This set of Thermal Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Centrifugal Compressor”.
1. A centrifugal compressor delivers a certain quantity of air at a pressure of 1.65bar and 87°C. The intake conditions are 17°C and 1bar. Which of the following is the isentropic efficiency of the compressor? Consider n = 1.625.
a) 87.98%
b) 100%
c) 74.36%
d) 42.36%
View Answer
Explanation:
Given: p2 = 1.65bar; T2 = 87°C = 87 + 273 = 360K;
T1 = 17°C = 17 + 273 = 290K; p1 = 1bar; n = 1.625;
Let T2‘ = Temperature of air at exit for isentropic compression,
We know that, \(\frac{T_2′}{T_1} = (\frac{p_2}{p_1})^{\frac{n-1}{n}} = (\frac{1.65}{1})^{\frac{1.625-1}{1.625}}\)
= (1.65)0.384 = 1.212
T2‘ = T1 × 1.212
= 290 × 1.212
T2‘ = 351.59K
We know that the isentropic efficiency is given by,
ηi = \(\frac{T_2′-T_1}{T_2-T_1}\)
= \(\frac{351.59-290}{60-290}\)
ηi = .8798 or 87.98%
2. A centrifugal compressor with 65% isentropic efficiency delivers certain amount of air at a pressure of 2.75bar and the temperature of air at the end of isentropic compression is given as 127.53°C. If compressor receives air at 27°C and at a pressure of 1bar, then which of the following is the actual temperature of the air at exit. Assuming γ and cp for air as 1.4 and 1kJ/kg-K respectively.
a) 18.16K
b) 181.66K
c) 181.66°C
d) 18.16°C
View Answer
Explanation:
Given: ηi = 65% = 0.65; p2 = 2.75; T1 = 27°C = 27 + 273 = 300K;
p1 = 1bar; T2‘ = 127.53°C = 127.53 + 273 = 400.53K; γ = 1.4; cp = 1kJ/kg-K;
Let, T2 = Actual temperature of the air at exit, and
T2‘ = Temperature of the air at exit of isentropic compression,
We know that isentropic efficiency is given by,
ηi = \(\frac{T_2′-T_1}{T_2-T_1}\)
0.65 = \(\frac{400.53-300}{T_2-300}\)
T2 – 300 = \(\frac{400.53-300}{.65}\)
T2 = 300 + 154.66 = 454.66K
∴ T2 = 454.66 – 273 = 181.66°C
3. A centrifugal compressor with 92% mechanical efficiency delivers 32kg of air per minute. If the compressor receives air at 27°C and the actual temperature of air at exit is 167°C, then which of the following is the power required to run the compressor. Consider γ and cp for air as 1.4 and 1kJ/kg-K respectively.
a) 811.5W
b) 811.5kW
c) 81.15kW
d) 81.15W
View Answer
Explanation:
Given: ηm = 92% = 0.92; m = 32kg/min; T1 = 27°C = 27 + 273 = 300K;
T2 = 167°C = 167 + 273 = 440K; cp = 1kJ/kg-K;
We know that the work done in compressing the air isentropically is given by,
Wisen = mcp (T2 – T1)
= 32 × 1 × (440 – 300)
= 32 × 140
= 4480kJ/min or \(\frac{4480}{60}\) = 74.66kW
We also know that the mechanical efficiency is given by;
ηm = \(\frac{W_{isen}}{W_{act}}\)
∴ Wact = \(\frac{W_{isen}}{η_m}\)
= \(\frac{74.66}{0.92}\) = 81.15kW
4. A centrifugal compressor compresses a certain amount of the air polytropically according to the law pv1.54 = constant. Which of the following is the polytropic efficiency of the compressor? Consider γ = 1.4.
a) 81.48%
b) 100%
c) 84.18%
d) 54.28%
View Answer
Explanation:
Given: n = 1.54; γ = 1.4;
We know that the polytropic efficiency is given by,
ηp = \((\frac{\gamma – 1}{\gamma} \frac{n}{n-1})\)
= \((\frac{1.4-1}{1.4} \frac{1.54}{1.54-1})\)
= 0.285 × 2.851
= 0.8148 or 81.48%
5. A centrifugal air compressor having blade velocity of 260m/s receives air at 27°C. Which of the following is the temperature rise of the air? Consider cp = 1kJ/kg-K.
a) 67.6°C
b) 27°C
c) 60.5°C
d) 122°C
View Answer
Explanation:
Given: Vb = 260m/s; T1 = 27°C = 27 + 273 = 300K; cp = 1kJ/kg-K;
We know that the work done by the compressor per kg of air is given by,
W = \(\frac{V_b^2}{1000}\)
= \(\frac{260^2}{1000}\) = 67.6kJ
We also know that work done by the compressor per kg of air is given by,
W = cp(T2 – T1)
67.6 = 1 × (T2 – T1)
(T2 – T1) = 67.6°C or K
6. A centrifugal air compressor compresses air from an initial pressure and temperature of 1bar and 15°C to some exit temperature by taking 185kJ/kg of work done. Which of the following is the delivery pressure of the air? Consider γ = 1.4 and cp = 1kJ/kg-K.
a) 6.28bar
b) 2.68bar
c) 5.68bar
d) 56.8bar
View Answer
Explanation:
Given: p1 = 1bar; T1 = 15°C = 15 + 273 = 288K;
W = 185kJ/kg; γ = 1.4; cp = 1kJ/kg-K;
We know that work done per kg of air is given by,
W = cp (T2 – T1)
185 = 1 × (T2 – 288)
T2 = 473K
We also know that,
\(\frac{T_2}{T_1} = (\frac{p_2}{p_1})^{\frac{\gamma-1}{\gamma}}\)
\(\frac{473}{288} = (\frac{p_2}{1})^{\frac{1.4-1}{1.4}}\)
1.642 = (p2)0.285
log(1.642) = 0.285 × log(p2)
0.215 = 0.285 × log(p2)
log(p2) = 0.754
p2 = 5.68bar
7. A centrifugal air compressor receives air at a pressure and temperature of 1bar and 27°C. The compressor polytropically compresses the air and delivers it at a temperature of 197°C. Which of the following is the index of compression if the heat transfer from the air to the jacket water is 140kJ? Consider and γ = 1.4 and cp = 1kJ/kg-K.
a) n = 1.07
b) n = 1.4
c) n = 1
d) n = 1.6
View Answer
Explanation:
Given: p1 = 1bar; T1 = 27°C = 27 + 273 = 300K; Q = 140kJ;
T2 = 197°C = 197 + 273 = 470K; γ = 1.4; cp = 1kJ/kg-K;
We know that work done per kg of air is given by,
W = cp (T2 – T1)
W = 1 × (470 – 300)
W = 170kJ
We also know that heat transfer in polytropic process is given by,
Q = \(\frac{\gamma-n}{\gamma – 1}\) × W
140 = \(\frac{1.4-n}{1.4-1}\) × 170
\(\frac{140}{170} = \frac{1.4-n}{1.4-1}\)
0.823 = \(\frac{1.4-n}{0.4}\)
0.823 × 0.4 = 1.4 – n
0.329 = 1.4 – n
n = 1.07
8. Which of the following is done in order to reduce the shock waves?
a) Prewhirl
b) Whirling
c) Post-whirl
d) Degree of reaction
View Answer
Explanation: Due to the higher r.p.m. of the compressor, the inlet velocity becomes high and because of this, the air steam tends to separate from the trailing face of the curved vane. This causes the shock waves to form and contributes to loss of energy. To reduce the shock waves, pre-rotation of air is done before it enters the impeller, also known as prewhirl.
9. A centrifugal air compressor having an internal diameter of 280mm compresses 36kg of air per minute while running at 4500rpm. The vane angles at inlet and outlet are 35° and 40° respectively. Which of the following is the velocity of flow at inlet?
a) 46.19m/s
b) 42.18m/s
c) 36m/s
d) 65.97m/s
View Answer
Explanation:
Given: D = 280mm = 0.28m; m = 36kg/min = 0.6kg/s;
N = 4500r.p.m.; θ = 35°; ϕ = 40°;
We know that impeller velocity at inlet is given by,
Vb = \(\frac{\pi D N }{60}\)
= \(\frac{\pi × 0.28 × 4500}{60}\)
f = Vb tan(θ)
= 65.97 × tan(35°)
= 46.19m/s
10. A centrifugal air compressor having internal diameters of 240mm compresses 24kg of air per minute while running at 3500rpm. The vane angles at inlet and outlet are 32° and 37° respectively. Which of the following is the value of thickness of the blade if the impeller contains 35 blades and the velocity of flow at inlet is given as 32m/s? Consider a specific volume of air as 0.8 m3/kg.
a) 212mm
b) 21.2mm
c) 2.12mm
d) 35.2mm
View Answer
Explanation:
Given: D = 240mm = 0.24m; m = 24kg/min = 0.4kg/s; N = 3500r.p.m.;
θ = 32°; ϕ = 37°; υs = 0.8m3/kg; Vf = 32m/s; n = 35;
We know that mass of air flowing through the impeller is given by,
m = \(\frac{(\pi D-nb) V_f}{υ_s}\);
0.4 = \(\frac{(\pi × 0.24 – 35b) 32}{0.8}\)
\(\frac{0.4 × 0.8}{32}\) = (π × 0.24 – 35b)
0.01 = 0.753 – 35b
b = \(\frac{0.753 – 0.01}{35}\)
= 0.0212m = 21.2mm
Sanfoundry Global Education & Learning Series – Thermal Engineering
To practice all areas of Thermal Engineering, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]
- Practice Chemical Engineering MCQs
- Apply for Mechanical Engineering Internship
- Practice Mechanical Engineering MCQs
- Check Chemical Engineering Books
- Check Mechanical Engineering Books
