Thermal Engineering Questions and Answers – Steam Nozzles – Super Saturated Flow and Wilson’s Line

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This set of Advanced Thermal Engineering Questions and Answers focuses on “Steam Nozzles – Super Saturated Flow and Wilson’s Line”.

1. What is Wilson line?
a) It is an isothermal line, at which the condensation completes
b) Saturation line of water is also called Wilson line
c) It represents the limiting condition of undercooling at which the condensation begins
d) It represents an Isobaric line, at which the condensation commences
View Answer

Answer: c
Explanation: At Wilson line the normal thermal equilibrium conditions are restored. It represents the limiting condition of undercooling at which the condensation begins. Wilson line is important while dealing with supersaturated expansion.
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2. Considering metastable expansion to be in effect, the flow of wet steam though a convergent-divergent nozzle shall result in a discharge slightly less than the calculated one.
a) True
b) False
View Answer

Answer: b
Explanation: The flow of wet steam through a convergent-divergent nozzle results in a discharge slightly greater than the calculated one. Converging part is small and steam velocity being high, eludes the possibility of normal condensation. This rapid expansion produces supersaturated state. Steam is undercooled to a temperature less than the saturation temperature corresponding to its pressure. This results in increase in density and hence increases in discharge.

3. The sudden expansion of dry saturated steam in the absence of dust, doesn’t lead to condensation until its density is about 8 times that of the saturated vapor of the same pressure.
a) True
b) False
View Answer

Answer: a
Explanation: Prof Wilson proved this fact though series of experiments. This delayed condensation reduces the enthalpy drop and also the final condition of the steam is improved. This type of expansion in steam nozzles is also called metastable expansion.
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4. Calculate the enthalpy drop steam enters a convergent-divergent nozzle at 11 bar and 250°C and leaves at 1 bar. The expansion is to be considered metastable.
a) 427031.34 kJ/kg
b) 55648.32 kJ/kg
c) 323654.25 kJ/kg
d) 158469.21 kJ/kg
View Answer

Answer: a
Explanation: Given, P1 = 11 bar, T1 = 250°C, P2 = 1 bar
At 11 bar and 250°C, from steam tables
vg1 = 0.2108 m3/kg
Since the steam entering is in superheated state,
n = 1.3
Enthalpy drop, hd = \(\frac{n}{n-1}\)(P1)(vg1){1-\((\frac{P_2}{P_1})^\frac{n-1}{n}\)}
hd = \(\frac{1.3}{1.3-1}\)(11*105)(0.2108)\(\big\{1-(\frac{1}{11})^\frac{1.3-1}{1.3}\big\} \)
hd = 427031.34 kJ/kg.

5. Which of the following the correct formula for calculating the exit velocity of steam through nozzle considering the expansion to be metastable.
a) C2 = \(\sqrt{2*\frac{n+1}{n-1}(P_1)(v_{g1}){1-(\frac{P_2}{P_1})^\frac{n-1}{n}}} \)
b) C2 = \(\sqrt{2*\frac{n-1}{n+1}(P_1)(v_{g1}){1-(\frac{P_2}{P_1})^\frac{n+1}{n}}}
\)
c) C2 = \(\sqrt{2*\frac{n}{n-1}(P_1)(v_{g1}){1-(\frac{P_2}{P_1})^\frac{n-1}{n}}}
\)
d) C2 = \(\sqrt{2*\frac{n}{n-1}(P_1)(v_{g1}){1-(\frac{P_2}{P_1})^\frac{n-1}{n}}}
\)
View Answer

Answer: c
Explanation: When the expansion of steam is metastable the relation Pvn = C is used. Some enthalpy drop is lost and hence we use the following formula for calculating the exit velocity of the steam.
C2 = \(\sqrt{2*\frac{n}{n-1}(P_1)(v_{g1}){1-(\frac{P_2}{P_1})^\frac{n-1}{n}}}
\)
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6. Which of the following is the correct formula for calculating the supercooled temperature of steam, when steam metastable expansion of steam takes place inside a steam nozzle. (T2 – Supercooled temperature, T1 – Inlet temperature of steam)
a) \(\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{n}{n-1} \)
b) \(\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{n+1}{n} \)
c) \(\frac{T_2}{T_1} = (\frac{P_1}{P_2})^\frac{n}{n-1} \)
d) \(\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{n}{n-1} \)
View Answer

Answer: d
Explanation: The correct formula for calculating the supercooled temperature is as follow –
\(\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{n}{n-1} \)
T2 is the supercooled temperature.

7. Steam at 12 bar and 300°C enters a convergent-divergent steam nozzle and leaves at 3 bar. If the expansion of the steam is metastable, calculate the supercooled temperature.
a) 165°C
b) 143°C
c) 154°C
d) 120°C
View Answer

Answer: b
Explanation: Given, P1 = 12 bar, T1 = 300°C or 573 K, P2 = 3 bar
We know that,
\(\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{n}{n-1} \)
Substituting the values, we get
\(\frac{T_2}{573} = (\frac{3}{12})^\frac{1.3-1}{1.3} \)
Therefore, T2 = 416.19 K or 143.19°C ≈ 143°C.
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8. Steam at 20 bar and 300°C enters a convergent-divergent steam nozzle and is discharged at 3 bar. Considering the effects of supersaturation calculate the degree of undercooling.
a) 36.65°C
b) 45.54°C
c) 23.32°C
d) 62.87°C
View Answer

Answer: a
Explanation: Given, P1 = 20 bar, T1 = 300°C or 573K, P2 = 3 bar
At 3 bar, from steam tables
Tsat = 133.5°C
n=1.3
Supercooled temperature, T2 = T1 \((\frac{P_2}{P_1})^\frac{n}{n-1}\) = 573\((\frac{3}{20})^\frac{1.3-1}{1.3}\) = 369.85 K or 96.85°C
Degree of undercooling = Tsat – T2 = 133.5 – 96.85 = 36.65°C.

9. Calculate the degree of supersaturation if steam at 10 bar 250°C is discharged at 2 bar by a convergent-divergent nozzle. Consider metastable steam expansion and negligible inlet velocity.
a) 5.65
b) 3.11
c) 2.15
d) 2.36
View Answer

Answer: b
Explanation: Given, P1 = 10 bar, T1 = 250°C or 523 K, P2 = 2 bar
Index of expansion, n = 1.3
Supercooled temperature, T2 = T1 \((\frac{P_2}{P_1})^\frac{n}{n-1}\) = 523\((\frac{2}{10})^\frac{1.3-1}{1.3}\) = 360.74 K or 87.74 °C
Saturation pressure corresponding to 87.74°C, Psat = 0.64309 bar (using interpolation)
Degree of supersaturation = \(\frac{P_2}{P_{sat}} = \frac{2}{0.64309}\) = 3.11.
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10. Steam at 15 bar pressure and 250°C enters a convergent-divergent steam nozzle and leaves at 4 bar. The mass flow rate is 5 kg/s. Calculate the exit area of the nozzle if the expansion is metastable.
a) 0.00542187 m2
b) 0.0029147 m2
c) 0.036521 m2
d) 0.003265 m2
View Answer

Answer: b
Explanation: Given, P1 = 15 bar, T1 = 250°C or 523 K, P2 = 4 bar, m = 5 kg/s
At 15 bar and 250°C, from steam tables
vg1 = 0.1520 m3/kg
Exit velocity, C2 = \(\sqrt{2*\frac{n-1}{n+1}(P_1)(v_{g1}){1-(\frac{P_2}{P_1})^\frac{n+1}{n}}} \)
C2 = \(\sqrt{2*\frac{1.3}{1.3-1}(15*10^5)(0.1520)\big\{1-(\frac{4}{15})^\frac{1.3-1}{1.3}\big\}}\)
C2 = 720.74 m/s
We know that,
vg2 = \((\frac{P_1}{P_2})^\frac{1}{n}\)*vg1 = \((\frac{15}{4})\frac{1}{1.3}\)*0.1520 = 0.42015 m3/kg
Mass flow rate, m = \(\frac{A_2*C_2}{v_{g2}} \)
5 = \(\frac{A2*720.74}{0.42015} \)
A2 = 0.0029147 m2

11. What is the value of index of expansion for supersaturated steam. (Pvn=C, n is the index of expansion)
a) 1.300
b) 1.013
c) 1.135
d) 1.000
View Answer

Answer: a
Explanation: In case of supersaturated steam the index of expansion is assumed to be same as that for superheated steam i.e. 1.3.
Therefore, supersaturated steam follows the following law –
Pv1.3 = Constant.

12. Which of the following statement about metastable expansion of steam through a steam nozzle is FALSE?
a) It increases the discharge through the nozzle
b) Steam is undercooled to a temperature less than that corresponding to its pressure
c) Density of steam is increased
d) Exit velocity of the steam is increased
View Answer

Answer: d
Explanation: Metastable expansion of steam produces a supersaturated state where steam is undercooled to a temperature less than that corresponding to its pressure. It has been observed after a series of experiments that discharge through the nozzle is increased as the density is increased. The exit velocity is reduced due to reduction in enthalpy drop.

13. The point on the h-s diagram at which the condensation beings and normal conditions of thermal equilibrium are restored is said to be on _____
a) wilson line
b) saturation vapor curve
c) constant dryness fraction line
d) constant pressure line
View Answer

Answer: a
Explanation: Wilson line is the locus of points at which the limit of undercooling is reached and the condensation commences. At this state the normal conditions of thermal equilibrium are also restored. Wilson line is not conventionally shown in a Mollier chart.

14. Which of the following is NOT an effect of supersaturation in steam flowing through a convergent-divergent nozzle?
a) Increase in specific volume of steam
b) Increase in the entropy of steam
c) Improvement in dryness fraction
d) Increase in the enthalpy drop
View Answer

Answer: d
Explanation: Enthalpy drop is reduced due to supersaturation in nozzles. Besides specific volume, entropy of the steam also increases due to supersaturation. The quality i.e. dryness fraction is also improved.

15. Determine the exit pressure of steam if the steam enters a nozzle at 11 bar and 300°C. At the outlet it is observed that the specific volume of the steam is 0.86804 m3/kg. The expansion of steam is metastable.
a) 4 bar
b) 3 bar
c) 2 bar
d) 1 bar
View Answer

Answer: c
Explanation: Given, P1 = 11 bar, T1 = 300°C, vg2 = 0.86804 m3/kg
At 11 bar and 300°C, from steam tables
vg1 = 0.2339 m3/kg
we know that,
P1vg1n = P2vg2n
Substituting the values
11(0.2339)1.3 = P2(0.86804)1.3
P2 = 2 bar.

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