# Thermal Engineering Questions and Answers – Draught – Definition and Classification

This set of Thermal Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Draught – Definition and Classification”.

1. What is draught?
a) The condition of a dry boiler is called draught
b) It’s a small pressure difference which causes the flow of gases
c) It is a boiler mounting
d) It is a boiler accessory

Explanation: Draught is the small pressure difference which causes the flow of gas to take place. Draught helps to carry air to the fire-box and then take away the gaseous products of combustion out. Sufficient amount of air is necessary for proper combustion.

2. Draught is classified into _____
a) Artificial and natural draught
b) Induced and forced draught
c) Steam jet and mechanical draught
d) Chimney draught and boiler draught

Explanation: Draught is classified into artificial draught and natural draught. Natural draught is also known as chimney draught since it makes use of a chimney of sufficient height to create the required pressure difference.

3. Artificial draught is classified into steam jet and mechanical draught.
a) True
b) False

Explanation: Artificial draught is classified into steam jet and mechanical draught. Steam jet draught is further classified into induced and forced draught. Mechanical draught is classified into induced fan, balanced (induced and forced fan) and forced fan draught.

4. Natural draught is obtained using _____
a) Economizer
b) Induced fan
c) Chimney
d) Induced and forced fan

Explanation: Chimney is used to obtain natural draught. Chimney creates the necessary pressure difference for draught. Use of chimney also helps in discharging the products of combustion to a height where they won’t be affecting the surroundings.

5. Which of the following is the correct formula for calculating the pressure difference causing the gas flow through the combustion chamber?
a) (ρa – ρg)gH
b) (ρa + ρg)gH
c) (ρg – ρa)gH
d) (ρg + ρa)gH

Explanation: The correct formula for calculating the pressure difference causing the gas flow though the combustion chamber (or draught) is ‘(ρa – ρg)gH’.
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6. Calculate the draught in mm of water if the chimney height is 30 m, the temperatures of hot gases inside chimney and outside air are 330°C and 30°C respectively. Also, the furnace is supplied with 15 kg of air per kg of coal burnt.
a) 14.67 mm of water
b) 17.64 mm of water
c) 16.22 mm of water
d) 16.19 mm of water

Explanation:
H = 30 m, Tg = 330°C or 603 K, Ta = 30°C or 303 K, ma = 15 kg of air/kg of coal
Draught in mm of water, hw = 353*H*$$\{\frac{1}{Ta}-\frac{1}{Tg}(\frac{ma +1}{ma})\}$$
Substituting the respective values
hw = 353*30*$$\{\frac{1}{303}-\frac{1}{603}(\frac{16}{15})\}$$
hw = 16.22 mm of water

7. A chimney is 25 m high. The temperature of hot gases inside chimney is 300°C and temperature of air outside chimney is 30°C. If 20 kg of air is supplied to the furnace per kg of coal find the equivalent column of hot gases that would produce the same draught pressure.
a) 20.02 m
b) 22.40 m
c) 24.42 m
d) 22.60 m

Explanation:
H = 25 m, Tg = 300°C or 573 K, Ta = 30°C or 303 K, ma = 20 kg of air/kg of coal
Height of equivalent column of hot gases (burnt gases), H1 =H*$$\{(\frac{ma}{ma +1})\frac{Tg}{Ta}-1\}$$
Substituting the respective values
H1 =25*$$\{(\frac{20}{21})\frac{573}{303}-1\}$$
H1 = 20.02 m

8. Calculate how much air is used per kg of coal burnt, if the chimney height is 35 m which creates a draught of 20 mm of water. The temperature of hot gases inside the chimney is 330°C and temperature of air outside is 27°C.
a) 29.52 kg of air per kg of coal
b) 21.34 kg of air per kg of coal
c) 22.45 kg of air per kg of coal
d) 19.54 kg of air per kg of coal

Explanation:
H = 35 m, Tg = 330°C or 603 K, Ta = 27°C or 300 K, hw = 20 mm of water
We know that,
hw = 353*H*$$\{\frac{1}{Ta}-\frac{1}{Tg}(\frac{ma +1}{ma})\}$$
After substituting the respective values
20 = 353*35*$$\{\frac{1}{300}-(\frac{ma +1}{ma})\frac{1}{603}\}$$
Solving for ma we get
ma = 29.52 kg of air per kg of coal

9. A 28 m chimney is full of hot gases at temperature 300°C. The air supplied for complete combustion of 1 kg of coal is 18 kg. calculate draught in N/m2 if the temperature of outside air is 25°C.
a) 140.32 N/m2
b) 187.54 N/m2
c) 150.32 N/m2
d) 146.76 N/m2

Explanation:
H = 28 m, Tg = 300°C or 573 K, Ta = 25°C or 298 K, ma = 18 kg of air/kg of coal burnt
draught (in N/m2) = 353*g*H*$$\{\frac{1}{Ta}-\frac{1}{Tg}(\frac{ma +1}{ma})\}$$
substituting the respective values
draught (in N/m2) = 353*9.81*28*$$\{\frac{1}{298}-\frac{1}{573}(\frac{19}{18})\}$$
draught = 146.76 N/m2

10. A chimney 30 m high produces a draught of 16 mm of water. The temperature of hot gases inside the chimney is 300°C. Calculate the temperature of outside air if 20 kg of air is supplied for complete combustion of 1 kg of fuel.
a) 30°C
b) 27°C
c) 25°C
d) 20°C

Explanation:
H = 30 m, hw = 16 mm of water, Tg = 300°C or 570 K, ma = 20 kg of air/kg of fuel
we know that,
hw = 353*H*$$\{\frac{1}{Ta}-\frac{1}{Tg}(\frac{ma +1}{ma})\}$$
Substituting the respective values, we get,
16 = 353*30*$$\{\frac{1}{Ta}-\frac{1}{573}(\frac{21}{20})\}$$
Solving for Ta we get,
Ta = 299.1 ≈ 300 K or 27°C

11. Find the temperature of the hot gases inside the chimney if the draught produced by a 25 m high chimney is 15 mm of water. The temperature of the air outside is 25°C and 18 kg of air is supplied for complete combustion of 1 kg of fuel.
a) 298°C
b) 330°C
c) 456°C
d) 364°C

Explanation:
H = 25, Ta = 25°C or 298 K, ma = 18 kg of air/kg of fuel, draught = 15 mm of water
We know that,
hw = 353*H*$$\{\frac{1}{Ta}-\frac{1}{Tg}(\frac{ma +1}{ma})\}$$
Substituting the values
15 = 353*25*$$\{\frac{1}{298}-\frac{1}{Tg}(\frac{19}{18})\}$$
Solving for Tg we get,
Tg = 637.42 K ≈ 637 K or 364°C

12. Find draught in terms of column of hot gases, if the temperature of hot gases inside the chimney is 350°C. The height of the chimney is 30 m. The temperature of outside air is 25°C. The air supplied for complete combustion of 1 kg of fuel is 20 kg.
a) 29.73 m
b) 32.46 m
c) 45.32 m
d) 30.25 m

Explanation:
H = 30 m, Tg = 350°C or 623 K, Ta = 25°C or 298 K, ma = 20 kg of air/kg of fuel
We know that, draught in terms of column of hot gases is given by –
H1 = H*$$\{(\frac{ma}{ma +1})\frac{Tg}{Ta}-1\}$$
Substituting the values H1 = 30*$$\{(\frac{20}{21})\frac{623}{298}-1\}$$
Therefore, H1 = 29.73 m

13. A 27 m high chimney produces a draught of 30 m in terms of Column of hot gases. The amount of air supplied for complete combustion of 1 kg of fuel is 20 kg. If the temperature of outside air is 27°C, determine the temperature of the hot gases inside the chimney.
a) 456°C
b) 392°C
c) 423°C
d) 398°C

Explanation:
H = 27 m, H1 = 30 m (column of hot gases), ma = 20 kg of air/kg of fuel, Ta = 27°C or 300 K
we know that,
H1 =H*$$\{(\frac{ma}{ma +1})\frac{Tg}{Ta}-1\}$$
Substituting the values, we get
30 =27*$$\{(\frac{20}{21}\frac{Tg}{300}-1\}$$
Solving for Tg, we get
Tg = 665 K or 392°C

14. How much air is used per kg of coal burnt in a boiler having a chimney of height 32 m to create draught of 170 N/m2. The temperature of outside air is 25°C and hot gases inside chimney is 330°C.
a) 10.16 kg of air per kg of coal burnt
b) 12.00 kg of air per kg of coal burnt
c) 14.32 kg of air per kg of coal burnt
d) 19.25 kg of air per kg of coal burnt

Explanation:
H = 32 m, draught = 170 N/m2, Ta = 25°C or 298 K, Tg = 330°C or 603 K
The following formula is used to calculate draught in N/m2
draught (in N/m2) = 353*g*H*$$\{\frac{1}{Ta}-\frac{1}{Tg}(\frac{ma +1}{ma})\}$$
Substituting the respective values, we get
170 = 353*9.81*32*$$\{\frac{1}{298}-\frac{1}{603}(\frac{ma +1}{ma})\}$$
Solving the above equation for ma, we get
ma = 10.16 kg of air per kg of coal burnt

15. Find out the temperature of outside air if a chimney 30 m high creates a draught of 25 m in terms of column if hot gases. The temperature of hot gases is 330°C. 18 kg of air is to be supplied for complete combustion of 1 kg of fuel.
a) 38°C
b) 28°C
c) 25°C
d) 32°C

Explanation:
H = 30 m, H1 = 25 m of hot gases, Tg = 330°C or 603 K, ma = 18 kg of air/kg of fuel
we know that,
H1 =H*$$\{\frac{ma}{ma +1}\frac{Tg}{Ta}-1\}$$
Substituting the values, we get
25 =30*$$\{\frac{18}{19}\frac{603}{Ta}-1\}$$
Solving for Tg, we get
Tg = 311.6 K ≈ 311 K or 38°C
Symbols
ρa – mass density of air
ma – mass of air supplied per kg of fuel
ρg – mass density of hot gases
hw – draught in terms of column of water
H1 – draught in terms of column of hot gases
Ta – absolute temperature of atmosphere
Tg – average absolute temperature of chimney gases
g – gravitational acceleration
H – chimney height

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