This set of Basic Thermal Engineering Questions and Answers focuses on “Evaporative Capacity and Equivalent Evaporation”.

1. Which of the following CANNOT be the unit of ‘evaporative capacity’?

a) kg of steam/h

b) kg of steam/h/m^{2} of heating surface

c) kg of steam/kg of air supplied

d) kg of steam/kg of fuel fired

View Answer

Explanation: Evaporative capacity is not expressed as kg of steam per kg of air supplied. It is expressed usually in kg of steam per hour. It is also expressed in kg of steam per hour per square meter of heating surface and kg of steam per kg of fuel fired.

2. Which of the following is the correct formula for factor of evaporation?

a) f = \(\frac{h-h_{f1}}{2257} \)

b) f = \(\frac{h_{f1}-h}{2257} \)

c) f = \(\frac{h-h_{f1}}{1257} \)

d) f = \(\frac{h_{f1}-h}{1257} \)

View Answer

Explanation: Factor of evaporation is the ratio of heat that is received by 1 kg of water under working conditions to that received by 1 kg of water evaporated from and at 100°C.

Factor of evaporation is calculated by the following formula –

f = \(\frac{h-h_{f1}}{2257} \)

3. Which of the correct formula for calculating equivalent evaporation?

a) m_{e}=\(\frac{m_a (h-h_{f1})}{2257} \)

b) m_{e}=\(\frac{m_a (h-h_{f1})}{1257} \)

c) m_{e}=\(\frac{m_a (h_{f1}-h)}{2257} \)

d) m_{e}=\(\frac{(h-h_{f1})}{2257} \)

View Answer

Explanation: Amount of water evaporated from water at 100°C to dry and saturated steam at 100°C is called equivalent evaporation.

It is calculated by the following formula

m

_{e}=\(\frac{m_a (h-h_{f1})}{2257} \)

4. 64 kg of steam is produced at 14 bar pressure having a dryness fraction of 0.82. The feed water temperature in the boiler is 39°C. Determine equivalent evaporation if mass of coal consumed is 8 kg.

a) 5.05 kg of steam/kg of fuel

b) 7.05 kg of steam/kg of fuel

c) 8.05 kg of steam/kg of fuel

d) 10.25 kg of steam/kg of fuel

View Answer

Explanation: Given,

m

_{s}= 64 kg, P = 14 bar, T = 39°C, x = 0.82, mf = 8 kg

We know that, m

_{a}= \(\frac{m_s}{m_f} \)

m

_{a}= \(\frac{64}{8} \)

m

_{a}= 8 kg steam/kg of coal

From steam tables at 14 bar

h

_{f}= 830.1 kJ/kg and h

_{fg}= 1957.7 kJ/kg

h = h

_{f}+ x(h

_{fg}) = 830.1 + 0.82(1957.7) = 2435.414 kJ/kg

h

_{f1}= 4.18(T – 0) = 4.18(39-0) = 163.02 kJ/kg

Equivalent evaporation, m

_{e}=\(\frac{m_a(h-h_{f1})}{2257} \)

m

_{e}= \(\frac{8(2435.414-163.02)}{2257} \)

m

_{e}= 8.05 kg of steam/kg of fuel.

5. What is the unit of equivalent evaporation?

a) kg of steam/kg of fuel fired

b) kg of steam/h/m^{2} of heating surface

c) kg of steam/h

d) kg of steam/kg of air supplied

View Answer

Explanation: The unit of equivalent evaporation is kg of steam per kg of fuel fired. It represents the quantity of steam that can be generated per kg of fuel fired at 100°C and 1 atm pressure. Equivalent evaporation is an important factor while determining the performance of the boiler.

6. A boiler generates 30 kg of steam at 11.5 bar in 1 hour with the consumption of 3 kg of coal. Feed water temperature is 40°C. Calculate equivalent evaporation if the steam is dry and saturated.

a) 10.78 kg of steam/kg of fuel

b) 11.58 kg of steam/kg of fuel

c) 6.32 kg of steam/kg of fuel

d) 5.65 kg of steam/kg of fuel

View Answer

Explanation: Given,

m

_{s}= 30 kg, P = 11.5 bar, m

_{f}= 3 kg, T = 40°C

From steam tables, at 11.5 bar

h

_{g}= 2781.3 kJ/kg

h = h

_{g}= 2781.3 KJ/kg ……………………………………………………………. (dry and saturated)

h

_{f1}= 4.18(T-0) = 4.18(40-1) = 167.2 kJ/kg

We know that, m

_{a}= \(\frac{ms}{mf} \)

m

_{a}= \(\frac{30}{3} \)

m

_{a}= 10 kg steam/kg of coal

Equivalent evaporation, m

_{e}=\(\frac{m_a (h-h_f1)}{2257} \)

m

_{e}= \(\frac{10(2781.3-167.2)}{2257} \)

m

_{e}= 11.58 kg of steam/kg of fuel.

7. A boiler produces 12 kg of steam per kg of coal burnt at a pressure of 11 bar. The steam generated is superheated to a temperature of 260°C. Determine the equivalent evaporation if the mean feed water temperature is 30°C.Take specific heat of superheated steam as 2.33 kJ/kg.

a) 10 kg of steam/kg of fuel

b) 12 kg of steam/kg of fuel

c) 15 kg of steam/kg of fuel

d) 17 kg of steam/kg of fuel

View Answer

Explanation: Given,

m

_{a}= 12 kg of steam/kg of fuel, P = 11 bar, T

_{sup}= 260°C, T = 30°C, c

_{p}= 2.33 kJ/kg-K

From steam tables at 11 bar

T

_{s}= 184.1°C

h

_{f}= 781.1 kJ/kg and h

_{fg}= 1998.5 kJ/kg

h = h

_{f}+ h

_{fg}+ c

_{p}(T

_{sup}-T

_{s}) = 781.1 + 1998.5 + 2.33(260-184.1) = 2956.45 kJ/kg

h

_{f1}= 4.18(T – 0) = 4.18(30-0) = 125.4 kJ/kg

Equivalent evaporation, m

_{e}=\(\frac{m_a (h-h_f1)}{2257} \)

m

_{e}= \(\frac{12(2956.45-125.4)}{2257} \)

m

_{e}= 15.05 kg of steam/kg of fuel

m

_{e}≈ 15 kg of steam/kg of fuel.

8. What is the latent heat of vaporization of water at 100°C and 1 atm pressure?

a) 2546 kJ/kg

b) 1254 kJ/kg

c) 1564 kJ/kg

d) 2257 kJ/kg

View Answer

Explanation: The latent heat of vaporization is the amount of heat energy required to convert a unit mass of a liquid (here water) into vapor without change in its temperature. The latent heat of vaporization of water is 2257 kJ/kg at 100°C and 1 atm pressure.

9. A boiler generates 600 kg of steam, consuming 40 kg of coal. Determine its evaporation capacity in kg of steam per hour if coal is fed to the furnace at the rate of 4 kg per hour.

a) 60 kg of steam/hour

b) 40 kg of steam/hour

c) 50 kg of steam/hour

d) 70 kg of steam/hour

View Answer

Explanation: Given,

Mass of steam, m

_{s}= 600 kg

Mass of coal, m

_{f}= 40 kg

Coal feeding rate, R = 4 kg/hr

Evaporation capacity =(m

_{s}/m

_{f}) * R

= (600/40) * 4

m

_{e}= 60 kg of steam/hour

10. A boiler produces dry and saturated steam at 12 bar pressure. Calculate the factor of evaporation if the feed water temperature is 45°C.

a) 0.95

b) 1.15

c) 1.10

d) 0.85

View Answer

Explanation: Given,

P = 12 bar, T = 45°C

From steam tables, at 12 bar pressure

h

_{f}= 798.4 kJ/kg, h

_{fg}= 1984.3 kJ/kg

h = h

_{f}+ h

_{fg}= 798.4 + 1984.3 = 2782.7 kJ/kg

h

_{f1}= 4.18(T-0) = 4.18(45-0) = 188.1 kJ/kg

Factor of evaporation, f=\(\frac{h-h_{f1}}{2257} \)

f=\(\frac{2782.7-188.1}{2257} \)

f=1.15.

11. A boiler generates steam at 12.5 bar pressure. Temperature of the feed water is 35°C. Determine the factor of evaporation if the dryness fraction of the steam is 0.85.

a) 1.32

b) 0.85

c) 1.56

d) 1.04

View Answer

Explanation: Given,

P = 12.5 bar, T = 35°C, x = 0.85

From steam tables, at 12.5 bar pressure

h

_{f}= 806.7 kJ/kg, h

_{fg}= 1977.4 kJ/kg

h = h

_{f}+ x(h

_{fg}) = 806.7 +0.85(1977.4) = 2487.49 kJ/kg

h

_{f1}= 4.18(T-0) = 4.18(35-0) = 146.3 kJ/kg

Factor of evaporation, f=\(\frac{h-h_{f1}}{2257} \)

f=\(\frac{2487.49-146.3}{2257} \)

f=1.04.

12. Steam generated by a boiler is superheated to a temperature of 230°C. The feed water temperature is 30°C. Calculate the factor of evaporation if the pressure of the steam generated is 14 bar. Specific heat of superheated steam is 2.34 kJ/kg-K

a) 1.22

b) 1.03

c) 1.52

d) 1.65

View Answer

Explanation: Given,

P = 14 bar, T = 30°C, T

_{sup}= 230°C, c

_{p}= 2.34 kJ/kg-K

From steam tables, at 14 bar pressure

h

_{f}= 830.1 kJ/kg, h

_{fg}= 1957.7 kJ/kg, T

_{sat}= °C

h = h

_{f}+ h

_{fg}+ c

_{p}(T

_{sup}– T

_{sat}) = 830.1 + 1957.7 + 2.34(230 – 195) = 2869.7 kJ/kg

h

_{f1}= 4.18(T-0) = 4.18(30-0) = 125.4 kJ/kg

Factor of evaporation, f=\(\frac{h-h_{f1}}{2257} \)

f=\(\frac{2869.7-125.4}{2257} \)

f=1.22.

13. A boiler generates dry and saturated steam at 10 bar pressure. The evaporative capacity of the boiler is 10 kg of steam per kg of fuel fired. The equivalent evaporation is 11.6 kg of steam per kg of fuel fired. Determine the temperature of feed water.

a) 30°C

b) 26°C

c) 38°C

d) 40°C

View Answer

Explanation: Given,

P = 10 bar, m

_{a}= 10 kg of steam/kg of fuel, m

_{e}= 11.6 kg of steam/kg of fuel

From steam tables, At 10 bar

h

_{g}= 2776.2 kJ/kg

h = h

_{g}= 2776.2 kJ/kg (Steam is dry and saturated)

We know that,

m

_{e}= \(\frac{ma(h-hf1)}{2257} \)

Substituting the values, we get

11.6 = \(\frac{10(2776.2-hf1)}{2257} \)

Solving for hf1, we get

h

_{f1}= 158.08 kJ/kg

but, h

_{f1}= 4.18(T-0)

158.08 = 4.18(T-0)

T = 37.82°C ≈ 38°C

14. The equivalent evaporation of a boiler is 11 kg of steam per kg of fuel fired. The evaporation capacity of the same boiler is 10 kg of steam per kg of fuel fired. Determine enthalpy of 1 kg of steam if the feed water temperature is 40°C.

a) 2750 kJ/kg

b) 2650 kJ/kg

c) 2250 kJ/kg

d) 2850 kJ/kg

View Answer

Explanation: m

_{e}= 11 kg of steam/kg of fuel, m

_{a}= 10 kg of steam/kg of fuel, T = 40°C

h

_{f1}= 4.18(T-0) = 4.18(40-0) = 167.2 kJ/kg

We know that,

m

_{e}= \(\frac{ma(h-hf1)}{2257} \)

Substituting the values, we get

11 = \(\frac{10(h-167.2)}{2257} \)

Solving for h, we get

h = 2649.9 kJ/kg ≈ 2650 kJ/kg.

15. Determine the temperature of feed water of a boiler having factor of evaporation equal to 1.14. The enthalpy of 1 kg of steam is 2700 kJ/kg.

a) 30°C

b) 40°C

c) 35°C

d) 45°C

View Answer

Explanation: f = 1.14, h = 2700 kJ/kg

We know that,

f=\(\frac{h-h_{f1}}{2257} \)

Substituting the values, we get

1.14=\(\frac{2700-h_{f1}}{2257} \)

Solving for h

_{f1}, we get

h

_{f1}= 127.02 kJ/kg

But, h

_{f1}= 4.18(T-0)

T = 127.02/4.18

T = 30.39°C ≈ 30°C.

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