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# Thermal Engineering MCQ’s – Steam Generators Performance – Boiler Efficiency

Which of the following is the correct formula for calculating boiler efficiency?

a) η = (h−hf1)C

b) η = ma(h−hf1)C

c) η = ma(hf1−hC

d) η = ma(h−hf1)2257

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Answer: bExplanation: The correct formula for calculating boiler efficiency is η = ma(h−hf1)C. The expression ma(h−hf1)2257 is used to calculate the equivalent evaporation of a boiler. Considering Superheater, boiler and economizer as a single unit, the calculated efficiency is called overall efficiency.

Which of the following is a fixed factor on which boiler efficiency depends?

a) Actual firing rate

b) Humidity of combustion air

c) Condition of heat absorbing surfaces

d) Properties of the fuel burnt

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Answer: dExplanation: The efficiency of a boiler depends on two types of factors – fixed factors and variable factors. Boiler design, Heat recovery equipment, built in loses, rated rate of firing and properties and characteristics of the fuel burnt are all fixed factors. Actual firing rate, humidity of combustion air, condition of heat absorbing surfaces are variable factors.

The following listed are some factors on which boiler efficiency depends. Which of the following is a variable factor?

a) Excess air fluctuations

b) Rated rate of firing

c) Boiler design

d) Heat recovery equipment

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Answer: aExplanation: Variable factors include actual firing rate, fuel condition as it is fired, Condition of heat absorbing surfaces, excess air fluctuations, change in draught due to atmospheric conditions and humidity and temperature of combustion air. Rated rate of firing, boiler design and heat recovery equipment are all fixed factors.

Boiler efficiency is the ratio of heat actually absorbed by water during generation of steam to the heat supplied in by the fuel in the same time period.

a) True

b) False

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Answer: aExplanation: The given statement is the correct definition of boiler efficiency. Mathematically, boiler efficiency = ma(h−hf1)C If economizer and superheater are also considered along with boiler, then the efficiency calculated is called overall efficiency.

Which of the following is not a heat recovery equipment?

a) Economizer

b) Air preheater

c) Feed water heater

d) Steam separator

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Answer: dExplanation: Steam separator is not a heat recovery equipment. Economizer, superheater, air preheater and feed water heater are all heat recovery equipments. Steam separator is a boiler accessory. Economizer, superheater, air preheater and feed water heater are also categorized under boiler accessories.

Following observations were made after the trial of a boiler for 24 hours.

Mass of coal burnt = 1300 kg

Mass of steam generated = 14000 kg

Mean effective pressure of steam = 10 bar

Temperature of the feed water = 40°C

Calorific value of coal = 30000 kJ/kg

Calculate the boiler efficiency if the dryness fraction of the steam generated is 0.95.

a) 90%

b) 80%

c) 70%

d) 60%

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Answer: a

Explanation: Given, mf = 1300 kg, ms = 14000 kg, P = 10 bar, T = 40°C, C = 30000 kJ/kg

ma = msmf

ma = 140001300

ma = 10.77 kg of steam/kg of fuel

At 10 bar, from steam tables

hf = 762.68 kJ/kg, hfg = 2014.44 kJ/kg

x = 0.95

h = hf + x(hfg) = 762.68 + 0.95(2014.44) = 2676.4 kJ/kg

hf1 = 4.18(T-0) = 4.18(40-0) = 167.2 kJ/kg

Boiler efficiency, η=ma(h−hf1)C=10.77(2676.4−167.2)30000 = 0.9 or 90%.

A boiler produces 4000 kg of dry and saturated steam in 6 hours. In the same time the mass of coal consumed is 300 kg of coal. The mean pressure of steam generated is 11 bar. The temperature of the feed water is 35°C. Determine the boiler efficiency if the calorific value of the fuel is 30000 kJ/kg.

a) 85.64%

b) 75.65%

c) 95.65%

d) 87.78%

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Answer: d

Explanation: Given, ms = 4000 kg, mf = 300 kg, P = 11 bar, T = 35°C, C = 30000 kJ/kg

At 11 bar, from steam tables

hf = 781.1 kJ/kg, hfg = 1998.5 kJ/kg

h = hf + hfg =781.1 + 1998.5 = 2779.6 kJ/kg

hf1 = 4.18 (T-0) = 4.18(35-0) = 146.3 kJ/kg

Boiler efficiency, η=ms(h−hf1mf∗C=3000(2779.6−146.3)300∗30000 = 0.8778 or 87.78%.

A boiler generates superheated steam having temperature 270°C and pressure 13 bar. The specific heat of the superheated steam 2.1 kJ/kg-K. The temperature of the feed water is 32°C. The evaporative capacity of the boiler is 8 kg of steam per kg of fuel. The calorific value of the coal is 26000 kJ/kg.

a) 86.65%

b) 70.32%

c) 65.87%

d) 90.80%

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Answer: a

Explanation: Given, Tsup = 270°C, P = 13 bar, T = 32°C, cp = 2.1 kJ/kg-K, ma = 8 kg of steam/kg of fuel, C = 26000 kJ/kg

At 13 bar, from steam tables

hf = 814.7 kJ/kg, hfg = 1970.7 kJ/kg, Ts = 191.6°C

h = hf + hfg + cp (Tsup – Ts) = 814.7 + 1970.7 + 2.1 (270 – 191.6)

h = 2950.04

hf1 = 4.18(T-0) = 4.18(32-0) = 133.76 kJ/kg

Boiler efficiency, η = ma(h−hf1)C=8(2950.04−133.76)26000 = 0.8665 or 86.65%.

A boiler produces steam at the rate of 1500 kg per hour at 12.5 bar. The water fed to the economizer is at a temperature of 35°C and it is raised to a temperature of 110°C before feeding it to the boiler. Coal is fired at a rate of 200 kg per hour. The dryness fraction of the steam generated is 0.87. If 15% of the coal remains unburnt and the calorific value of the coal being fired is 36200 kJ/kg, determine the overall efficiency.

a) 58%

b) 65%

c) 78%

d) 82%

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Answer: a

Explanation: Given, ms = 1500 kg/h, P = 12.5 bar, T1 = 35°C, T2 = 110°C, mf = 200 kg/h, x = 0.87, C = 36200 kJ/kg

Mass of coal actually burnt per hour = 200 85100 = 170 kg/h

ma = msmf=1500170 = 8.82 kg of steam/kg of fuel

At 12.5 bar, from steam tables

hf = 806.7 kJ/kg, hfg = 1977.4 kJ/kg

h = hf + x(hfg) = 806.7 + 0.87(1977.4) = 2527.04 kJ/kg

hf1 = 4.18(T-0) = 4.18(35-0) = 146.3 kJ/kg

Overall efficiency = \frac{m_a (h-h_{f1}}{C} = \frac{8.2(2527.04-146.3)}{36200} = 0.58 or 58%.

A boiler having evaporative capacity 8 kg of steam per kg of fuel produces dry and saturated steam at 13 bar pressure. The feed water temperature is 35°C. Determine the calorific value of coal if the efficiency of the boiler is 70%.

a) 35000 kJ/kg

b) 29560 kJ/kg

c) 31000 kJ/kg

d) 30160 kJ/kg

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Answer: d

Explanation: Given, ma = 8 kg of steam/kg of fuel, P = 13 bar, T = 35°C, η = 0.7

At 13 bar, from the steam tables

hg = 2785.4 kJ/kg

h = hg = 2785.4 kJ/kg

hf1 = 4.18(T-0) = 4.18(35-0) = 146.3 kJ/kg

we know that,

η = ma(h−hf1)C

Substituting the values, we get

0.7 = 8(2785.4−146.3)C

Solving for ‘C’, we get

C = 30161.14 kJ/kg ≈ 30160 kJ/kg.

A boiler is fired with fuel having calorific value 30000 kJ/kg. It produces steam at 5 bar. The dryness fraction of the steam generated 0.85. The efficiency of the boiler is 77% and the evaporative e capacity of the boiler is 10 kg of steam per kg of fuel. Determine the temperature of feed water.

a) 35°C

b) 29°C

c) 25°C

d) 37°C

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Answer: b

Explanation: Given, C = 30000 kJ/kg, P = 5 bar, x = 0.85, η = 0.77, ma = 10 kg of steam/kg of fuel

At 5 bar, from steam tables

hf = 640.1 kJ/kg, hfg = 2170.4 kJ/kg

h = hf + x(hfg) = 640.1 + 0.85(2170.4) = 2431.39 kJ/kg

we know that,

η = ma(h−hf1)C

Substituting the values, we get

0.77 = 10(2431.39−hf1)30000

Solving for hf1, we get

hf1 = 121.39 kJ/kg

But, hf1 = 4.18(T-0)

121.39 = 4.18(T)

Therefore, T = 29°C.

Determine the evaporative capacity of a boiler in kg of steam per kg of fuel fired if the feed water temperature is 40°C, the efficiency of the boiler is 95%, calorific value of the coal fried is 33000 kJ/kg and the steam generated is dry and saturated at 12 bar pressure.

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) 15 kg of steam/kg of fuel

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Answer: c

Explanation: Given, T = 40°C, η = 0.95, C = 33000 kJ/kg, P = 12 bar

At 12 bar, from steam tables

hg = 2782.7 kJ/kg

h = hg = 2782.7 kJ/kg

hf1 = 4.18(T-0) = 4.18(40-0) = 167.2 kJ/kg

we know that,

η = ma(h−hf1)C

Substituting the values, we get

0.95 = ma(2782.7−167.2)33000

Solving for ma, we get

ma = 11.98 kg of steam/kg of fuel ≈ 12 kg of steam/kg of fuel.

Determine the enthalpy of 1 kg of steam if the boiler efficiency is 79%, the evaporative capacity of the boiler is 9 kg of steam per kg of fuel fired, the feed water temperature is 40°C and the calorific value of coal is 30000 kJ/kg.

a) 2500 kJ/kg

b) 3000 kJ/kg

c) 2800 kJ/kg

d) 2200 kJ/kg

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Answer: c

Explanation: Given, η = 0.79, ma = 9 kg of steam/ kg of fuel, T = 40°C, C = 30000 kJ/kg

hf1 = 4.18(T-0) = 4.18(40) = 167.2 kJ/kg

We know that

η = ma(h−hf1)C

Substituting the values, we get

0.79 = 9(h−167.2)30000

Solving for ‘h’, we get

h = 2800 kJ/kg.

Boiler efficiency doesn’t depend on _____

a) calorific value of fuel fired

b) specific heat of steam generated

c) boiler design

d) operation time

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Answer: dExplanation: Boiler efficiency does not depend on operation time. It depends on boiler design, heat recovery equipments, calorific value of fuel fired, specific heat of steam generated, fuel condition, firing rate etc.

In a boiler, some part of heat is lost to flue gases.

a) True

b) False

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Answer: aExplanation: The flue gases generated as product of combustion contain dry products of combustion. They also contain steam generated by the combustion of hydrogen in fuel. This heat loss can be reduced by passing the flue gases through the air preheater and economizer.