Engineering Questions with Answers - Multiple Choice Questions
Thermal Engineering MCQ’s – Evaporative Capacity and Equivalent Evaporation
1 - Question
Which of the following CANNOT be the unit of ‘evaporative capacity’?
a) kg of steam/h
b) kg of steam/h/m2 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 - Question
Which of the following is the correct formula for factor of evaporation?
a) f = h−hf12257
b) f = hf1−h2257
c) f = h−hf11257
d) f = hf1−h1257
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 = h−hf12257
3 - Question
Which of the correct formula for calculating equivalent evaporation?
a) me=ma(h−hf1)2257
b) me=ma(h−hf1)1257
c) me=ma(hf1−h)2257
d) me=(h−hf1)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 me=ma(h−hf1)2257
4 - Question
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, ms = 64 kg, P = 14 bar, T = 39°C, x = 0.82, mf = 8 kg We know that, ma = msmf ma = 648 ma = 8 kg steam/kg of coal From steam tables at 14 bar hf = 830.1 kJ/kg and hfg = 1957.7 kJ/kg h = hf + x(hfg) = 830.1 + 0.82(1957.7) = 2435.414 kJ/kg hf1 = 4.18(T – 0) = 4.18(39-0) = 163.02 kJ/kg Equivalent evaporation, me=ma(h−hf1)2257 me = 8(2435.414−163.02)2257 me = 8.05 kg of steam/kg of fuel.
5 - Question
What is the unit of equivalent evaporation?
a) kg of steam/kg of fuel fired
b) kg of steam/h/m2 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 - Question
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
Answer: b
Explanation: Given,
ms = 30 kg, P = 11.5 bar, mf = 3 kg, T = 40°C
From steam tables, at 11.5 bar
hg = 2781.3 kJ/kg
h = hg = 2781.3 KJ/kg ……………………………………………………………. (dry and saturated)
hf1 = 4.18(T-0) = 4.18(40-1) = 167.2 kJ/kg
We know that, ma = msmf
ma = 303
ma = 10 kg steam/kg of coal
Equivalent evaporation, me=ma(h−hf1)2257
me = 10(2781.3−167.2)2257
me = 11.58 kg of steam/kg of fuel.
7 - Question
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, ma = 12 kg of steam/kg of fuel, P = 11 bar, Tsup = 260°C, T = 30°C, cp = 2.33 kJ/kg-K From steam tables at 11 bar Ts = 184.1°C hf = 781.1 kJ/kg and hfg = 1998.5 kJ/kg h = hf + hfg + cp(Tsup-Ts) = 781.1 + 1998.5 + 2.33(260-184.1) = 2956.45 kJ/kg hf1 = 4.18(T – 0) = 4.18(30-0) = 125.4 kJ/kg Equivalent evaporation, me=ma(h−hf1)2257 me = 12(2956.45−125.4)2257 me = 15.05 kg of steam/kg of fuel me ≈ 15 kg of steam/kg of fuel.
8 - Question
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 - Question
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, ms = 600 kg Mass of coal, mf = 40 kg Coal feeding rate, R = 4 kg/hr Evaporation capacity =(ms/mf) * R = (600/40) * 4 me = 60 kg of steam/hour
10 - Question
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 hf = 798.4 kJ/kg, hfg = 1984.3 kJ/kg h = hf + hfg = 798.4 + 1984.3 = 2782.7 kJ/kg hf1 = 4.18(T-0) = 4.18(45-0) = 188.1 kJ/kg Factor of evaporation, f=h−hf12257 f=2782.7−188.12257 f=1.15.
11 - Question
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 hf = 806.7 kJ/kg, hfg = 1977.4 kJ/kg h = hf + x(hfg) = 806.7 +0.85(1977.4) = 2487.49 kJ/kg hf1 = 4.18(T-0) = 4.18(35-0) = 146.3 kJ/kg Factor of evaporation, f=h−hf12257 f=2487.49−146.32257 f=1.04.
12 - Question
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, Tsup = 230°C, cp = 2.34 kJ/kg-K From steam tables, at 14 bar pressure hf = 830.1 kJ/kg, hfg = 1957.7 kJ/kg, Tsat = °C h = hf + hfg + cp (Tsup – Tsat) = 830.1 + 1957.7 + 2.34(230 – 195) = 2869.7 kJ/kg hf1 = 4.18(T-0) = 4.18(30-0) = 125.4 kJ/kg Factor of evaporation, f=h−hf12257 f=2869.7−125.42257 f=1.22.
13 - Question
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
Answer: c
Explanation: Given,
P = 10 bar, ma = 10 kg of steam/kg of fuel, me = 11.6 kg of steam/kg of fuel
From steam tables, At 10 bar
hg = 2776.2 kJ/kg
h = hg = 2776.2 kJ/kg (Steam is dry and saturated)
We know that,
me = ma(h−hf1)2257
Substituting the values, we get
11.6 = 10(2776.2−hf1)2257
Solving for hf1, we get
hf1 = 158.08 kJ/kg
but, hf1 = 4.18(T-0)
158.08 = 4.18(T-0)
T = 37.82°C ≈ 38°C
14 - Question
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: me = 11 kg of steam/kg of fuel, ma = 10 kg of steam/kg of fuel, T = 40°C hf1 = 4.18(T-0) = 4.18(40-0) = 167.2 kJ/kg We know that, me = ma(h−hf1)2257 Substituting the values, we get 11 = 10(h−167.2)2257 Solving for h, we get h = 2649.9 kJ/kg ≈ 2650 kJ/kg.
15 - Question
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=h−hf12257 Substituting the values, we get 1.14=2700−hf12257 Solving for hf1, we get hf1 = 127.02 kJ/kg But, hf1 = 4.18(T-0) T = 127.02/4.18 T = 30.39°C ≈ 30°C.