This set of Thermal Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Rotary Compressor”.
1. A roots blower compressor compresses 0.08m3 of air from 1bar to 1.8bar per revolution. Which of the following is the actual work done per revolution?
a) 6.25kJ
b) 5.6kJ
c) 5.4kJ
d) 6.4kJ
View Answer
Explanation:
Given: v1 = 0.08m3; p1 = 1bar = 1 × 105 N/m2; p2 = 1.8bar = 1.8 × 105 N/m2;
We know that the actual work done per revolution is given by,
Wact = v1 (p2 – p1)
Wact = 0.08(1.8 × 105 – 1 × 105)
Wact = 0.08 × 0.8 × 105
Wact = 6400J or Wact = 6.4kJ
2. A roots blower compressor compresses 0.06m3 of air from 1bar to 1.6bar per revolution. Which of the following is the ideal work done per revolution?
a) 3288.59J
b) 3000.13J
c) 1518.13J
d) 3018.13J
View Answer
Explanation:
Given: v1 = 0.06m3; p1 = 1bar = 1 × 105 N/m2; p2 = 1.6bar = 1.6 × 105 N/m2;
We know that the ideal work done per revolution is given by,
Wideal = \(\frac{\gamma}{\gamma – 1}\) × p1 v1 \(\bigg[(\frac{p_2}{p_1} )^{\frac{\gamma – 1}{\gamma}} – 1\bigg] \)
= \(\frac{1.4}{1.4-1}\) × 1 × 105 × 0.06\(\bigg[(\frac{1.6}{1})^{\frac{1.4-1}{1.4}} – 1\bigg] \)
= 21000 × [1.143 – 1]
Wideal = 3018.13J
3. A roots blower compressor consumes 4386N-m of actual work done. The required ideal work done is 3500N-m. Which of the following is the efficiency of the compressor in percentage?
a) 79.79%
b) 72.52%
c) 40.79%
d) 100%
View Answer
Explanation:
Given: Wact = 4386J; Wideal = 3500J;
We know that the efficiency of the compressor is given by,
η = \(\frac{W_ideal}{W_act}\)
η = \(\frac{3500}{4386}\)
η = 0.7979 or 79.79%
4. A centrifugal compressor delivers 45kg per minute of air at a pressure of 3.2bar and 83°C. The intake pressure and temperature of the air is 1bar and 27°C. Which of the following is the index of compression? Consider R = 287J/kg-K.
a) 1.171
b) 1.4
c) 1.2
d) 2
View Answer
Explanation:
Given: m = 45kg/min; p1 = 1bar; T1 = 27°C = 273 + 27 = 300K;
p2 = 3.2bar; T2 = 83°C = 83 + 273 = 356K; R = 287J/kg-K;
Let n = index of compression,
We know that, \(\frac{T_2}{T_1} = (\frac{p_2}{p_1})^{\frac{n-1}{n}}\)
\(\frac{356}{300} = (\frac{3.2}{1})^{\frac{n-1}{n}}\)
1.186 = (3.2)\(^{\frac{n-1}{n}}\)
log(1.186) = \(\frac{n-1}{n}\) log(3.2)
0.074 = \(\frac{n-1}{n}\) × 0.505
0.074n = 0.505n – 0.505
0.431n = 0.505
n = 1.171
5. A centrifugal compressor delivers 32kg per minute of air at a pressure of 2.75bar and 87°C. The intake pressure and temperature of the air is 1bar and 15°C. Which of the following is the power required, if the compression is isothermal? Consider R = 287J/kg-K.
a) 44.56kW
b) 40.39kW
c) 56.85kW
d) 26.73kW
View Answer
Explanation:
Given: m = 32kg/min; p1 = 1bar; T1 = 15°C = 273 + 15 = 288K;
p2 = 2.75bar; T2 = 87°C = 87 + 273 = 360K; R = 287J/kg-K;
We know that work done by the compressor if the compression is isothermal,
W = 2.3 mRT1 log(r)
Where, r = \(\frac{p_2}{p_1}\) = \(\frac{2.75}{1}\) = 2.75
W = 2.3 × 32 × 287 × 288 × log(2.75)
W = 6083481.6 × 0.4395
W = 2673690.163J/min = 2673.69kJ/min
∴ Power required, P = \(\frac{2673.69}{60}\) = 44.56kW
6. A centrifugal air compressor having a pressure compression ratio of 4.75 compresses air at the rate of 7.5kg/s. If the initial pressure and temperature of the air is 1bar and 17°C, then which of the following is the final temperature of the gas? Consider γ = 1.4 and cp = 1kJ/kg-K.
a) 259.62°C
b) 279.62°C
c) 179.62°C
d) 156.25°C
View Answer
Explanation:
Given: \(\frac{p_2}{p_1}\) = 4.75; m = 7.5kg/sec; p1 = 1bar; T1 = 17°C = 17 + 273 = 290K;
γ = 1.4; cp = 1kJ/kg-K;
Let T2 = Final temperature of the gas,
We know that, \(\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{\frac{\gamma – 1}{\gamma}}\)
\(\frac{T_2}{T_1}\) = (4.75)\(^{\frac{1.4-1}{1.4}}\) = 1.56
∴ T2 = T1 × 1.56
= 290 × 1.56
T2 = 452.62K = 179.62°C
7. In a rotary air compressor the work done by the compressor is 358kJ. Which of the following is the heat exchanged with the jacket water? The compression follows the relation pv1.6 = constant.
a) -179kJ
b) +179kJ
c) -259kJ
d) +259kJ
View Answer
Explanation:
Given: W = 358kJ; n = 1.6;
We know that the heat exchanged in a polytropic process,
Q = \(\frac{\gamma – n}{\gamma – 1}\) × W
Q = \(\frac{1.4-1.6}{1.4-1}\) × 358
Q = -179kJ
8. A centrifugal compressor running at 1500r.p.m. the outer diameter of the blade tip is 540mm. Which of the following is the tangential velocity of the outer blade tip?
a) 72.41m/s
b) 65.75m/s
c) 49.29m/s
d) 42.41m/s
View Answer
Explanation:
Given: N = 1500r.p.m.; D1 = 540mm = 0.54m;
We know that tangential velocity of the outer blade tip,
Vb1 = \(\frac{\pi × D_1 × N}{60}\)
= \(\frac{\pi × 0.54 × 1500}{60}\)
= 42.41m/s
9. A centrifugal compressor running at 1200r.p.m. the tangential velocity of the outer blade tip is 52m/sec. Which of the following is the work done per kg of air?
a) 2704kW
b) 27.04kW
c) 4504kW
d) 2.704kW
View Answer
Explanation:
Given: N = 1200r.p.m.; Vb1 = 52m/s;
We know that work done per kg of air is given by,
W = m(Vb1)2
= 1 × (52)2
= 2704W
= 2.704kW
10. A centrifugal compressor running at 1000r.p.m. receives air at 25°C. If the work done per kg of air is 5.675kW, then which of the following is the temperature of the air leaving the compressor? Consider cp =1kJ/kg K.
a) 303.675°C
b) 30.675°C
c) 306.75°C
d) 3.0675°C
View Answer
Explanation:
Given: N = 1000r.p.m.; T1 = 25°C = 25 + 273 = 298K; W = 5.675kW;
We know that work done is given by,
W = mcp (T2 – T1)
5.675 = 1 × 1 × (T2 – 298)
T2 = 5.675 + 298
T2 = 303.675K
T2 = 30.675°C
Sanfoundry Global Education & Learning Series – Thermal Engineering
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