Thermal Engineering Questions and Answers – Nozzle Efficiency

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

1. The final velocity obtained after passing the steam through a nozzle is less than the calculated one. Which of the following is NOT a valid reason for the same?
a) Friction between steam and nozzle surface
b) Steam not being superheated
c) Shock loses
d) Internal friction of steam
View Answer

Answer: b
Explanation: The final velocity of the steam after passing it through a nozzle is reduced due to the internal friction of the steam, friction between steam and nozzle surface. Shock loses are also responsible for reduction of final velocity. The quality of steam being passed has no effect on the enthalpy drop.
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2. Which of the following is NOT an effect of frictional loses in a convergent-divergent nozzle?
a) Enthalpy drop is increased
b) The expansion is not isentropic
c) The final dryness fraction of the steam is increased
d) The specific volume of steam is increased
View Answer

Answer: a
Explanation: The enthalpy drop is reduced due to the frictional loses in a convergent-divergent nozzle. The expansion is not isentropic. The dryness fraction of the steam is increased i.e. quality of the steam is improved because the steam absorbs heat generated by friction. As the steam becomes more dry, the specific volume of the steam is increased.

3. Presence of friction in a convergent-divergent nozzle, decreases the final velocity of the steam and increases the dryness fraction of the steam.
a) True
b) False
View Answer

Answer: a
Explanation: Final velocity of the steam is reduced due to shock loses and presence of friction in the nozzle. Friction between the nozzle surface and steam and the internal friction of the steam generates heat. This heat is absorbed by the steam and the dryness fraction of the steam is increased.
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4. Determine the velocity coefficient if the efficiency of a convergent-divergent steam nozzle is 92%. The inlet velocity of steam is negligible.
a) 0.959
b) 0.120
c) 0.465
d) 0.542
View Answer

Answer: a
Explanation: Given, η = 0.92
When inlet velocity is negligible, the velocity coefficient is the square root of the nozzle efficiency.
Velocity coefficient = \(\sqrt{η} = \sqrt{0.92}\) = 0.959.

5. Dry and saturated steam at 12 bar expands through a convergent-divergent nozzle having 90% efficiency to 2 bar. Determine the actual exit velocity of the steam. Neglect the initial velocity of steam.
a) 451.20 m/s
b) 754.62 m/s
c) 650.32 m/s
d) 856.96 m/s
View Answer

Answer: b
Explanation: P1 = 12 bar, η = 0.90, P2 = 2 bar
At 12 bar, from steam tables
h1 = hg1 = 2782.7 kJ/kg, s1 = sg1 = 6.519 kJ/kg-K
At 2 bar, from steam tables
hf2 = 504.7 kJ/kg, hfg2 = 2201.6 kJ/kg
sf2 = 1.530 kJ/kg-K, sfg2 = 5.597 kJ/kg-K
s1 = s2
s1 = sf2 + x2(sfg2)
6.519 = 1.530 + x2(5.597)
x2 = 0.891
h2 = hf2 + x2(hfg2)
= 504.7 +0.891(2201.6)
= 2466.32 kJ/kg
C2 = 44.72\(\sqrt{η(h_1-h_2)}\)
C2 = 44.72\(\sqrt{0.90(2782.7-2466.32)}\)
C2 = 754.62 m/s.
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6. Initial velocity of the steam entering a convergent-divergent steam nozzle is 60 m/s. Steam expands from 15 bar and 250°C to 1 bar. Determine the exit velocity of steam if nozzle has an efficiency of 95%.
a) 973.08 m/s
b) 654.32 m/s
c) 785.45 m/s
d) 900.75 m/s
View Answer

Answer: a
Explanation: P1 = 15 bar, T1 = 250°C, η = 0.95, P2 = 1 bar, C1 = 60 m/s
At 15 bar and 200°C, from steam tables
h1 = 2923.5 kJ/kg, s1 = 6.710 kJ/kg-K
At 1 bar, from steam tables
hf2 = 417.5 kJ/kg, hfg2 = 2257.9 kJ/kg
sf2 = 1.303 kJ/kg-K, sfg2 = 6.057 kJ/kg-K
s1 = s2
s1 = sf2 + x2(sfg2)
6.70 = 1.303 + x2(6.057)
x2 = 0.89
h2 = hf2 + x2(hfg2)
= 417.5 + 0.89(2257.9)
= 2427.03 kJ/kg
C2 = \(\sqrt{2*1000*η(h_1-h_2)+C_1^2} \)
C2 = \(\sqrt{2*1000*0.95(2923.5-2427.03)+60^2} \)
C2 = 973.08 m/s.

7. Dry and saturated steam enters a convergent-divergent steam nozzle at 11 bar and leaves at 3 bar. Determine the nozzle efficiency if the actual enthalpy drop is 200 kJ/kg.
a) 78.56%
b) 96.36%
c) 90.35%
d) 85.27%
View Answer

Answer: d
Explanation: P1 = 11 bar, P2 = 3 bar
Actual enthalpy drop = 200 kJ/kg
At 11 bar, from steam tables
h1 = hg = 2779.7 kJ/kg
s1 = sg = 6.550 kJ/kg-K
At 3 bar, from steam tables
hf2 = 561.5 kJ/kg, hfg2 = 2163.2 kJ/kg
sf2 = 1.672 kJ/kg-K, sfg2 = 5.319 kJ/kg-K
s1 = s2
s1 = sf2 + x2(sfg2)
6.550 = 1.672 + x2(5.319)
x2 = 0.917
h2 = hf2 + x2(hfg2)
= 561.5 + 0.917(2163.2)
= 2545.15 kJ/kg
Isentropic enthalpy drop = h1 – h2 = 2779.7 – 2545.15 = 234.55 kJ/kg
Nozzle efficiency, η = \(\frac{Actual \, enthalpy \, drop}{Isentropic \, enthalpy \, drop} = \frac{200}{234.55} \)= 0.8527 or 85.27%.
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8. Dry and saturated steam at 10 bar enters a convergent-divergent nozzle and leaves at 1 bar. If 8% of heat drop is lost in friction find the percentage decrease in final velocity. Neglect initial velocity.
a) 5.65%
b) 3.87%
c) 2.14%
d) 4.08%
View Answer

Answer: d
Explanation: P1 = 10 bar, P2 = 1 bar, η = 1.0 – 0.08 = 0.92
At 10 bar, from steam tables
h1 = hg = 2776.2 kJ/kg
s1 = 6.583 kJ/kg-K
At 1 bar, from steam tables
hf2 = 417.5 kJ/kg, hfg2 = 2257.9 kJ/kg
sf2 = 1.303 kJ/kg-K, sfg2 = 6.057 kJ/kg-K
s1 = s2
s1 = sf2 + x2(sfg2)
6.583 = 1.303 + x2(6.057)
x2 = 0.872
h2 = hf2 + x2(hfg2)
h2 = 417.5 + 0.872(2257.9)
h2 = 2386.39 kJ/kg
Isentropic Enthalpy drop, hd = h1 – h2 = 2776.2 – 2386.39 = 389.81 kJ/kg
C2 = 44.72\(\sqrt{h_d}\) = 44.72\(\sqrt{389.81}\) = 882.93 m/s
C2’ = 44.72\(\sqrt{η(h_d)}\) = 44.72\(\sqrt{0.92(389.81)}\) = 846.88 m/s
% decrease in exit velocity = \(\frac{C_2-C_2′}{C_2}\) *100 = \(\frac{882.93- 846.88}{882.93}*100\) = 4.08%.

9. Steam at 10 bar superheated to 200°C enters a convergent-divergent steam nozzles and leaves at 2 bar. Determine the dryness fraction of the steam discharged if the nozzle efficiency is 93%.
a) 0.931
b) 0.865
c) 0.758
d) 0.963
View Answer

Answer: a
Explanation: P1 = 10 bar, T1= 200°C, P2 = 2 bar, η = 0.93
At 10 bar and 200°C, from steam tables
h1 = 2826.8 kJ/kg
s1 = 6.692 kJ/kg-K
At 2 bar, from steam tables
hf2 = 504.7 kJ/kg, hfg2 = 2201.6 kJ/kg
sf2 = 1.530 kJ/kg-K, sfg2 = 5.597 kJ/kg-K
s1 = s2
s1 = sf2 + x2(sfg2)
6.692 = 1.530 + x2(5.597)
x2 = 0.922
h2 = hf2 + x2(hfg2)
h2 = 504.7 + 0.922(2201.6)
h2 = 2534.58 kJ/kg
We know that,
η=\(\frac{h_1-h_2′}{h_1-h_2}\)
Substituting the values
0.93=\(\frac{2826.8-h_2”}{2826.8-2534.58}\)
h2’ = 2555.04 kJ/kg
But, h2’ = hf2 + x2(hfg2)
2555.04 = 504.7 + x2’ (2201.6)
x2’ = 0.931.
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10. Find the mass flow rate if dry and saturated steam at 12 bar pressure is discharged at 2 bar by a convergent-divergent nozzle. The nozzle efficiency and exit area of the nozzle are 95% and 0.00256 m2.
a) 1.5 kg/s
b) 2.5 kg/s
c) 3.0 kg/s
d) 4.0 kg/s
View Answer

Answer: b
Explanation: P1 = 12 bar, P2 = 2 bar, η = 0.95, A2 = 0.00256 m2
At 12 bar, from steam tables
h1 = hg1 = 2782.7 kJ/kg, s1 = sg1 = 6.519 kJ/kg-K
At 2 bar, from steam tables
hf2 = 504.7 kJ/kg, hfg2 = 2201.6 kJ/kg
sf2 = 1.530 kJ/kg-K, sfg2 = 5.597 kJ/kg-K
vg2 = 0.88540 m3/kg
s1 = s2
s1 = sf2 + x2(sfg2)
6.519 = 1.530 + x2(5.597)
x2 = 0.891
h2 = hf2 + x2(hfg2)
= 504.7 +0.891(2201.6)
= 2466.32 kJ/kg
C2 = 44.72\(\sqrt{η(h_1-h_2)}\)
C2 = 44.72\(\sqrt{0.95(2782.7-2466.32)}\)
C2 = 775.30 m/s
We know that,
η=\(\frac{h_1-h_2′}{h_1-h_2}\)
Substituting the values
0.95=\(\frac{2782.7-h_2”}{2782.7-2466.32} \)
h2’ = 2482.14 kJ/kg
But, h2’ = hf2 + x2(hfg2)
2482.14 = 504.7 + x2’ (2201.6)
x2’ = 0.898
Mass flow rate, m = \(\frac{A_2*C_2}{x_2’*v_2g} = \frac{0.00256*775.30}{0.898*0.88540}\) = 2.5 kg/s.

11. Nozzle efficiency is the ratio of isentropic enthalpy drop to actual enthalpy drop, between the same pressures.
a) True
b) False
View Answer

Answer: b
Explanation: Nozzle efficiency is defined as the ratio of actual enthalpy drop to isentropic enthalpy drop between the same pressures.
Mathematically,
η=\(\frac{Actual \, Enthalpy \, drop}{Isentropic \, Enthapy \, drop} = \frac{h_d’}{h_d}\)

12. A convergent-divergent steam nozzle is supplied with steam at 15 bar. It discharges the steam at 1 bar. Determine the exit velocity if the velocity coefficient of the given nozzle is 0.95.
a) 785.42 m/s
b) 654.32 m/s
c) 908.90 m/s
d) 854.65 m/s
View Answer

Answer: c
Explanation: P1 = 15 bar, P2 = 1 bar, Velocity coefficient = 0.95
At 15 bar, from steam tables
h1 = 2789.9 kJ/kg, s1 = 6.441 kJ/kg-K
At 1 bar, from steam tables
hf2 = 417.5 kJ/kg, hfg2 = 2257.9 kJ/kg
sf2 = 1.303 kJ/kg-K, sfg2 = 6.057 kJ/kg-K
s1 = s2
s1 = sf2 + x2(sfg2)
6.441= 1.303 + x2(6.057)
x2 = 0.848
h2 = hf2 + x2(hfg2)
= 417.5 + 0.848(2257.9)
= 2332.2 kJ/kg
C2 = 44.72\(\sqrt{(h_1-h_2)}\) = 44.72\(\sqrt{(2789.9-2332.2)}\) = 956.74 m/s
The actual exit velocity, C2’ = Velocity coefficient * C2
C2’ = 0.95 * 956.74
C2’ = 908.90 m/s.

13. Steam at 14 bar and 250°C enters a convergent-divergent steam nozzle and is discharged at 2 bar. Determine the nozzle efficiency if the velocity coefficient is 0.95. Neglect the initial velocity of steam.
a) 95.65%
b) 90.25%
c) 99.65%
d) 85.32%
View Answer

Answer: b
Explanation: Velocity coefficient is the square root of nozzle efficiency, if the inlet velocity is negligible.
Using this relation,
Nozzle efficiency = (Velocity coefficient)2
= 0.952
= 0.9025 or 90.25%.

14. Choose the most appropriate statement regarding velocity coefficient.
a) It can be zero
b) It can be greater than one
c) It should strictly lie between zero and one
d) It is square of nozzle efficiency
View Answer

Answer: c
Explanation: Velocity coefficient is the ratio of actual exit velocity to exit velocity considering the flow isentropic.
Mathematically,
Velocity Coeffiecint=\(\frac{Actual \, Exit \, Velocity}{Exit \,Velocity(Considering \, flow \, to \, be \, isentropic)} = \frac{C_2′}{C_2} \)

Since the actual exit velocity can never be equal to the exit velocity considering flow to be isentropic, velocity coefficient lies strictly between zero and one.

15. Dry and saturated steam at 11 bar pressure enters a convergent-divergent nozzle and is discharged at 2 bar. Determine the actual enthalpy drop if he efficiency of the nozzle is 96%.
a) 254.56 kJ/kg
b) 321.32 kJ/kg
c) 288.14 kJ/kg
d) 245.63 kJ/kg
View Answer

Answer: c
Explanation: Given, P1 = 11 bar, P2 = 2 bar, η = 0.96
At 11 bar, from steam tables
h1 = hg = kJ/kg
s1 = sg = 6.550 kJ/kg-K
At 2 bar, from steam tables
hf2 = 504.7 kJ/kg, hfg2 = 2201.6 kJ/kg
sf2 = 1.530 kJ/kg-K, sfg2 = 5.597 kJ/kg-K
s1 = s2
s1 = sf2 + x2(sfg2)
6.550 = 1.530 + x2(5.597)
x2 = 0.897
h2 = hf2 + x2(hfg2)
h2 = 504.7 + 0.897(2201.6)
h2 = kJ/kg
Enthalpy drop, hd = h1 – h2 = 2779.7 – 2479.54 = 300.16 kJ/kg
Actual enthalpy drop, hd’ = ηhd = 0.96 * 300.15 = 288.14 kJ/kg.

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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