This set of Thermal Engineering Interview Questions and Answers for Experienced people focuses on “Relationship Between Area, Velocity and Pressure in Nozzle Flow”.

1. Which of the following expressions correctly represents the relationship between cross-sectional area of a nozzle, fluid velocity and specific volume?

a) \(\frac{dA}{A}+\frac{dC}{C}=\frac{dv}{v} \)

b) \(\frac{dA}{A}-\frac{dC}{C}=\frac{dv}{v} \)

c) \(\frac{dA}{A}+\frac{dC}{C}+\frac{dv}{v}\) = 0

d) \(\frac{dA}{A}+\frac{dv}{v}=\frac{dC}{C} \)

View Answer

Explanation: The correct relationship between cross-sectional area, fluid velocity and specific volume is represented by the following equation –

\(\frac{dA}{A}+\frac{dC}{C}=\frac{dv}{v} \)

The nature of change (positive or negative) in any two parameters in the equation above predicts the nature of change in the third parameter.

2. Which of the following expressions is correct? (P – fluid pressure, M- Mach number, A – cross-sectional area of nozzle/diffuser)

a) \(\frac{dA}{A}=\frac{1}{γ}\frac{dP}{P}\big\{\frac{M^2}{1-M^2} \big\} \)

b) \(\frac{dA}{A}=\frac{1}{γ}\frac{dP}{P}\big\{\frac{M^2-1}{M^2}\big\} \)

c) \(\frac{dA}{A}=\frac{1}{γ}\frac{dP}{P}\big\{\frac{1-M^2}{M^2} \big\} \)

d) \(\frac{1}{γ}\frac{dA}{A}=\frac{dP}{P}\big\{\frac{1-M^2}{M^2} \big\} \)

View Answer

Explanation: The correct expression relating the pressure, cross-sectional area of nozzle/diffuser and Mach number is –

\(\frac{dA}{A}=\frac{1}{γ} \frac{dP}{P}\big\{\frac{1-M^2}{M^2} \big\} \)

or

\(\frac{dA}{A}= \frac{1}{γ} \frac{dP}{P}{\frac{Cs^2}{C^2 -1}} \)

where, C

_{s}– sonic velocity

C – fluid velocity

3. What is Mach number?

a) It is the ratio of sonic velocity of a fluid at N.T.P. to the local sonic velocity of the same fluid

b) It is the ratio of fluid velocity to sonic velocity of the same fluid at N.T.P.

c) It is the ratio of local sonic velocity to fluid velocity

d) It is the ratio of fluid velocity to local sonic velocity

View Answer

Explanation: Mach number is the ratio of fluid velocity to the local sonic velocity. It is the ratio of the same quantity and hence is dimensionless. The fluid velocity is called subsonic is Mach number is less than one and supersonic if the Mach number is greater than one.

4. In case of accelerated flow, when the pressure decreases along the flow direction and Mach number is less than one, it corresponds to _____

a) Convergent part of a nozzle

b) Divergent part of a nozzle

c) Throat of a nozzle

d) Convergent part of a diffuser

View Answer

Explanation: The flow is accelerated, hence it’s a nozzle. Since \(\frac{dP}{P}\) is negative and Mach number is less than one, for the following equation to hold –

\(\frac{dA}{A} = \frac{1}{γ} \frac{dP}{P} \big\{\frac{1-M^2}{M^2} \big\} \)

L.H.S should also be negative. This implies that \(\frac{dA}{A}\) should be negative, which corresponds to convergent part of the nozzle.

5. Which of the following statements regarding the Mach number is TRUE, when the fluid reaches the throat of a nozzle?

a) It becomes unity

b) It is less than one

c) It is greater than one

d) Mach number is not defined at throat of a nozzle

View Answer

Explanation: At the throat of the nozzle, there is no change in cross-sectional are of the nozzle i.e. \(\frac{dA}{A}\)=0. Therefore, from the following equation –

\(\frac{dA}{A}=\frac{1}{γ} \frac{dP}{P} \big\{\frac{1-M^2}{M^2} \big\} \)

It is evident that the R.H.S. should also be zero. This implies that the Mach number must be one. The fluid velocity attains the value of local sonic velocity at the throat.

6. Which of the following conditions corresponds to divergent part of a nozzle?

a) M < 1 and \(\frac{dP}{P}\) > 0

b) M < 1 and \(\frac{dP}{P}\) < 0

c) M > 1 and \(\frac{dP}{P}\) < 0

d) M > 1 and \(\frac{dP}{P}\) > 0

View Answer

Explanation: Consider the following equation –

\(\frac{dA}{A}=\frac{1}{γ}\frac{dP}{P} \big\{\frac{1-M^2}{M^2} \big\} \)

For a nozzle \(\frac{dP}{P}\)<0, and for divergent part, \(\frac{dA}{A}\)>0. For the above equation to hold M > 1.

Therefore, M > 1 and \(\frac{dP}{P}\) < 0 is the correct answer.

7. A decelerated flow, having fluid velocity greater than the local sonic velocity corresponds to _____

a) Convergent part of a nozzle

b) Divergent part of a nozzle

c) Convergent part of a diffuser

d) Divergent part of a diffuser

View Answer

Explanation: For a diffuser, \(\frac{dP}{P}\)>0. It is given that fluid velocity is greater than the local sonic velocity i.e. M > 1. Therefore, according to the following equation –

\(\frac{dA}{A}=\frac{1}{γ}\frac{dP}{P} \big\{\frac{1-M^2}{M^2} \big\} \)

8. Which of the following conditions correspond to divergent type diffuser?

a) M < 1 and \(\frac{dA}{A}\)>0

b) M < 1 and \(\frac{dA}{A}\)>0

c) M < 1 and \(\frac{dA}{A}\)<0

d) M > 1 and \(\frac{dA}{A}\)<0

View Answer

Explanation: Diffuser promotes decelerated flow, \frac{dP}{P}>0. Consider the following equation –

\(\frac{dA}{A}=\frac{1}{γ}\frac{dP}{P} \big\{\frac{1-M^2}{M^2} \big\} \)

For the divergent part, \frac{dA}{A}>0. For the above equation to hold under the listed conditions, the Mach number must be less than one. Hence, M < 1 and \(\frac{dA}{A}\)>0 is the correct answer.

9. The purpose of a steam injector is to force water into the boiler under pressure.

a) True

b) False

View Answer

Explanation: A steam injector is used to force water into the boiler under pressure. It employees the principle of steam nozzles. It makes use of the kinetic energy of a steam jet for increasing the pressure and velocity of a corresponding amount of water.

10. Air at 18 bar and 100°C enters a convergent nozzle. Assume the flow to be isentropic and calculate the sonic velocity. Take adiabatic index equal to 1.4.

a) 353.40 m/s

b) 321.56 m/s

c) 360.87 m/s

d) 400.32 m/s

View Answer

Explanation: P = 18 bar, T = 100°C or 373 K, γ = 1.4

Critical pressure, P* = P\((\frac{2}{γ+1})^\frac{γ}{γ-1}\) = 18\((\frac{2}{1.4+1})^\frac{1.4}{1.4-1}\) = 9.51 bar

T

_{1}= T\((\frac{P*}{P})^\frac{γ-1}{γ}\) = (373)\((\frac{9.51}{18})^\frac{1.4-1}{1.4}\) = 310.83 K

Sonic velocity, C

_{s}= \(\sqrt{γR(T_1)} = \sqrt{1.4*287*310.83}\) = 353.40 m/s.

11. Air enters a frictionless adiabatic horizontal nozzle at 12 bar and 167°C with inlet velocity 50 m/s and leaves at 3 bar. Take adiabatic index equal to 1.4 and c_{p} = 1.005 kJ/kg-K.

a) 654.78 m/s

b) 321.75 m/s

c) 552.45 m/s

d) 456.87 m/s

View Answer

Explanation: P

_{1}= 13 bar, T

_{1}= 167°C or 440 K, C

_{1}= 50 m/s, P

_{2}= 3 bar, c

_{p}= 1.005 kJ/kg-K, γ = 1.4

We know that, T

_{2}= T

_{1}\((\frac{P_2}{P_1})^\frac{γ-1}{γ}\) = 440 \((\frac{3}{13})^\frac{1.4-1}{1.4}\) = 289.40 K

Applying the energy equation at inlet and outlet of the nozzle, we get

m[h

_{1}+\(\frac{c_1^2}{2}\)+Z

_{1}*g]+Q=m[h

_{2}+\(\frac{c_2^2}{2}\)+Z

_{2}*g]+W

Q = 0, W = 0, Z

_{1}= Z

_{2}

h

_{1}+\(\frac{c_1^2}{2}\)=h

_{2}+\(\frac{c_2^2}{2}\)

c

_{2}

^{2}=2(h

_{1}-h

_{2})+c

_{1}

^{2}

C

_{2}= \(\sqrt{2(h_1-h_2)+c_1^2} \)

C

_{2}= \(\sqrt{2c_p(T_1-T_2)+c_1^2} \)

C

_{2}= \(\sqrt{2*1.005*10^3 (440-289.40)+50^2} \)

C

_{2}= 552.45 m/s.

**Sanfoundry Global Education & Learning Series – Thermal Engineering**

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