This set of Rocket Propulsion Questions and Answers for Freshers focuses on “Nozzle Theory – Isentropic Flow Through Nozzles”.
1. Which of the following will happen if there is a flow obstruction in the nozzle inner wall?
a) Local conversion of kinetic to thermal energy
b) Local drop in pressure around the region of obstruction
c) Formation of normal shock waves at the point of obstruction
d) Rapid decrease in mass flow rate
View Answer
Explanation: In the presence of an obstruction, a stagnation zone develops locally, leading to the conversion of useful kinetic energy to thermal energy. It decreases the overall performance of the engine. In the stagnation zone, the local pressure is the stagnation pressure whose value is larger than the static pressure. Point of obstruction may or may not have normal shock waves and it depends on the flow velocity, the shape of the obstruction, nature of the gas flowing through the nozzle, etc.
2. If at a point along the isentropic rocket nozzle, the temperature of the gas is 718 K, and the flow velocity is 150 m/s, determine the nozzle exit velocity, given the exhaust gas temperature is 460 K (Assume Cp = 1005 J/kg K)
a) 840.25 m/s
b) 735.58 m/s
c) 913.07 m/s
d) 374.95 m/s
View Answer
Explanation: Since the process is isentropic and since no work is done in the nozzle, stagnation enthalpy (ho) is constant.
ho = h + v2/2 = const.
Then, h1 + v12/2 = he + ve2/2
Writing h = CpT, we have
ve = \(\sqrt{(2C_p(T_1-T_e) +v_1^2)}\)
= 735.58 m/s.
3. If the rocket exhaust gases are calorically perfect and have a mean molecular mass of 28 kg/kmol, and a constant specific heat ratio (γ) of 1.33, determine the nozzle exit velocity. Assume the nozzle inlet conditions to be the following: P1 = 5 MPa, V1 = 240 m/s, T1 = 750 K and exit pressure (Pe) to be 1 MPa.
a) 204 m/s
b) 699.21 m/s
c) 712.32 m/s
d) 805.56 m/s
View Answer
Explanation: Using the adiabatic relation, Pe / P1 = (Te / T1)γ/γ-1,
Te = T1(Pe / P1)γ/γ-1
Te = 750 x (1/5)0.33/1.33 = 503 K
R = Ru / Mmol = 8314/28 = 297 J / kg K.
Cp = γR/(γ-1) = 1.33 x 297 / 0.33 = 1197 J / kg K.
ho is constant. So, h1 + v12/2 = he + ve2/2.
Using this, ve = \(\sqrt{(2C_p(T_1-T_e) +v_1^2)}\)
= \(\sqrt{(2*1197*( 750 – 503) + 240^2)}\)
= 805.56 m/s.
4. The area of a de Laval Nozzle is directly proportional to the ratio v/V, where v is the flow velocity and V is the specific volume.
a) True
b) False
View Answer
Explanation: For a de Laval Nozzle, the area is inversely proportional to the ratio v/V. This follows from the continuity equation.
5. What is the correct definition of the nozzle area expansion ratio?
a) The ratio of nozzle exit area to the throat area
b) The ratio of the nozzle area at the inflection point to the throat area
c) The ratio of throat area to the nozzle inlet area
d) The ratio of throat area to the area at the point of inflection
View Answer
Explanation: Nozzle area expansion ratio is the ratio of the nozzle exit area to the nozzle’s throat area. The throat area is the smallest area and lies in between the convergent and divergent portions of a de Laval nozzle.
6. Which of the following will improve the performance of a rocket engine?
a) Decrease in gas temperature
b) Increase in the molecular mass of the propellant
c) Decrease in the pressure ratio
d) Increase in the chamber pressure
View Answer
Explanation: Increase of chamber pressure P1 will help in performance improvement of the rocket engine. With the increase in pressure ratio Prc = P1/P2, performance increases (here P2 denotes the exit pressure). From the expression for exhaust velocity v2 = \(\sqrt{(2\gamma R_uT_0(1-P_{rc}^{1-\gamma/\gamma})/(\gamma-1)M_{mol})}\), it can be seen that decrease in propellant molecular mass will aid in increased exhaust jet velocity and hence enhanced performance.
7. What are the standardized chamber pressure and exit pressure values used for comparing the specific impulse values or various design parameters of different rocket engines?
a) 1000 psia, 1 atm
b)1000 atm, 1 psia
c) 100 psia, 1 atm
d)100 atm, 1 psia
View Answer
Explanation: Specific impulse and other design parameters of various rocket engines are compared with one another while assuming some general standard values for the chamber pressure and exit pressure. A chamber pressure of 1000 psia and an exit pressure of 1 atm are typically in use today. (1 psia = 6894.76 Pa and 1 atm = 101.325 Pa).
8. What is the expression for the maximum theoretical value of the nozzle outlet velocity (ve) with an infinite expansion, if γ represents the ratio of specific heats, To represents the chamber temperature, R is the gas constant?
a) ve = \(\sqrt{(2\gamma RT_o(\gamma-1))}\)
b) ve = \(\sqrt{(2\gamma RT_o/(\gamma-1))}\)
c) ve = \(\sqrt{(2(\gamma+1)RT_o/(\gamma-1))}\)
d) ve = \(\sqrt{(2(\gamma-1)RT_o(\gamma+1))}\)
View Answer
Explanation: ve = \(\sqrt{(2R\gamma T_o/(\gamma-1))}\) is the correct expression. It is obtained from the bigger expression v2 = \(\sqrt{(2\gamma R_uT_0(1-P_{rc}^{1-\gamma/\gamma})/(\gamma-1)M_{mol})}\). Here Prc = P1/P2, where P2 is the exit pressure and P1 is the chamber pressure. The maximum theoretical value is possible when P2/P1 = 0, i.e. when the jet is exhausted to vacuum.
9. What is the pressure ratio for the maximum theoretical value of nozzle outlet velocity?
a) Infinite
b) Zero
c) Finite value < 1
d) Finite value > 1
View Answer
Explanation: Pressure ratio Prc = P1 / P2, where P2 is the jet exit pressure and P1 is the chamber pressure. For the maximum theoretical value of nozzle outlet velocity, the flow has to be exhausted to vacuum. So P2 = 0 or in other words, Prc → ∞.
10. What will happen if the maximum theoretical nozzle exhaust velocity tends to a very large value?
a) Working fluid ceases to be a gas
b) The temperature of the working fluid increases exponentially
c) The thermal energy content of the fluid becomes infinite
d) The flow undergoes rapid compression
View Answer
Explanation: The maximum theoretical exhaust velocity is finite because it is limited by the thermal energy content of the fluid. If the exhaust velocity tends to a very large value, then the thermal energy content (and hence the temperature) of the working fluid rapidly decreases and may lead to the species to fall under their liquefaction or freezing point. When that happens, the working fluid changes its phase from the gaseous state.
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