Rocket Propulsion Questions and Answers – Nozzle Theory – Four Performance Parameters

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This set of Rocket Propulsion Multiple Choice Questions & Answers (MCQs) focuses 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
View Answer

Answer: a
Explanation: 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.
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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
View Answer

Answer: a
Explanation: Nozzle area ratio is defined as Ae/A*, where Ae denotes exit area and A* denotes throat area. Nozzle contraction area ratio is Ai/A*, where Ai 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 Cp = 1005 J/(kgK) and γ = 1.33.
a) 0.133 m2
b) 0.210 m2
c) 0.295 m2
d) 0.225 m2
View Answer

Answer: d
Explanation: Choking occurs if Pe/Po < {2/(γ+1)}γ/γ-1
{2/(γ+1)}γ/γ-1 = (2/2.33)(1.33/0.33) = 0.54
Here Pe/Po = 0.4. Hence its a C-D nozzle.
Te = To(Pe/Po)γ-1/γ = 648 x (101/250)0.33/1.33 = 517.5 K
ve = \(\sqrt{(2C_p(T_t-T_e))} = \sqrt{(2 * 1005 * (648 – 517.5))}\) = 512.16 m/s
ρe = Pe/RTe = 0.78 kg/m3 (Take Cp = γR/(γ-1) to get approximate value of R)
Using m = ρeAeve,
Ae = m/ρeve = 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
View Answer

Answer: b
Explanation: 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
View Answer

Answer:
Explanation: 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.
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6. A gas tank discharges faster if it is thermally isolated than if it is kept isothermal.
a) True
b) False
View Answer

Answer: b
Explanation: 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
View Answer

Answer: d
Explanation: The required distance x = 0.67Do√(P0/P1), where Do is the nozzle diameter, Po is the flow pressure and P1 is the atmospheric pressure.
Do = \(\sqrt{(4A_e/\pi)} = \sqrt{(4 * 0.110/\pi)}\) = 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
View Answer

Answer: c
Explanation: Ae/A* = 9.
So de2/d*2 = 9 (Since A = πd2/4).
Which means de = 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
View Answer

Answer: d
Explanation: Thrust coefficient CF = T/PoA*.
A* = πd2/4 = 78.53 cm2.
∴ CF = 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 Cp = 1005 J/kgK)
a) 1591 m/s
b) 2041 m/s
c) 991 m/s
d) 781 m/s
View Answer

Answer: a
Explanation: Using Cp = γR/γ-1, we get R = 1005 x 0.4/1.4 = 287 J/kgK.
ue = Me\(\sqrt{\gamma RT} = 3 * \sqrt{(1.4 * 287 * 700)}\) = 1591 m/s.
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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn