Engineering Questions with Answers - Multiple Choice Questions

# MCQs on Jet Propulsion Engine Thrust

1 - Question

What forces are exerted on the body by a fluid flow?
a) Pressure force
b) Shear stress
c) Pressure and shear stress
d) Lift force

c
Explanation: When a body is kept in a stream of fluid flowing over it, there are two forces exerted. One is the pressure distribution and the other is the shear stress distribution. This leads to formation of lift or drag.

2 - Question

Thrust produced by engine is due to the shear stress distribution from the fluid.
a) True
b) False

b
Explanation: When the gas flows through the engine, there is a distribution of pressure and shear stress over all the components of the engine – compressor blades, turbine blades, combustion chamber and nozzle. This leads to the thrust formation.

3 - Question

What is the formula for net thrust of the engine?
a) F = -∯pdS
b) F = ∯pdS
c) F = -∯dSp
d) F = -∯p2 dS

a
Explanation: Usually the pressure distribution is more dominant than the effect of shear stress thus for computing thrust, pressure force is only considered. The net thrust is given by the formula
F = -∯pdS
Where, p is the pressure distribution
dS is the outer surface over which it is integrated and thus vector points away from the surface
The negative sign is present because thrust acts in the opposite direction of the acceleration.

4 - Question

What is the static thrust for an engine as derived from the integral momentum equation?
a) T = m˙ue + Ae(pe – p0)
b) T = m˙Ae(pe – p0)
c) T = Ae(pe – p0)
d) T = m˙Ae

a
Explanation: Considering an engine in a control volume with ambient pressure p0, exit pressure pe and exit velocity of the fluid ue, the net force is given by:
∑Fx = T + Ae(p0 – pe)
Fx = m˙ue
Thus, solving the equations we get,
T + Ae(p0 – pe) = m˙ue
On rearranging the terms we get,
T = m˙ue + Ae(pe – p0)

5 - Question

What is the thrust of an engine if the inlet and exhaust mass flow rate are 150 kg/s and 190 kg/s respectively with the pressure difference between the exhaust and ambient air is 1.85 × 104? The exhaust area is 2m2, the ambient and exhaust velocities are 280 m/s and 400 m/s.
a) 50 kN
b) 62 kN
c) 71 kN
d) 84 kN

c
Explanation: Given, m˙i = 150 kg/s, m˙e = 190 kg/s, pe – p∞= 1.85 × 104, Ae = 2m2, V∞ = 280 m/s
Ve = 400 m/s.
The thrust equation is given by:
T = m˙eVe – m˙iV∞ + Ae(pe – p∞)
Substituting the values, we get
T = 190 × 400 – 150 × 280 + 2(1.85 × 104)
T = 76,000 – 42,000 + 37,000 = 71 kN

6 - Question

The integral form of the conservative equation is used for computing thrust of an engine.
a) True
b) False

a
Explanation: The integral form of conservative equation finds use in many practical scenarios such as finding thrust of a jet propulsion device – engine. It can be used to find out the thrust of turbojet, turboprop, ramjet, turbofan, rocket engine and many more.

7 - Question

Integral form of momentum equation cannot replace the need for theoretical and experimental measurements.
a) True
b) False

b
Explanation: In real life scenario, computing the thrust of the jet engine using theoretical and experimental results can be very exhaustive. The data obtained is not just extensive but also complex. Thus, to replace this, an integral form of momentum equation is used which gives the result much more accurately.

8 - Question

What happens to the integral ∫dS when the area of exhaust is less than the area of inlet?
a) Positive
b) Negative
c) Zero
d) Infinity

b
Explanation: The sign of the integral ∫dS in the momentum equation is determined by the area of exhaust and inlet. When the exit area is less than the inlet area, the sum of negative components of pressure is more than the sum of positive components of the pressure. Thus, the sign becomes negative.

9 - Question

What causes thrust of an engine?
a) Change in momentum
b) Difference in lift distribution
c) Difference in drag distribution
d) Mass flow entering the engine