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

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.

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

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

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.

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

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

Explanation: P

_{1}= 12 bar, η = 0.90, P

_{2}= 2 bar

At 12 bar, from steam tables

h

_{1}= h

_{g1}= 2782.7 kJ/kg, s

_{1}= s

_{g1}= 6.519 kJ/kg-K

At 2 bar, from steam tables

h

_{f2}= 504.7 kJ/kg, h

_{fg2}= 2201.6 kJ/kg

s

_{f2}= 1.530 kJ/kg-K, s

_{fg2}= 5.597 kJ/kg-K

s

_{1}= s

_{2}

s

_{1}= s

_{f2}+ x

_{2}(s

_{fg2})

6.519 = 1.530 + x

_{2}(5.597)

x

_{2}= 0.891

h

_{2}= h

_{f2}+ x

_{2}(h

_{fg2})

= 504.7 +0.891(2201.6)

= 2466.32 kJ/kg

C

_{2}= 44.72\(\sqrt{η(h_1-h_2)}\)

C

_{2}= 44.72\(\sqrt{0.90(2782.7-2466.32)}\)

C

_{2}= 754.62 m/s.

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

Explanation: P

_{1}= 15 bar, T

_{1}= 250°C, η = 0.95, P

_{2}= 1 bar, C

_{1}= 60 m/s

At 15 bar and 200°C, from steam tables

h

_{1}= 2923.5 kJ/kg, s

_{1}= 6.710 kJ/kg-K

At 1 bar, from steam tables

h

_{f2}= 417.5 kJ/kg, h

_{fg2}= 2257.9 kJ/kg

s

_{f2}= 1.303 kJ/kg-K, s

_{fg2}= 6.057 kJ/kg-K

s

_{1}= s

_{2}

s

_{1}= s

_{f2}+ x

_{2}(s

_{fg2})

6.70 = 1.303 + x

_{2}(6.057)

x

_{2}= 0.89

h

_{2}= h

_{f2}+ x

_{2}(h

_{fg2})

= 417.5 + 0.89(2257.9)

= 2427.03 kJ/kg

C

_{2}= \(\sqrt{2*1000*η(h_1-h_2)+C_1^2} \)

C

_{2}= \(\sqrt{2*1000*0.95(2923.5-2427.03)+60^2} \)

C

_{2}= 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

Explanation: P

_{1}= 11 bar, P

_{2}= 3 bar

Actual enthalpy drop = 200 kJ/kg

At 11 bar, from steam tables

h

_{1}= h

_{g}= 2779.7 kJ/kg

s

_{1}= s

_{g}= 6.550 kJ/kg-K

At 3 bar, from steam tables

h

_{f2}= 561.5 kJ/kg, h

_{fg2}= 2163.2 kJ/kg

s

_{f2}= 1.672 kJ/kg-K, s

_{fg2}= 5.319 kJ/kg-K

s

_{1}= s

_{2}

s

_{1}= s

_{f2}+ x

_{2}(s

_{fg2})

6.550 = 1.672 + x

_{2}(5.319)

x

_{2}= 0.917

h

_{2}= h

_{f2}+ x

_{2}(h

_{fg2})

= 561.5 + 0.917(2163.2)

= 2545.15 kJ/kg

Isentropic enthalpy drop = h

_{1}– h

_{2}= 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%.

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

Explanation: P

_{1}= 10 bar, P

_{2}= 1 bar, η = 1.0 – 0.08 = 0.92

At 10 bar, from steam tables

h

_{1}= h

_{g}= 2776.2 kJ/kg

s

_{1}= 6.583 kJ/kg-K

At 1 bar, from steam tables

h

_{f2}= 417.5 kJ/kg, h

_{fg2}= 2257.9 kJ/kg

s

_{f2}= 1.303 kJ/kg-K, s

_{fg2}= 6.057 kJ/kg-K

s

_{1}= s

_{2}

s

_{1}= s

_{f2}+ x

_{2}(s

_{fg2})

6.583 = 1.303 + x

_{2}(6.057)

x

_{2}= 0.872

h

_{2}= h

_{f2}+ x

_{2}(h

_{fg2})

h

_{2}= 417.5 + 0.872(2257.9)

h

_{2}= 2386.39 kJ/kg

Isentropic Enthalpy drop, h

_{d}= h

_{1}– h

_{2}= 2776.2 – 2386.39 = 389.81 kJ/kg

C

_{2}= 44.72\(\sqrt{h_d}\) = 44.72\(\sqrt{389.81}\) = 882.93 m/s

C

_{2}’ = 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

Explanation: P

_{1}= 10 bar, T

_{1}= 200°C, P

_{2}= 2 bar, η = 0.93

At 10 bar and 200°C, from steam tables

h

_{1}= 2826.8 kJ/kg

s

_{1}= 6.692 kJ/kg-K

At 2 bar, from steam tables

h

_{f2}= 504.7 kJ/kg, h

_{fg2}= 2201.6 kJ/kg

s

_{f2}= 1.530 kJ/kg-K, s

_{fg2}= 5.597 kJ/kg-K

s

_{1}= s

_{2}

s

_{1}= s

_{f2}+ x

_{2}(s

_{fg2})

6.692 = 1.530 + x

_{2}(5.597)

x

_{2}= 0.922

h

_{2}= h

_{f2}+ x

_{2}(h

_{fg2})

h

_{2}= 504.7 + 0.922(2201.6)

h

_{2}= 2534.58 kJ/kg

We know that,

η=\(\frac{h_1-h_2′}{h_1-h_2}\)

Substituting the values

0.93=\(\frac{2826.8-h_2”}{2826.8-2534.58}\)

h

_{2}’ = 2555.04 kJ/kg

But, h

_{2}’ = h

_{f2}+ x

_{2}(h

_{fg2})

2555.04 = 504.7 + x

_{2}’ (2201.6)

x

_{2}’ = 0.931.

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 m^{2}.

a) 1.5 kg/s

b) 2.5 kg/s

c) 3.0 kg/s

d) 4.0 kg/s

View Answer

Explanation: P

_{1}= 12 bar, P

_{2}= 2 bar, η = 0.95, A

_{2}= 0.00256 m

^{2}

At 12 bar, from steam tables

h

_{1}= h

_{g1}= 2782.7 kJ/kg, s

_{1}= s

_{g1}= 6.519 kJ/kg-K

At 2 bar, from steam tables

h

_{f2}= 504.7 kJ/kg, h

_{fg2}= 2201.6 kJ/kg

s

_{f2}= 1.530 kJ/kg-K, s

_{fg2}= 5.597 kJ/kg-K

v

_{g2}= 0.88540 m

^{3}/kg

s

_{1}= s

_{2}

s

_{1}= s

_{f2}+ x

_{2}(s

_{fg2})

6.519 = 1.530 + x

_{2}(5.597)

x

_{2}= 0.891

h

_{2}= h

_{f2}+ x

_{2}(h

_{fg2})

= 504.7 +0.891(2201.6)

= 2466.32 kJ/kg

C

_{2}= 44.72\(\sqrt{η(h_1-h_2)}\)

C

_{2}= 44.72\(\sqrt{0.95(2782.7-2466.32)}\)

C

_{2}= 775.30 m/s

We know that,

η=\(\frac{h_1-h_2′}{h_1-h_2}\)

Substituting the values

0.95=\(\frac{2782.7-h_2”}{2782.7-2466.32} \)

h

_{2}’ = 2482.14 kJ/kg

But, h

_{2}’ = h

_{f2}+ x

_{2}(h

_{fg2})

2482.14 = 504.7 + x

_{2}’ (2201.6)

x

_{2}’ = 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

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

Explanation: P

_{1}= 15 bar, P

_{2}= 1 bar, Velocity coefficient = 0.95

At 15 bar, from steam tables

h

_{1}= 2789.9 kJ/kg, s

_{1}= 6.441 kJ/kg-K

At 1 bar, from steam tables

h

_{f2}= 417.5 kJ/kg, h

_{fg2}= 2257.9 kJ/kg

s

_{f2}= 1.303 kJ/kg-K, s

_{fg2}= 6.057 kJ/kg-K

s

_{1}= s

_{2}

s

_{1}= s

_{f2}+ x

_{2}(s

_{fg2})

6.441= 1.303 + x

_{2}(6.057)

x

_{2}= 0.848

h

_{2}= h

_{f2}+ x

_{2}(h

_{fg2})

= 417.5 + 0.848(2257.9)

= 2332.2 kJ/kg

C

_{2}= 44.72\(\sqrt{(h_1-h_2)}\) = 44.72\(\sqrt{(2789.9-2332.2)}\) = 956.74 m/s

The actual exit velocity, C

_{2}’ = Velocity coefficient * C

_{2}

C

_{2}’ = 0.95 * 956.74

C

_{2}’ = 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

Explanation: Velocity coefficient is the square root of nozzle efficiency, if the inlet velocity is negligible.

Using this relation,

Nozzle efficiency = (Velocity coefficient)

^{2}

= 0.95

^{2}

= 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

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

Explanation: Given, P

_{1}= 11 bar, P

_{2}= 2 bar, η = 0.96

At 11 bar, from steam tables

h

_{1}= h

_{g}= kJ/kg

s

_{1}= s

_{g}= 6.550 kJ/kg-K

At 2 bar, from steam tables

h

_{f2}= 504.7 kJ/kg, h

_{fg2}= 2201.6 kJ/kg

s

_{f2}= 1.530 kJ/kg-K, s

_{fg2}= 5.597 kJ/kg-K

s

_{1}= s

_{2}

s

_{1}= s

_{f2}+ x

_{2}(s

_{fg2})

6.550 = 1.530 + x

_{2}(5.597)

x

_{2}= 0.897

h

_{2}= h

_{f2}+ x

_{2}(h

_{fg2})

h

_{2}= 504.7 + 0.897(2201.6)

h

_{2}= kJ/kg

Enthalpy drop, h

_{d}= h

_{1}– h

_{2}= 2779.7 – 2479.54 = 300.16 kJ/kg

Actual enthalpy drop, h

_{d}’ = ηhd = 0.96 * 300.15 = 288.14 kJ/kg.

**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]**

**Related Posts:**

- Practice Chemical Engineering MCQs
- Practice Mechanical Engineering MCQs
- Apply for Chemical Engineering Internship
- Check Chemical Engineering Books
- Apply for Mechanical Engineering Internship