250+ TOP MCQs on Nozzle Theory – Four Performance Parameters and Answers

Rocket Propulsion Multiple Choice Questions on “Nozzle Theory – Four Performance Parameters”.

1. How are delivered performance values estimated for a propulsion system?
a) Static tests or flight tests of full-scale models
b) Dimensional analysis and similitude of models
c) A theoretical analysis using the known relations
d) Using computational simulation
Answer: a
Clarification: Theoretical and actual (or delivered) performance values of a propulsion system may not be equal at all times. For evaluating delivered performance parameter values, static tests or flight test of full-scale models are used.

2. What is another name for the nozzle area ratio?
a) Nozzle expansion ratio
b) Nozzle contraction ratio
c) Nozzle convergence ratio
d) Nozzle divergence ratio
Answer: a
Clarification: Nozzle area ratio is defined as A_e/A*, where A_e denotes exit area and A* denotes throat area. Nozzle contraction area ratio is A_i/A*, where A_i is the inlet area of the nozzle.

3. If isentropic efficiency for expansion in a nozzle having a mass flow rate of 90 kg/s is 0.95, find the exit area. Given the low speed entrance pressure is 250 kPa, temperature is 375 °C and outside atmospheric pressure is 101 kPa. Assume C_p = 1005 J/(kgK) and γ = 1.33.
a) 0.133 m²
b) 0.210 m²
c) 0.295 m²
d) 0.225 m²
Answer: d
Clarification: Choking occurs if P_e/P_o < {2/(γ+1)}^γ/γ-1
{2/(γ+1)}^γ/γ-1 = (2/2.33)(1.33/0.33) = 0.54
Here P_e/P_o = 0.4. Hence its a C-D nozzle.
T_e = T_o(P_e/P_o)^γ-1/γ = 648 x (101/250)^0.33/1.33 = 517.5 K
v_e = (sqrt{(2C_p(T_t-T_e))} = sqrt{(2 * 1005 * (648 – 517.5))}) = 512.16 m/s
ρe = P_e/RT_e = 0.78 kg/m³ (Take C_p = γR/(γ-1) to get approximate value of R)
Using m = ρ_eA_ev_e,
A_e = m/ρ_ev_e = 90/(0.78×512.16) = 0.225 m².

4. Find the critical pressure ratio for a nozzle flow from a chamber of pressure 300 kPa to the atmospheric pressure of 100 kPa. Assume γ = 1.4.
a) 0.637
b) 0.528
c) 0.981
d) 0.243
Answer: b
Clarification: Critical pressure ratio = {2/(γ+1)}^γ/γ-1
Using γ=1.4, we have the value equal to (2/2.4)^1.4/0.4 = 0.528.

5. Determine the acoustic speed in chamber conditions if the chamber pressure is 500 kPa, density is 0.8 kg/m³ and γ=1.4.
a) 900.2 m/s
b) 879.5 m/s
c) 935.4 m/s
d) 1014.2 m/s
Answer:
Clarification: a = (sqrt{(gamma RT)}). Assuming the gas to be ideal, P/ρ = RT. So, a = (sqrt{(gamma P_o/rho_o)} = sqrt{(1.4 * 500 * 1000/0.8)}) = 935.4 m/s.

6. A gas tank discharges faster if it is thermally isolated than if it is kept isothermal.
a) True
b) False
Answer: b
Clarification: A thermally isolated gas tank discharges slowly than an isothermal one. This is because in the isothermal gas tank, in order to maintain the temperature constant, any kind of heat addition will result in an increase of tank pressure thereby resulting in a faster ejection.

7. For a slightly over-expanded supersonic exhaust from a nozzle with exit area of 0.110 m², flow pressure of 180 kPa and an atmospheric pressure of 100 kPa, determine the approximate distance between the nozzle and the first shock diamond.
a) 30.2 cm
b) 25.9 cm
c) 28.5 cm
d) 33.6 cm
Answer: d
Clarification: The required distance x = 0.67D_o√(P₀/P₁), where D_o is the nozzle diameter, P_o is the flow pressure and P₁ is the atmospheric pressure.
D_o = (sqrt{(4A_e/pi)} = sqrt{(4 * 0.110/pi)}) = 0.374 m.
x = 0.67 * 0.374 * (sqrt{(180/100)}) = 0.336 m or 33.6 cm.

8. If a nozzle has an optimum expansion ratio of 9, determine the nozzle cone exit diameter for a throat diameter of 15 mm.
a) 30 mm
b) 135 mm
c) 45 mm
d) 90 mm
Answer: c
Clarification: A_e/A* = 9.
So d_e²/d*2 = 9 (Since A = πd²/4).
Which means d_e = d* x 3 = 45 mm.

9. Determine the rocket thrust coefficient for a total thrust of 20 kN, chamber pressure of 500 kPa and a throat diameter of 10 cm.
a) 2.39
b) 6.51
c) 4.97
d) 5.09
Answer: d
Clarification: Thrust coefficient C_F = T/P_oA*.
A* = πd²/4 = 78.53 cm².
∴ C_F = 20,000/(500 x 1000 x 78.53 x 10^-4)
= 5.09.

10. Determine the jet exhaust velocity, if the flow has an exit Mach number of 3 and temperature of 700 K. (Assume γ= 1.4 and C_p = 1005 J/kgK)
a) 1591 m/s
b) 2041 m/s
c) 991 m/s
d) 781 m/s
Answer: a
Clarification: Using C_p = γR/γ-1, we get R = 1005 x 0.4/1.4 = 287 J/kgK.
u_e = M_e(sqrt{gamma RT} = 3 * sqrt{(1.4 * 287 * 700)}) = 1591 m/s.