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BEE Paper-1 — Chapter 5: Material & Energy Balance

167 questions — 68 objective (1 mark), 67 short (5 marks), 32 long (10 marks). Every answer is checked against the 2014 BEE guidebook and carries its book section reference plus an explanation.
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Objective questions (1 mark) — 68

📖 §5.5 Example 5.5 — concentrations (mole fraction)

1. 54 kg of water is mixed with 0.34 moles of salt to make a solution. The mole fraction of the solution is ____.

  1. 0.1
  2. 18.36
  3. 158.8
  4. none of the above
Answer: A) 0.1
Confirmed vs Book-1 §5.5 Ex.5.5: Book computes moles as (mass)/(mol. wt), mol. wt of water = 18. Moles of water = 54/18 = 3. Mole fraction of salt = moles salt/(total moles) = 0.34/(3 + 0.34) = 0.34/3.34 = 0.102 ≈ 0.1. Hence option (a).
Source: Sep 2021
📖 §5.3 Basic principles — element (stoichiometric) balance

2. C2H4 + xO2 ----> 2CO2 + yH2O, what is the value of x + y?

  1. 2
  2. 3
  3. 5
  4. 8
Answer: C) 5
Confirmed vs Book-1 §5.3 (mass of each element is conserved): C2H4 + xO2 → 2CO2 + yH2O. Hydrogen: 4 = 2y → y = 2. Oxygen: 2x = (2×2) + 2 = 6 → x = 3. Therefore x + y = 3 + 2 = 5, option (c).
Source: Sep 2021
📖 §5.5 Example 5.5 — weight/weight concentration

3. A solution of common salt is prepared by adding 25 kg of salt to 100 kg of water. The weight fraction of solution is ____.

  1. 20%
  2. 25%
  3. 4%
  4. none of the above
Answer: A) 20%
Confirmed vs Book-1 §5.5 Ex.5.5: weight fraction = weight of solute / total weight of solution = 25/(25 + 100) = 25/125 = 0.20, i.e. % w/w = 20%. (Book's own case: 20/(100+20) = 16.7%.) Option (a).
Source: Sep 2021
📖 §5.3 Master balance — Raw Materials = Products + Waste + Stored + Losses

4. A system uses 100 kg of raw material A, 200 kg of B and 220 kg of C. The mix is heated to 220 deg C. Air carries away on average 60% of A and 30% of B through the chimney. The output product would be

  1. 520 kg
  2. 400 kg
  3. 312 kg
  4. 208 kg
Answer: B) 400 kg
Confirmed vs Book-1 §5.3: Total raw material in = 100 + 200 + 220 = 520 kg. Waste carried away by air = 60% of A + 30% of B = (0.60×100) + (0.30×200) = 60 + 60 = 120 kg. With no storage, Products = 520 − 120 = 400 kg. Option (b).
Source: Nov 2009
📖 §5.5 Example 5.6 — evaporator solids (tie-component) balance

5. If feed of 15 tonnes per hour at 6% concentration is fed to an evaporator, the product obtained at 30% concentration is equal to ____ tonnes per hour.

  1. 3
  2. 9
  3. 0.9
  4. 4.5
Answer: A) 3
Confirmed vs Book-1 §5.5 Ex.5.6: Solids are conserved. Solids in feed = 15 × 0.06 = 0.9 t/h. Product at 30% solids = 0.9/0.30 = 3 t/h. (Water evaporated = 15 − 3 = 12 t/h.) Option (a).
Source: 2019
📖 §5.5 Material balance — moisture + water formed from hydrogen

6. 1 kg of wood contains 15% moisture and 5% hydrogen by weight. How much water is evaporated during complete combustion of 1kg of wood?

  1. 0.6 kg
  2. 200 g
  3. 0.15 kg
  4. none of the above
Answer: A) 0.6 kg
Confirmed vs Book-1 §5.5 (component mass balance): Free moisture = 1 × 0.15 = 0.15 kg. Hydrogen burns as H2 + ½O2 → H2O, so 2 kg H gives 18 kg water, i.e. 9 kg water per kg of hydrogen: 9 × 0.05 = 0.45 kg. Total water evaporated = 0.15 + 0.45 = 0.60 kg. Option (a).
Source: 2019
📖 §5.5 Material balance procedure — bone-dry solids balance

7. In a drying process product moisture is reduced from 60% to 30%. Inlet weight of the material is 200 kg. Calculate the weight of the outlet product.

  1. 80
  2. 120.5
  3. 114.3
  4. none of the above
Answer: C) 114.3
Confirmed vs Book-1 §5.5 (dry-solids balance, as in Ex.5.11): Bone-dry solids = 200 × (1 − 0.60) = 80 kg and are unchanged. Outlet product at 30% moisture is 70% solids, so outlet = 80/0.70 = 114.3 kg. Option (c).
Source: 2016
📖 §5.2 Components of material and energy balance (Fig 5.1)

8. A mass balance for energy conservation does not consider which of the following

  1. Steam
  2. water
  3. Lubricating oil
  4. Raw material
Answer: C) Lubricating oil
Confirmed vs Book-1 §5.2 Fig 5.1: the streams counted are raw materials, chemicals, water/air, energy/power (inputs) and products, by-products, emissions, wastewater and wastes (outputs). Steam, water and raw material are all such process streams; lubricating oil is a maintenance consumable and is not taken in the mass balance. Option (c).
Source: 2016
📖 §5.8 Energy analysis and the Sankey diagram (Fig 5.9)

9. Diagrammatic representation of input and output energy streams of an equipment or system is known as

  1. mollier diagram
  2. sankey diagram
  3. psychrometric chart
  4. balance diagram
Answer: B) sankey diagram
Confirmed vs Book-1 §5.8: 'The Sankey diagram is a very useful tool to represent an entire input and output energy flow in any energy equipment or system', the width of each arrow being proportional to the flow. Hence it is the diagrammatic representation of input/output energy streams. Option (b).
Source: 2016
📖 §5.1 Purpose of Material and Energy Balance (box)

10. Which of the following tool is made use of to assess the input, conversion efficiency, output, losses, quantification of all material, energy and waste streams in a process or system?

  1. material balance
  2. energy balance
  3. material and energy balance
  4. Sankey diagram
Answer: C) material and energy balance
Confirmed vs Book-1 §5.1 purpose box: material AND energy balance is used 'to assess the input, conversion efficiency, output and losses' and 'to quantify all material, energy and waste streams in a process or a system'. Only the combined material and energy balance covers all of these. Option (c).
Source: 2013
📖 §5.5 Example 5.6 — evaporator solids balance

11. If feed of 100 tonnes per hour at 10% concentration is fed to an evaporator, the product obtained at 25% concentration is equal to ____ tonnes per hour.

  1. 25
  2. 40
  3. 50
  4. 62.5
Answer: B) 40
Confirmed vs Book-1 §5.5 Ex.5.6: Solids in feed = 100 × 0.10 = 10 t/h and are conserved. Product at 25% concentration = 10/0.25 = 40 t/h. (Water evaporated = 60 t/h.) Option (b).
Source: 2013
📖 §5.5 Material balance — moisture + water formed from hydrogen

12. 1 kg of wood contains 15% moisture and 7% hydrogen by weight. How much water is evaporated from wood during complete combustion of 1 kg of wood ?

  1. 0.78 kg
  2. 0.22 kg
  3. 0.15 kg
  4. 0.63 kg
Answer: A) 0.78 kg
Confirmed vs Book-1 §5.5 (component mass balance): Free moisture = 0.15 kg. Water from hydrogen = 9 × 0.07 = 0.63 kg (9 kg water per kg H, from H2 + ½O2 → H2O). Total water evaporated = 0.15 + 0.63 = 0.78 kg. Option (a).
Source: 2012
📖 §5.1 Purpose of Material and Energy Balance (box)

13. Material and energy balance is used to quantify

  1. material and energy losses
  2. profit
  3. cost of production
  4. all of the above
Answer: A) material and energy losses
Confirmed vs Book-1 §5.1 purpose box: the M&E balance quantifies all material, energy and waste streams and assesses input, conversion efficiency, output and losses — i.e. it quantifies material and energy losses. Profit and cost of production are financial, not balance, outputs. Option (a).
Source: 2012
📖 §5 End-of-chapter Objective Q1 (§5.1 purpose box)

14. The objective of material and energy balance is to assess the

  1. input-output
  2. conversion efficiency
  3. losses
  4. all the above
Answer: D) all the above
Confirmed vs Book-1 §5.1 purpose box: the objective is to assess the input, conversion efficiency, output AND losses; hence all three listed items apply. Option (d) all the above.
Source: Guidebook
📖 §5.2 Components of M&E balance, Fig 5.1 (EOC Objective Q2)

15. In the material balance of a process or unit operation, which component will not be considered on the input side?

  1. chemicals
  2. water/air
  3. recycle
  4. by product
Answer: D) by product
Confirmed vs Book-1 §5.2 Fig 5.1: inputs are raw materials, chemicals, water/air, energy/power and the RECYCLE stream ('recycle stream is shown along with input side'). By-products appear on the output side with products, emissions and wastes. Option (d).
Source: Guidebook
📖 §5.2 Components of M&E balance, Fig 5.1 (EOC Objective Q3)

16. In material balance of a process, recycle product is always considered as

  1. input to process
  2. output to process
  3. both (a) and (b)
  4. none of them
Answer: A) input to process
Confirmed vs Book-1 §5.2 Fig 5.1 note: 'It may be noted that recycle stream is shown along with input side.' The recycle is fed back into the process, so in the material balance it is always taken as an input. Option (a).
Source: Guidebook
📖 §5.3 Conservation of mass (EOC Objective Q4)

17. In a chemical process two reactants A (200 kg) and B (200 kg) are used as reactants. If conversion is 50% and A and B react in equal proportion, calculate the weight of the product formed.

  1. 150 kg
  2. 200 kg
  3. 250 kg
  4. 400 kg
Answer: B) 200 kg
Confirmed vs Book-1 §5.3 (mass is conserved): total reactants charged = 200 + 200 = 400 kg; A and B react in equal proportion so they are consumed together. At 50% conversion the reacted mass = 0.50 × 400 = 200 kg, and by conservation of mass this appears as 200 kg of product. Option (b).
Source: Guidebook
📖 §5.6 Heat balance, Q = m·Cp·ΔT (EOC Objective Q5)

18. In a heat treatment furnace the material is heated up to 800 degC from ambient temperature of 30 degC. Considering the specific heat of material as 0.13 kcal/kg degC, what is the energy content in one kg of material after heating?

  1. 50 kcal
  2. 250 kcal
  3. 350 kcal
  4. 100 kcal
Answer: D) 100 kcal
Confirmed vs Book-1 §5.6 heat balance (Q = m·Cp·ΔT, as in Ex.5.10): Q = 1 kg × 0.13 kcal/kg°C × (800 − 30)°C = 0.13 × 770 = 100.1 ≈ 100 kcal per kg. Option (d) 100 kcal.
Source: Guidebook
📖 §5.6 Energy balance — radiation losses / sinks (EOC Objective Q6)

19. In a utility steam boiler, heat loss due to radiation normally is in the range of

  1. 10%
  2. 14%
  3. 1%
  4. 8%
Answer: C) 1%
Confirmed vs Book-1 §5.6: sinks are 'depositories of leakage or rejected energy... usually low-grade heat, as in radiation losses from boilers'. For a large utility steam boiler this radiation and convection loss is small — of the order of 1% of input (it is significant only in small boilers). Option (c).
Source: Guidebook
📖 §5.6 Energy balance — electrical-to-heat conversion (EOC Objective Q7)

20. Energy supplied by electricity, Q in kcal is equal to

  1. kWh x 8.6
  2. kWh x 86
  3. kWh x 860
  4. none
Answer: C) kWh x 860
Confirmed vs Book-1 §5.6: to compare energy streams they are converted to equivalent heat energy. 1 kWh = 860 kcal, so Q (kcal) = kWh × 860. Option (c).
Source: Guidebook
📖 §5.8 Sankey diagram (EOC Objective Q8)

21. Sankey diagram is a useful tool to represent

  1. financial strength of the company
  2. management philosophy
  3. input and output energy flow
  4. human resource strength of the company
Answer: C) input and output energy flow
Confirmed vs Book-1 §5.8: the Sankey diagram 'is very useful tool to represent an entire input and output energy flow in any energy equipment or system', arrow width proportional to the flow. Option (c).
Source: Guidebook
📖 §5.6 Heat balances — heat duty formula (EOC Objective Q9)

22. Which of the following formula is useful to determine the heat duty in conducting heat balance?

  1. Q = M Cp deltaT
  2. Q = AV
  3. PV = nRT
  4. none of the above
Answer: A) Q = M Cp deltaT
Confirmed vs Book-1 §5.6/Ex.5.10: heat duty (sensible heat) = mass × specific heat × temperature change, Q = M·Cp·ΔT. Q = AV is a flow (continuity) relation and PV = nRT is the gas law used for gas concentrations (Ex.5.7/5.8), not for heat duty. Option (a).
Source: Guidebook
📖 §5.6 Energy balance — electrical energy (EOC Objective Q10)

23. A 230V, 100 W rated incandescent bulb is operated at a constant voltage of 250V. The approximate power consumption of bulb is

  1. 100 W
  2. 118 W
  3. 85 W
  4. none of the above
Answer: B) 118 W
Confirmed vs Book-1 §5.6: for a fixed-resistance (incandescent) load R = V²/P = 230²/100 = 529 Ω, so P ∝ V². P = 250²/529 = 62500/529 = 118.1 W, i.e. P = 100 × (250/230)² ≈ 118 W. Option (b).
Source: Guidebook
📖 §5.5 Example 5.7 — mole fraction of a gas mixture

24. A gaseous mixture contains 7.50 gms of H2 and 3.25 gms of O2 and 5.55gms of N2. The Mole fraction of N2 is ____.

  1. 0.34
  2. 0.03
  3. 0.0294
  4. 0.049
Answer: D) 0.049
Confirmed vs Book-1 §5.5 Ex.5.7 method (moles = mass/mol. wt): H2 = 7.50/2 = 3.75; O2 = 3.25/32 = 0.1016; N2 = 5.55/28 = 0.1982. Total = 4.0498 moles. Mole fraction of N2 = 0.1982/4.0498 = 0.0489 ≈ 0.049. Option (d).
Source: Mar 2023
📖 §5.5 Material balance procedure — bone-dry solids balance

25. A dry feed contains 7% moisture was feed to a water spray chamber to increase the moisture content to 35% in the dry feed. The output feed quantity coming from the spray chamber is ____.

  1. 0.5 kg/kg of input feed
  2. 1.43 kg/kg of input feed
  3. 1.48 kg/kg of input feed
  4. 2.66 kg/kg of input feed
Answer: B) 1.43 kg/kg of input feed
Confirmed vs Book-1 §5.5 (dry solids are unchanged): per 1 kg of input feed, bone-dry solids = 1 × (1 − 0.07) = 0.93 kg. In the output the moisture is 35%, so solids are 65%: output = 0.93/0.65 = 1.43 kg per kg of input feed. Option (b).
Source: Mar 2023
📖 §5.3 Basic principles — master material balance equation

26. Law of conservation of mass can be expressed by the following equation:

  1. Products = Raw Materials + Waste Products + Stored Products + Losses
  2. Products = Raw Materials + Waste Products + Stored Products - Losses
  3. Raw Material = Products + Waste Products + Stored Products - Losses
  4. Raw Materials = Products + Waste Products + Stored Products + Losses
Answer: D) Raw Materials = Products + Waste Products + Stored Products + Losses
Confirmed vs Book-1 §5.3: 'Raw Materials = Products + Waste Products + Stored Products + Losses', where Losses are the unidentified materials. Input equals the sum of all outputs plus what is stored plus unaccounted losses. Option (d).
Source: Mar 2023
📖 §5.7 Example 5.11 / §5.5 — evaporation of moisture into air

27. When the evaporation of water from a wet substance is zero, the relative humidity of air is likely to be ____.

  1. 0%
  2. 10%
  3. 50%
  4. 100%
Answer: D) 100%
Confirmed vs Book-1 §5.5–§5.7 (drying material balance): moisture leaves the wet substance only while the surrounding air can still take up water vapour. When the air is saturated — relative humidity 100% — its moisture-carrying capacity is exhausted and the evaporation rate falls to zero. Option (d).
Source: Mar 2023
📖 §5.3 Basic principles — element (stoichiometric) balance

28. 1 mole of sulphur react with X moles of H2SO4 to form Y moles of H2O and Z moles of SO2 then (X/Y) + Z is ____.

  1. 2
  2. 4
  3. 1
  4. 3
Answer: B) 4
Confirmed vs Book-1 §5.3 (element balance): S + 2H2SO4 → 3SO2 + 2H2O. So 1 mole of sulphur reacts with X = 2 moles of H2SO4 giving Y = 2 moles of H2O and Z = 3 moles of SO2. Therefore (X/Y) + Z = (2/2) + 3 = 1 + 3 = 4. Option (b).
Source: Mar 2023
📖 §5.5 Material balance procedure — bone-dry solids balance

29. In a drying process, moisture is reduced from 50% to 30%. Initial weight of the material is 100 kg. Calculate the weight of the final product in kg.

  1. 80
  2. 86
  3. 71.4
  4. 74.3
Answer: C) 71.4
Confirmed vs Book-1 §5.5 (dry solids unchanged, as in Ex.5.11): Bone-dry solids = 100 × (1 − 0.50) = 50 kg. Final product at 30% moisture is 70% solids, so final weight = 50/0.70 = 71.4 kg. Option (c).
Source: Jul 2022
📖 §5.5 Example 5.6 — solids balance (crystallizer/evaporator)

30. If feed of 100 tons per hour at 5% concentration is fed to a crystallizer, the rate in tons per hour of the product obtained at 25% concentration is equal to:

  1. 40
  2. 20
  3. 25
  4. 100
Answer: B) 20
Confirmed vs Book-1 §5.5 Ex.5.6 method: Solids in feed = 100 × 0.05 = 5 t/h and are conserved. Product at 25% concentration = 5/0.25 = 20 t/h. Option (b).
Source: Jul 2022
📖 §5.3 Basic principles — Raw Materials = Products + Waste + Stored + Losses

31. A process receives 1000 kg/hr of raw material. The hourly outputs are 700 kg of product, 200 kg of waste, and 50 kg stored. What is the unaccounted loss?

  1. 100 kg/hr
  2. 150 kg/hr
  3. 200 kg/hr
  4. 50 kg/hr
Answer: D) 50 kg/hr
Confirmed vs Book-1 §5.3 master equation: Losses = Raw Materials − (Products + Waste + Stored) = 1000 − (700 + 200 + 50) = 1000 − 950 = 50 kg/hr of unidentified (unaccounted) loss. Option (d).
Source: Sep 2025
📖 §5.5 Example 5.6 — evaporator water evaporated

32. Calculate the quantity of water evaporated when 100 kg of feed containing 6% solids is concentrated to 30% solids.

  1. 600 kg
  2. 180 kg
  3. 80 kg
  4. 800 kg
Answer: C) 80 kg
Confirmed vs Book-1 §5.5 Ex.5.6 (book's own numbers): Solids in feed = 100 × 0.06 = 6 kg and are conserved. Output = 6/0.30 = 20 kg. Water evaporated = 100 − 20 = 80 kg. Option (c).
Source: Sep 2025
📖 §5.5 Example 5.5 — moles = mass/molecular weight

33. Moles of water in 54 grams:

  1. 3
  2. 4
  3. 5
  4. 6
Answer: A) 3
Confirmed vs Book-1 §5.5 Ex.5.5 (mol. wt of water = 18): moles = 54/18 = 3 moles. Option (a).
Source: Sep 2025
📖 §5.5 Example 5.7(a) — mean molecular weight of air

34. Mean molecular weight of air (77% N2, 23% O2 by weight) is ___________ grams.

  1. 26.8
  2. 27.8
  3. 28.8
  4. 29.8
Answer: C) 28.8
Confirmed vs Book-1 §5.5 Ex.5.7(a): basis 100 kg air contains 77/28 = 2.75 moles N2 and 23/32 = 0.72 moles O2; total = 3.47 moles. Mean molecular weight = 100/3.47 = 28.8. Option (c).
Source: Sep 2025
📖 §5.5 Example 5.5 — moles = mass/molecular weight

35. The number of moles of water contained in 72 grams of water is

  1. 2
  2. 3
  3. 4
  4. 5
Answer: C) 4
Confirmed vs Book-1 §5.5 Ex.5.5 (mol. wt of water = 18): moles = 72/18 = 4 moles. Option (c).
Source: Sep 2024
📖 §5.5 Example 5.5 — weight/weight concentration

36. A solution of common salt in water is prepared by adding 25 kg of salt to 100 kg of water. The concentration of salt in this solution as a weight fraction shall be

  1. 10%
  2. 15%
  3. 20%
  4. 25%
Answer: C) 20%
Confirmed vs Book-1 §5.5 Ex.5.5: weight fraction = solute/(solute + solvent) = 25/(25 + 100) = 0.20 = 20% w/w. Option (c).
Source: Sep 2024
📖 Book-1 §5.3 — Basic Principles of Material and Energy Balance (Figure 5.2)

37. The general material balance equation for a process can be written as:

  1. Raw Materials = Products + Waste Products + Stored Products + Losses
  2. Raw Materials = Products − Losses
  3. Products = Raw Materials + Losses
  4. Raw Materials + Losses = Products
Answer: A) Raw Materials = Products + Waste Products + Stored Products + Losses
Confirmed vs Book-1 §5.3 — the balance box gives Mass In = Mass Out + Mass Stored, i.e. Raw Materials = Products + Wastes + Stored Materials. When part of the input cannot be accounted for, the book writes the full form: Raw Materials = Products + Waste Products + Stored Products + Losses, where the losses are the unidentified materials. Rearranging this equation is what lets you solve for any unknown stream.
Source: AI practice
📖 Book-1 §5.2 — Components of Material and Energy Balance; §5.4 (Figure 5.3)

38. In a process flow diagram, which of the following is NOT on the INPUT side of the process box?

  1. Raw material
  2. Recycle stream
  3. By-product
  4. Combustion air
Answer: C) By-product
Confirmed vs Book-1 §5.2 — the book states explicitly that the RECYCLE stream is shown along with the input side; inputs also include raw materials, water/air, chemicals and energy. Figure 5.3 in §5.4 shows desired products, by-products, wastes and energy coming OUT of the box, so a by-product is an output. This is exactly the chapter's own objective Q2 (which component is not considered on the input side — by-product).
Source: AI practice
📖 Book-1 §5.8 — Energy Analysis and the Sankey Diagram (Figure 5.9)

39. A diagram that visually represents the input and output energy flows of a system, where the width of each arrow is proportional to the magnitude of the flow, is called a:

  1. Gantt chart
  2. Sankey diagram
  3. Pie chart
  4. Fishbone diagram
Answer: B) Sankey diagram
Confirmed vs Book-1 §5.8 — flows in a Sankey diagram are represented by arrows whose width is proportional to the size of the actual flow, so input, useful output and losses can be compared at a glance. Figure 5.9 for an IC engine shows 25% effective power, 5% friction and parasitic losses, 30% coolant and 40% exhaust gas — making exhaust heat recovery the priority.
Source: AI practice
📖 Book-1 §5.5 — Example 5.9 (dust balance) / Short Question S-1

40. Dust-laden gas enters a bag filter at 200,000 m^3/hr carrying 6 g/m^3 of dust. The cleaned gas leaves at 210,000 m^3/hr carrying 0.1 g/m^3. The dust collected in the hopper is:

  1. 1179 kg/hr
  2. 1200 kg/hr
  3. 21 kg/hr
  4. 1221 kg/hr
Answer: A) 1179 kg/hr
Confirmed vs Book-1 §5.5 — component (dust) balance: In = Out + hopper. Dust in = 200,000 x 6 / 1000 = 1200 kg/hr; dust out = 210,000 x 0.1 / 1000 = 21 kg/hr; collected in the bin = 1200 - 21 = 1179 kg/hr. Identical method to Example 5.9, where 777.7 - 10.6 = 767.1 kg/hr.
Source: AI practice
📖 Book-1 §5.5 — Example 5.6 (evaporator)

41. A liquor containing 8% solids is concentrated to 40% solids. Starting with 100 kg of feed, the mass of water evaporated is:

  1. 32 kg
  2. 60 kg
  3. 80 kg
  4. 20 kg
Answer: C) 80 kg
Confirmed vs Book-1 §5.5 — the dry solids are conserved. Solids = 100 x 0.08 = 8 kg; concentrated output = 8 / 0.40 = 20 kg; water evaporated = 100 - 20 = 80 kg. Example 5.6 works the same case (6% to 30% solids on 100 kg feed) and also gives 80 kg of water evaporated.
Source: AI practice
📖 Book-1 §5.7 — Example 5.10 (furnace cooling water)

42. A furnace shell (mass 2000 kg, specific heat 0.2 kcal/kg°C) is cooled from 90°C to 55°C by water whose temperature rises by 5°C. The mass of cooling water required is (Cp_water = 1 kcal/kg°C):

  1. 1400 kg
  2. 2800 kg
  3. 14000 kg
  4. 700 kg
Answer: B) 2800 kg
Confirmed vs Book-1 §5.7 — heat to be removed from the shell Q = m x Cp x dT = 2000 x 0.2 x (90 - 55) = 14,000 kcal. Heat picked up by the water = X x 1 x 5 = 5X kcal. Setting stream 1 = stream 2 gives X = 14,000 / 5 = 2800 kg. This is Example 5.10 verbatim (water enters at 28 degC and may rise only to 33 degC).
Source: AI practice
📖 Book-1 §5.5 — Short Question S-4 (boiler blow-down)

43. Boiler feed water has a TDS of 120 mg/l and the maximum permissible TDS in the boiler drum is 3500 mg/l. The required blow-down (as % of feed) is approximately:

  1. 29.2%
  2. 0.34%
  3. 3.43%
  4. 12.0%
Answer: C) 3.43%
Confirmed vs Book-1 §5.5 — a steady-state solids (TDS) balance gives blow-down as % of feed = feed TDS / maximum permissible TDS x 100 = 120 / 3500 x 100 = 3.43%. All dissolved solids entering with the feed water must leave with the blow-down to hold the drum at the IS limit of 3500 mg/l for boilers operating up to 2 MPa.
Source: AI practice
📖 Book-1 §5.5 — Example 5.7 (air composition)

44. Air is approximately 77% N2 and 23% O2 by weight. Taking a 100 kg basis (N2 = 28, O2 = 32), the mean molecular weight of air is about:

  1. 30.0
  2. 28.8
  3. 32.0
  4. 28.0
Answer: B) 28.8
Confirmed vs Book-1 §5.5 — on a basis of 100 kg of air: moles N2 = 77/28 = 2.75 and moles O2 = 23/32 = 0.72, total 3.47 moles. Mean molecular weight of air = 100 / 3.47 = 28.8. Cross-check: mole fraction of oxygen = 0.72 / 3.47 = 0.21, the familiar 21% by volume.
Source: AI practice
📖 Book-1 §5.7 — Short Question S-5 (heat rejected to cooling water)

45. Water flowing at 200 m^3/hr is heated through a temperature rise of 7°C. The heat duty is (1 m^3 water = 1000 kg, Cp = 1 kcal/kg°C):

  1. 140,000 kcal/hr
  2. 1,400,000 kcal/hr
  3. 14,000 kcal/hr
  4. 1400 kcal/hr
Answer: B) 1,400,000 kcal/hr
Confirmed vs Book-1 §5.7 — 200 m3/hr of water = 200 x 1000 = 200,000 kg/hr (1 m3 of water = 1000 kg). Q = m x Cp x dT = 200,000 x 1 x 7 = 14,00,000 kcal/hr. The usual slip is applying Q = m Cp dT to the volume without converting to mass.
Source: AI practice
📖 Book-1 §5.7 — Example 5.11 (dryer heat balance)

46. A solid is dried from 55% moisture to 10% moisture. For 60 kg/hr of wet feed, the moisture removed is:

  1. 27 kg/hr
  2. 33 kg/hr
  3. 30 kg/hr
  4. 45 kg/hr
Answer: C) 30 kg/hr
Confirmed vs Book-1 §5.7 — the bone-dry cloth is the tie component: 60 x (1 - 0.55) = 27 kg/hr, unchanged through the dryer. Final product = 27 / 0.90 = 30 kg/hr, so moisture removed = 60 - 30 = 30 kg/hr. Example 5.11 gets the same 30 kg/hr (33 kg initial moisture less 3 kg final).
Source: AI practice
📖 Book-1 §5.6 — Heat Balances (Chapter-5 objective Q7)

47. Electrical energy of 1 kWh is equivalent to how much heat energy?

  1. 860 kcal
  2. 632 kcal
  3. 1000 kcal
  4. 427 kcal
Answer: A) 860 kcal
Confirmed vs Book-1 §5.6 — the chapter's own objective question states that energy supplied by electricity, Q in kcal = kWh x 860. The book advises converting every energy stream (oil, gas, coal, steam, electricity) to equivalent heat energy before balancing, and this is the electrical conversion factor. 632 kcal/kg in Example 5.12 is the enthalpy of evaporated moisture, a different quantity.
Source: AI practice
📖 Book-1 §5.5 — Short Question S-3 (unburnt carbon in refuse)

48. Coal contains 67.2% carbon and 22.3% ash. The refuse contains 7.1% carbon (rest is ash). Taking 100 kg of coal as basis, the fraction of original carbon left unburnt in the refuse is about:

  1. 7.1%
  2. 2.5%
  3. 1.7%
  4. 24%
Answer: B) 2.5%
Confirmed vs Book-1 §5.5 — ash is inert and is therefore the tie component that fixes the refuse mass: refuse = 22.3 / (1 - 0.071) = 22.3 / 0.929 = 24.0 kg per 100 kg of coal. Carbon in the refuse = 24.0 x 0.071 = 1.70 kg. As a percentage of the original carbon = 1.70 / 67.2 x 100 = 2.5%.
Source: AI practice
📖 §5.3 Basic principles of material and energy balance

49. Which of the following will not be a major component of mass balance

  1. steam
  2. water
  3. raw materials
  4. lubricating oil
Answer: D) lubricating oil
A material balance is built on the streams that matter in mass and cost: raw materials in, products and by-products out, plus the major utility streams such as water and steam. Lubricating oil is a consumable used in trivial quantity relative to the process streams, so it is not a major component. The book's own guidance is that costly and high-volume materials are the ones worth balancing.
Source: Sep 2015
📖 §5.3 Basic principles of material and energy balance

50. The number of moles of water contained in 54 kg of water is ------------

  1. 2
  2. 3
  3. 4
  4. 5
Answer: B) 3
Moles = mass / molecular weight, and water's molecular weight is 18. The printed options only work if the quantity is read as 54 GRAMS: 54/18 = 3 moles. As literally printed, 54 kg = 54,000/18 = 3,000 moles. Carry both facts: the formula, and the habit of checking whether the paper's kg should have been g — 1 kmol of water is 18 kg, 1 mol is 18 g.
Source: Sep 2015
📖 §5.5 Material balance

51. In a drying process, moisture is reduced from 60% to 30%. Initial weight of the material is 200 kg. Calculate the weight of the product

  1. 104
  2. 266.6
  3. 130
  4. 114.3
Answer: D) 114.3
Solids are conserved: 200 x (1-0.60) = 80 kg. In the product the solids are (1-0.30) = 70% of the mass, so product = 80/0.70 = 114.3 kg. Option (c) 130 is what you get by subtracting the moisture percentages, and (a) 104 by anchoring on the water instead of the solids. Always anchor on the component that does NOT leave.
Source: Sep 2015
📖 §5.3 Basic principles of material and energy balance

52. If air consists of 77% by weight of nitrogen and 23% by weight of oxygen, the mean molecular weight of air is

  1. 11.9
  2. 28.8
  3. 17.7
  4. insufficient data
Answer: B) 28.8
For a mass-basis composition, mean molecular weight = 100 / sum(weight% / molecular weight) = 100 / (77/28 + 23/32) = 100/3.469 = 28.8. The trap is treating 77/23 as MOLE percentages and computing 0.77 x 28 + 0.23 x 32 = 28.9 — it happens to be close for air, but the method is wrong and will fail on any other mixture. Rule: mole fractions get a weighted average, mass fractions get a harmonic-style reciprocal sum.
Source: Sep 2015
📖 §5.3 Basic principles of material and energy balance

53. An oil-fired boiler operates at an excess air of 6%. If the stoichiometric air fuel ratio is 14 then for an oil consumption of 100 kg per hour, the flue gas liberated in kg/hr would be

  1. 1484
  2. 1584
  3. 106
  4. 114
Answer: B) 1584
Actual air = stoichiometric x (1 + excess) = 14 x 1.06 = 14.84 kg air per kg fuel. By conservation of mass everything that goes in comes out as flue gas, so flue gas = fuel + air = 1 + 14.84 = 15.84 kg per kg fuel, and for 100 kg/h of oil that is 1,584 kg/h. The mark is lost by forgetting to ADD the fuel mass (giving 1,484) — mass in equals mass out, and the fuel does not disappear.
Source: Sep 2015
📖 §5.3 Basic principles of material and energy balance

54. A process requires 10 Kg of fuel with a calorific value of 5000 kcal/kg. The system efficiency is 80% and the losses will be

  1. 10000 kcal
  2. 45000 kcal
  3. 500 kcal
  4. 2000 kcal
Answer: A) 10000 kcal
Input = 10 x 5,000 = 50,000 kcal. At 80% efficiency the useful heat is 40,000 kcal, so the loss is the other 20% = 10,000 kcal. Read the question wording: 'the losses will be' asks for the wasted fraction, and 45,000/40,000 are planted for anyone who computes the useful heat instead. Loss = input x (1 - efficiency).
Source: Sep 2015
📖 §5.5 Material balance

55. A centrifugal pump draws 12 m3/hr. Due to leakages from the body of the pump a continuous flow of 2 m3/hr is lost. The efficiency of the pump is 55%. The flow at the discharge side would be

  1. 12 m3/hr
  2. 10 m3/hr
  3. 5.5 m3/hr
  4. 6.6 m3/hr
Answer: B) 10 m3/hr
Mass in = mass out: 12 - 2 (leakage) = 10 m3/h at the discharge. The 55% efficiency is a red herring — it governs the POWER the pump draws, not the volume of water it moves. Spotting irrelevant data is half the skill in these balance questions; efficiency never appears in a mass balance.
Source: Sep 2015
📖 §5.3 Basic principles of material and energy balance

56. 20 m3 of water is mixed with 30 m3 of another liquid with a specific gravity of 0.9. The volume of the mixture would be

  1. 47 m3
  2. 48 m3
  3. 50 m3
  4. 53 m3
Answer: C) 50 m3
Volumes of miscible liquids are taken as additive in this balance, so 20 + 30 = 50 m3. The specific gravity of 0.9 is supplied only to tempt you into a mass calculation — it would matter if the question asked for the MASS of the mixture (20,000 + 27,000 = 47,000 kg) or its mean density, which is exactly why 47 is offered as an option. Read whether the question asks for volume or mass.
Source: Sep 2015
📖 §5.3 Basic principles of material and energy balance

57. 1 kg of wood contains 15% moisture and 7% hydrogen by weight. How much water is evaporated during complete combustion of 1 kg of wood

  1. 0.78 kg
  2. 220 grams
  3. 0.15 kg
  4. 0.63 kg
Answer: A) 0.78 kg
Two sources of water. The free moisture is 15% x 1 kg = 0.15 kg. The hydrogen burns to water in the ratio 9 kg of water per kg of H2 (from 2H2 + O2 -> 2H2O, i.e. 4 kg H2 gives 36 kg H2O), so 0.07 x 9 = 0.63 kg. Total = 0.78 kg. That factor of 9 is worth memorising — it is the same factor that creates the whole GCV-minus-NCV difference.
Source: Sep 2015
📖 §5.3 Basic principles of material and energy balance

58. A process requires 120 kg of fuel with a calorific value of 4800 kcal/kg for heating with a system efficiency of 82 %. The loss would be____________.

  1. 576000 kcal
  2. 472320 kcal
  3. 103680 kcal
  4. 480000 kcal
Answer: C) 103680 kcal
Input = 120 x 4,800 = 576,000 kcal. Loss = input x (1 - efficiency) = 576,000 x 0.18 = 103,680 kcal. Option (b) 472,320 is the USEFUL heat and (a) 576,000 is the input, both planted for candidates who stop one step early. Read whether the question wants input, output or loss.
Source: Sep 2019
📖 §5.3 Basic principles of material and energy balance

59. 54 kg of water is mixed with 0.34 moles of salt to make a solution. The mole fraction of the solution is

  1. 0.1
  2. 18.36
  3. 158.8
  4. none of the above
Answer: A) 0.1. The printed answer key is (a), which is obtained only if the water quantity is read as 54 grams: moles of water = 54/18 = 3; mole fraction of salt = 0.34/(3 + 0.34) = 0.102 ~ 0.1. [Note: as printed, 54 kg of water = 3000 moles would give a mole fraction of 0.000113, i.e. option (d). The paper itself prints kg where g is meant - the same slip occurs in Q.50 of this paper.] Answer key printed in the question paper.
Mole fraction of a component = its moles / total moles. Taking the water as 54 GRAMS (which is what the printed options require): moles of water = 54/18 = 3, so mole fraction of salt = 0.34/(3 + 0.34) = 0.102, about 0.1. As printed in kg the answer would be negligible. Keep the definitions apart: mole fraction uses MOLES, weight fraction uses MASS — mixing them is what generates the other three options.
Source: Mar 2021
📖 §5.3 Basic principles of material and energy balance

60. C2H4 + xO2 -----> 2CO2 + yH2O, what is the value of x + y?

  1. 2
  2. 3
  3. 5
  4. 8
Answer: C) 5. Balancing C2H4 + 3O2 -> 2CO2 + 2H2O: carbon 2 = 2, hydrogen 4 = 2x2, oxygen 3x2 = 6 = (2x2) + 2. Hence x = 3, y = 2 and x + y = 5. [OCR: the formula is printed as 'CoH4 + xO', read as 'C2H4 + xO2'.] Derived; the printed answer-key column was not legible in the scan for this question.
Balance the equation element by element: C2H4 + 3O2 -> 2CO2 + 2H2O. Carbon 2 = 2; hydrogen 4 = 2 x 2; oxygen 3 x 2 = 6 = (2 x 2) + 2. So x = 3, y = 2 and x + y = 5. Balance carbon first, then hydrogen, and let oxygen fall out last — that order works for every hydrocarbon combustion equation in the paper.
Source: Mar 2021
📖 §5.3 Basic principles of material and energy balance

61. A solution of common salt is prepared by adding 25 kg of salt to 100 kg of water. The weight fraction of solution is

  1. 20%
  2. 25%
  3. 4%
  4. none of the above
Answer: A) 20%. Total solution = 25 + 100 = 125 kg. Weight fraction of salt = 25/125 = 0.20 = 20%. [The printed answer-key mark for this question was illegible in the scan; the value is computed.] Derived; the printed answer-key column was not legible in the scan for this question.
Weight fraction is on the TOTAL solution, not on the solvent: total = 25 + 100 = 125 kg, so the fraction is 25/125 = 0.20 = 20%. Option (b) 25% comes from dividing by the 100 kg of water alone, which is the intended trap. Rule: the denominator of a weight fraction is always the whole mixture.
Source: Mar 2021
📖 §5.3 Basic principles of material and energy balance

62. The number of moles in 90 kg of water is

  1. 5
  2. 18
  3. 2
  4. none of the above
Answer: A) 5. The printed answer key is (a), which follows only if the quantity is read as 90 GRAMS: moles = 90/18 = 5. [As printed, 90 kg of water = 90,000/18 = 5000 moles, i.e. option (d). The paper prints kg where g is intended - the same slip as in Q.7 of this paper.] Answer key printed in the question paper.
Moles = mass / molecular weight, with water at 18. The printed key of 5 follows only if the quantity is read as 90 GRAMS (90/18 = 5); as printed, 90 kg gives 5,000 moles, or 5 kmol. Note that 5 kmol is the same number with a different prefix, which is probably how the discrepancy arose — carry both 1 mol = 18 g and 1 kmol = 18 kg.
Source: Mar 2021
📖 §5.3 Basic principles of material and energy balance

63. A process requires 100 kg of fuel with a calorific value of 5000 kcal/kg for heating with a system efficiency of 83%. The loss would be

  1. 235,000 kcal
  2. 85,000 kcal
  3. 103680 kcal
  4. 415,000 kcal
Answer: B) 85,000 kcal. Heat input = 100 x 5,000 = 500,000 kcal. Useful heat at 83% efficiency = 415,000 kcal. Loss = 500,000 - 415,000 = 85,000 kcal (i.e. 17% of input). Derived - the question paper carries no printed answer key for Section-I.
Input = 100 x 5,000 = 500,000 kcal; useful at 83% = 415,000 kcal; loss = 500,000 - 415,000 = 85,000 kcal (the 17% not converted). The distractors are the input-related figures 415,000 (useful heat) and 235,000, so identify what is asked before computing. Loss = input x (1 - efficiency) is the one-line route.
Source: Jul 2022
📖 §5.3 Basic principles of material and energy balance

64. The number of moles of water contained in 27 kg of water is

  1. 5
  2. 3
  3. 4
  4. 1.5
Answer: D) 1.5. The options are consistent only with 27 GRAMS of water: moles = mass / molecular weight = 27/18 = 1.5. [As printed, 27 kg of water would be 27,000/18 = 1,500 moles. This paper series repeatedly prints kg where grams are intended.] Derived - the question paper carries no printed answer key for Section-I.
Moles = mass / molecular weight = 27/18 = 1.5, which is what the options are built for — the quantity has to be read as 27 GRAMS. As printed, 27 kg of water is 1,500 moles (1.5 kmol). The recurring lesson across this whole family of mole questions: check the unit against the magnitude of the options before dividing.
Source: Jul 2022
📖 §5.5 Material balance

65. In a drying process, moisture is reduced from 50% to 30%. Initial weight of the material is 100 kg. Calculate the weight of the final product in kg

  1. 80
  2. 86
  3. 71.4
  4. 74.3
Answer: C) 71.4 kg. Bone-dry solids = 100 x (1 - 0.50) = 50 kg and this quantity is unchanged by drying. In the final product the solids form (1 - 0.30) = 70% of the mass, so final weight = 50 / 0.70 = 71.4 kg. Derived - the question paper carries no printed answer key for Section-I.
Bone-dry solids = 100 x (1-0.50) = 50 kg and they do not change. In the product the solids form (1-0.30) = 70% of the mass, so the product = 50/0.70 = 71.4 kg. Option (a) 80 comes from subtracting the moisture percentages (50-30 = 20 kg removed), which is wrong because the two percentages are on different totals. Always anchor on the solids.
Source: Jul 2022
📖 §5.3 Basic principles of material and energy balance

66. A gaseous mixture contains 7.50 gms of H2 and 3.25 gms of O2 and 5.55 gms of N2. The Mole fraction of N2 is

  1. 0.34
  2. 0.03
  3. 0.0294
  4. 0.049
Answer: D) 0.049. Moles: H2 = 7.50/2 = 3.750; O2 = 3.25/32 = 0.1016; N2 = 5.55/28 = 0.1982. Total = 4.0498 moles. Mole fraction of N2 = 0.1982/4.0498 = 0.0489 ~ 0.049. Derived - the question paper carries no printed answer key for Section-I.
Convert each mass to moles with its own molecular weight: H2 = 7.50/2 = 3.750; O2 = 3.25/32 = 0.1016; N2 = 5.55/28 = 0.1982. Total = 4.0498 mol, so the N2 mole fraction = 0.1982/4.0498 = 0.049. Hydrogen dominates because its molecular weight is tiny — which is exactly the point of the question. Taking mass fractions instead (5.55/16.3 = 0.34) produces the planted wrong option.
Source: Mar 2023
📖 §5.5 Material balance

67. A dry feed containing 7% moisture was fed to a water spray chamber to increase the moisture content to 35% in the dry feed. The output feed quantity coming from the spray chamber is

  1. 0.5 kg/kg of input feed
  2. 1.43 kg/kg of input feed
  3. 1.48 kg/kg of input feed
  4. 2.66 kg/kg of input feed
Answer: B) 1.43 kg/kg of input feed. Basis 1 kg of feed: bone-dry solids = 1 x (1 - 0.07) = 0.93 kg, and this is unchanged by spraying. In the product the solids form (1 - 0.35) = 65%, so output = 0.93/0.65 = 1.43 kg per kg of input feed. Derived - the question paper carries no printed answer key for Section-I.
Bone-dry solids are conserved even when water is ADDED. Basis 1 kg of feed: solids = 1 x (1-0.07) = 0.93 kg. In the product the solids are (1-0.35) = 65%, so product = 0.93/0.65 = 1.43 kg per kg of input feed. Same technique as the drying questions, run in reverse — the solids anchor works whether moisture is being removed or sprayed on.
Source: Mar 2023
📖 §5.3 Basic principles of material and energy balance

68. 1 mole of sulphur reacts with X moles of H2SO4 to form Y moles of H2O and Z moles of SO2, then (X/Y) + Z is

  1. 2
  2. 4
  3. 1
  4. 3
Answer: B) 4. The balanced reaction is S + 2H2SO4 -> 3SO2 + 2H2O, so X = 2, Y = 2 and Z = 3. Therefore (X/Y) + Z = (2/2) + 3 = 1 + 3 = 4. Derived - the question paper carries no printed answer key for Section-I.
The balanced reaction is S + 2H2SO4 -> 3SO2 + 2H2O, so X = 2, Y = 2 and Z = 3, giving (X/Y) + Z = 1 + 3 = 4. Verify by counting: sulphur 1 + 2 = 3; hydrogen 4 = 4; oxygen 8 = 6 + 2. Balance the element that appears in fewest species first, then check oxygen last.
Source: Mar 2023

Short questions (5 marks) — 67

📖 §5.1 Purpose of Material and Energy Balance (box)

1. State the purpose of carrying out a material and energy (M&E) balance for an industrial process.

Model answer: A material and energy balance accounts for all mass and energy entering and leaving a process. Its purpose is: (i) to assess the input, conversion efficiency, output and losses; (ii) to quantify all material, energy and waste streams in a process or system; and (iii) it is a powerful tool for establishing the basis for improvement and identifying potential savings. It rests on the conservation principle (Mass In = Mass Out + Stored; Energy In = Energy Out + Stored) and exposes losses/leaks that are otherwise invisible.
Guidebook box 'Purpose of Material and Energy Balance' lists exactly these three points. Objective-Q1 answer is 'all the above' (input-output, conversion efficiency, losses).
Source: Year not recorded
📖 §5.1 Introduction — laws of conservation of mass and energy

2. State the law of conservation of mass and the law of conservation of energy on which M&E balances are based.

Model answer: Law of conservation of mass: matter is neither created nor destroyed; matter may flow through a control volume and may react to form another species, but no matter is ever lost or gained. Law of conservation of energy: energy is neither created nor destroyed, it is simply converted from one form into another. Together these laws lead to the mass (material) balance and the energy balance: if there is no accumulation, what goes into a process must come out — true for both batch and continuous operation over any chosen time interval.
Direct from §5.1 Introduction.
Source: Year not recorded
📖 §5.3 Basic principles — master material balance equation (sugar-plant losses)

3. Write the general material balance equation for a unit operation and explain why a 'Losses' term is included.

Model answer: Treating the unit operation as a box, mass in must balance mass out: Mass In = Mass Out + Mass Stored, i.e. Raw Materials = Products + Waste Products + Stored Products. In practice the measured products, waste and stored material do not fully account for the input; the unidentified shortfall is added as Losses, so the working equation becomes: Raw Materials = Products + Waste Products + Stored Products + Losses. Losses are the unidentified materials (e.g. material chemically changed, or going unnoticed down a drain) that must be tracked down.
§5.3, sugar-plant example: m_Au is the unknown loss; equation extended to include Losses.
Source: Year not recorded
📖 §5.3 Basic principles — energy balance equation; §5.6

4. Write the general energy balance equation for a unit operation and state why energy balances are more complicated than mass balances.

Model answer: Energy In = Energy Out + Energy Stored, where energy entering with raw materials plus energy added in the plant equals energy leaving with products, energy leaving with waste, energy lost to surroundings, plus energy stored. Energy balances are more complicated than mass balances because energy exists in many inter-convertible forms — kinetic, potential, heat, chemical, electrical, mechanical — and these forms can be converted from one to another during processing (e.g. mechanical energy converted by friction into heat). It is the sum total of all forms that is conserved, so the quantities must still balance overall.
§5.3 energy balance + §5.6 conservation of energy.
Source: Year not recorded
📖 §5.2 Components of M&E balance, Fig 5.1 — recycle on input side

5. In the material balance of a process, where are the recycle stream and the by-product placed — on the input side or the output side?

Model answer: The recycle stream is shown on the INPUT side (it re-enters the process and is treated as an input along with raw materials, chemicals, water/air and energy/power). The by-product is an OUTPUT of the process (along with products, gaseous emissions, wastewater, liquid/solid waste). This is why, in the objective questions, recycle is 'always considered as input to process' and the by-product is the component NOT considered on the input side.
Fig 5.1 note states recycle stream is shown along with input side; Objective Q2 (by-product not on input) and Q3 (recycle = input to process).
Source: Year not recorded
📖 §5.4 Classification of processes (steady/unsteady; continuous/batch)

6. Distinguish between a steady-state process and an unsteady-state process, and between a continuous and a batch process.

Model answer: A steady-state process is one in which none of the process variables change with time (flows and quantities held in vessels are constant), so balances can be written per unit time. An unsteady-state process is one in which the process variables change with time. Based on how the process is built to operate: a continuous process has feed and product streams moving into and out of the process all the time (e.g. oil refinery, distillation). A batch process has feed charged in to start, processed through steps with no mass added or removed during the cycle (parameters monitored/controlled), and finished products taken out at specific times.
§5.4 Classification of Processes (A by time, B by operation).
Source: Year not recorded
📖 §5.5 Levels of material balance

7. Describe the different levels at which a material balance can be developed.

Model answer: Material balances can be developed at three levels: (1) Overall material balance — covers input and output streams for the complete plant. (2) Section-wise material balance — made for each section/department/cost centre; helps prioritise focus areas for efficiency improvement. (3) Equipment-wise material balance — made for key equipment; helps assess equipment performance and identify energy and material losses. The choice depends on the reason for the balance; the major factor is cost of materials, so costly materials and products are considered more readily than cheaper ones and waste materials.
§5.5 'Levels of Material Balance'.
Source: Year not recorded
📖 §5.5 Material balance procedure — steps (a), (b), (c)

8. List the typical steps in carrying out a material balance for a process.

Model answer: First identify materials in, materials out and material stored, and decide whether to treat each material as a whole (gross balance) or treat individual constituents separately (e.g. dry solids vs water). Typical steps are: (a) Define basis & units — choose a basis as quantity (mass for batch) or flow rate (mass/hr for continuous) of one process stream, with convenient units (w/w, w/v, molar concentration, mole fraction). (b) Draw a flowchart — establish a boundary so flow streams in/out can be identified, showing inputs, process steps, intermediates, by-products, recycle and outputs with operating parameters and flow rates. (c) Write the material balance equations and solve.
§5.5 'Material Balance Procedure' steps a, b, c.
Source: Year not recorded
📖 §5.5 step (b) Draw a flowchart; Fig 5.4 Typical process flowchart

9. What is a process flow diagram and what should it represent when prepared for a material and energy balance?

Model answer: A process flow (flow chart) is a schematic representation of the production process showing the various input resources, conversion steps, outputs and recycle streams; a boundary is established so flow streams in and out can be identified. It should be drawn stepwise from raw material to finished product, showing intermediates and by-products. It must represent the operating process parameters (temperature, pressure, % concentration, etc.) and the flow rate of each stream in appropriate units (m³/h or kg/h; for batch processes the total cycle time is included). Inputs include raw materials, water, steam and energy; wastes/by-products include solids, water, chemicals and energy. The output is the final product.
§5.5 step (b) Draw a flowchart and the bullets that follow; Fig 5.4 typical arrangement.
Source: Year not recorded
📖 §5.5 step (a) Define basis & units — w/w, w/v, molar, mole fraction

10. Define the four ways of expressing material concentration used when choosing units for a material balance.

Model answer: (1) Weight/weight (w/w) concentration — weight of solute divided by total weight of solution (fractional form of % composition by weight). (2) Weight/volume (w/v) concentration — weight of solute in the total volume of the solution. (3) Molar concentration (M) — the number of molecular weights of solute expressed in kg in 1 m³ of solution. (4) Mole fraction — the ratio of the number of moles of solute to the total number of moles of all species present in the solution. (For gases, concentration is measured as weight per unit volume or as partial pressures.)
§5.5 step (a) Define basis & units.
Source: Year not recorded
📖 §5.6 Heat balances — enthalpy and datum

11. What is a heat balance, and why must enthalpy always be referred to a datum?

Model answer: A heat balance is an energy balance that considers only heat energy (enthalpy), ignoring internal energies — the most common and important practical form of energy balance. In heating and drying operations enthalpy (total heat) is conserved, so enthalpy balances can be written around items of equipment, process stages or the whole plant, assuming no appreciable heat is converted to work. Enthalpy (H) is always referred to a reference level or datum so that quantities are relative to that datum; working out the balance is then a matter of considering the masses involved, their specific heats, and their changes in temperature or state (latent heat).
§5.6 'Heat Balances'.
Source: Year not recorded
📖 §5.6 'How can energy be lost in a system?'

12. How can energy be 'lost' in a system, and what are the main causes of this loss of usable energy?

Model answer: Because of inter-conversions it is not easy to isolate the separate forms of energy (heat, kinetic, chemical, potential). When energy is said to be 'lost' it really means it has changed into a form that is not counted — most often work done against friction, which changes other forms into heat and wear. No energy is actually destroyed; only the forms have changed. The causes of loss of usable energy are: in mechanical systems — friction; in electrical systems — resistance; in fluid systems — turbulence, viscosity or mixing. Practically, energy balances take into account only heat balances, ignoring internal energies.
§5.6 'How can energy be lost in a system?'
Source: Year not recorded
📖 §5.6 Energy balance in power plant cycle, Fig 5.6 (ΣQ + ΣW = ΔU)

13. Apply the law of conservation of energy to a power plant (Rankine) cycle and state the cycle energy balance.

Model answer: If a system undergoes a process by heat and work transfer, the net heat supplied ΣQ plus the net work input ΣW equals the change in internal energy ΔU: ΣQ + ΣW = ΔU. For a complete thermodynamic cycle ΔU = 0, so ΣQ + ΣW = 0. For the power plant: heat Q_in is supplied to water in the boiler, feed-pump work W_in is added, the steam drives the turbine producing useful work W_out, and heat Q_out is rejected in the condenser. ΣQ = Q_in − Q_out. The working fluid (feed water) changes state from water to steam and back to condensate.
§5.6 Fig 5.6 Energy Balance in Power Plant Cycle; sign convention: Q supplied (+) / rejected (−), W on system (+) / by system (−).
Source: Year not recorded
📖 §5.6 Efficiency, Fig 5.7; sinks (rejected low-grade energy)

14. Define the thermodynamic efficiency of a process and explain what 'sinks' represent in an energy balance.

Model answer: The thermodynamic efficiency of a process is the ratio of useful output to input, and is always less than 100% (Efficiency = Output / Input). For example, with input = 15 + 185 = 200 units and useful output = 45, efficiency = 45/200 = 0.225 = 22.5%, so energy lost = 100 − 22.5 = 77.5%. Sinks are the depositories of leakage or rejected energy — usually low-grade heat such as radiation losses from boilers or heat carried away by cooling water. The outputs represent the useful work; when storage does not change, inputs must equal outputs.
§5.6 Efficiency, Fig 5.7; sinks defined just before Fig 5.6.
Source: Year not recorded
📖 §5.6 Conservation of energy, Fig 5.5 a/b/c — storage change

15. Using the conservation of energy, explain how a change in stored energy is inferred when the input does not equal the output.

Model answer: If the storage in a system does not change, the ingoing and outgoing energy must be equal. If the storage changes, this is reflected in the balance and the energy input need not equal the output. Example: if 75 units enter but only 60 units leave, the first law requires energy be conserved, so the system must have gained (stored) 15 units — storage increased. Conversely, if the input is short by 15 units relative to the output, the system must have lost (released) 15 units of stored energy — storage decreased.
§5.6 Fig 5.5 a/b/c (No storage / Storage increased / Storage decreased).
Source: Year not recorded
📖 §5.7 Facility as an Energy System, Fig 5.8 Plant Energy Systems

16. Explain the concept of a 'facility as an energy system' and the three areas into which energy/utility systems are classified.

Model answer: In a production facility, primary energy as coal, oil, gas and electricity enters and is converted into more convenient forms such as steam, compressed air and chilled water; the outgoing energy is usually heat and motion. Energy usage splits into: electricity (purchased HT, converted to LT, or self-generated by DG sets/captive plants), fuels (furnace oil, coal converted to steam or electricity), boilers (steam for heating/drying), cooling towers/cooling-water systems (cooling demand) and air compressors (compressed-air needs). For a system approach and energy analysis, all energy/utility systems are classified into three areas: generation, distribution and utilisation.
§5.7 'Facility as an Energy System', Fig 5.8 Plant Energy Systems.
Source: Year not recorded
📖 §5.8 Sankey diagram, Fig 5.9 (IC engine) — EOC Short Q S-2

17. Explain the Sankey diagram with an example.

Model answer: A Sankey diagram is a flow diagram that represents the entire input and output energy flow of an energy equipment or system (boiler, fired heater, furnace, etc.) after an energy balance has been carried out. Flows are shown as arrows, and the WIDTH of each arrow is proportional to the magnitude of the actual flow. Better than numbers or tables, it visually shows outputs (benefits) and losses so that energy managers can prioritise improvements. Example: for an internal combustion engine, 100% fuel energy enters and splits into ~25% effective power, ~5% friction/parasitic losses, ~30% coolant loss and ~40% exhaust gas loss. The wide exhaust-gas arrow flags it as the priority area — pointing to a waste-heat-recovery device.
§5.8 + Fig 5.9 (IC engine). Objective Q8: Sankey = input and output energy flow.
Source: Year not recorded
📖 §5.8 Energy analysis and the Sankey diagram

18. What basic data is needed for an energy analysis, and why is the Sankey diagram preferred to present it?

Model answer: The basic data needed for an energy analysis is an energy balance of each process section. The objective is to define in detail the energy input, the energy utilised, and the energy dissipated or wasted. This is best represented by a Sankey diagram because, rather than numbers, tables or descriptions, it visually represents the various outputs (benefits) and losses — with arrow width proportional to flow — so the energy manager can immediately see the largest loss and focus on improvements in a prioritized manner. For an IC engine it makes clear that exhaust flue-gas loss is the key area for attention, justifying a waste-heat boiler.
§5.8 opening paragraph + IC-engine discussion.
Source: Year not recorded
📖 §5.6 Heat balances; EOC Objective Q7 (kWh × 860) and Q9 (Q = M·Cp·ΔT)

19. State the heat-duty formula used in a heat balance and the relation converting electrical energy to heat (kcal).

Model answer: The heat duty (sensible heat) is found from Q = m · Cp · ΔT, where m = mass, Cp = specific heat and ΔT = temperature change; for a phase change the latent-heat form Q = m · h_latent is used. Electrical energy supplied, Q in kcal, equals kWh × 860 (since 1 kWh = 860 kcal). These, with PV = nRT for gases, are the core relations for Chapter-5 heat and material balances.
Objective Q9 (Q = m·Cp·ΔT) and Q7 (Q kcal = kWh × 860).
Source: Year not recorded
📖 §5.5–§5.8 constants used in Examples 5.5–5.12 and the Solved Example

20. List the key physical constants used in Chapter-5 material and energy balance calculations.

Model answer: Latent heat of evaporation of water ≈ 540 kcal/kg (≈ 2257 kJ/kg). Specific heat of water Cp = 1 kcal/kg°C. Electrical energy: 1 kWh = 860 kcal. Gas law: PV = nRT with R = 0.08206 m³·atm/mol·K; 1 mole of gas at STP occupies 22.4 litres. Common molecular weights: water = 18, NaCl (common salt) = 58.5, CO₂ = 44, N₂ = 28, O₂ = 32; mean molecular weight of air ≈ 28.8. Density of water = 1000 kg/m³.
Drawn from Examples 5.5–5.12 and the Solved Example (water 540 kcal/kg, mol wts, 22.4 L, R value).
Source: Year not recorded
📖 §5.5 Material balance procedure — dry-solids (tie-component) balance

21. Explain the method (tie-component / dry-solids balance) for solving a drying or evaporation material balance.

Model answer: Use a constituent (tie-component) balance: identify a component whose mass is conserved through the process — in drying/evaporation the bone-dry solids do not change. Steps: (1) From the feed, compute the dry-solid mass = feed × (1 − moisture fraction). (2) This same dry-solid mass appears in the product, so product mass = dry solids ÷ (1 − final moisture fraction). (3) Water (or vapour) removed = feed − product. The solids act as the tie component linking inlet and outlet, letting you find the unknown stream without needing the water balance directly.
§5.5 'do material balance for dry solids alone... separating into non-water and water'; method underlies Examples 5.3, 5.6, 5.11, 5.12.
Source: Year not recorded
📖 §5.7 Example 5.10 — furnace shell cooling water

22. Explain how to compute the cooling-water requirement to cool a hot body, using a furnace shell as an example.

Model answer: Use an energy (heat) balance: heat lost by the hot body = heat gained by the cooling water. Heat to be removed = m·Cp·ΔT of the body. Heat gained by water = m_water·Cp_water·ΔT_water (Cp_water = 1 kcal/kg°C). Equate the two and solve for m_water. Example: furnace shell m = 2000 kg, Cp = 0.2 kcal/kg°C, cooled 90→55°C, so heat removed = 2000 × 0.2 × 35 = 14,000 kcal. Water enters 28°C, leaves 33°C (ΔT = 5°C), so 14,000 = m_water × 1 × 5, giving m_water = 2800 kg.
Example 5.10 (furnace); verified numbers.
Source: Year not recorded
📖 §5 EOC Short Q S-1 — dust balance (method of Example 5.9)

23. During an air pollution monitoring study, the inlet gas stream to a bag filter was 200,000 m³/hr and the outlet stream was 210,000 m³/hr. Dust load at the inlet was 6 g/m³ and at the outlet 0.1 g/m³. How much dust (kg/hr) was collected in the bag filter bin?

Model answer: Apply a dust mass balance: Dust in = Dust out + Dust collected in hopper. Inlet dust = 200,000 m³/hr × 6 g/m³ = 1,200,000 g/hr = 1200 kg/hr. Outlet dust = 210,000 m³/hr × 0.1 g/m³ = 21,000 g/hr = 21 kg/hr. Dust collected in bin = 1200 − 21 = 1179 kg/hr.
Short-Q S-1; method from Example 5.9 dust balance. Verified arithmetic.
Source: Year not recorded
📖 §5 EOC Short Q S-3 — unburnt carbon by ash (tie-component) balance

24. A coal sample from the mine contains 67.2% carbon and 22.3% ash. The refuse after combustion contains 7.1% carbon and the rest is ash. Compute the % of the original carbon left unburnt in the refuse.

Model answer: Basis 100 kg coal: carbon = 67.2 kg, ash = 22.3 kg. Ash is inert and is conserved, so ash in refuse = 22.3 kg. Refuse is 7.1% carbon and 92.9% ash, so total refuse = 22.3 / 0.929 = 24.0 kg. Carbon (unburnt) in refuse = 7.1% of 24.0 = 1.70 kg. % of original carbon unburnt = (1.70 / 67.2) × 100 ≈ 2.54%.
Short-Q S-3; ash is the tie (conserved) component. Verified.
Source: Year not recorded
📖 §5 EOC Short Q S-4 — boiler blowdown from TDS balance

25. A boiler is fed with soft water containing 120 mg/l dissolved solids. As per IS standards the dissolved solids in the boiler should not exceed 3500 mg/l (boilers up to 2 MPa). A continuous blowdown system is adopted. Find the percentage of feed water that is blown down.

Model answer: Dissolved solids are conserved: solids entering with feed water must leave with the blowdown (steam carries no solids). At steady state, % blowdown (as fraction of feed water) = Feed-water TDS / Maximum boiler TDS × 100 = 120 / 3500 × 100 = 3.43%. So about 3.4% of the feed water must be blown down to hold the boiler TDS at 3500 mg/l.
Short-Q S-4; rewritten cleanly (existing answer was self-contradictory). Method per master notes: %blowdown = feed TDS / max TDS.
Source: Year not recorded
📖 §5 EOC Short Q S-5 — Q = m·Cp·ΔT for cooling water

26. A shell-and-tube heat exchanger is cooled with a stream of demineralized water. Evaluate the total heat rejected to the cooling water (kcal/hr) if the water flow rate is 200 m³/hr and the temperature rise is 7°C.

Model answer: Mass flow of water = 200 m³/hr × 1000 kg/m³ = 200,000 kg/hr. Specific heat of water Cp = 1 kcal/kg°C. Heat rejected Q = m · Cp · ΔT = 200,000 × 1 × 7 = 1,400,000 kcal/hr (1.4 × 10⁶ kcal/hr).
Short-Q S-5; verified.
Source: Year not recorded
📖 §5 EOC Long Q L-2 — autoclave cooling-water requirement

27. An autoclave holds 1000 cans of pea soup heated to 100°C; the cans must be cooled to 40°C. Each can weighs 60 g (Cp 0.50 kJ/kg°C) and holds 0.45 kg pea soup (Cp 4.1 kJ/kg°C). Heat content of the autoclave walls above 40°C is 1.6×10⁴ kJ. Cooling water enters at 15°C and leaves at 35°C. How much cooling water is required (no wall heat loss)?

Model answer: Heat to be removed (cooling 100→40°C, ΔT = 60°C): Soup = 1000 × 0.45 × 4.1 × 60 = 110,700 kJ. Cans = 1000 × 0.060 × 0.50 × 60 = 1,800 kJ. Walls = 1.6×10⁴ = 16,000 kJ. Total heat removed = 110,700 + 1,800 + 16,000 = 128,500 kJ. Cooling water (Cp = 4.187 kJ/kg°C, ΔT = 35−15 = 20°C) gains this heat: m_w = 128,500 / (4.187 × 20) ≈ 1534 kg of cooling water.
Long-Q L-2 framed as a short worked balance. Numbers from OCR; water Cp 4.187 kJ/kg°C assumed.
Source: Year not recorded
📖 §5.5 Example 5.1 — mixing two solutions (solids balance)

28. A 10% solids solution (entering at 5.0 kg/s) is mixed with a 25% solids solution; a single output of 20% solids is removed. Find the other flow rates (no accumulation).

Model answer: Let B (25% solids) = X kg/s and output C (20% solids) = Y kg/s. Total balance: 5 + X = Y. Solids balance: 0.10×5 + 0.25X = 0.20Y → 0.5 + 0.25X = 0.20(5 + X) → 0.05X = 0.5 → X = 10 kg/s. Then Y = 5 + 10 = 15 kg/s. So the 25% stream is 10 kg/s and the 20% output is 15 kg/s.
Example 5.1; verified.
Source: Year not recorded
📖 §5.5 Example 5.6 — evaporator, water evaporated per 100 kg feed

29. In a textile mill an evaporator concentrates a liquor containing 6% solids (w/w) to an output containing 30% solids (w/w). Calculate the water evaporated per 100 kg of feed.

Model answer: Solids are conserved. Feed = 100 kg → solid content = 100 × 0.06 = 6 kg. Outlet solid content = 6 kg (mass in = mass out). Output (thick liquor) = 6 / 0.30 = 20 kg. Water evaporated = feed − output = 100 − 20 = 80 kg.
Example 5.6; verified.
Source: Year not recorded
📖 §5.5 Example 5.4 — continuous centrifuge (milk) balance

30. In continuous centrifuging, 35,000 kg of whole milk containing 4% fat is separated over 6 hours into skim milk (0.45% fat) and cream (45% fat). Find the flow rates of the two output streams.

Model answer: Feed per hour = 35,000 / 6 = 5833 kg/hr. Let skim milk = Y and cream = Z. Mass balance: 5833 = Y + Z (Z = 5833 − Y). Fat balance: 0.04 × 5833 = 0.0045Y + 0.45Z. Substituting: 233.3 = 0.0045Y + 0.45(5833 − Y) → solving gives Y ≈ 5369 kg/hr (skim milk) and Z = 5833 − 5369 ≈ 464 kg/hr (cream).
Example 5.4; verified.
Source: Year not recorded
📖 §5.5 Example 5.3 — constituent balance, skim vs whole milk

31. Skim milk made by removing fat from whole milk contains 90.5% water, 3.5% protein, 5.1% carbohydrate, 0.1% fat and 0.8% ash. If the original whole milk contained 4.5% fat (only fat removed, no losses), find the whole-milk composition.

Model answer: Basis 100 kg skim milk → it contains 0.1 kg fat. Let removed fat = x kg. Original fat = (x + 0.1) kg; original mass = (100 + x) kg. Since original fat was 4.5%: (x + 0.1)/(100 + x) = 0.045 → x = 4.6 kg. So whole milk = 104.6 kg. Composition: fat = 4.5%, water = 90.5/104.6 = 86.5%, protein = 3.5/104.6 = 3.3%, carbohydrate = 5.1/104.6 = 4.9%, ash = 0.8/104.6 = 0.76% (the book prints 0.8%). Whole milk total = 104.6 kg from 100 kg skim milk + 4.6 kg fat.
Example 5.3; verified.
Source: Year not recorded
📖 §5.7 Example 5.11 — textile dryer heat balance / thermal efficiency

32. A textile dryer consumes 4 m³/hr of natural gas (CV 800 kJ/mole) and dries 60 kg/hr of wet cloth from 55% to 10% moisture. Estimate its overall thermal efficiency (latent heat of evaporation only).

Model answer: Bone-dry cloth = 60 × (1 − 0.55) = 27 kg; initial moisture = 33 kg. Final product at 10% moisture: dry cloth is 90%, so product = 27/0.9 = 30 kg, final moisture = 3 kg. Moisture removed = 33 − 3 = 30 kg/hr. Heat used = 30 × 2257 = 6.8×10⁴ kJ/hr. Gas at STP: 1 mole = 22.4 L, so 4 m³/hr = 4000/22.4 = 179 mole/hr. Heat available = 179 × 800 = 14.3×10⁴ kJ/hr. Thermal efficiency = 6.8×10⁴ / 14.3×10⁴ ≈ 48%.
Example 5.11; verified.
Source: Year not recorded
📖 §5.7 Example 5.12 — paper machine evaporation rate and steam

33. A paper machine produces 340 TPD; inlet/outlet dryness is 40%/95%; evaporated moisture at 80°C (enthalpy 632 kcal/kg); steam supplied at 35 kg/cm² (latent heat 513 kcal/kg). Find (a) moisture evaporated and (b) steam required per hour.

Model answer: Production = 340 TPD = 14.16 TPH. Paper (bone-dry) in product = 14.16 × 0.95 = 13.45 TPH. Moisture before dryer = [(100−40)/40] × 13.45 = 20.175 TPH. Moisture after dryer = 14.16 × 0.05 = 0.707 TPH. (a) Evaporated moisture = 20.175 − 0.707 = 19.468 TPH. (b) Heat in moisture = 632 × 19,468 = 12,303,776 kcal/hr; steam required = 12,303,776 / 513 = 23,984 kg ≈ 23.98 MT/hour.
Example 5.12; verified.
Source: Year not recorded
📖 §5.8 Solved Example — evaporator vapour and steam (enthalpy balance)

34. An evaporator is fed 10,000 kg/hr of 1% solids solution (feed at 38°C), concentrated to 2% solids. Steam enters at enthalpy 640 kcal/kg, condensate leaves at 100°C; enthalpies: feed 38.1, product 100.8, vapour 640 kcal/kg. Find the vapour formed and the steam used per hour.

Model answer: Solids balance: solids = 10,000 × 0.01 = 100 kg/hr; product = 100/0.02 = 5000 kg/hr; vapour formed = 10,000 − 5000 = 5000 kg/hr. Heat balance: heat in steam + heat in feed = heat in vapour + heat in thick liquor. M×(640−100) + 38.1×10,000 = 640×5000 + 100.8×5000 → 540M + 381,000 = 3,200,000 + 504,000 → 540M = 3,323,000 → M = 6153.7 kg steam/hr.
Solved Example before the Questions section; verified (steam gives 640−100 = 540 kcal/kg).
Source: Year not recorded
📖 §5.5 Example 5.5 — salt solution expressed four ways

35. 20 kg of salt is dissolved in 100 kg of water giving a solution of density 1323 kg/m³. Express the salt concentration as (a) weight fraction, (b) weight/volume fraction, (c) mole fraction, (d) molar concentration. (Mol wt: salt 58.5, water 18.)

Model answer: (a) Weight fraction = 20/(100+20) = 0.167 = 16.7% w/w. (b) 1 m³ solution = 1323 kg, of which salt = 20×1323/120 = 220.5 kg/m³; weight/volume fraction = 220.5/1000 = 0.2205 ≈ 22.1%. (c) Moles water = 100/18 = 5.56, moles salt = 20/58.5 = 0.34; mole fraction salt = 0.34/(5.56+0.34) = 0.058. (d) Molar concentration M = 220.5/58.5 = 3.77 mol/m³.
Example 5.5; verified.
Source: Year not recorded
📖 §5 EOC Objective Q4 — conversion and mass conservation

36. In a chemical process two reactants A (200 kg) and B (200 kg) react in equal proportion. If the conversion is 50%, calculate the weight of product formed.

Model answer: Total reactants charged = 200 + 200 = 400 kg. They react in equal proportion, so all of A and B can react together. At 50% conversion, only half of the reactant mass is converted to product: product formed = 0.50 × 400 = 200 kg. (By conservation of mass the product equals the converted reactant mass.)
Objective Q4; correct option = 200 kg.
Source: Year not recorded
📖 §5 EOC Objective Q5 — sensible heat Q = m·Cp·ΔT

37. In a heat-treatment furnace, material is heated from an ambient 30°C to 800°C. Taking the specific heat as 0.13 kcal/kg°C, what is the energy content of one kg of material after heating?

Model answer: Sensible heat Q = m·Cp·ΔT = 1 × 0.13 × (800 − 30) = 0.13 × 770 = 100.1 ≈ 100 kcal per kg.
Objective Q5; correct option = 100 kcal.
Source: Year not recorded
📖 §5.6 Energy balance — milk pasteurizer example (whole and part process)

38. Using the milk pasteurizer as an example, show how an energy (heat) balance can be applied to a whole process and to only a part of a process.

Model answer: In pasteurizing, milk is pumped through a heat exchanger, first heated then cooled; the energy of interest is heat energy in the milk. Whole process: heat energy leaving in the milk = initial heat energy + heat added by pump + heat added in the heating section − heat removed in the cooling section − heat lost to surroundings. Part of a process: considering only the heating section, heat lost by the hot water = heat gained by the milk + heat lost from the heat exchanger to its surroundings. Thus the conservation of energy applies equally to the whole plant or to any chosen sub-section.
§5.6 pasteurizer example; verified.
Source: Year not recorded
📖 §5.1 Introduction — planning → pilot → commissioning → production

39. At what stages of a process is a material balance used, from concept to production?

Model answer: A material balance can be determined from the conceptual stage to the final production stage. Initially it is estimated during the planning stage of a new process or equipment. This estimate is improved while carrying out pilot-scale tests on the new process. The material balance is then verified during the commissioning stage, and finally used as a control measure during actual production. Material balances are fundamental to the control of processing, particularly the control of product yields.
§5.1 Introduction (planning → pilot → commissioning → production).
Source: Year not recorded
📖 §5.3 Component balance with no chemical change — sugar plant

40. If no chemical changes occur in a plant, how does the law of conservation of mass apply to each individual component? Illustrate with a sugar plant.

Model answer: If no chemical changes occur, the conservation of mass applies to each component separately: mass of component A in entering materials = mass of A in exit materials + mass of A stored in the plant. Example (sugar plant): if total sugar (mA) entering is not equalled by purified sugar (mAp) + sugar in waste liquors (mAw) + sugar accumulated (mAs), something is wrong — sugar is either being burned (chemically changed) or going unnoticed down a drain. The unidentified loss mAu must be found, giving: Raw materials = Products + Waste + Stored + Losses.
§5.3 component balance + sugar example.
Source: Year not recorded
📖 §5.5 Example 5.7 (a) and (b) — air composition

41. Air consists of 77% by weight nitrogen and 23% by weight oxygen. Calculate (a) the mean molecular weight of air and (b) the mole fraction of oxygen. (Mol wt: N₂ = 28, O₂ = 32.)

Model answer: Basis 100 kg air: moles N₂ = 77/28 = 2.75, moles O₂ = 23/32 = 0.72. Total moles = 2.75 + 0.72 = 3.47. (a) Mean molecular weight of air = 100/3.47 = 28.8. (b) Mole fraction of oxygen = 0.72/3.47 = 0.21.
Example 5.7 (a) and (b); verified.
Source: Year not recorded
📖 §5 EOC Short Q S-3 (exam variant) — ash tie-component method

42. A coal sample contains 64% carbon and 24% ash. The refuse after combustion contains 8% carbon and the rest ash. Compute the % of the original carbon left unburnt in the refuse.

Model answer: Basis 100 kg coal: carbon = 64 kg, ash = 24 kg. Ash is conserved, so ash in refuse = 24 kg. Refuse is 8% carbon and 92% ash, so total refuse = 24/0.92 = 26.087 kg. Unburnt carbon in refuse = 8% × 26.087 = 2.087 kg. % of original carbon unburnt = (2.087/64) × 100 = 3.26%.
Exam variant of S-3; same ash-tie method. Verified arithmetic.
Source: Jul 2022 Exam
📖 §5.6 Heat balance — Q = m·Cp·ΔT, latent heat; EOC Objective Q7 (kWh × 860)

43. Steam heats 5 kL/hr of furnace oil from 30°C to 90°C. Furnace-oil Cp = 0.22 kcal/kg°C, sp. gravity 0.95. (a) Steam per hour needed if steam latent heat is 510 kcal/kg. (b) If steam costs Rs 3.40/kg and electricity Rs 6/kWh, which is more economical?

Model answer: (a) Mass of oil = 5×1000×0.95 = 4750 kg/hr. Heat required = m·Cp·ΔT = 4750 × 0.22 × (90−30) = 62,700 kcal/hr. Steam = 62,700/510 = 123 kg/hr. (b) Steam cost = 123 × 3.40 = Rs 417.9/hr. Electricity = 62,700/860 = 72.9 kWh; cost = 72.9 × 6 = Rs 437.4/hr. Steam heating is more economical.
Exam item; uses Q=mCpΔT, latent heat, kWh×860. Verified arithmetic.
Source: 24th Exam Sep 2024
📖 §5.4 Classification of processes, Fig 5.3 — process as a control box

44. Explain how a process or unit operation is represented as a 'box', and what must be true of the mass crossing the box.

Model answer: A process can be viewed overall or as a series of units, each a unit operation represented by a box (control volume). Raw materials and energy go into the box, and desired products, by-products, wastes and energy come out. The mass into and out of the control box must be equal (Mass In = Mass Out when there is no accumulation). The equipment within the box makes the required changes with as little waste and energy use as possible. The box may sit between previous unit operations (feeding it) and subsequent unit operations (receiving its product), with wasted energy shown leaving.
§5.4 + Fig 5.3 Representation of Process.
Source: Year not recorded
📖 Book-1 §5.5 — Material Balance Procedure (gross vs constituent balance)

45. Distinguish between a gross (overall) material balance and a constituent (component) material balance.

Model answer: A gross (overall) material balance treats all the material in each stream as a whole — total mass in = total mass out (+ stored). A constituent (component) balance separates the material into individual constituents and balances each one separately; for example, splitting a stream into non-water (dry solids) and water, then writing a balance on the dry solids alone. The constituent approach is essential when one component is conserved (a 'tie component') and is used to find unknown stream rates in drying, evaporation and separation problems.
Synthesised from §5.5 'Material Balance Procedure' (gross vs individual constituents); concept not given as a single Q in OCR.
Source: n/a
📖 Book-1 §5.5 (levels of material balance) & §5.8 (energy analysis); SEC as energy intensity per §4.6

46. What is specific energy consumption (SEC), and how does an energy balance help establish it?

Model answer: Specific energy consumption is the energy used per unit of output — for example kcal (or kWh) per kg or per tonne of product. It normalises energy use against production so that performance can be compared over time or against benchmarks regardless of output level. An energy balance, by quantifying the total energy input and the useful output of a facility, section or equipment, provides exactly the numbers needed to compute SEC (energy input ÷ production). Tracking SEC highlights deteriorating efficiency and the scope for energy-saving measures.
Concept flagged in the task brief; supported by §5.5 levels of balance and §5.8 energy analysis but not defined verbatim in OCR.
Source: n/a
📖 §5.6 Energy balance

47. In a heat exchanger the inlet and outlet temperatures of the cooling water are 30 oC and 36 oC. The flow rate of cooling water is 400 litres/hr. The process fluid enters the heat exchanger at 60 oC and leaves at 45 oC. Find out the flow rate of the process fluid? (Cp of process fluid is 0.8 kCal/kg oC).

Model answer: Heat transferred to cooling water = m x Cp x dT = 400 x 1 x (36-30) = 2400 kcal/hour. Flow rate of process fluid = 2400/((60-45) x 0.8) = 200 kgs/hr.
Heat picked up by the cooling water = 400 x 1 x (36-30) = 2,400 kcal/h. All of it came from the process fluid, so m = Q / (Cp x dT) = 2,400 / (0.8 x (60-45)) = 200 kg/h. The heat-exchanger balance is always 'heat lost by hot = heat gained by cold'; the errors that cost marks are using the water's Cp of 1 on the oil side, and mixing up which stream has which dT.
Source: Oct 2011
📖 §5.5 Material balance

48. A cotton mill dries 1200 kg of wet fabric in a drier from 54% initial moisture to 9% final moisture. How many kilograms of water are removed during drying operation?

Model answer: Basis: 1200 kg/hr of wet fabric. Dry fabric = 1200 x 0.46 = 552 kg. Weight of final fabric = 552/0.91 = 606.6 kg. Water removed = 1200 - 606.6 = 593.4 kg.
Work on BONE-DRY solids, which do not change during drying: 1200 x (1-0.54) = 552 kg. In the product, solids are (1-0.09) = 91% of the mass, so final mass = 552/0.91 = 606.6 kg and water removed = 1200 - 606.6 = 593.4 kg. Never subtract the moisture percentages (54 - 9 = 45% of 1200 = 540 kg is wrong) — the percentages are on different total masses. Hook: fix the dry solids, then re-inflate.
Source: Oct 2011
📖 §5.6 Energy balance

49. In a heat exchanger the inlet and outlet temperatures of the cooling water are 30 oC and 36 oC. The flow rate of cooling water is 500 litres/hr. The process fluid enters the heat exchanger at 60 oC and leaves at 45 oC. Find out the flow rate of the process fluid? (Cp of process fluid is 0.8 kCal/kg oC).

Model answer: Heat transferred to cooling water = m x Cp x dT = 500 x 1 x (36-30) = 3000 kcal/hour. Flow rate of process fluid = 3000/((60-45) x 0.8) = 250 kgs/hr.
Water side: 500 x 1 x (36-30) = 3,000 kcal/h. Process fluid: m = 3,000 / (0.8 x 15) = 250 kg/h. Same structure as its 400 L/h twin — set heat gained by the cold stream equal to heat lost by the hot stream, and keep each stream's own Cp with its own dT.
Source: Oct 2011
📖 §5.5 Material balance

50. A cotton mill dries 2000 kg of wet fabric in a drier from 54% initial moisture to 9% final moisture. How many kilograms of water are removed during drying operation?

Model answer: Basis: 2000 kg/hr of wet fabric. Dry fabric = 2000 x 0.46 = 920 kg. Weight of final fabric = 920/0.91 = 1011 kg. Water removed = 2000 - 1011 = 989 kg.
Bone-dry solids = 2000 x 0.46 = 920 kg, unchanged by drying. Product mass = 920/0.91 = 1,011 kg, so water removed = 2000 - 1011 = 989 kg. Percentages quoted on a WET basis always need this solids-anchored route; subtracting 54% - 9% directly gives 900 kg and loses the mark.
Source: Oct 2011
📖 §5.6 Energy balance

51. A Diesel Generator performance trial gives specific generation of 3.5 kWh per liter of diesel. The cooling water loss and exhaust flue gas loss as percentage of fuel input are 28% and 32% respectively. The calorific value of diesel is 10,200 kcal/kg. The specific gravity of Diesel is 0.85. Calculate unaccounted loss as percentage of input energy.

Model answer: CV of Diesel = 10,200 kcal/kg. Heat in input diesel = 10,200 x 0.85 = 8670 kcal/litre. Heat in kWh energy output = 3.5 x 860 = 3010 kcal/litre. % of heat used for kWh output = 3010/8670 = 34.72 %. Unaccounted loss = 100 - (34.72 + 28 + 32) = 5.28 %.
Per litre of diesel: input = 10,200 kcal/kg x 0.85 kg/L = 8,670 kcal/L. Useful electrical output = 3.5 kWh x 860 = 3,010 kcal/L, i.e. 3,010/8,670 = 34.7% — that is the DG's thermal efficiency. Cooling water 28% + exhaust 32% = 60%. Unaccounted (radiation, lube oil, generator loss) = 100 - 34.7 - 60 = 5.3%. The two traps: forgetting the specific gravity when converting kcal/kg to kcal/litre, and forgetting to convert the kWh output to kcal with the 860 factor.
Source: Aug 2013
📖 §5.6 Energy balance

52. A Diesel Generator performance trial gives specific generation of 3.5 kWh per liter of diesel. The cooling water loss and exhaust flue gas loss as percentage of fuel input are 29% and 31% respectively. The calorific value of diesel is 10,200 kcal/kg. The specific gravity of Diesel is 0.85. Calculate unaccounted loss as percentage of input energy.

Model answer: CV of Diesel = 10,200 kcal/kg. Heat in input diesel = 10,200 x 0.85 = 8670 kcal/litre. Heat in kWh energy output = 3.5 x 860 = 3010 kcal/litre. % of heat used for kWh output = 3010/8670 = 34.72 %. Unaccounted loss = 100 - (34.72 + 29 + 31) = 5.28 %.
Input = 10,200 x 0.85 = 8,670 kcal/L; output = 3.5 x 860 = 3,010 kcal/L = 34.7%. Cooling 29% + exhaust 31% = 60%, so unaccounted = 100 - 34.7 - 60 = 5.3%. Note the answer is the same as the 28/32 variant because only the split between the two named losses changed — a useful check that you have set the balance up correctly.
Source: Aug 2013
📖 §5.6 Energy balance

53. A gas fired water heater heats water flowing at a rate of 20 litres per minute from 25 0 C to 85oC. If the GCV of the gas is 9200 kcal/kg, what is the rate of combustion of gas in kg/min (assume efficiency of water heater as 82%)

Model answer: Volume of water heated = 20 liters/min Mass of water heated = 20 Kg/min Heat supplied by gas * efficiency = Heat required by water. … 1 mark Mass of gas Kg/min * 9200 * 0.82 = 20 Kg/min * 1 kcal/Kg/oC)* (85-25)oC … 1 mark Mass of gas Kg/min = (20*1*60)/ (9200*0.82) = 0.159 Kg/ min. …. 3 marks
Water side: 20 L/min = 20 kg/min; Q = 20 x 1 x (85-25) = 1,200 kcal/min. Gas = Q / (GCV x efficiency) = 1,200 / (9,200 x 0.82) = 0.159 kg/min. Divide by the efficiency — multiplying by 0.82 is the standard sign error, and it makes the burner look better than it is. Write the balance as 'gas x GCV x efficiency = water heat duty' and the algebra takes care of itself.
Source: Sep 2015
📖 §5.5 Material balance

54. In a process plant, an evaporator concentrates a liquor containing solids of 6% by w/w (weight by weight) to produce an output containing 30% solids w/w. calculate the evaporation of water per 500 kgs of feed to the evaporator.

Model answer: Inlet solid contents = 6 % Output solid contents = 30% Feed = 500 kgs Inlet solid content in kg in feed = 500 x 0.06 = 30 kg …… 1 mark Outlet solid content in kg = 30 kg …… 1 mark Quantity of water evaporated = [500 – {(30 / 30) x 100}] = 400 kg. …… 3 marks
Solids balance: 500 x 0.06 = 30 kg of solids, and they leave in an output that is 30% solids, so output = 30/0.30 = 100 kg. Water evaporated = 500 - 100 = 400 kg per 500 kg of feed. General shortcut for evaporators: output = feed x (inlet solids %) / (outlet solids %), so raising the concentration five-fold cuts the mass to a fifth.
Source: Sep 2015
📖 §5.6 Energy balance

55. In a chemical factory where dyes are made, wet cake at 30 OC consisting of 60% moisture is put in a dryer to obtain an output having only 5% moisture, at atmospheric pressure. In each batch about 120 kgs of material is dried. a. The quantity of moisture removed per batch. b. What is the total quantity ( sensible & latent) of heat required to evaporate the moisture, if the latent heat of water is 540 kcal/kg at atmospheric conditions, Ignore heat absorbed by the solids c. Find the quantity of steam required for the drying process (per batch), if steam at 4 kg/cm2 is used for generating hot air in the dryer and the dryer efficiency is 80%. Latent heat of steam at 4 kg/cm2 is 520 kcal/kg.

Model answer: Given that  Qty of material dried per batch - 120 Kgs  Moisture at inlet - 60% a. The quantity of moisture removed per batch.  Water quantity in a wet batch - 120 x 0.6 = 72 Kgs.  Quantity of bone dry material - 120 – 72 = 48 Kgs.  Moisture at outlet - 5%  Total weight of dry batch output - 48/0.95 = 50.5 Kgs.  Equivalent water in a dry batch - 50.5 - 48 = 2.5 Kgs.  Total water removed in drying - 72 – 2.5 = 69.5 Kgs./batch …………………….1.5 marks b. The total quantity of heat required to evaporate the moisture. To evaporate the moisture at atmospheric pressure, the material has to be first heated up to 100 OC. The total heat required would be; Sensible heat - 72 x 1 x (100 – 30) = 5040 Kcal/batch Latent heat - 69.5 x 540 = 37530 Kcal/batch Total heat required - 5040 + 37530 = 42570 Kcal/batch …………………….2 marks c. The quantity of steam required for the drying process Dryer Efficiency - 80% Heat input to dryer - 42570/0.8 = 53212.50 Kcal/batch Latent heat in 4 Kg/cm2 steam - 520 Kcal/Kg Steam quantity required - 53212.50 / 520 = 102.3 Kgs / batch …………………….1.5 marks
(a) Solids = 120 x 0.40 = 48 kg, conserved; product = 48/0.95 = 50.53 kg; moisture removed = 120 - 50.53 = 69.47 kg. (b) The heat has TWO parts: sensible, to take that water from 30 deg C to 100 deg C = 69.47 x 1 x 70 = 4,863 kcal, plus latent, to evaporate it = 69.47 x 540 = 37,514 kcal; total about 42,377 kcal per batch. Marks are routinely lost by giving only the latent term — the question says 'sensible AND latent' for exactly that reason.
Source: Sep 2017
📖 §5.6 Energy balance

56. In a heat treatment shop, steel components are heat-treated in batches of 80 Tons. The heat treatment cycle is as follows;  Increase temperature from 30 OC to 850 OC in 3 hours.  Maintain 850 OC for 1 hour (soaking time).  Cool the material to 60 OC in 4 hours. a) Calculate the efficiency of the furnace, if the specific heat of steel is 0.12 kcal/kg OC and fuel oil consumption per batch is 1400 litres.  GCV of fuel oil - 10200 kcal/kg,  Cost of fuel oil - Rs. 46,000/kL,  Sp. gr. of fuel oil - 0.92. b) Due to high cost of oil, the plant management decides to convert to a lower operating cost LPG fired furnace lined on the inside with ceramic fibre insulation and with an operating efficiency of 80%, for same requirement. The investment towards installation of the new furnace is Rs. 50 lakhs. Calculate the Return on Investment, if the plant operates two batches per day and 250 days in a year.  Cost of LPG - Rs. 75/kg,  GCV of LPG - 12500 kcal/kg.

Model answer: Quantity of steel treated per batch - 80 Tons a. Efficiency of Furnace: Useful heat supplied to steel - 80000 x 0.12 x (850 – 30) = 7872000 kcal/batch …………………….1 mark Total heat supplied by fuel - 1400 x 0.92 x 10200 = 13137600 kcal/batch Efficiency of Furnace - 7872000/12067824 = 59.9% …………………….1 mark b. Return on Investment (RoI): Cost of operating fuel oil furnace - 1400 x 46 = Rs. 64400/batch Efficiency of new LPG furnace - 80% Heat supplied in new LPG furnace - 7872000/0.8 = 9840000 kcal/batch Equivalent LPG consumption - 9840000/12500 = 787.2 kg/batch …………………….1 mark Cost of operating LPG Furnace - 787.2 x 75 =Rs. 59040/batch Cost saving per batch - 64400 – 59040 =Rs. 5360/- Annual cost saving - 5360 x 2 x 250 =Rs. 26,80,000/- …………………….1 mark Investment for new furnace - Rs. 50 Lakhs Return on Investment (RoI) - (26.8/50)*100 = 53.6% …………………….1 mark
Useful heat to the steel = 80,000 kg x 0.12 x (850-30) = 7,872,000 kcal per batch. Heat input = 1,400 litres x density x 10,200 kcal/kg (convert litres to kg with the specific gravity given in the paper — omitting it is the standard error). Efficiency = useful/input x 100. Note the soaking hour and the cooling leg add nothing to the USEFUL heat: only the 30 to 850 deg C rise is charged to the furnace, and the heat given up during cooling is a heat-recovery opportunity, not an input.
Source: Sep 2017
📖 §5.5 Material balance

57. In a 100 TPD Sponge Iron plant, the sponge iron is fed to the Induction melting furnace, producing molten steel at 88% yield. The Energy consumption details are as follows: Coal Consumption : 130 TPD GCV of coal : 4500 kcal/kg Power Purchased from Grid : 82400 kWh / day Specific Energy consumption for Kiln producing Sponge Iron: 120 kWh / ton sponge iron 82400 kWh/day from Grid Factory Boundary Electricity for Induction Melting 130 TPD Coal Furnace 4500 kcals/kg 100 TPD Sponge Induction Melting Iron Ore Iron Molten Sponge Iron Kiln Furnace steel Grid Electricity Yield: 88% 120 kWh/t of Sponge iron Calculate the following 1. Specific Energy Consumption of Induction melting furnace in terms of kWh/ton of molten steel 2. Specific Energy Consumption of the entire plant, in terms of kcal/kg of molten steel (product). 3. Total Energy Consumption of Plant in Tons of Oil Equivalent (TOE )

Model answer: a) Specific Energy Consumption of Induction Melting Furnace Molten Steel Production from the Induction melting furnace per day = 100 x 88/100 = 88 TPD Total Energy Consumption of the Plant = 82400 kWh Electrical Energy Consumption in Sponge Iron Making = 120 x 100 = 12000 kWh per day Electrical Energy Consumption in Induction Melting Furnace = 82400-12000 = 70400 kWh/day …………………….1 mark Specific Energy Consumption of Induction Melting Furnace= 70400 / 88 = 800 kWh/ton of molten steel …………………….1 mark b)Total Energy Consumption of the Plant: (82400x860) + (130x1000x4500) = (70864000+585000000) = 655864000 kcal/day …………………….1 mark Specific Energy Consumption in terms of kcal/kg of Molten metal =655864000/88000 =7453 kcal/kg of molten metal …………………….1 mark c) Total Energy consumption of Plant in ToE = 655864000/107 = 65.586 ToE …………………….1 mark
Split the plant into two boxes at the factory boundary. Kiln: 100 TPD of sponge iron at 120 kWh/t = 12,000 kWh/day, plus the coal 130 t x 4,500 kcal/kg = 5.85 x 10^8 kcal/day. Induction furnace: molten steel = 100 x 0.88 = 88 TPD, and its electricity = total grid 82,400 - kiln 12,000 = 70,400 kWh/day, so SEC = 70,400/88 = 800 kWh per tonne of molten steel. Draw the boundary box before calculating — subtracting the kiln's share from the total purchase is the step the question is really testing.
Source: Sep 2017
📖 §5.5 Material balance

58. In a 100 TPD Sponge Iron plant, the sponge iron is fed to the Induction melting furnace, producing molten steel at 86% yield. The Energy consumption details are as follows: Coal Consumption : 130 TPD GCV of coal : 4500 kcal/kg Power Purchased from Grid : 82400 kWh / day Specific Energy consumption for Kiln producing Sponge Iron: 120 kWh / ton sponge iron 82400 kWh/day from Grid Factory Boundary Electricity for Induction Melting 130 TPD Coal Furnace 4500 kcals/kg 100 TPD Sponge Induction Melting Iron Ore Iron Molten Sponge Iron Kiln Furnace steel Grid Electricity Yield: 86% 120 kWh/t of Sponge iron Calculate the following 1. Specific Energy Consumption of Induction melting furnace in terms of kWh/ton of molten steel. 2. Specific Energy Consumption of the entire plant, in terms of kcal/kg of molten steel (product). 3. Total Energy Consumption of Plant in Tons of Oil Equivalent (TOE ).

Model answer: a) Specific Energy Consumption of Induction Melting Furnace Molten Steel Production from the Induction melting furnace per day = 100 x 86/100 = 86 TPD Total Energy Consumption of the Plant = 82400 kWh Electrical Energy Consumption in Sponge Iron Making = 120 x 100 = 12000 kWh per day Electrical Energy Consumption in Induction Melting Furnace = 82400-12000 = 70400 kWh/day …………………….1 mark Specific Energy Consumption of Induction Melting Furnace= 70400/86 = 818.6 kWh/ton of molten steel …………………….1 mark b)Total Energy Consumption of the Plant: = (82400x860) + (130x1000x4500) = (70864000+585000000) = 655864000 kcal/day …………………….1 mark Specific Energy Consumption in terms of kcal/kg of Molten metal =655864000/86000 =7626.3 kcal/kg of molten metal …………………….1 mark c) Total Energy consumption of Plant in ToE = 655864000/107 = 65.586 ToE …………………….1 mark
Kiln electricity = 100 TPD x 120 kWh/t = 12,000 kWh/day, so the induction furnace gets 82,400 - 12,000 = 70,400 kWh/day. Molten steel at 86% yield = 86 TPD, giving an SEC of 70,400/86 = 819 kWh per tonne of molten steel. Compare with the 88% variant (800 kWh/t) and the lesson is visible: a two-point drop in yield costs about 19 kWh per tonne, because the same energy is spread over less saleable product.
Source: Sep 2017
📖 §5.6 Energy balance

59. In a chemical factory where dyes are made, wet cake at 30 OC consisting of 60% moisture is put in a dryer to obtain an output having only 8% moisture, at atmospheric pressure. In each batch about 120 kgs of material is dried. a. The quantity of moisture removed per batch. b. What is the total quantity (sensible & latent) of heat required to evaporate the moisture, if the latent heat of water is 540 kcal/kg at atmospheric conditions, Ignore heat absorbed by the solids c. Find the quantity of steam required for the drying process (per batch), if steam at 4 kg/cm2 is used for generating hot air in the dryer and the dryer efficiency is 70%. Latent heat of steam at 4 kg/cm2 is 520 Kcal/kg.

Model answer: Given that  Qty of material dried per batch - 120 kgs  Moisture at inlet - 60% a. The quantity of moisture removed per batch.  Water quantity in a wet batch - 120 x 0.6 = 72 kgs.  Quantity of bone dry material - 120 – 72 = 48 kgs.  Moisture at outlet - 8%  Total weight of dry batch output - 48/0.92 = 52.2 kgs.  Equivalent water in a dry batch - 52.2 - 48 = 4.2 kgs.  Total water removed in drying - 72 – 4.2 = 67.8 kgs./batch …………………….1.5 marks b. The total quantity of heat required to evaporate the moisture. To evaporate the moisture at atmospheric pressure, the material has to be first heated up to 100 OC. The total heat required would be; Sensible heat - 72 x 1 x (100 – 30) = 5040 kcal/batch Latent heat - 67.8 x 540 = 36612 kcal/batch Total heat required - 5040 + 36612 = 41652 kcal/batch …………………….2 marks c. The quantity of steam required for the drying process Dryer Efficiency - 70% Heat input to dryer - 41652/0.7 = 59502.86 kcal/batch Latent heat in 4 Kg/cm2 steam - 520 kcal/kg Steam quantity required - 59502.86 / 520 = 114.4 kgs / batch …………………….1.5 marks
(a) Solids = 120 x 0.40 = 48 kg; product = 48/0.92 = 52.17 kg; moisture removed = 67.83 kg. (b) Sensible heat 30 to 100 deg C = 67.83 x 1 x 70 = 4,748 kcal, plus latent = 67.83 x 540 = 36,628 kcal, total about 41,376 kcal. Note how little the answer moves between the 5% and 8% final-moisture versions — the mass of water is dominated by the 60% inlet moisture, which is why drying the last few percent is disproportionately expensive per kilogram of product.
Source: Sep 2017
📖 §5.6 Energy balance

60. In a heat treatment shop, steel components are heat-treated in batches of 80 Tons. The heat treatment cycle is as follows;  Increase temperature from 30 OC to 850 OC in 3 hours.  Maintain 850 OC for 1 hour (soaking time).  Cool the material to 60 OC in 4 hours. a) Calculate the efficiency of the furnace, if the specific heat of steel is 0.12 kcal/kg OC and fuel oil consumption per batch is 1400 litres. GCV of fuel oil - 10200 kcal/kg, Cost of fuel oil - Rs. 46,000/kL, Sp. gr. of fuel oil - 0.92. b) Due to high cost of oil, the plant management decides to convert to a lower operating cost LPG fired furnace lined on the inside with ceramic fibre insulation and with an operating efficiency of 75%, for same requirement. The investment towards installation of the new furnace is Rs. 50 lakhs. Calculate the Return on Investment, if the plant operates two batches per day and 270 days in a year. Cost of LPG - Rs. 75/kg, GCV of LPG - 12500 kcal/kg.

Model answer: Quantity of steel treated per batch - 80 Tons a. Efficiency of Furnace: Useful heat supplied to steel - 80000 x 0.12 x (850 – 30) = 7872000 kcal/batch …………………….1 mark Total heat supplied by fuel - 1400 x 0.92 x 10200 = 13137600 kcal/batch Efficiency of Furnace - 7872000/12067824 = 59.9% …………………….1 mark b. Return on Investment (RoI): Cost of operating fuel oil furnace - 1400 x 46 = Rs. 64400/batch Efficiency of new LPG furnace - 75% Heat supplied in new LPG furnace - 7872000/0.75 = 10496000 kcal/batch Equivalent LPG consumption - 10496000/12500 = 839.68 kg/batch …………………….1 mark Cost of operating LPG Furnace - 839.68 x 75 =Rs. 62976/batch Cost saving per batch - 64400 – 62976 =Rs. 1424/- Annual cost saving - 1424 x 2 x 270 = Rs. 768960/- …………………….1 mark Investment for new furnace - Rs. 50 Lakhs Return on Investment (RoI) - (7.69/50)*100 = 15.38% …….…….
Useful heat = 80,000 x 0.12 x (850-30) = 7,872,000 kcal per batch; input = 1,400 L x specific gravity x 10,200 kcal/kg; efficiency = useful/input. Only the heating leg counts as useful output; the soaking period covers standing losses and the controlled cooling from 850 to 60 deg C is a recoverable heat stream, not an input. Convert the fuel volume to mass before applying the GCV.
Source: Sep 2017
📖 §5.6 Energy balance

61. A paint drier requires 75.4 m3/min of air at 93°C, which is heated in a steam-coil unit. How many kg of steam at 4 bar does this unit require per hour? The density of air is 1.2 kg/m3 and specific heat of air is 0.24 kcal/kg°C. The ambient temperature is 32°C. Steam table data: Pressure 4 bar, Temperature 143 °C, Enthalpy of water 143 kcal/kg, Enthalpy of evaporation 510 kcal/kg, Enthalpy of steam 653 kcal/kg.

Model answer: Air flow rate = 75.4 m3/min x 60 = 4524 m3/hr Mass flow rate of air = 4524 x 1.2 = 5428.8 kg/hr Sensible heat of air = m x Cp x dT = 5428.8 x 0.24 x (93 - 32) = 79477.6 kcal/hr Latent heat of steam at 4 bar = 510 kcal/kg Steam required = 79477.6 / 510 = 156 kg/hr
Air: 75.4 m3/min x 60 = 4,524 m3/h, x 1.2 kg/m3 = 5,428.8 kg/h. Heat = 5,428.8 x 0.24 x (93-32) = 79,477 kcal/h. Steam = heat / LATENT heat = 79,477/510 = 156 kg/h. The steam table gives three numbers and only one is right: use the enthalpy of EVAPORATION (510), because a steam coil condenses the steam and returns the condensate at 143 deg C carrying its 143 kcal/kg back. Dividing by the 653 total enthalpy is the standard error.
Source: Sep 2018
📖 §5.5 Material balance

62. A continuous centrifuge separates 36,000 kg of whole milk containing 4% fat in 6-hour period into skim milk with 0.40% fat and cream with 40% fat. Find out the flow rates of whole milk, cream and skim milk using mass balance.

Model answer: MASS IN: Total mass flow of whole milk = 36000/6 = 6000 kg per hour Fat per hour = 6000 x 0.04 = 240 kg/hr Therefore water plus solids other than fat = (6000 - 240) = 5760 kg per hr MASS OUT: Let the mass of cream be X kg/hr; its total fat content is 0.40X. The mass of skim milk is (6000 - X) and its total fat content is 0.0040 (6000 - X). Material balance on fat: Fat in = Fat out 6000 x 0.04 = 0.0040 (6000 - X) + 0.40X Solving, X = 545 kg/hr So the flow of whole milk is 6000 kg/hr, the flow of cream is 545 kg/hr and the flow of skim milk is (6000 - 545) = 5455 kg/hr.
Total balance: 36,000/6 = 6,000 kg/h of whole milk, carrying 6,000 x 0.04 = 240 kg/h of fat. Let C = cream and S = skim, with C + S = 6,000 and 0.40C + 0.004S = 240. Substituting, 0.396C = 216, so C = 545.5 kg/h and S = 5,454.5 kg/h. Two balances — total mass and the KEY COMPONENT (fat) — is the standard method for any separator, and checking that 0.4 x 545.5 + 0.004 x 5454.5 = 240 confirms the arithmetic.
Source: Sep 2019
📖 §5.3 Basic principles of material and energy balance

63. A conveyor delivers coal with a width of 0.9 m and coal bed height of 0.15 m at a speed of 0.8 m/s. Determine the coal delivery in tons per hour considering the coal density as 1.1 ton/m3.

Model answer: Volume of coal delivered = Cross sectional area x Length travelled per second = 0.9 m x 0.15 m x 0.8 m/s = 0.108 m3/s = 0.108 x 3600 = 388.8 m3/hr Coal delivery rate = 388.8 m3/hr x 1.1 t/m3 = 427.7 tonnes/hr
Volume rate = cross-section x belt speed = 0.9 x 0.15 x 0.8 = 0.108 m3/s = 388.8 m3/h. Mass = 388.8 x 1.1 = 427.7 tonnes/h. Note the density is already in tonnes per cubic metre, so no further conversion is needed — the usual slip is treating 1.1 as kg/m3 or forgetting the 3,600 seconds. This is the standard field method for checking a coal feed rate against the weigh-feeder reading.
Source: Sep 2019
📖 §5.6 Energy balance

64. In a textile industry, 25,000 kg/hr water is currently being heated from 28 °C to 80 °C by indirect heating of steam in dyeing machines. It is proposed to recover heat from the hot effluent and generate hot water at 45 °C which would be further raised to 80 °C by steam. Estimate the reduction in steam in kg/hr considering the latent heat of steam as 520 kcal/kg in both the cases.

Model answer: WITHOUT HEAT RECOVERY: Heating required (Q1) = m x Cp x dT = 25000 x 1 x (80 - 28) = 13,00,000 kcal/hr Steam required = 13,00,000 / 520 = 2500 kg/hr AFTER HEAT RECOVERY: Heating required (Q2) = 25000 x 1 x (80 - 45) = 8,75,000 kcal/hr Steam required = 8,75,000 / 520 = 1682.7 kg/hr Reduction in steam required = 2500 - 1682.7 = 817.3 kg/hr
Without recovery: Q = 25,000 x 1 x (80-28) = 1,300,000 kcal/h, steam = 1,300,000/520 = 2,500 kg/h. With recovery the effluent lifts the water to 45 deg C, so steam covers only 80-45 = 35 deg C: Q = 875,000 kcal/h, steam = 1,682.7 kg/h. Reduction = 817.3 kg/h, i.e. 32.7%. The saving is proportional to the share of the temperature rise that the recovered heat covers — 17 of 52 degrees — which is the quick sanity check.
Source: Sep 2019
📖 §5.5 Material balance

65. A sample of coal is found to contain 64% carbon and 24% ash. The refuse obtained at the end of combustion is analyzed and found to contain 8% carbon and the rest is ash. Compute the percentage of the original carbon unburnt in the refuse. (5 Marks)

Model answer: Coal: carbon = 64%, ash = 24%. Refuse: carbon = 8%, ash = 92%. Basis: 100 kg of coal. The key is that the ash is inert - the same quantity of ash appears in the refuse as was in the coal. Mass of carbon in coal = 64 kg; mass of ash in coal = 24 kg = mass of ash in refuse. Mass of refuse = 24 x (100/92) = 26.087 kg. Mass of carbon in refuse = 26.087 x (8/100) = 2.087 kg. Percentage of the original carbon remaining unburnt in the refuse = (2.087 / 64) x 100 = 3.26%.
Basis 100 kg of coal, and anchor on the INERT ASH: 24 kg of ash leaves in refuse that is (100-8) = 92% ash, so refuse = 24/0.92 = 26.09 kg, of which 8% is carbon = 2.09 kg. Unburnt fraction of the original carbon = 2.09/64 x 100 = 3.26%. Do not compare the 8% in the refuse against the 64% in the coal directly — they are percentages of very different masses, and the ash tie-component is what converts between them.
Source: Jul 2022
📖 §5.5 Material balance

66. A continuous centrifuge separates 36,000 kg of whole milk containing 4% fat in a 6-hour period into skim milk with 0.40% fat and cream with 40% fat. Find out the flow rates of whole milk, cream and skim milk using mass balance. (5 Marks)

Model answer: MASS INLET Total mass flow of whole milk = 36,000/6 = 6,000 kg/hr Fat entering = 6,000 x 0.04 = 240 kg/hr Water plus non-fat solids = 6,000 - 240 = 5,760 kg/hr MASS OUTLET Let the mass of cream be X kg/hr; its fat content = 0.40X. Skim milk = (6,000 - X) kg/hr; its fat content = 0.0040 x (6,000 - X). FAT BALANCE (fat in = fat out) 6,000 x 0.04 = 0.0040 (6,000 - X) + 0.40X 240 = 24 - 0.004X + 0.40X 216 = 0.396X, so X = 545 kg/hr Result: whole milk = 6,000 kg/hr, cream = 545 kg/hr, skim milk = 6,000 - 545 = 5,455 kg/hr.
Whole milk = 36,000/6 = 6,000 kg/h with 240 kg/h of fat. Total balance C + S = 6,000; fat balance 0.40C + 0.004S = 240. Substituting S = 6,000 - C gives 0.396C = 216, so cream = 545.5 kg/h and skim milk = 5,454.5 kg/h. Present it as two labelled balances — total mass and key component — because that structure is where the marks sit, and check by recomputing the fat.
Source: Sep 2025
📖 §5.6 Energy balance

67. A food dryer processes 1,000 kg/hr of wet material with an initial moisture content of 55% (wet basis) and dries it to a final moisture content of 10% (wet basis). Steam Flow: 2,500 kg/hr at 3.5 bar (enthalpy = 660 kcal/kg); Latent heat of water vaporization = 540 kcal/kg; Specific heat of dry material = 0.45 kcal/kg deg C; Drying temperature rise = 60 deg C; Ignore heat loss and assume 100% steam use for moisture removal and solid heating. (Each 1 Mark) a) Calculate the mass of bone-dry solid in the feed b) Calculate the mass of water removed per hour c) Estimate the energy required to evaporate the moisture d) Estimate the energy required to heat the dry solids e) Calculate the total energy input from steam

Model answer: a) Dry matter fraction = 1 - 0.55 = 0.45. Bone-dry solid = 1000 x 0.45 = 450 kg/hr. b) Initial water = 1000 - 450 = 550 kg/hr. At 10% final moisture (wet basis) the dry solid is 90% of the product: 0.90M = 450, so final product M = 500 kg/hr. Final water = 500 - 450 = 50 kg/hr. Water removed = 550 - 50 = 500 kg/hr. c) Q(evaporation) = water removed x latent heat = 500 x 540 = 2,70,000 kcal/hr. d) Q(solids) = mass of dry solid x specific heat x temperature rise = 450 x 0.45 x 60 = 12,150 kcal/hr. e) Total energy demand = 2,70,000 + 12,150 = 2,82,150 kcal/hr. Energy available from steam = 2,500 x 660 = 16,50,000 kcal/hr. The steam supply (16,50,000 kcal/hr) far exceeds the drying demand (2,82,150 kcal/hr), indicating substantial scope to reduce steam consumption.
(a) Bone-dry solids = 1,000 x 0.45 = 450 kg/h. (b) Initial water = 550 kg/h; at 10% final moisture the product = 450/0.90 = 500 kg/h, so water removed = 1,000 - 500 = 500 kg/h and 50 kg/h of water remains in the product. (c) Heat = latent 500 x 540 = 270,000 kcal/h plus sensible heating of the solids 450 x 0.45 x 60 = 12,150 kcal/h, total about 282,150 kcal/h. (d) Steam actually needed = 282,150/660 = about 428 kg/h against 2,500 kg/h supplied, so the dryer is heavily over-steamed — the audit finding. Do not forget the solids-heating term: it is small here but it is where the marks separate.
Source: Sep 2025

Long questions (10 marks) — 32

📖 §5.8 Solved Example — evaporator (book worked example)

1. An evaporator is fed with 10,000 kg/hr of a solution having 1% solids. The feed is at 38 degC and is to be concentrated to 2% solids. Steam enters at a total enthalpy of 640 kcal/kg and the condensate leaves at 100 degC. Enthalpy of feed = 38.1 kcal/kg, enthalpy of product (thick liquor) = 100.8 kcal/kg and enthalpy of vapour = 640 kcal/kg. Find (i) the mass of vapour formed per hour and (ii) the mass of steam used per hour.

Model answer: STEP 1 - Mass (solids) balance to get product and vapour. Solids in feed = 10,000 x 1/100 = 100 kg/hr (solids are conserved). Product (thick liquor) is 2% solids: Product x 2/100 = 100 -> Product = 100/0.02 = 5000 kg/hr. Vapour formed = Feed - Product = 10,000 - 5000 = 5000 kg/hr. STEP 2 - Heat (enthalpy) balance to get steam. Heat in with feed = 10,000 x 38.1 = 3,81,000 kcal/hr. Heat out in thick liquor = 5000 x 100.8 = 5,04,000 kcal/hr. Heat out in vapour = 5000 x 640 = 32,00,000 kcal/hr. Steam gives up (640 - 100) = 540 kcal/kg (enthalpy of steam minus condensate at 100 degC). Balance: Heat by steam + Heat in feed = Heat in vapour + Heat in thick liquor M x 540 + 3,81,000 = 32,00,000 + 5,04,000 M x 540 = 37,04,000 - 3,81,000 = 33,23,000 M (steam) = 33,23,000 / 540 = 6153.7 kg/hr. ANSWER: Vapour formed = 5000 kg/hr; Steam used = 6153.7 kg/hr.
Canonical two-part evaporator problem. Part (a) is a pure solids balance (solids unchanged, water leaves as vapour). Part (b) is an enthalpy balance where steam contributes latent heat = h_steam - h_condensate = 640 - 100 = 540 kcal/kg.
Source: Year not recorded
📖 §5.7 Example 5.12 — paper machine (book worked example)

2. Production rate from a paper machine is 340 tonnes per day (TPD). Inlet and outlet dryness to the paper machine are 40% and 95% respectively. Evaporated moisture temperature is 80 degC. To evaporate the moisture, steam is supplied at 35 kg/cm2 (latent heat = 513 kcal/kg). Assume 24 hours/day operation and enthalpy of evaporated moisture = 632 kcal/kg. Estimate (a) the quantity of moisture to be evaporated per hour and (b) the input steam quantity required for evaporation per hour.

Model answer: Production = 340 TPD = 340/24 = 14.16 TPH (tonnes per hour) of paper. STEP 1 - Moisture to be evaporated (solids balance; bone-dry paper is conserved). Bone-dry paper in final product (95% dry) = 14.16 x 0.95 = 13.45 TPH. Weight of moisture BEFORE dryer (inlet 40% dry means 60% moisture on the dry solids): = [(100-40)/40] x 13.45 = (60/40) x 13.45 = 20.175 TPH. Weight of moisture AFTER dryer (outlet 95% dry -> 5% moisture) = 14.16 x 0.05 = 0.707 TPH. Evaporated moisture = 20.175 - 0.707 = 19.468 TPH = 19,468 kg/hr. STEP 2 - Steam required (heat balance). Heat to be carried away in the moisture (sensible + latent) = 632 x 19,468 = 1,23,03,776 kcal/hr. This heat is supplied by the latent heat of the steam (513 kcal/kg): Steam required = 1,23,03,776 / 513 = 23,984 kg/hr = 23.98 MT/hr. ANSWER: Moisture evaporated = 19.468 TPH (19,468 kg/hr); Steam required = 23,984 kg/hr (~24 MT/hr).
Drying + steam heat balance. Track bone-dry paper (unchanged). Moisture before/after found from dryness %, evaporated = difference. Steam = heat in evaporated moisture / latent heat of steam.
Source: Year not recorded
📖 §5 EOC Long Q L-1 — wet scrubber make-up water balance

3. A wet scrubber removes fine dust from an inlet gas stream with a spray of water so that the outlet gas meets emission standards. Stream 1 (recirculation liquid back to the scrubber) = 4.54 m3/hr. The liquid withdrawn for treatment and disposal (stream 4) = 0.454 m3/hr. The inlet gas (stream 2) is completely dry and the outlet gas (stream 6) carries 272.16 kg/hr of moisture evaporated in the scrubber. Make-up water is added as stream 5. How much make-up water must be continually added to keep the unit running?

Model answer: Apply an OVERALL WATER BALANCE across the scrubber system boundary: Water IN = Water OUT. The recirculation stream (stream 1 = 4.54 m3/hr) is internal to the loop and cancels out - it does not cross the system boundary. Water LEAVES the system by two routes: (a) Liquid withdrawn for treatment/disposal (stream 4) = 0.454 m3/hr = 454 kg/hr (1 m3 water = 1000 kg). (b) Moisture evaporated and carried away with the outlet gas (stream 6) = 272.16 kg/hr. Total water leaving = 454 + 272.16 = 726.16 kg/hr. Water ENTERS the system only as make-up water (stream 5). By balance: Make-up water (stream 5) = Water leaving = 726.16 kg/hr ~ 0.726 m3/hr. ANSWER: Make-up water to be added continually = 726.16 kg/hr (approximately 0.73 m3/hr).
Classic make-up water balance. Key insight: ignore the internal recirculation loop; balance only what crosses the boundary. Make-up replaces water lost to disposal plus water evaporated into the gas.
Source: Year not recorded
📖 §5 EOC Long Q L-2 — autoclave cooling-water requirement

4. An autoclave contains 1000 cans of pea soup heated to an overall temperature of 100 degC. The cans are to be cooled to 40 degC before leaving the autoclave. How much cooling water is required if it enters at 15 degC and leaves at 35 degC? The specific heats of the pea soup and the can metal are 4.1 and 0.50 kJ/kg degC respectively. Each can weighs 60 g and contains 0.45 kg of pea soup. The heat content of the autoclave walls above 40 degC is 1.6 x 10^4 kJ, and there is no heat loss through the walls.

Model answer: HEAT TO BE REMOVED (cooling all hot components from 100 to 40 degC, delta_T = 60 degC). Use Q = m x Cp x delta_T for each component: 1) Pea soup: m = 1000 x 0.45 = 450 kg; Q = 450 x 4.1 x 60 = 1,10,700 kJ. 2) Can metal: m = 1000 x 0.060 = 60 kg; Q = 60 x 0.50 x 60 = 1,800 kJ. 3) Autoclave walls (heat content above 40 degC, given) = 16,000 kJ. Total heat to be removed = 1,10,700 + 1,800 + 16,000 = 1,28,500 kJ. HEAT ABSORBED BY COOLING WATER (15 -> 35 degC, delta_T = 20 degC; Cp water = 4.186 kJ/kg degC). Heat gained = m_water x 4.186 x 20 = 83.72 x m_water kJ. ENERGY BALANCE: Heat removed = Heat gained by water 1,28,500 = 83.72 x m_water m_water = 1,28,500 / 83.72 = 1535 kg. ANSWER: Cooling water required ~ 1535 kg (about 1530-1540 kg depending on the Cp of water used).
Multi-component sensible-heat balance plus stored wall heat. Total heat released by soup + cans + walls equals heat picked up by cooling water (m Cp delta_T).
Source: Year not recorded
📖 §5.5 Example 5.9 — bag filter dust balance (book worked example)

5. A bag filter is used to remove dust from a gas stream in a cement plant. Inlet gas to the bag filter is 1,69,920 m3/hr with a dust loading of 4577 mg/m3. Outlet gas from the bag filter is 1,85,040 m3/hr with a dust loading of 57 mg/m3. What is the maximum quantity of ash that will have to be removed per hour from the bag filter hopper?

Model answer: Apply a DUST (mass) BALANCE: Mass in = Mass out. Inlet gas dust = Outlet gas dust + Hopper ash. STEP 1 - Inlet dust quantity. = 1,69,920 m3/hr x 4577 mg/m3 x (1/1,000,000) kg/mg = 7,77,72,... /1,000,000 = 777.7 kg/hr. STEP 2 - Outlet dust quantity. = 1,85,040 m3/hr x 57 mg/m3 x (1/1,000,000) kg/mg = 10.6 kg/hr. STEP 3 - Hopper ash (removed). Hopper ash = Inlet dust - Outlet dust = 777.7 - 10.6 = 767.1 kg/hr. ANSWER: Ash removed from the hopper = 767.1 kg/hr.
Straight dust mass balance: convert loading (mg/m3) x flow (m3/hr) to kg/hr for inlet and outlet; collected ash = inlet - outlet.
Source: Year not recorded
📖 §5.7 Example 5.10 — furnace shell cooling water (book worked example)

6. A furnace shell has to be cooled from 90 degC to 55 degC. The mass of the furnace shell is 2 tonnes and its specific heat is 0.2 kcal/kg degC. Water is available at 28 degC and the maximum allowed increase in water temperature is 5 degC. Calculate the quantity of water required to cool the furnace. Neglect heat loss.

Model answer: ENERGY STREAM 1 - Heat to be removed from the furnace shell. m = 2 tonnes = 2000 kg; Cp = 0.2 kcal/kg degC; T1 = 90 degC, T2 = 55 degC. Q = m x Cp x (T1 - T2) = 2000 x 0.2 x (90 - 55) = 2000 x 0.2 x 35 = 14,000 kcal. ENERGY STREAM 2 - Heat picked up by cooling water. Cp water = 1 kcal/kg degC; inlet 28 degC, outlet (max) 33 degC, so delta_T = 5 degC. Heat removed by water = X x 1 x (33 - 28) = 5X kcal, where X = mass of water. ENERGY BALANCE: Stream 1 = Stream 2 14,000 = 5X X = 14,000 / 5 = 2800 kg. ANSWER: Quantity of cooling water required = 2800 kg.
Sensible-heat balance: heat lost by shell (m Cp delta_T) = heat gained by water (m Cp delta_T). Solve for water mass.
Source: Year not recorded
📖 §5.7 Example 5.11 — textile dryer thermal efficiency (book worked example)

7. A textile dryer consumes 4 m3/hr of natural gas with a calorific value of 800 kJ/mole. The throughput of the dryer is 60 kg of wet cloth per hour, drying it from 55% moisture to 10% moisture. Estimate the overall thermal efficiency of the dryer, taking into account the latent heat of evaporation only. (1 mole of gas occupies 22.4 litres at STP; latent heat of evaporation of water = 2257 kJ/kg.)

Model answer: STEP 1 - Moisture removed (solids/moisture balance). Initial moisture in wet cloth = 60 x 0.55 = 33 kg. Bone-dry cloth = 60 x (1 - 0.55) = 27 kg (unchanged). Final product at 10% moisture: dry solids are 90% -> total final mass = 27/0.90 = 30 kg, so final moisture = 30 - 27 = 3 kg. Moisture removed per hour = 33 - 3 = 30 kg/hr. STEP 2 - Heat usefully used for drying (latent heat only). Q_used = 30 kg/hr x 2257 kJ/kg = 67,710 kJ/hr ~ 6.8 x 10^4 kJ/hr. STEP 3 - Heat available from combustion of the gas. Gas rate = 4 m3/hr = 4 x 1000 litres/hr; at STP 1 mole = 22.4 litres, so moles = 4000/22.4 = 179 moles/hr. Q_available = 179 x 800 = 1,43,200 kJ/hr ~ 14.3 x 10^4 kJ/hr. STEP 4 - Thermal efficiency. eta = Q_used / Q_available = 6.8 x 10^4 / 14.3 x 10^4 = 0.48 = 48%. ANSWER: Overall thermal efficiency of the dryer ~ 48%.
Energy balance on a dryer. Efficiency = heat used to evaporate moisture (m x latent heat) divided by heat available from fuel (moles/hr x CV per mole). Sensible heat neglected as latent heat dominates.
Source: Year not recorded
📖 §5.5 Example 5.4 — continuous centrifuge (book worked example)

8. In a continuous centrifuging of milk, 35,000 kg of whole milk containing 4% fat is to be separated in a 6 hour period into skim milk with 0.45% fat and cream with 45% fat. What is the flow rate of the two output streams from the continuous centrifuge?

Model answer: Work on a per-hour basis (steady state). BASIS: Total mass input per hour = 35,000 / 6 = 5833 kg/hr. Let Y = skim milk (kg/hr), Z = cream (kg/hr). EQ-1 (Total mass balance): 5833 = Y + Z -> Z = 5833 - Y. EQ-2 (Fat balance): 0.04 x 5833 = 0.0045 x Y + 0.45 x Z. Substitute Z from EQ-1 into EQ-2: 0.04 x 5833 = 0.0045 Y + 0.45 (5833 - Y) 233.3 = 0.0045 Y + 2624.85 - 0.45 Y 0.4455 Y = 2624.85 - 233.3 = 2391.55 Y = 2391.55 / 0.4455 = 5369 kg/hr (skim milk). Z = 5833 - 5369 = 464 kg/hr (cream). ANSWER: Skim milk = 5369 kg/hr; Cream = 464 kg/hr.
Continuous two-output split. Two equations: total mass balance and the tracked-component (fat) balance. Substitute and solve for the two output flow rates.
Source: Year not recorded
📖 §5.5 Example 5.3 — constituent balance (book worked example)

9. Skim milk is prepared by removing some fat from whole milk. The skim milk contains 90.5% water, 3.5% protein, 5.1% carbohydrate, 0.1% fat and 0.8% ash. If the original whole milk contained 4.5% fat, calculate the composition of the whole milk, assuming that only fat was removed and there are no processing losses.

Model answer: BASIS: 100 kg of skim milk (which therefore contains 0.1 kg fat). Let x = mass of fat removed to make the skim milk. Total original fat = (x + 0.1) kg. Total original (whole milk) mass = (100 + x) kg. The original fat content was 4.5%, so: (x + 0.1) / (100 + x) = 0.045 x + 0.1 = 0.045 (100 + x) = 4.5 + 0.045 x x - 0.045 x = 4.5 - 0.1 0.955 x = 4.4 -> x = 4.6 kg. So total whole milk = 100 + 4.6 = 104.6 kg. Compositions of the whole milk (each skim-milk component mass now divided by 104.6): - Fat = 4.5% (given / by construction). - Water = 90.5 / 104.6 = 86.5%. - Protein = 3.5 / 104.6 = 3.3%. - Carbohydrate = 5.1 / 104.6 = 4.9%. - Ash = 0.8 / 104.6 = 0.8%. ANSWER: Whole milk = 4.5% fat, 86.5% water, 3.3% protein, 4.9% carbohydrate, 0.8% ash (total 104.6 kg from 100 kg skim).
Constituent balance: fix a 100 kg basis on the known (skim) stream, track the single component that changed (fat), set up the % equation, solve for added/removed mass, then re-express all components on the new total mass.
Source: Year not recorded
📖 §5.5 Example 5.2 — multi-component balance (book worked example)

10. A solution which is 80% oil, 15% usable by-products and 5% impurities enters a refinery. One output is 92% oil and 6% usable by-products. The other output is 60% oil and flows at 1000 lit/hr (assume no accumulation, percentages by volume). Find (a) the flow rate of the input, (b) the percent composition of the 1000 lit/hr output, and (c) what percent of the original impurities are in the 1000 lit/hr output.

Model answer: BASIS: Input A = X lit/hr; Output B = Y lit/hr (92% oil, 6% UBP, 2% impurities); Output C = 1000 lit/hr (60% oil, V fraction UBP, W fraction impurities). Equations: EQ-1 Total: X = Y + 1000. EQ-2 Oil: 0.80 X = 0.92 Y + 0.60 x 1000. EQ-3 UBP: 0.15 X = 0.06 Y + V x 1000. EQ-4 Impurities: 0.05 X = 0.02 Y + W x 1000. (a) Solve EQ-1 and EQ-2. Substitute X = Y + 1000 into EQ-2: 0.80 (Y + 1000) = 0.92 Y + 600 0.80 Y + 800 = 0.92 Y + 600 200 = 0.12 Y -> Y = 1666.7 lit/hr. X = 1666.7 + 1000 = 2666.7 lit/hr = flow rate of input. (b) Composition of the 1000 lit/hr stream: UBP (EQ-3): V x 1000 = 0.15 x 2666.7 - 0.06 x 1666.7 = 400.0 - 100.0 = 300 -> V = 0.30 (30%). Impurities (EQ-4): W x 1000 = 0.05 x 2666.7 - 0.02 x 1666.7 = 133.3 - 33.3 = 100 -> W = 0.10 (10%). So the 1000 lit/hr output = 60% oil, 30% usable by-products, 10% impurities. (Check: 60+30+10 = 100.) (c) Impurities in input = 0.05 x 2666.7 = 133.3 lit/hr; impurities in the 1000 lit/hr stream = 100 lit/hr. Percent of original impurities in this stream = 100 / 133.3 x 100 = 75%. ANSWER: (a) Input = 2666.7 lit/hr; (b) 60% oil, 30% UBP, 10% impurities; (c) 75% of the original impurities.
Multi-component (three constituents) material balance with two outputs. Write total + component balances, solve the two-unknown pair first, then back out the unknown output fractions and the impurity ratio.
Source: Year not recorded
📖 §5.8 Sankey diagram + §5.6 energy balance — DG set application

11. Prepare the energy balance of a Diesel Generator and draw a Sankey diagram. Given: calorific value of diesel = 10,000 kcal/litre; average energy generated = 4.07 kWh/litre; alternator efficiency = 96%; stack (flue gas) losses = 33%; coolant losses = 24%; and the balance is radiation losses. (1 kWh = 860 kcal.)

Model answer: Take fuel input = 100% thermal energy (basis: 1 litre diesel = 10,000 kcal). STEP 1 - Electrical output %. Energy generated = 4.07 kWh/litre = 4.07 x 860 = 3500.2 kcal/litre. Electrical output = (3500.2 / 10,000) x 100 = 35%. STEP 2 - Alternator losses. Alternator efficiency = 96%, so alternator loss = 100 - 96 = 4% (of the mechanical energy fed to it; taken as ~4% of input for the Sankey split). STEP 3 - Given loss splits. Stack (flue gas) loss = 33%; Coolant loss = 24%. STEP 4 - Radiation (balance) loss. Radiation loss = 100 - (Electrical 35 + Alternator 4 + Stack 33 + Coolant 24) = 100 - 96 = 4%. ENERGY BALANCE (of 100% input): 35% electrical output + 4% alternator + 33% stack + 24% coolant + 4% radiation = 100%. SANKEY DIAGRAM: a single 100% input arrow (fuel energy) branches into a 35% useful electrical output arrow and loss arrows of 33% (stack/flue gas), 24% (coolant), 4% (alternator) and 4% (radiation), the arrow widths drawn proportional to each percentage. Total losses = 65%.
Energy balance expressed as % of fuel input. Electrical output from kWh x 860 / CV. Radiation is the closing (balance) term. Sankey arrow widths are proportional to the magnitude of each stream.
Source: Year not recorded
📖 §5.6 Energy balance — specific energy consumption; EOC Objective Q7 (kWh × 860)

12. A foundry has an induction furnace of 5 TPH with a specific energy consumption of 620 kWh/tonne of liquid metal. The yield of foundry castings is 60%. The castings are heat treated in an oil-fired furnace consuming 75 kg oil/tonne of castings. Find the energy consumption per tonne of finished product in terms of oil equivalent. (GCV of oil = 10,000 kcal/kg; 1 kWh = 860 kcal.)

Model answer: STEP 1 - Melting energy per tonne of finished (heat-treated) product. SEC of melting = 620 kWh per tonne of liquid metal. Yield of castings = 60%, so per tonne of castings: 620 / 0.60 = 1033.3 kWh/tonne. Convert to heat units: 1033.3 x 860 = 8,88,667 kcal/tonne. STEP 2 - Heat-treatment energy per tonne of castings. Oil used = 75 kg oil/tonne x 10,000 kcal/kg = 7,50,000 kcal/tonne. STEP 3 - Total energy per tonne of finished product. Total = 8,88,667 + 7,50,000 = 16,38,667 kcal/tonne. STEP 4 - Express as oil equivalent. Oil equivalent = 16,38,667 / 10,000 = 163.87 kg oil/tonne of finished product. ANSWER: ~163.9 kg (163.87 kg) oil equivalent per tonne of finished product.
Combine electrical melting energy (adjusted for 60% yield, converted kWh->kcal) with oil heat-treatment energy, then divide the total kcal by the oil GCV to get kg oil equivalent.
Source: Year not recorded
📖 §5.5 Example 5.9 (dust balance) + §5 EOC Short Q S-3 (ash tie-component)

13. (a) A fuel used in a boiler contains 40% carbon and 23% ash. The refuse obtained after combustion is analysed and found to contain 7% carbon and the rest ash. Compute the percentage of the original carbon in the fuel that remains unburnt in the refuse. (5 marks) (b) During an ESP performance study, the inlet gas stream to the ESP is 2,89,920 Nm3/hr with a dust loading of 5500 mg/Nm3, and the outlet gas stream is 3,01,100 Nm3/hr with a dust loading of 110 mg/Nm3. How much fly ash is collected in the system in kg/hr? (5 marks)

Model answer: PART (a) - Unburnt carbon (use ash as the conserved/inert tracer). Basis: 100 kg of refuse -> unburnt carbon = 7 kg, ash = 93 kg. All the ash in the fuel reports to the refuse, and ash = 23% of the fuel. So 93 kg ash corresponds to 23% of the fuel: quantity of raw fuel = 93 / 0.23 = 404.35 kg. Original carbon in the fuel = 0.40 x 404.35 = 161.74 kg. Unburnt carbon (in refuse) = 7 kg. % of original carbon unburnt = (7 / 161.74) x 100 = 4.33%. PART (b) - Fly ash collected (dust mass balance): Inlet dust = Outlet dust + Fly ash collected. Inlet dust = 2,89,920 x 5500 / 1,000,000 = 1594.56 kg/hr. Outlet dust = 3,01,100 x 110 / 1,000,000 = 33.12 kg/hr. Fly ash collected = 1594.56 - 33.12 = 1561.44 kg/hr. ANSWER: (a) 4.33% of the original carbon remains unburnt; (b) 1561.44 kg/hr of fly ash collected.
Part (a): ash is inert and conserved, so use it to back-calculate the fuel mass, then compare unburnt carbon to original carbon. Part (b): dust mass balance, collected = inlet loading x flow - outlet loading x flow.
Source: Year not recorded
📖 §5.5 Example 5.6 — evaporator solids balance (improvement case)

14. In a Chlor-Alkali plant, an evaporator was designed to concentrate 500 kg of liquor containing 7% w/w solids to 45% solids w/w in the output. Presently the output from the evaporator has 30% solids w/w. The energy manager suggested overhauling the evaporator to achieve the design solids in the output. Calculate the percentage improvement in water removal in the evaporator after overhauling.

Model answer: Feed = 500 kg with 7% solids. Solids in feed = 500 x 7/100 = 35 kg (conserved through the evaporator). PRESENT case (output 30% solids): Output (thick liquor) = solids / 0.30 = 35 / 0.30 = 116.7 kg. Water removed = 500 - 116.7 = 383.3 kg. DESIGN case (output 45% solids, after overhaul): Output = 35 / 0.45 = 77.8 kg. Water removed = 500 - 77.8 = 422.2 kg. IMPROVEMENT: Incremental water removal = 422.2 - 383.3 = 38.9 kg. % improvement in water removal = 38.9 / 383.3 x 100 = 10.14%. ANSWER: About 10.14% improvement in water removal after overhauling.
Solids are conserved; output mass = solids / (solids fraction). Compute water removed = feed - output for both present and design cases, then percentage improvement relative to the present case.
Source: Year not recorded
📖 §5.6 Heat balances — enthalpy balance on a steam mixing point

15. Saturated steam at 1 atm is discharged from a turbine at 1200 kg/h. Superheated steam at 300 degC and 1 atm is required as feed to a heat exchanger. To produce it, the turbine discharge is mixed with superheated steam at 400 degC, 1 atm (specific volume 3.11 m3/kg). Calculate the amount of superheated steam at 300 degC produced and the volumetric flow rate of the 400 degC steam. (Enthalpies: saturated steam at 1 atm = 2676 kJ/kg; 400 degC steam = 3278 kJ/kg; 300 degC steam = 3074 kJ/kg.)

Model answer: Let m1 = mass flow of 400 degC steam (kg/h), m2 = mass flow of 300 degC product steam (kg/h). STEP 1 - Mass balance of water: 1200 + m1 = m2 ... (1) STEP 2 - Energy (enthalpy) balance: (1200)(2676) + m1(3278) = m2(3074) ... (2) Solve (1) and (2) simultaneously. Substitute m2 = 1200 + m1 into (2): 32,11,200 + 3278 m1 = (1200 + m1)(3074) = 36,88,800 + 3074 m1 3278 m1 - 3074 m1 = 36,88,800 - 32,11,200 204 m1 = 4,77,600 m1 = 2341.2 kg/h. m2 = 1200 + 2341.2 = 3541.2 kg/h (superheated steam at 300 degC produced). STEP 3 - Volumetric flow rate of the 400 degC steam (specific volume 3.11 m3/kg): = 2341.2 kg/h x 3.11 m3/kg = 7281.1 m3/h. ANSWER: 300 degC steam produced = 3541.2 kg/h; volumetric flow of 400 degC steam = 7281.1 m3/h.
Mixing problem solved with a mass balance and an enthalpy balance (two equations, two unknowns). Volumetric flow = mass flow x specific volume.
Source: Year not recorded
📖 §5.2 Fig 5.1 / §5.5 — material balance with a recycle stream (dry-solids balance)

16. In a drying operation, the moisture content of the feed to a calciner must be held at 15% (w/w) to prevent lumping and sticking. This is achieved by mixing the fresh feed (30% moisture w/w) with a recycle stream of dried material (3% moisture w/w). What fraction of the dried product must be recycled?

Model answer: Let F = fresh feed, R = recycle, P = product (all mass units). Track SOLIDS (dry matter): feed is 70% solids, recycle and product are 97% solids, mixed stream to dryer is 85% solids (15% moisture). STEP 1 - Solids balance at the MIXER: 0.70 F + 0.97 R = 0.85 (F + R) 0.70 F + 0.97 R = 0.85 F + 0.85 R 0.12 R = 0.15 F R = 1.25 F ... (1) STEP 2 - Solids balance at the DRYER (mixed stream in; product + recycle out, both at 97% solids): 0.85 (F + R) = 0.97 (P + R) 0.85 (F + 1.25 F) = 0.97 P + 0.97 x 1.25 F 0.85 x 2.25 F = 0.97 P + 1.2125 F 1.9125 F = 0.97 P + 1.2125 F 0.70 F = 0.97 P F = 1.386 P ... (2) STEP 3 - Recycle in terms of product. Substitute (2) into (1): R = 1.25 x 1.386 P = 1.7325 P ... (3) Product + Recycle = P + 1.7325 P = 2.7325 P. STEP 4 - Recycle as a fraction of dried product stream: R / (P + R) = 1.7325 P / 2.7325 P = 0.634 = 63.4%. ANSWER: About 63.4% of the dried product must be recycled.
Recycle problem solved by two solids balances (mixer and dryer), because dry solids are conserved. Express R and F in terms of P, then the recycle fraction of the dried stream.
Source: Year not recorded
📖 §5.7 Example 5.12 — paper drying machine (method)

17. A paper drying machine has a production capacity of 500 TPD and currently operates at an output of 480 TPD. The dryness of the paper is 60% at the inlet and 95% at the outlet. Steam is supplied at 4 kg/cm2 with a latent heat of 510 kcal/kg. The evaporated moisture is at about 100 degC with an enthalpy of 640 kcal/kg. The plant operates 24 hours per day. Assuming only the latent heat of steam is used for drying and neglecting the enthalpy of moisture in the wet paper, estimate (i) the quantity of moisture to be evaporated per hour and (ii) the input steam quantity required per hour.

Model answer: Output = 480 TPD at 95% dryness. STEP 1 - Bone-dry paper (conserved). Bone-dry mass at output = 480 x 0.95 = 456 TPD. STEP 2 - Total wet paper at the inlet (60% dryness = 60% bone-dry). Total wet paper in = 456 / 0.60 = 760 TPD. STEP 3 - Moisture evaporated per hour. Moisture evaporated = (inlet wet - outlet) = (760 - 480) = 280 TPD. Per hour = 280 / 24 = 11.67 TPH. STEP 4 - Steam required (heat balance). Heat to evaporate = moisture x enthalpy of evaporated moisture = 11.67 x 640 kcal (per T basis). Steam = (11.67 x 640) / 510 = 14.6 TPH. ANSWER: Moisture evaporated ~ 11.67 TPH; Steam required ~ 14.6 TPH.
Bone-dry paper is conserved; find total inlet wet mass from outlet dry mass and inlet dryness. Moisture evaporated = inlet - outlet. Steam = heat carried by evaporated moisture / latent heat of steam.
Source: Year not recorded
📖 §5.7 Example 5.12 — paper machine evaporation and steam

18. The production through a paper machine is 300 tonnes per day (TPD). The inlet and outlet dryness to the paper machine are 50% and 95% respectively. The evaporated moisture temperature is 80 degC. Steam is supplied at 3.5 kg/cm2 (latent heat 513 kcal/kg). Assuming 24 hours/day operation and enthalpy of evaporated moisture = 632 kcal/kg, estimate (i) the quantity of moisture to be evaporated in kg/hr and (ii) the steam quantity required for evaporation in kg/hr.

Model answer: Production = 300 TPD = 300/24 = 12.5 TPH of paper. STEP 1 - Moisture to be evaporated (bone-dry paper conserved). Bone-dry paper in product (95% dry) = 12.5 x 0.95 = 11.875 TPH. Moisture after dryer (5% of product) = 12.5 - 11.875 = 0.625 TPH. At the inlet the paper is 50% dry, so on the same 11.875 TPH of dry solids the total inlet wet mass = 11.875 / 0.50 = 23.75 TPH, and moisture before dryer = 23.75 - 11.875 = 11.875 TPH. Evaporated moisture = 11.875 - 0.625 = 11.25 TPH = 11,250 kg/hr. STEP 2 - Steam required (heat balance). Heat carried by evaporated moisture = 632 x 11,250 = 71,10,000 kcal/hr. Steam required = 71,10,000 / 513 = 13,859.6 kg/hr. ANSWER: Moisture evaporated = 11,250 kg/hr; Steam required = 13,859.6 kg/hr.
Same method as the book Example 5.12 but with 50% inlet dryness. Track bone-dry paper, find moisture before and after, evaporated = difference; steam = heat in moisture / latent heat.
Source: Year not recorded
📖 §5.8 Solved Example — evaporator mass + enthalpy balance (method)

19. An evaporator is fed with 5000 kg/hr of a solution having 0.5% solids at 38 degC, and is concentrated to 1% solids. Steam enters at total enthalpy 640 kcal/kg and condensate leaves at 100 degC. Enthalpy of feed = 38.1 kcal/kg, product solution = 100.8 kcal/kg, vapour = 640 kcal/kg. Find the mass of vapour formed per hour and the mass of steam used per hour.

Model answer: STEP 1 - Mass (solids) balance. Solids in feed = 5000 x 0.5/100 = 25 kg/hr (conserved). Product (thick liquor) at 1% solids: mass x 1/100 = 25 -> product = 25 / 0.01 = 2500 kg/hr. Vapour formed = 5000 - 2500 = 2500 kg/hr. STEP 2 - Heat (enthalpy) balance. Steam gives up (640 - 100) = 540 kcal/kg. Heat by steam + heat in feed = heat in vapour + heat in thick liquor M x 540 + 38.1 x 5000 = 640 x 2500 + 100.8 x 2500 M x 540 + 1,90,500 = 16,00,000 + 2,52,000 = 18,52,000 M x 540 = 18,52,000 - 1,90,500 = 16,61,500 M (steam) = 16,61,500 / 540 = 3076.8 kg/hr. ANSWER: Vapour formed = 2500 kg/hr; Steam used = 3076.8 kg/hr.
Same structure as the book Solved Example: solids balance for vapour/product, then enthalpy balance with steam latent heat = 640 - 100 = 540 kcal/kg.
Source: Year not recorded
📖 §5.6 Energy balance

20. An evaporator is to be fed with 10,000 kg/hr of a solution having 1% solids. The feed is at 38 oC. It is to be concentrated to 2% solids. Steam is entering at a total enthalpy of 640 kCal/kg and the condensate leaves at 100 oC. Enthalpies of feed are 38.1 kcal/kg, product solution is 100.8 kCal/kg and that of the vapour is 640 kCal/kg. Find the mass of vapour formed per hour and the mass of steam used per hour.

Model answer: Mass of vapour: Feed = 10,000 kg/hr @ 1% solids; Solids = 10,000 x 1/100 = 100 kg/hr; Mass_out x 2/100 = 100, so Mass_out = 10,000/2 = 5000 kg/hr. Vapour formed = 10,000 - 5000 = 5000 kg/hr. Thick liquor = 5000 kg/hr. Steam consumption: Enthalpy of feed = 10,000 x 38.1 = 38.1 x 10^4 kCal; Enthalpy of the thick liquor = 100.8 x 5000 = 5,04,000 kCal; Enthalpy of the vapour = 640 x 5000 = 32,00,000 kCal. Heat balance: Heat input by steam + heat in feed = heat out in vapour + heat out in thick liquor: [M x (640-100) + 38.1 x 10,000] = (32,00,000 + 5,04,000); M x 540 = 33,23,000; Mass of steam required = 33,23,000/540 = 6153.7 kg/hr.
Mass first, by a SOLIDS balance: solids = 10,000 x 0.01 = 100 kg/h, and they leave in a product that is 2% solids, so product = 100/0.02 = 5,000 kg/h and vapour = 10,000 - 5,000 = 5,000 kg/h. Then the energy balance: heat out - heat in = (5,000 x 640) + (5,000 x 100.8) - (10,000 x 38.1) = 3,323,000 kcal/h. Each kg of steam gives up (640 - 100) = 540 kcal because the condensate leaves at 100 deg C carrying 100 kcal/kg, so steam = 3,323,000/540 = about 6,154 kg/h. The classic error is dividing by the steam's TOTAL enthalpy of 640 instead of the heat actually released, 640 minus the condensate enthalpy.
Source: Oct 2011
📖 §5.6 Energy balance

21. An evaporator is to be fed with 6000 kg/hr of a solution having 1% solids. The feed is at 38 oC. It is to be concentrated to 2% solids. Steam is entering at a total enthalpy of 640 kCal/kg and the condensate leaves at 100 oC. Enthalpies of feed are 38.1 kcal/kg, product solution is 100.8 kCal/kg and that of the vapour is 640 kCal/kg. Find the mass of vapour formed per hour and the mass of steam used per hour.

Model answer: Mass of vapour: Feed = 6000 kg/hr @ 1% solids; Solids = 6000 x 1/100 = 60 kg/hr; Mass_out x 2/100 = 60, so Mass_out = 6000/2 = 3000 kg/hr. Vapour formed = 6000 - 3000 = 3000 kg/hr. Thick liquor = 3000 kg/hr. Steam consumption: Enthalpy of feed = 6000 x 38.1 = 22.8 x 10^4 kCal; Enthalpy of the thick liquor = 100.8 x 3000 = 3,02,400 kCal; Enthalpy of the vapour = 640 x 3000 = 19,20,000 kCal. Heat balance: [M x (640-100) + 38.1 x 6000] = (19,20,000 + 3,02,400); M x 540 = 21,99,540; Mass of steam required = 21,99,540/540 = 4073 kg/hr.
Solids = 6,000 x 0.01 = 60 kg/h, so product = 60/0.02 = 3,000 kg/h and vapour = 3,000 kg/h. Energy: (3,000 x 640) + (3,000 x 100.8) - (6,000 x 38.1) = 1,993,800 kcal/h; steam = 1,993,800 / (640-100) = about 3,692 kg/h. Note the useful shortcut in this family of problems: doubling the solids concentration always halves the outgoing mass, so the vapour equals half the feed.
Source: Oct 2011
📖 §5.5 Material balance — Example 5.9 (dust balance)

22. A bag house is being used to remove dust from an air exhaust stream flowing at 100 m3/min. The dirty air contains 15 g/m3 of particles, while the cleaned air from the bag house contains 0.02 g/m3. The industry's operating permit allows the exhaust stream to contain as much as 0.9 g/m3. For various operating reasons, the industry wishes to bypass some of the dirty air around the bag house and blend it back into the cleaned air so that the total exhaust stream meets the permissible limit. Assume no air leakage and negligible change in pressure or temperature of the air throughout the process. Draw a schematic diagram and calculate the flow rate of air through the bag house and the mass of dust collected per day in kg. [refers to a figure in the original paper]

Model answer: Draw a flow diagram of the process. In this problem two balances can be made, namely, flow rate of dust in g/m3 and flow rate of air in m3/min. Balancing of flow rate of air in m3/min is possible because the temperature and pressure of air remain constant in the system. Balance for dust around the total system: Input = Output from bag house + Output in the mixed exhaust. Dust removed from bag house (Z) = 100 m3/min x 15 g/m3 - 100 m3/min x 0.90 g/m3 = 1410 g/min. Daily dust output = 1410 g/min x 24 h/1 d x 60 min/1 h x 1 kg/1000 g = 2030 kg. Balance for airflow: 100 = X + Y, where X and Y are the bypass stream and the flow through bag house respectively. Balance for dust around B: 15X + 0.02Y = 0.9 x 100. Solving the last two equations: X, the bypass stream = 5.9 m3/min; Y, the flow through bag house = 94.1 m3/min.
Two balances on the same control volume. Let x be the bypassed dirty air in m3/min. Air: cleaned stream = 100 - x, and the blended total is still 100 m3/min. Dust: 0.02(100 - x) + 15x = 0.9 x 100, so 2 + 14.98x = 90 and x = 5.87 m3/min bypassed, leaving 94.13 m3/min through the bag house. Set up the CONCENTRATION balance in g/min (concentration x volume flow), not in g/m3 — adding concentrations directly is the standard error.
Source: Aug 2013
📖 §5.6 Energy balance

23. a) A furnace heating steel ingots is fired with oil having a calorific value of 10,500 kCal/kg and efficiency of 75%. Calculate the oil consumption per hour when the throughput of the furnace is 50 TPH and the temperature of the finished product is 600 oC. Take ambient temperature as 30 oC and Specific Heat of Steel as 0.12 kCal/kg oC b) In Steel industry, different types of gases are generated during steel making process. Volumetric Flow rate and Calorific Values of each gases are: Type of Gas Flow (SM3/hr) CV (kCal/SM3) Coke Oven Gas 75,000 4,000 COREX Gas 50,000 2,000 BOF Gas 55,000 1,500 Blast Furnace Gas 80,000 700 All these gases are mixed in the gas mixer before combustion. Find out the Calorific Value (in kCal/SM3) of mix gas.

Model answer: a) Oil Consumption / hr 50 (TPH) x 0.12 (kCal/kg oC) x (600 – 30) (oC) = ------------------------------------------------------------------ 0.75 (%) x 10,500 (kCal/kg) = 0.43 TPH (5 marks) b) Total flow of Mix Gas = 75,000 + 50,000 + 55,000 + 80,000 = 2,60,000 SM3/hr (1 mark) CV of Mix Gas = [(75,000 x 4,000) + (50,000 x 2,000) + (55,000 x 1,500) + (80,000 x 700)] / 2,60,000 = 2,071 kCal/SM3 (4 marks)
(a) Useful heat = m x Cp x dT = 50,000 kg/h x 0.12 x (600-30) = 3,420,000 kcal/h. Fuel heat needed = useful/efficiency = 3,420,000/0.75 = 4,560,000 kcal/h. Oil = 4,560,000/10,500 = 434 kg/h. Convert TPH to kg/h first and divide by the efficiency, never multiply. (b) For the by-product gases, compute each gas's heat as flow (Nm3/h) x calorific value (kcal/Nm3) and compare with the fuel it can displace — the balance is again just energy in equals energy usefully used plus losses.
Source: Sep 2015
📖 §5.5 Material balance

24. The production capacity of a paper drying machine is 500 TPD and is currently operating at an output of 480 TPD. To find out the steam requirement for drying, the Energy Manager measures the dryness of the paper both at inlet and outlet of the paper drying machine which found to be 60% and 95% respectively. The steam is supplied at 4 kg/cm2, having a latent heat of 510 kCal/kg. The evaporated moisture temperature is around 100 0C having enthalpy of 640 kCal/kg. Plant operates 24 hours per day. Assume only latent heat of steam is being used for drying the paper and neglect the enthalpy of the moisture in the wet paper. i) Estimate the quantity of moisture to be evaporated per hr. ii) Input steam quantity required for evaporation per hr.

Model answer: Output of the drying machine = 480 TPD with 95% dryness. Bone dry mass of paper at the output = 480 x 0.95 = 456 TPD …. (2 marks) Since the dryness at the inlet is 60%, Total mass of wet paper at the inlet = (456 x 100) / 60 = 760 TPD …..(2 marks) Moisture evaporated per hour = (760 – 480)/ 24 = 11.67TPH ….(3 marks) Mass of Steam, m = (11.67 x 640)/ 510 = 14.6 TPH …..(3 marks)
Anchor on BONE-DRY paper, which passes through unchanged: 480 x 0.95 = 456 TPD. At the inlet the sheet is 60% dry, so inlet mass = 456/0.60 = 760 TPD. Moisture evaporated = 760 - 480 = 280 TPD (about 11,667 kg/h). Steam = heat to evaporate that moisture / 510 kcal/kg of latent heat. Note the machine's 500 TPD capacity is not used in the calculation — it is there for the utilisation comment only. Dryness fraction here means dry solids percentage, not steam quality.
Source: Sep 2015
📖 §5.6 Energy balance

25. Saturated steam at 1 atm is discharged from a turbine at 1200 kg/h. Superheated steam at 300 0C and 1 atm is needed as a feed to a heat exchanger. To produce it, the turbine discharge stream is mixed with superheated steam at 400 0C, 1 atm and specific volume of 3.11 m3/kg. Calculate the amount of superheated steam at 300 0C produced and the volumetric flow rate of the 400 0C steam.

Model answer: Solution 1. Mass balance of water 1200 + m1 = m2 ………………………………………… (1) …………………….1 mark 2. Energy balance (1200 kg/h)(2676 kJ/kg) + m1(3278 kJ/kg) = m2(3074 kJ/kg) ……………………………………. …. (2) …………………….1 mark Eqs. (1) and (2) are solved simultaneously 3211200 + 3278m1 = (1200 + m1)3074 m1 = 2341.2 kg/h m2 = 1200 + 2341.2 = 3541.2 kg/h (superheated steam produced) …………………….4 marks o 3. Volumetric flow rate of 400 C steam The specific volume of steam at 400 C and 1 atm is 3.11 m3/kg. The volumetric flow rate is calculated as follows: (2341.2 kg/h)(3.11 m3/kg) = 7281.1 m3/h …………………….4 marks
Two simultaneous equations. Mass: 1,200 + m1 = m2. Energy: (1,200 x 2,676) + m1 x 3,278 = m2 x 3,074, using the enthalpies of saturated steam at 1 atm, and of superheated steam at 400 deg C and at 300 deg C. Substituting gives m1 = about 2,341 kg/h of 400 deg C steam and m2 = about 3,541 kg/h of 300 deg C steam. The volumetric flow of the hot stream = m1 x 3.11 m3/kg = about 7,280 m3/h. Specific volume converts mass flow to volume flow — multiply, do not divide.
Source: Sep 2017
📖 §5.6 Energy balance

26. Saturated steam at 1 atm is discharged from a turbine at 1000 kg/h. Superheated steam at 300 0C and 1 atm is needed as a feed to a heat exchanger. To produce it, the turbine discharge stream is mixed with superheated steam at 400 0C, 1 atm and specific volume of 3.11 m3/kg Calculate the amount of superheated steam at 300 0C produced and the volumetric flow rate of the 400 0C steam.

Model answer: Solution 1. Mass balance of water 1000 + m1 = m2 ………………………………………… (1) …………………….1 mark 2. Energy balance (1000 kg/h)(2676 kJ/kg) + m1(3278 kJ/kg) = m2(3074 kJ/kg) ……………………………………. …. (2) …………………….1 mark Eqs. (1) and (2) are solved simultaneously 2676000 + 3278m1 = (1000 + m1)3074 m1 = 1950.98 kg/h m2 = 1000 + 1950.98 = 2950.98 kg/h (superheated steam produced) …………………….4 marks o 3. Volumetric flow rate of 400 C steam The specific volume of steam at 400 o C and 1 atm is 3.11 m3/kg. The volumetric flow rate is calculated as follows: (1950.98 kg/h)(3.11 m3/kg) = 6067.55 m3/h …………………….4 marks
Mass: 1,000 + m1 = m2. Energy: (1,000 x 2,676) + m1 x 3,278 = m2 x 3,074 (enthalpies of saturated steam at 1 atm, superheated at 400 deg C and at 300 deg C). Solving, m1 = about 1,951 kg/h and m2 = about 2,951 kg/h; volumetric flow of the 400 deg C stream = 1,951 x 3.11 = about 6,067 m3/h. Two equations, two unknowns — always write the mass balance first, then substitute it into the energy balance.
Source: Sep 2017
📖 §5.5 Material balance

27. In a Chlor-Alkali plant, an evaporator was designed to concentrate 500 kg of liquor containing solids of 7% w/w (weight by weight) to 45% solids w/w in the output. Presently the output from evaporator has 30% solids w/w. The energy manager suggested overhauling the evaporator to achieve the design rate of solids w/w in the output. Calculate the percentage improvement in water removal in the evaporator after overhauling of the evaporator.

Model answer: Amount of feed (input) to the evaporator = 500 kg Concentration of solids in feed = 7 wt% Amount of solids in feed (input) = 500 x 7/100 = 35 kg PRESENT SCENARIO: Concentration of solids in product (output) = 30 wt% = 0.3 Mass balance across the evaporator: Amount of product (output) from the evaporator = 35 / 0.3 = 116.7 kg Water vapour removed from the evaporator = 500 - 116.7 = 383.3 kg DESIGN SCENARIO: Concentration of solids in product (output) = 45 wt% = 0.45 Amount of product (output) from the evaporator = 35 / 0.45 = 77.8 kg Water vapour removed from the evaporator = 500 - 77.8 = 422.2 kg Incremental water removal achieved = 422.2 - 383.3 = 38.9 kg % increase in water removal = 38.9 / 383.3 x 100 % improvement in water removal after overhaul = 10.14 %
Solids = 500 x 0.07 = 35 kg, conserved in both cases. PRESENT: output = 35/0.30 = 116.7 kg, so water removed = 500 - 116.7 = 383.3 kg. DESIGN: output = 35/0.45 = 77.8 kg, water removed = 500 - 77.8 = 422.2 kg. Improvement = (422.2 - 383.3)/383.3 x 100 = about 10.1%. Take the percentage on the PRESENT water removal (the base being improved), not on the feed — that choice of base is what the mark hangs on.
Source: Sep 2018
📖 §5.6 Energy balance

28. A medium size chemical plant receives electricity from grid and also generates electricity from coal based Captive Power Plant (CPP). Coal is also used for process requirements. The fine coal from CPP is sold to neighboring plant. The annual energy details are given below: Electricity purchased from grid 5 MU; Electricity exported to grid 11 MU; Power generation from CPP 36 MU; Power supplied from CPP to Process plant 25 MU; Fine coal sold to neighboring unit 1000 ton; Coal used for process plant 5000 ton; GCV of coal 4500 kcal/kg; Heat rate of CPP 3500 kcal/kWh; Annual Operating Hours 7200. Calculate a. Energy usage in TOE (Tons of oil equivalent) (5 Marks) b. Coal used in CPP (3 Marks) c. Calculate the CPP operating power in MW. (2 Marks)

Model answer: ENERGY USAGE IN TOE (TONS OF OIL EQUIVALENT): Grid electricity imported = (5 x 10^6 kWh) x (860 kcal/kWh) = (+) 43 x 10^8 kcal/year Power generated from CPP = (36 x 10^6 kWh) x (3500 kcal/kWh) = (+) 1260 x 10^8 kcal/year Coal imported for process = (5000 x 10^3 kg) x (4500 kcal/kg) = (+) 225 x 10^8 kcal/year Power exported to grid = (11 x 10^6 kWh) x (3500 kcal/kWh) = (-) 385 x 10^8 kcal/year Coal fines exported to neighbour = (1000 x 10^3 kg) x (4500 kcal/kg) = (-) 45 x 10^8 kcal/year Net annual energy consumption = (43 + 1260 + 225) - (385 + 45) = (+) 1098 x 10^8 kcal/year a. Energy usage in TOE = (1098 x 10^8 kcal/year) / 10^7 = 10980 MTOE (1 MTOE = 10^7 kcal) b. Coal used in CPP = ((36 x 10^6 kWh) x (3500 kcal/kWh)) / (4500 kcal/kg) = 28 x 10^6 kg coal/year = 28000 tons of coal/year c. CPP operating power = (36 x 10^6 kWh/year) / (7200 hrs/year) = 5000 kW = 5 MW
Draw the plant boundary and give every stream a sign: imports and generation are positive, exports and sales are negative. Convert electricity with 860 kcal/kwh and coal with its GCV, then divide by 10^7 for toe. Grid import 5 MU = 43 x 10^8 kcal; export of 11 MU and the fine coal SOLD to the neighbour are both deducted because they leave the boundary; the CPP's coal enters as fuel while the CPP's power to the process is an internal transfer that must NOT be counted twice. Double counting the CPP (once as coal, once as kWh) is the single biggest mark-loser in this question — count the fuel entering, not the electricity it makes.
Source: Sep 2018
Also uses Ch 3 · Basics of Energy & Its Forms — see that chapter
📖 §5.5 Material balance

29. a) A sample of fuel being used in a boiler is found to contain 40% carbon and 23% ash. The refuse obtained after combustion is analyzed and found to contain 7% carbon & the rest is ash. Compute the percentage of the original carbon in fuel which remains as unburnt in the refuse. (5 Marks) b) During an ESP performance evaluation study, the inlet gas stream to ESP is 2,89,920 Nm3/hr and the dust loading is 5,500 mg/Nm3. The outlet gas stream from ESP is 3,01,100 Nm3/hr and the dust loading is 110 mg/Nm3. How much fly ash is collected in the system in kg/hr? (5 Marks)

Model answer: a) Let the quantity of refuse sample = 100 kg Amount of unburnt carbon in refuse = 7 kg Amount of ash in the refuse = 93 kg Total ash in the fuel that has come into the refuse = 23 % of fuel 93 kg of ash corresponds to 23 % ash in the fuel Therefore quantity of total raw fuel = 93 / 0.23 = 404.35 kg Quantity of original carbon in the fuel = 0.40 x 404.35 = 161.74 kg Quantity of unburnt carbon in refuse = 7 kg % of the original carbon unburnt in the refuse = (7 / 161.74) x 100 = 4.32 % b) Based on mass balance: Inlet gas stream dust = Outlet gas stream dust + Fly ash collected i) Inlet gas stream flow = 2,89,920 Nm3/hr; dust concentration = 5500 mg/Nm3 Inlet dust quantity = 289920 x 5500 / 1000000 = 1594.56 kg/hr ii) Outlet dust quantity = 301100 (Nm3/hr) x 110 (mg/Nm3) / 1000000 = 33.12 kg/hr iii) Fly ash collected = Inlet gas stream dust - Outlet gas stream dust = 1594.56 - 33.12 = 1561.44 kg/hr
(a) Basis 100 kg of coal. ASH is inert and is fully conserved, which is the key that unlocks the whole problem: 23 kg of ash leaves in a refuse that is (100-7) = 93% ash, so refuse = 23/0.93 = 24.73 kg, of which 7% is carbon = 1.73 kg. Percentage of the original carbon left unburnt = 1.73/40 x 100 = 4.3%. (b) ESP efficiency = (inlet loading - outlet loading)/inlet x 100, and the dust collected = flow x (inlet - outlet) concentration, converted from mg to kg. Anchoring on the inert ash, not on the carbon, is the technique to carry into every unburnt-carbon question.
Source: Sep 2019
📖 §5.5 Material balance

30. The production through a paper machine is 300 tonnes per day (TPD). Inlet and outlet dryness to paper machine is 50% and 95% respectively. Evaporated moisture temperature is 80 deg C. To evaporate moisture, the steam is supplied at 3.5 kg/cm2. Latent heat of steam at 3.5 kg/cm2 is 513 kcal/kg. Assume 24 hours/day operation and estimate the following (Each 5 Marks): i) Quantity of moisture to be evaporated in kg/hr ii) Steam quantity required for evaporation in kg/hour. Note: Consider enthalpy of evaporated moisture as 632 kcal/kg.

Model answer: Production = 300 TPD = 12.5 TPH. Inlet dryness = 50%, outlet dryness = 95%. i) Moisture to be evaporated: Paper (dry fibre) weight in the final product = 12.5 x 0.95 = 11.875 TPH Moisture leaving with the product = 12.5 - 11.875 = 0.625 TPH Dry fibre entering the dryer = 11.875 TPH; at 50% dryness the wet weight in = 11.875 x (100/50) = 23.75 TPH Moisture entering = 23.75 - 11.875 = 11.875 TPH Moisture evaporated = 11.875 - 0.625 = 11.25 TPH = 11,250 kg/hr ii) Steam quantity: Heat carried away by the evaporated moisture = 632 kcal/kg x 11,250 kg/hr = 7,110,000 kcal/hr Latent heat available in the supply steam at 3.5 kg/cm2 = 513 kcal/kg Steam required = 7,110,000 / 513 = 13,859.6 kg/hr = 13.86 TPH
Production = 300 TPD = 12.5 TPH at 95% dryness, so bone-dry fibre = 12.5 x 0.95 = 11.875 TPH. At the inlet the sheet is only 50% dry, so inlet mass = 11.875/0.50 = 23.75 TPH. (i) Moisture evaporated = 23.75 - 12.5 = 11.25 TPH = 11,250 kg/h. (ii) Steam = heat required to raise and evaporate that moisture, divided by the latent heat 513 kcal/kg. Anchor on the bone-dry fibre, which is the only quantity that passes through the machine unchanged.
Source: Jul 2022
📖 §5.5 Material balance

31. A) In an industry, de-humidified air containing 0.0089 kg H2O/kg air is required to be fed to a dryer for textile drying. The atmospheric air has a specific humidity of 0.02 kg H2O/kg air. The atmospheric air is to be passed through a de-humidifying system that reduces the absolute humidity to the desired level. The de-humidifying system installed has the capacity to reduce the absolute humidity to 0.0005 kg H2O/kg air. Therefore, to get the desired humidity of 0.0089 kg H2O/kg air at the de-humidifying system outlet, a part of fresh atmospheric air by-passes and mixes at the outlet of the de-humidification system. Find out the mass of water removed per 100 kg of air fed to the dehumidifier and the percentage of by-passed air. (4 Marks) B) The waste acid contains 30% H2SO4, 35% HNO3 and 35% water. The waste acid is to be concentrated to contain 39% H2SO4 and 42% HNO3 by addition of concentrated Sulphuric acid containing 98% H2SO4 and concentrated Nitric acid containing 72% HNO3. Calculate the quantities of three acids to be mixed to get 1000 kg of desired mixed acid. (6 Marks) [refers to a figure in the original paper]

Model answer: A) Basis: 100 kg of air fed to the dehumidifier. Moisture in = 100 x 0.02 = 2 kg Moisture out of the dehumidifier = 100 x 0.0005 = 0.05 kg Water removed = 2 - 0.05 = 1.95 kg per 100 kg of air fed. Let x kg of atmospheric air by-pass the dehumidifier. Moisture balance at the mixing point: 0.05 + 0.02x = 0.0089 x (100 + x) 0.05 + 0.02x = 0.89 + 0.0089x 0.0111x = 0.84, so x = 75.68 kg Total air to the dryer = 100 + 75.68 = 175.68 kg Percentage of by-passed air = 75.68/175.68 x 100 = 43.08% B) Basis: 1000 kg of desired mixed acid (39% H2SO4, 42% HNO3). Let X = waste acid (30% H2SO4, 35% HNO3), Y = concentrated sulphuric acid (98% H2SO4), Z = concentrated nitric acid (72% HNO3). Overall mass balance: X + Y + Z = 1000 ... (1) H2SO4 balance: 0.30X + 0.98Y = 0.39 x 1000 = 390 ... (2) HNO3 balance: 0.35X + 0.72Z = 0.42 x 1000 = 420 ... (3) Solving (1), (2) and (3) simultaneously: X = 90.1 kg of waste acid; Y = 370.4 kg of concentrated sulphuric acid; Z = 539.5 kg of concentrated nitric acid.
A blending (mixing) balance on the moisture. Let y be the fraction of atmospheric air bypassed around the dehumidifier: 0.0005(1 - y) + 0.02y = 0.0089, so 0.0195y = 0.0084 and y = 0.431, i.e. about 43% of the air is bypassed and 57% is dehumidified. The balance is written on the moisture MASS carried by each stream (kg H2O per kg of dry air x kg of dry air) — the dry air itself is the tie component and is conserved throughout.
Source: Mar 2023
📖 §5.6 Energy balance

32. An evaporator is to be fed with 5000 kg/hr of a solution having 0.5% solids. The feed is at 38 deg C and is to be concentrated to 1% solids. Steam is entering at a total enthalpy of 640 kcal/kg and the condensate leaves at 100 deg C. Enthalpies of feed are 38.1 kcal/kg, product solution is 100.8 kcal/kg and that of the vapour is 640 kcal/kg. Find the mass of vapour formed per hour and the mass of steam used per hour. (10 Marks)

Model answer: MASS OF VAPOUR (solids balance) Feed = 5000 kg/hr at 0.5% solids, so solids = 5000 x 0.5/100 = 25 kg/hr The solids all leave in the thick liquor at 1% concentration: Mass(out) x 1/100 = 25, so Mass(out) = 2500 kg/hr Vapour formed = 5000 - 2500 = 2500 kg/hr Thick liquor = 2500 kg/hr STEAM CONSUMPTION (heat balance) Enthalpy of feed = 5000 x 38.1 = 190,500 kcal/hr Enthalpy of thick liquor = 2500 x 100.8 = 252,000 kcal/hr Enthalpy of vapour = 2500 x 640 = 16,00,000 kcal/hr Heat given up per kg of steam = 640 - 100 = 540 kcal/kg (condensate leaves at 100 deg C) Heat input by steam + heat in feed = heat out in vapour + heat out in thick liquor M x 540 + 190,500 = 16,00,000 + 252,000 M x 540 = 16,61,500 M = 3,076.8 kg/hr of steam
Solids = 5,000 x 0.005 = 25 kg/h; at 1% solids the product = 25/0.01 = 2,500 kg/h, so vapour = 2,500 kg/h. Energy: (2,500 x 640) + (2,500 x 100.8) - (5,000 x 38.1) = 1,661,500 kcal/h; steam = 1,661,500 / (640-100) = about 3,077 kg/h. The steam gives up only (total enthalpy - condensate enthalpy) = 540 kcal/kg, because the condensate leaves at 100 deg C still carrying 100 kcal/kg.
Source: Sep 2024
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