Engineering Questions with Answers - Multiple Choice Questions
Rocket Propulsion – Definitions and Fundamentals – Energy and Efficiencies
1 - Question
Which form of energy is most useful for propulsion?
a) The kinetic energy of ejected matter
b) The chemical energy of combustion of propellants
c) The potential energy of the rocket at some altitude
d) The electrical energy of the rocket control system
Explanation: At the nozzle exit, the exhaust gases are ejected with some kinetic energy. This is the most useful form of energy crucial to propulsion. Chemical energy produced from the combustion of propellants is of no use unless it can be harnessed for propelling the rocket forward by some means.
2 - Question
For a constant gas ejection velocity (v) of 1440 m/s and a mass flow rate (m) of 15 kg/s, find the power of the jet (Pjet).
a) 7.78 MW
b) 15.55 MW
c) 31.10 MW
d) 62.21 MW
Explanation: Power of the jet is 1/2mv2. It is the time rate of expenditure of ejected jet’s kinetic energy. Pjet = 1/2 x 15 x 14402 = 15.55 MW.
3 - Question
If the power of the jet is given to be 25 MW for a chemical rocket having a weight flow rate of 147 N/s, find out the vehicle’s specific impulse.
a) 98.2 s
b) 372.6 s
c) 186.3 s
d) 243.5 s
Explanation: Power of the jet can be expressed in terms of weight flow rate and specific impulse. It is given by 1/2wogoIsp2. Pjet = 1/2 wo go Isp2 Isp = (2Pjet/ẇ ogo)−−−−−−−−−−−√ = (2∗25∗106/(147∗9.8))−−−−−−−−−−−−−−−−−−−√ = 186.3 s.
4 - Question
If a chemical rocket has a jet power of 18 MW, inert mass of 50 kg, a propellant mass of 1000 kg, then what is its specific power?
a) 34.28 kW/kg
b) 17.14 kW/kg
c) 8.91 kW/kg
d) 25.43 kW/kg
Explanation: Specific power is jet power divided by the total mass of that the propulsion system prior to its launch. It is lower for electric propulsion system when compared to chemical rockets. This is because of the heavier and relatively inefficient power source that it has to carry. Pspec = Pjet / mtotal mtotal = minert + mprop = 50 + 1000 = 1050 kg ∴ Pspec = 18 x 106 / 1050 = 17.14 kW/kg.
5 - Question
What is the way of production of energy in a chemical rocket?
a) Ionization of propellants
b) Combustion of propellants
c) Expulsion of propellants
d) Atomization of propellants
Explanation: Propellants are mixed together and combusted to create energy for propulsion. Expulsion of propellants after combustion converts this energy into a more useful form (kinetic energy).
6 - Question
Which of the following Is the maximum energy available per unit mass of chemical propellants?
a) heat of combustion
b) enthalpy of vaporization
c) enthalpy of atomization
d) enthalpy of fusion
Explanation: The heat of combustion is the maximum energy per unit mass that can be extracted from chemical propellants. It can be of two types – higher or lower heating value. Higher heating value is determined by taking all the propellants to their pre-combustion temperature and letting all the vapor produced in the reaction to condense.
7 - Question
For a propulsive device, if power available is denoted by P1, power input by P2, combustion efficiency by e, then which of the following expression is correct?
Explanation: P1=P2*e is the correct expression. Input power will be reduced by a factor of combustion efficiency to become power available.
8 - Question
If a propulsion system produces a thrust of 5 kN, exhaust velocity of 4500 m/s, and vehicle velocity of 3000 m/s, what is the power transmitted to the vehicle?
a) 225 MW
b) 75 MW
c) 10 MW
d) 15 MW
Explanation: Power transmitted to the vehicle is the product of vehicle velocity and the thrust produced by the propulsion system. It is not the same as exhaust jet power. Ptrans = T v = 5 x 103 x 3000 = 15 MW.
9 - Question
Which of the following is not a way of losing energy in a chemical rocket?
a) Poor mixing
b) Incomplete burning
c) Heat loss to the walls
d) Radial combustion propagation
Explanation: Axial and radial combustion propagation are two different kinds of burning propagation. Poor mixing and incomplete burning constitute combustion losses.
10 - Question
For a rocket having 95% combustion efficiency, total chemical power of 17MW, a mass flow rate of propellants of 15 kg/s, and effective jet exhaust velocity of 1443 m/s, find the internal efficiency of rocket propulsion.
Explanation: Internal efficiency of rocket propulsion is the ratio of kinetic power in the jet to the available chemical power. It is an indicator of how much the system is capable of converting input energy into useful kinetic energy. Pav chem = ηcomb x Ptot chem = 0.95 x 17 x 106 W ηint = Pkinetic / Pav chem = 1/2μe2 / Pchem = 0.5 x 15 x 14432 / (0.95 x 17 x 106) ≅ 0.966.