# Gas Dynamics Questions and Answers – De Laval Nozzle

This set of Gas Dynamics Multiple Choice Questions & Answers (MCQs) focuses on “De Laval Nozzle”.

1. A De Laval Nozzle is the only means to produce supersonic flow.
a) True
b) False

Explanation: A De Laval Nozzle is the nozzle consisting of a convergent-divergent section. Now as per the stream tube area-velocity relation when a subsonic flow passes through the convergent section it accelerates as it moves downstream, and as it passes through divergent section its Mach number further increases resulting into the supersonic flow. Hence to produce a supersonic flow a nozzle must have an expanding section, i.e. a nozzle must be a De Laval Nozzle.

2. At which condition the flow reaches to supersonic velocity inside the De Laval Nozzle.
a) pth/p0 ≤ 0.663
b) pth/p0 ≤ 0.354
c) pth/p0 ≤ 0.528
d) pth/p0 ≤ 0.721

Explanation: For flow to be a supersonic velocity it must obtain M = 1 at the throat. Now from isentropic flow relations at M = 1, the critical pressure ratio at the throat is given by;
$$(\frac {p^*}{p_0} )=(\frac {T^*}{T_0 })$$γ/(γ-1)
$$(\frac {p^*}{p_0} )$$ = [1+($$\frac {\gamma -1}{2}$$)M2]1/(γ-1)
$$(\frac {p^*}{p_0} )$$ = [($$\frac {2}{\gamma +1}$$)]1/(γ-1)
$$(\frac {p^*}{p_0} )$$ = 0.528
Hence at p*/p0 = 0.528, the flow chocks at throat. Thus pth/p0 should be ≤ 0.528 for the flow to keep accelerating after the throat.

3. What will be the area ratio for the flow of air having M = 2.3 inside a divergent section?
a) 1.24
b) 2.19
c) 2.37
d) 4.80

Explanation: The area ratio for any given section having local Mach number is given by;
$$(\frac {A}{A^*} )^2=\frac {1}{M^2} [ \frac {2}{γ+1} ( 1+\frac {γ-1}{2} M^2 ) ]$$(γ+1)/(γ-1)
Therefore, for the section having local M = 2.3, the area ratio is given by;
$$(\frac {A}{A^*} )^2=\frac {1}{(2.3)^2} [ \frac {2}{1.4+1} ( 1 + \frac {1.4-1}{2} (2.3)^2 ) ]$$(1.4+1)/(1.4-1)
$$(\frac {A}{A^*} )$$=2.19

4. Calculate the temperature at the nozzle exit, for a C-D nozzle having a throat to exit area ratio of 5.9 and reservoir temperature 288 K.
a) 723.49 K
b) 764.26 K
c) 890.42 K
d) 933.12 K

Explanation: From the Area-Mach number relation,
$$(\frac {A}{A^*} )^2=\frac {1}{M^2} [ \frac {2}{γ+1} ( 1+\frac {γ-1}{2} M^2 ) ]$$(γ+1)/(γ-1)
for area ratio Ae/A* = 5.9, the exit Mach number is Me = 3.35
Now temperature ratio $$(\frac {T_e}{T_0} ) = [ 1+(\frac {γ-1}{2}) M^2 ]$$
$$(\frac {T_e}{T_0} )$$ = 3.24
Thus, Te = 3.24*288 = 933.12 K

5. The nozzle test conditions are determined on the basis of which characteristic?
a) Ae/A*
b) Pe/P*
c) Te/T*
d) ρe/ρ*

Explanation: It is evident from area-Mach number relation, that the nozzle exit Mach number depends on the nozzle area ratio. Now since the throat section of the nozzle geometry is where Mach number becomes a unity, and also the nozzle geometry is designed according to required exit Mach number, we can say that the nozzle test conditions are determined on basis of Ae/A*.

6. The De Laval Nozzle transform pressure energy into __________
a) Thermal energy
b) Potential energy
c) Kinetic energy
d) Chemical energy

Explanation: A De Laval nozzle is the convergent-divergent nozzle where the pressure decreases as the fluid flows downstream and velocity increases. Therefore the decreases in pressure are converted into kinetic energy that creates change in momentum of fluid.

7. If the temperature of the reservoir is increased by four times, then what will happen to the mass flow rate?
a) Increases by the same amount
b) Decreases by the same amount
c) Increases by half
d) Decreases by half

Explanation: The mass flow of fluid through the nozzle is expressed as;
$$\dot{m} = \frac {p_0 A^*}{\sqrt {T_0 }} \sqrt {\frac {γ}{R} \frac {2}{γ+1}^{(γ+1)/(γ-1)} }$$
I.e. the mass flow rate varies with temperature as; $$\dot{m}$$ ∝ $$\frac {1}{\sqrt {T_0} }$$
Hence by increasing the temperature by 4 times, the mass flow rate decreases by half.

8. For a nozzle area ratio, A must be ≥ A*.
a) True
b) False

Explanation: By differentiating the nozzle area ratio A/A* twice, it shows that the second derivative of A/A* is positive. Hence the location where the area ratio is minimum is the choked location, thus A should be ≥ A*, as A < A* is physically impossible.

9. What is the maximum mass flow rate that can be achieved for gas with γ = 1.4?
a) $$\dot{m} = \frac {p_0 A^*}{\sqrt {RT_0 }}$$*0.7486
b) $$\dot{m} = \frac {p_0 A^*}{\sqrt {RT_0 }}$$*0.6847
c) $$\dot{m} = \frac {p_0 A^*}{\sqrt {RT_0 }}$$*0.9329
d) $$\dot{m} = \frac {p_0 A^*}{\sqrt {RT_0 }}$$*0.4265

Explanation: Based on the expression for the mass flow rate of fluid through nozzle; $$\dot{m} = \frac {p_0 A^*}{\sqrt {T_0 }} \sqrt {\frac {γ}{R} \frac {2}{γ+1}^{(γ+1)/(γ-1)} }$$
For gases with γ = 1.4: $$\dot{m} = \frac {p_0 A^*}{\sqrt {RT_0 }}$$*0.6847

10. Which of the following C-D nozzle has minimum divergence losses?
a) Conical nozzle
b) Aerospike nozzle
c) Annular nozzle
d) Bell nozzle

Explanation: The Bell nozzle out of other nozzles have more divergence angle at the throat that moderately decreases up to exit nearly to about 2° to 5°, while the conical nozzle has a constant divergence angle. As a result of this, the divergence losses at throat decreases compared to other nozzles.

Sanfoundry Global Education & Learning Series – Gas Dynamics.

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