Engineering Questions with Answers - Multiple Choice Questions

# Aerodynamics – Governing Equations

1 - Question

What causes the flow properties to vary in quasi – one – dimensional flow?
a) Cross – sectional area
b) Normal shock
d) Frictional drag
Explanation: In case of quasi – one dimensional flow the flow properties keep changing along the distance x due to the varying cross – sectional area (A) which is in contrast to the one – dimensional flow. In that, the area remains constant thus, the flow properties change fir to the presence of shock wave friction and heat addition or removal.

<img class="alignnone size-medium wp-image-9959" src="https://mocktestpro.in/wp-content/uploads/2021/08/sds-300x163.png" alt="" width="300" height="163"

2 - Question

What is the momentum equation for a quasi – one dimensional flow?
a) p1A1 + ρ1u21A1 + ∫A2A1pdA = p2A2 + ρ2u22A2
b) p1A1u1+ ρ1u21A1 + ∫A2A1pdA = p2A2u2+ ρ2u22A2
c) p1A1 + ρ1u21A1 = p2A2 + ρ2u22A2
d) p1A1u1+ ρ1u21A1 = p2A2u2+ ρ2u22A2
Explanation: The integral form of the momentum equation is given by: ∯S(ρV.dS)V = -∯SpdS In order to find the x – component of this the equation becomes: ∯S(ρV.dS)u = -∯SpdSx Where, pdSx is the x component of pressure u is the velocity On the control surfaces of the streamtube, V.dS = 0 because they are streamlines. At 1, A1, V, dS are in opposite direction thus they are negative. This results in the left side of the equation to be – ρ1u21A1 + ρ2u22A2. For the right side of the equation, it is –(-p1A1 + p2A2). Negative sign is because at A1, dS points to the left and is negative. For the upper and lower surfaces of the control volume, pressure integral becomes: – ∫A2A1 – pdA = ∫A2A1pdA Where the negative sign is because the dS points to the left. This results In the equation to be: – ρ1u21A1 + ρ2u22A2 = -(-p1A1 + p2A2 ) + ∫A2A1pdA On rearranging the terms we get: p1A1 + ρ1u21A1 + ∫A2A1pdA = p2A2 + ρ2u22A2

3 - Question

Which is the Euler’s equation for the quasi – one dimensional flow?
a) dp = ρudu
b) dp = –uρdu
c) dp = -ρudu
d) dp = ρudu

Explanation: We consider a duct of variable cross sectional area with two stations 1 and 2, having properties as given in the figure. Using the momentum equation:
p1A1 + ρ1u21A1 + A2A1pdA = p2A2 + ρ2u22A2
We get,
pA + ρu2A + pdA = (p + dp)(A + dA) + (ρ + dρ)(u + du)2 (A + dA)
pA + ρu2A + pdA = pA + pdA + Adp + dpdA + ρu2 + (ρdu2 + 2ρudu + dρu2 + dρdu2 + 2dρudu)(A + dA)
Since the conditions at station 1 are: p, ρ, A and conditions at station 2 are (p + dp), (ρ + dρ), (A + dA).
The product of differentials dPdA, dρ(du)2(A + dA) are negligible thus are ignored.
The resulting equation is:
The continuity equation is given by:
d(uρA) = 0
Expanding this we get,
ρudA + ρAdu + Audρ = 0
Multiplying the above equation by u on both sides we get:
ρu2dA + ρuAdu + Au2dρ = 0 ➔ eqn 2
Subtracting eqn 2 from eqn 1, we get the differential equation for the quasi one – dimensional flow:
dp = -ρudu

4 - Question

What is the differential form of momentum equation for the quasi one – dimensional flow known as?
a) Froude equation
b) Euler’s equation
c) Kelvin’s equation
d) Bernoulli’s equation
Explanation: The differential form of momentum equation for the quasi one – dimensional flow is given by: dp = – ρudu This is known as Euler’s equation which is derived from the momentum equation.

5 - Question

What is the differential form of energy equation for quasi one – dimensional flow?
a) dh – u2du = 0
b) dh – udu = 0
c) dh + u2du = 0
d) dh + udu = 0
Explanation: The energy equation for quasi one – dimensional flow is given by: h + u22 = constant On differentiating the above equation we arrive at the differential energy equation for the quasi one – dimensional flow: dh + udu = 0

6 - Question

What is the value of surface integral between surface 1 and 2 of the control volume while finding the continuity equation? a) V.dS
b) dS
c) Zero
d) V2.dS

Explanation: The continuity equation for quasi one – dimensional flow is obtained by integrating conservative equation in a control volume which is given by: ρ1u1A1 = ρ2u2A2 Where subscript 1 and 2 correspond to the cross – sectional area in the control volume. The surface integral between 1 and 2 is considered to be zero as V is along the surface of the streamtube and hence V.dS = 0 and there is no fluid flow across the boundary of the stream tube.

7 - Question

According to the energy equation, which of these properties remain constant along the flow?
a) Total enthalpy
b) Total entropy
c) Kinetic energy
d) Potential energy