Thermal Engineering Questions and Answers – Natural Draught

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This set of Thermal Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Natural Draught”.

1. A 25 m high chimney is filled with hot gases of temperature 330°C. The temperature of outside air is 27°C. If 20 kg of air is supplied for complete combustion of 1 kg of fuel, calculate the velocity of hot gases though the chimney.
a) 21.18 m/s
b) 25.65 m/s
c) 18.56 m/s
d) 15.65 m/s
View Answer

Answer: a
Explanation: H = 25 m, Tg = 330°C or 603 K, Ta = 27°C or 300 K, ma = 20 kg of air/kg of fuel
we know that,
Draught in terms of column of hot gases, H1 = H*\(\{(\frac{m_a}{m_a+1})\frac{T_g}{T_a} -1\}\)
Substituting the values
H1 = 25*\(\{(\frac{20}{21})\frac{603}{300}-1\}\)
H1 = 22.86 m
The velocity of hot gases through chimney is given by, V = \(\sqrt{2*g*H_1}\)
V = \(\sqrt{(2*9.81*22.86)}\)
V = 21.18 m/s.
Symbols:
h – Actual draught in terms of column of hot gases
H1 – Theoretical draught in terms of column of hot gases
ma – mass of air supplied per kg of fuel
Ta – absolute temperature of atmosphere
Tg – average absolute temperature of chimney gases
H – chimney height.
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2. A chimney has a height 30 m. The ambient temperature and the temperature of hot gases inside the chimney are 27°C and 300°C respectively. Find the velocity of the flue gases passing through the chimney if 50% of the theoretical draught is lost in friction at the grate and passage. It is given that 20 kg of air is required for complete combustion of 1 kg of fuel.
a) 21.35 m/s
b) 20.65 m/s
c) 15.52 m/s
d) 10.32 m/s
View Answer

Answer: c
Explanation: H = 30 m, Ta = 27°C or 300 K, Tg = 300°C or 573 K, ma = 20 kg of air/kg of fuel
we know that,
Draught in terms of column of hot gases, H1 = H*\(\{(\frac{m_a}{m_a+1})\frac{T_g}{T_a} -1\}\)
Substituting the values, we get
H1 = 30*\(\{(\frac{20}{21})\frac{573}{300}-1\}\)
H1 = 24.57 m
Available head, h = 0.5*H1
h = 0.5 * 24.57
h = 12.28 m
Velocity of the flue gases , V = \(\sqrt{2*g*h}\)
V = \(\sqrt{2*9.81*12.28}\)
V = 15.52 m/s.
Symbols:
h – Actual draught in terms of column of hot gases
H1 – Theoretical draught in terms of column of hot gases
ma – mass of air supplied per kg of fuel
Ta – absolute temperature of atmosphere
Tg – average absolute temperature of chimney gases
H – chimney height
g – gravitational acceleration

3. The velocity of flue gases though chimney is 23 m/s. The chimney height is 35 m. The temperature of the flue gases is 300°C. The flue gases formed per kg of fuel burnt is 23 kg. Find out the ambient temperature.
a) 30°C
b) 27°C
c) 40°C
d) 37°C
View Answer

Answer: d
Explanation: V = 23 m/s, H = 35 m, Tg = 300°C or 573 K, ma +1 = 23 kg of flue gases/kg of fuel
ma = 22 kg of air/kg of fuel burnt
we know that,
Velocity of flue gases though chimney, V = \(\sqrt{2*g*H_1}\)
Substituting the values
23 = \(\sqrt{2*9.81*H_1}\)
Solving for H1, we get
H1 = 26.96 m
Also,
H1 = H*\(\{(\frac{m_a}{m_a+1})\frac{T_g}{T_a} -1\}\)
Substituting the values
26.96 = 35*\(\{(\frac{22}{23}) \frac{573}{T_a}-1\}\)
Solving for Ta, we get
Ta = 309.6 K ≈ 310 K or 37°C.
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4. The velocity and temperature of flue gases passing through a chimney are 20 m/s and 330°C. 18 kg of air is supplied for complete combustion of 1 kg of fuel. The ambient temperature is 27°C. Determine the chimney height 40% of the theoretical draught is lost in friction at the grate and passage.
a) 58.32 m
b) 37.57 m
c) 29.65 m
d) 35.26 m
View Answer

Answer: b
Explanation: V = 20 m/s, Tg = 330°C or 603 K, ma = 18 kg of air/kg of fuel, Ta = 27°C or 300 K
we know that,
V = \(\sqrt{2*g*h}\)
Substituting the respective values
20 = \(\sqrt{2*9.81*h}\)
Solving for h, we get
h = 20.38 m
Theoretical draught in terms of column of hot gases, H1 = \(\frac{h}{60}\)*100
H1 = \(\frac{20.38}{60}\)*100
H1 = 33.97 m
Also
H1 = H*\(\{(\frac{m_a}{m_a+1})\frac{T_g}{T_a} -1\}\)
Substituting the respective values
33.97 = H*\(\{(\frac{18}{19})\frac{603}{300}-1\}\)
Solving for ‘H’, we get
H = 37.57 m.

5. Which of the following expressions represents the correct ratio of absolute temperature of the chimney gases and absolute temperature of outside air when the discharge through the chimney is maximum.
a) \(\frac{T_g}{T_a} = \frac{m_a+1}{m_a} \)
b) \(\frac{T_g}{T_a} = \frac{m_a}{m_a+1} \)
c) \(\frac{T_g}{T_a} = 2(\frac{m_a+1}{m_a }) \)
d) \(\frac{T_g}{T_a} = 2(\frac{m_a+1}{m_a }) \)
View Answer

Answer: c
Explanation: The natural draught or the chimney draught is more effective when the maximum weight of hot gases is discharged in a given time.
At this condition the ratio of absolute temperature of hot gases to absolute temperature of outside air bears a specific value, which is
\(\frac{T_g}{T_a} = 2(\frac{m_a+1}{m_a }) \)
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6. What is the ratio of draught in terms of column of hot gases to the chimney height when the discharge though chimney is maximum?
a) 2.0
b) 1.5
c) 1.0
d) 0.5
View Answer

Answer: c
Explanation: When the discharge though chimney is maximum, the ratio of draught in terms of column of hot gases to chimney height becomes 1.
That is, H1 = H
It is also the maximum value of H1.

7. The frictional resistance offered to the flow of flue gases by the grate and gas passages result in draught losses.
a) True
b) False
View Answer

Answer: a
Explanation: Apart from resistance offered by grate and gas flow passage, draught is also lost near the bends in the gas flow circuit. Also presence of friction head in equipments like superheater, economizer etc. causes loss in a draught.
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8. The draught produced by the chimney is due to the difference in densities between the column of hot gases inside the chimney and the cold air present outside.
a) True
b) False
View Answer

Answer: a
Explanation: Draught is classified into natural draught and artificial draught. Natural draught is produced by chimney. It produces draught by virtue of density difference between the hot air column inside the chimney and the cold air outside.

9. A chimney of height 25 m is used to provide natural draught. The ambient temperature is 25°C and the temperature of the hot gases inside the chimney is 352°C. If the discharge though the chimney is maximum determine the mass of air supplied per kg of fuel burnt.
a) 20.55 kg of air/kg of fuel burnt
b) 18.54 kg of air/kg of fuel burnt
c) 15.23 kg of air/kg of fuel burnt
d) 25.21 kg of air/kg of fuel burnt
View Answer

Answer: a
Explanation: Ta = 25°C or 298 K, Tg = 352°C or 625 K
It is given that the discharge is maximum, therefore
\(\frac{T_g}{T_a} =2(\frac{m_a+1}{m_a }) \)
Substituting the respective values
\(\frac{625}{298}=2(\frac{m_a+1}{m_a }) \)
Solving for ma, we get
ma = 20.55 kg of air/kg of fuel burnt.
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10. Determine the temperature of hot gases inside the chimney if the temperature of cold air outside is 27°C and 20 kg of air is required for complete combustion of 1 kg of fuel. It is given that the discharge though the chimney is maximum.
a) 335°C
b) 300°C
c) 318°C
d) 357°C
View Answer

Answer: d
Explanation: Ta = 27°C or 300 K, ma = 20 kg of air/kg of fuel
According to the given condition
\(\frac{T_g}{T_a} = 2(\frac{m_a+1}{m_a }) \)
Substituting the respective values
\(\frac{T_g}{300} = 2(\frac{21}{20}) \)
Solving for Tg, we get
Tg = 630 K or 357°C.

11. A chimney of height 45 m is full of hot gases at temperature 330°C. If the discharge through the chimney is maximum, determine the draught in terms of column of hot gases. The ambient is 25°C.
a) 45 m
b) 35 m
c) 25 m
d) 39 m
View Answer

Answer: a
Explanation: H = 45 m, Tg = 330°C or 603 K, Ta = 25°C or 298 K
Since it is given that the discharge though the chimney is maximum.
Draught in terms of column of hot gases, H1 = H (Chimney height)
H1 = 45 m.

12. A boiler has a chimney of height 45 m. The ambient temperature is 25°C. The temperature of the hot gases inside chimney is 330°C. Calculate the draught in terms of column of water if it is given that discharge through the chimney is maximum.
a) 40.23 m
b) 35.45 m
c) 26.65 m
d) 24.32 m
View Answer

Answer: c
Explanation: H = 45 m, Ta = 25°C or 298 K, Tg = 330°C or 603 K
It is given that discharge though the chimney is maximum, therefore
hw = 176.5 \(\frac{H}{T_a} \)
Substituting the values, we get
hw = 176.5 \(\frac{45}{298} \)
hw = 26.65 m.
Symbols –
hw – draught in terms of column of water
Ta – absolute temperature of atmosphere
Tg – average absolute temperature of chimney gases
H – chimney height.

13. The temperature of flue gases was observed to be 350°C with a chimney (natural draught) of 45 m. The same draught was developed using an induced draught fan and the temperature of the flue gases was measured to be 150°C. The mass of air supplied for complete combustion of 1 kg of fuel is 20 kg. The ambient temperature is 30°C. Assuming cp = 1.004 kJ/kg for flue gases calculate the efficiency of chimney.
a) 0.21%
b) 0.56%
c) 1.23%
d) 0.11%
View Answer

Answer: a
Explanation: H = 45 m, Tg1 = 350°C or 623 K, Tg2 = 150°C or 423 K, ma = 20 kg of air/kg of fuel
Ta = 30°C or 303 K
Chimney efficiency is given by-
η = \(\frac{H\big\{(\frac{m_a}{m_a+1})\frac{T_{g1}}{T_a}-1\big\}*g}{cp(T_{g1}-T_{g2})*1000 }*100 \)
Substituting the respective values
η = \(\frac{45\big\{(\frac{20}{21})\frac{623}{303}-1\big\}*9.81}{1.004(623-423)*1000}*100 \)
η = 0.21%.
Symbols –
ma – mass of air supplied per kg of fuel
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.

14. A chimney is used to develop natural draught of 20 mm of water. The ambient temperature is 30°C. Determine the chimney height if the discharge though the chimney is maximum.
a) 25.36 m
b) 34.33 m
c) 45.32 m
d) 41.32 m
View Answer

Answer: b
Explanation: hw = 20 m of water, Ta = 30°C or 303 K
Since the discharge through the chimney is maximum
hw = 176.5 \(\frac{H}{T_a} \)

Substituting the respective values
20 = \(176.5\frac{H}{303} \)
Solving for H, we get
H = 34.33 m.

15. Determine the ambient temperature if the temperature of flue gases inside the chimney is 363°C and the 20 kg of air is supplied for complete combustion of 1 kg of fuel. The discharge though the chimney is maximum.
a) 27°C
b) 35°C
c) 30°C
d) 25°C
View Answer

Answer: c
Explanation: Tg = 363°C or 636 K, ma = 20 kg of air/kg of fuel
Since the discharge though the chimney is maximum
\(\frac{T_g}{T_a} = 2(\frac{m_a+1}{m_a }) \)

Substituting the values
\(\frac{636}{Ta} = 2(\frac{21}{20}) \)

Solving for Ta, we get
Ta = 302.86 K ≈ 303 K or 30°C.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter