This set of Rocket Propulsion Questions and Answers for Campus interviews focuses on “Definitions and Fundamentals – Exhaust Velocity”.
1. A rocket produces a total thrust of 20,000 N. If the mass flow rate of propellants is 16 kg/s, what will be effective exhaust velocity?
a) 1250 m/s
b) 2500 m/s
c) 625 m/s
d) 5000 m/s
Explanation: Total Thrust = mass flow rate x effective exhaust velocity. Here we assume that the mass flow rate is constant, flow is axially directed and frictional effects are neglected.
So, 20,000 = 16 x ueff
ueff = 20,000/16 = 1250 m/s.
2. What is the characteristic velocity of a rocket if the chamber pressure is 5 MPa, the nozzle throat diameter is 75 mm and the mass flow rate is 25 kg/s?
a) 884 m/s
b) 1768 m/s
c) 3536 m/s
d) 442 m/s
Explanation: Characteristic velocity = Chamber pressure x throat area / mass flow rate. It is an important performance parameter of rocket engines.
So, Characteristic velocity = [5 x 106 x π x (75 x 10-3)2/4] / 25 = 884 m/s.
3. Characteristic velocity is a performance parameter that relates to the efficiency of _____
b) Combustion chamber
Explanation: Characteristic velocity can be used to compare the performance of combustion chambers in chemical rocket engines. It can also be used to compare different propellants and propulsion systems.
4. Which of the following variables is not useful in the determination of rocket effective exhaust velocity?
a) Mass flow rate
b) Exhaust velocity
c) Rocket velocity
d) Exit area
Explanation: In the formulation of effective exhaust velocity, mass flow rate, exhaust velocity, exit area, nozzle exit pressure, ambient pressure are all relevant variables. But the rocket velocity doesn’t come into picture in the final expression.
5. If C denotes effective exhaust velocity and Isp is specific impulse, then which of them is a better performance parameter?
c) Both of them are equal
d) Both of them are equivalent
Explanation: Isp = C/go, where go is the standard acceleration due to gravity at sea level. Since Isp and c differ only by a constant factor, they are equivalent.
6. If mo denotes initial mass, mb denotes burnout mass, then what is the expression for mass ratio?
Explanation: Correct expression of mass flow rate is mb/mo (according to Sutton). It is a measure of rocket efficiency and it gives us an idea about how massive the rocket is with and without propellants.
7. For a rocket traveling at 700 m/s having an exhaust velocity of 1.4 km/s, exhaust gas density of 0.3 kg/m3, nozzle throat diameter of 0.14 cm and nozzle exit diameter of 25 cm, find the total mass flow rate of the propellants.
a) 6.46 kg/s
b) 103.1 kg/s
c) 20.62 kg/s
d) 32.33 kg/s
Explanation: Mass flow rate of the propellants is the same as the mass flow rate of the rocket exhaust jet exiting the nozzle. That will be equal to the product of exhaust gas density, nozzle exit area, and exhaust velocity.
m = ρ Ae ue
= 0.3 x π/4 x (25 x 10-2)2 x 1400
= 20.62 kg/s.
8. What constitutes the exhaust gas at the nozzle exit of a chemical rocket engine?
a) combustion products of the propellants
b) A mixture of combustion products and surrounding air
c) Surrounding air
Explanation: The exhaust gas at the nozzle exit of a chemical rocket engine consists of combustion products of the propellants. If the combustion were inefficient, it would also contain unburnt propellants. Downstream of the nozzle exit, the exhaust jet entrains the surrounding air and becomes a mixture of both.
9. Which of the following is not part of a rocket nozzle?
a) Convergent section
b) Divergent section
c) Throat section
d) Staggering section
Explanation: Main elements of a nozzle include convergent, divergent and throat sections. Staggered nozzle sections are uncommon and unfavorable for the smooth flow of exhaust gases.
10. What is the nozzle throat diameter of a rocket having chamber pressure (Pc) of 3 MPa, mass flow rate (m) of 20 kg/s and characteristic velocity (c*) of 1500 m/s?
a) 10 cm
b) 11.3 cm
c) 12.8 cm
d) 25 cm
Explanation: From the expression c* = Pc At / m, throat area At can be determined.
At = 1500 x 20/(3 x 106) = 0.01 m2
At = πd2/4
d = √(4At/π)
= √(4 x 0.01/*π)
= 0.112 m or 11.2 cm.
Sanfoundry Global Education & Learning Series – Rocket Propulsion.
To practice all areas of Rocket Propulsion for Campus Interviews, here is complete set of 1000+ Multiple Choice Questions and Answers.