Thermal Engineering Questions and Answers – Steam Flow Through Nozzles

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This set of Thermal Engineering Questions and Answers for Experienced people focuses on “Steam Flow Through Nozzles”.

1. A steam nozzle is a passage of varying cross section through which the kinetic energy of steam is converted into heat energy.
a) True
b) False
View Answer

Answer: b
Explanation: Steam nozzle is a passage of varying cross section, through which the heat energy of the steam is converted into kinetic energy. The velocity of steam in increased its pressure is decreased. It is usually used to produce high steam jets to drive the steam turbines.
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2. Which of the following statements about steam nozzles is FALSE?
a) It converts the heat energy of the steam into kinetic energy
b) It has a varying cross section
c) The smallest section is called throat
d) The pressure at the outlet is more than at the inlet
View Answer

Answer: d
Explanation: Pressure at the inlet of the steam nozzle is greater than at the outlet. Steam nozzles increase the velocity of the steam passing through by converting the heat energy of the steam into kinetic energy. It has a varying cross section and the smallest section is called throat.

3. The smallest section of a steam nozzle is called _____
a) maw
b) neck
c) throat
d) muzzle
View Answer

Answer: c
Explanation: Steam nozzle has varying cross section; the smallest section is called throat. Based upon the variation of cross-section nozzles are categorized into diverging nozzles, converging nozzles and converging-diverging nozzles.
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4. Which type of nozzle does the following picture represent?

a) Convergent
b) Divergent
c) Convergent-Divergent
d) Subsonic
View Answer

Answer: b
Explanation: This picture represents a convergent-divergent nozzle. The smallest section of the nozzle is called throat. A convergent-divergent nozzle converges to throat and diverges afterwards. A convergent-divergent nozzle is able to impart higher velocity to steam than a convergent nozzle.

5. The steam flow though nozzle is considered to be _____
a) adiabatic
b) isobaric
c) isothermal
d) isochoric
View Answer

Answer: a
Explanation: During the expansion no heat is added not rejected from the system hence, the steam flow though through nozzle is considered to be adiabatic. Steam while passing through nozzle loses its pressure and temperature, also the nozzle is of varying cross section, therefore, the flow cannot be isobaric, isothermal or isochoric.
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6. Steam at a temperature 200°C and pressure 4 bar enters a steam nozzles and is discharged at 1 bar pressure. Determine the exit velocity of steam if the inlet velocity is negligible.
a) 500 m/s
b) 720 m/s
c) 580 m/s
d) 750 m/s
View Answer

Answer: b
Explanation: Given, P1 = 4 bar, T1 = 200°C, P2 = 1 bar
From steam tables, At 4 bar and 200°C
h1 = 2860.5 kJ/kg, s1 = 7.171 kJ/kg/K
From steam tables, At 1 bar
hf = 417.5 kJ/kg, hfg = 2257.9 kJ/kg
sf = 1.3027 kJ/kg-K, sfg = 6.0657 kJ/kg-K
Since, s1 = s2
s1 = sf +x(sfg)
7.171 = 1.3027 + x(6.0657)
x = 0.967
therefore, h2 = hf + x(hfg)
h2 = 417.5 + 0.967(2257.9)
h2 = 2601.92 kJ/kg
Exit velocity, C2 = \(\sqrt{2*1000(h_1-h_2)}\)
C2 = \(\sqrt{2*1000(2860.5-2601.92)}\)
C2 = 719.14 m/s ≈ 720 m/s.

7. Steam entering a nozzle at 15 bar and 350°C is discharged at 1 bar. If the inlet velocity of the steam is 50 m/s, determine the exit velocity.
a) 785 m/s
b) 365 m/s
c) 565 m/s
d) 475 m/s
View Answer

Answer: c
Explanation: Given, P1 = 15 bar, T1 = 350°C, P2 = 1 bar, C1 = 50m/s
Using Mollier chart,
h1 = 3150 kJ/kg and h2 = 2992 kJ/kg
Exit velocity, C2 = \(\sqrt{2*1000(h_1-h_2)+C_1^2}\)
C2 = \(\sqrt{2*1000(3150-2992)+50^2}\)
C2 = 564.36 m/s ≈ 565 m/s.
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8. Dry and saturated steam enters a nozzle at 11 bar and leaves at 2 bar. What minimum exit area should the nozzle have if the steam flow of 2 kg/s is to be maintained?
a) 0.00365 m2
b) 0.02156 m2
c) 0.00312 m2
d) 0.00228 m2
View Answer

Answer: d
Explanation: Given, P1 = 11 bar, P2 = 2 bar, m = 2 kg/s
Using Mollier chart,
h1 = 2780 kJ/kg, h2 = 2480 kJ/kg, vg2 = 0.885 m3
We know that,
C2 = \(\sqrt{2*1000(h_1-h_2)}\)
C2 = \(\sqrt{2*1000(2780-2480)}\)
C2 = 774.6 m/s
Exit area, A2 = \(\frac{m*v_{g2}}{C_2} = \frac{2*0.885}{774.6}\) = 0.00228 m2

9. Final velocity of steam passing thorough a nozzle is found out to be 586 m/s. The pressure of steam at inlet is 15 bar and is dry and saturated. The exit pressure is 2 bar. Determine the dryness fraction of the steam when it leaves the nozzle.
a) 0.79
b) 0.96
c) 0.88
d) 0.81
View Answer

Answer: b
Explanation: Given, C2 = 586 m/s, P1 = 15 bar, P2 = 2 bar
From Steam tables,
At 15 bar:
h1 = hg = 2789.9 kJ/kg
At 2 bar:
hf2 = 504.7 kJ/kg, hfg2 = 2201.6 kJ/kg
We know that,
C2 = 44.72\(\sqrt{(h_1-h_2)} \)
Substituting the values
586 = 44.72\(\sqrt{(2789.9-h_2)} \)
Solving for ‘h2’, we get
h2 = 2618.19 kJ/kg
we know that
h2 = hf2 + x(hfg2)
2618.19 = 504.7 + x2(2201.6)
Solving for x2, we get
x2 = 0.96.
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10. Dry saturated steam enters a steam nozzle at 12 bar and 200°C and leaves at 2 bar. If the exit area is 0.00197 m2, determine the mass flow rate of steam.
a) 1 kg/s
b) 2 kg/s
c) 3 kg/s
d) 4 kg/s
View Answer

Answer: b
Explanation: Given, P1 = 12 bar, T1 = 200°C, P2 = 2 bar, A2 = 0.00197 m2
From steam tables, at 12 bar and 200°C
h1 = 2814.4 kJ/kg, s1 = 6.587 kJ/kg-K
From steam tables, at 2 bar
hf2 = 504.7 kJ/kg, hfg2 = 2201.6 kJ/kg
sf2 = 1.530 kJ/kg-K, sfg2 = 5.597 kJ/kg-K
vg2 = 0.8854 m3/kg
We know that
s1 = s2
s1 = sf2 + x2(sfg2)
6.587 = 1.530 + x2(5.597)
x2 = 0.90
h2 = hf2 + x2(hfg2)
= 504.7 + 0.9(2201.6)
= 2486.14 kJ/kg
We know that,
C2 = 44.72\(\sqrt{(h_2)}\)
Substituting the values
C2 = 44.72\(\sqrt{(2814.4-2486.14)}\)
C2 = 810.23 m/s
Mass flow rate, m = \(\frac{A_2*C_2}{xv_g}\)
m = \(\frac{0.00197*810.23}{0.9*0.8854}\)
m = 2 kg/s.

11. Which of the following equations represent the correct critical pressure ratio?
a) \(\frac{P_1}{P_2} = (\frac{2}{n+1})^\frac{n}{n-1} \)
b) \(\frac{P_2}{P_1} = (\frac{2}{n-1})^\frac{n}{n-1} \)
c) \(\frac{P_2}{P_1} = (\frac{2}{n+1})^\frac{n}{n-1} \)
d) \(\frac{P_2}{P_1} = (\frac{2}{n+1})^\frac{n}{n+1} \)
View Answer

Answer: c
Explanation: Critical pressure ratio represents the ratio of throat pressure to inlet pressure for maximum discharge through nozzle. The correct equation for critical pressure ratio is-
\(\frac{Throat \, Pressure}{Inlet \, Pressure} = \frac{P_2}{P_1} = (\frac{2}{n+1})^\frac{n}{n-1} \)

12. Calculate the throat pressure if the dry and saturated steam enters a nozzle at 15 bar. Consider the discharge through the nozzle maximum.
a) 8.66 bar
b) 7.58 bar
c) 5.65 bar
d) 2.36 bar
View Answer

Answer: a
Explanation: P1 = 15 bar, x2 = 0.95
n = 1.135 (Dry and saturated steam)
According to the given condition,
\(\frac{P_2}{P_1} = (\frac{2}{n+1})^\frac{n}{n-1} \)
\(\frac{P_2}{15} = (\frac{2}{1.135+1})^\frac{1.135}{1.135-1} \)
Therefore, P2 = 8.66 bar.

13. Steam enters a nozzle at 14 bar and 200°C. Determine the discharge if the discharge through the nozzle is maximum. Take the area of throat as 0.002 m2.
a) 3.95 kg/s
b) 4.18 kg/s
c) 2.36 kg/s
d) 1.65 kg/s
View Answer

Answer: b
Explanation: Given, P1 = 14 bar, T1 = 200°C, x2 = 0.85, A2 = 0.002 m2
n = 1.3 (Supersaturated steam)
From steam tables, at 14 bar and 200°C
vg1 = 0.1429 m3/kg
According to the given condition
mmax = A\(\sqrt{n(\frac{P_1}{v_1})(\frac{2}{n+1})^\frac{n+1}{n-1}} \)
mmax = 0.002\(\sqrt{1.3{14*10^5}{0.1429})(\frac{2}{1.3+1})^\frac{1.3+1}{1.3-1}} \)
mmax = 4.18 kg/s.

14. The maximum mass flow through a steam nozzle is independent of _____
a) Initial pressure
b) Initial density
c) Final pressure
d) Throat area
View Answer

Answer: c
Explanation: The equation below represents the maximum mass flow through a nozzle.
mmax = A\(\sqrt{n(\frac{P_1}{v_1})(\frac{2}{n+1})^\frac{n+1}{n-1}} \)
It can be observed from the above equation that the maximum mass flow depends in Initial conditions i.e. initial pressure and initial density (or initial specific volume), quality and throat area. It is independent of final pressure and quality (dryness fraction) of steam.

15. Determine the velocity of steam at throat if the discharge through the nozzle is maximum. Steam enters the nozzle in dry and saturated state and at 11 bar.
a) 455.45 m/s
b) 625.32 m/s
c) 545.87 m/s
d) 325.65 m/s
View Answer

Answer: a
Explanation: Given, P1 = 11 bar
n = 1.135 (Dry and saturated steam)
From steam tables, at 11 bar
vg1 = 0.17739 m3/kg
According to the given conditions
C2 = \(\sqrt{2(\frac{n}{n+1})(P_1)(v_{g1})}\)
C2 = \(\sqrt{2(\frac{1.135}{1.135+1})(11*10^5)(0.17739)}\)
C2 = 455.45 m/s.

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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