This set of Aerodynamics Multiple Choice Questions & Answers (MCQs) focuses on “Governing Equations for Quasi One Dimensional Flow”.

1. Which of these parameters in quasi – one dimensional flow varies w.r.t. x as opposed to the one – dimensional flow?

a) Cross – sectional area

b) Pressure

c) Density

d) Temperature

View Answer

Explanation: One – dimensional flows are those flows that have a constant crossectional area and only the flow variables i.e. pressure, velocity, density, temperature etc vary wrt x. What separates Quasi one – dimensional flow from this is the variable cross – sectional area wrt x.

2. What is the momentum equation for a quasi – one dimensional flow?

a) p_{1}A_{1} + ρ_{1}u\(_1 ^2\)A_{1} + \(\int _{A_1} ^{A_2}\)pdA = p_{2}A_{2} + ρ_{2}u\(_2 ^2\)A_{2}

b) p_{1}A_{1}u_{1} + ρ_{1}u\(_1 ^2\)A_{1} + \(\int _{A_1} ^{A_2}\)pdA = p_{2}A_{2}u_{2} + ρ_{2}u\(_2 ^2\)A_{2}

c) p_{1}A_{1} + ρ_{1}u\(_1 ^2\)A_{1} = p_{2}A_{2} + ρ_{2}u\(_2 ^2\)A_{2}

d) p_{1}A_{1}u_{1} + ρ_{1}u\(_1 ^2\)A_{1} = p_{2}A_{2}u_{2} + ρ_{2}u\(_2 ^2\)A_{2}

View Answer

Explanation: The integral form of the momentum equation is given by:

∯

_{s}(ρ

**= – ∯**

*V.dS)V*_{s}p

*dS*In order to find the x – component of this the equation becomes:

∯

_{s}(ρ

**= – ∯**

*V.dS)u*_{s}p

*dS*_{x}Where, p

**is the x component of pressure**

*dS*_{x}u is the velocity

On the control surfaces of the streamtube,

**= 0 because they are streamlines. At 1, A**

*V.dS*_{1},

**are in opposite direction thus they are negative. This results in the left side of the equation to be – ρ**

*V,dS*_{1}u\(_1 ^2\)A

_{1}+ ρ

_{2}u\(_2 ^2\)A

_{2}.

For the right side of the equation, it is – (-p

_{1}A

_{1}+ p

_{2}A

_{2}). Negative sign is because at A

_{1}, dS points to the left and is negative.

For the upper and lower surfaces of the control volume, pressure integral becomes:

– \(\int _{A_1} ^{A_2}\) – pdA = \(\int _{A_1} ^{A_2}\)pdA

Where the negative sign is because the dS points to the left.

This results In the equation to be:

– ρ

_{1}u\(_1 ^2\)A

_{1}+ ρ

_{2}u\(_2 ^2\)A

_{2}= – ( – p

_{1}A

_{1}+ p

_{2}A

_{2}) + \(\int _{A_1} ^{A_2}\)pdA

On rearraging the terms we get:

p

_{1}A

_{1}+ ρ

_{1}u\(_1 ^2\)A

_{1}+ \(\int _{A_1} ^{A_2}\)pdA = p

_{2}A

_{2}+ ρ

_{2}u\(_2 ^2\)A

_{2}

3. Which is the Euler’s equation for the quasi – one dimensional flow?

a) dp = \(\frac {ρ}{u}\)du

b) dp = – \(\frac {u}{ρ}\)du

c) dp = – ρudu

d) dp = ρudu

View Answer

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:

p

_{1}A

_{1}+ ρ

_{1}u\(_1 ^2\)A

_{1}+ \(\int _{A_1} ^{A_2}\)pdA = p

_{2}A

_{2}+ ρ

_{2}u\(_2 ^2\)A

_{2}

We get,

pA + ρu

^{2}A + pdA = (p + dp)(A + dA) + (ρ + dρ)(u + du)

^{2}(A + dA)

pA + ρu

^{2}A + pdA = pA + pdA + Adp + dpdA + ρu

^{2}+ (ρdu

^{2}+ 2ρudu + dρu

^{2}+ dρdu

^{2}+ 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:

AdP + Adρu

^{2}+ ρu

^{2}dA + 2ρuAdu = 0 ➔ eqn 1

The continuity equation is given by:

d(uρA) = 0

Explanding this we get,

ρudA + ρAdu + Audρ = 0

Multiplying the above equation by u on both sides we get:

ρu

^{2}dA + ρuAdu + Au

^{2}dρ = 0 ➔ eqn 2

Subtracting eqn 2 from eqn 1, we get the differential equation for the quasi one – dimensional flow:

dp = – ρudu

4. 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

View Answer

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. For which of these flows do we need a convergent duct to increase the velocity of the flow?

a) Subsonic flow

b) Supersonic flow

c) Hypersonic flow

d) Sonic flow

View Answer

Explanation: The area – velocity relation is given by:

\(\frac {dA}{A}\) = (M

^{2}– 1)\(\frac {du}{u}\)

According to this formula, for subsonic flows the value of M

^{2}– 1 becomes negative since 0 < M < 1. Thus, with decreasing cross – sectional area, the velocity increases. Decreasing area is achieved by convergent duct.

6. What happens to the velocity of the supersonic flow in the divergent duct?

a) Decreases

b) Increases

c) Remains the same

d) Changes periodically

View Answer

Explanation: Supersonic flows have M > 1 which results in M

^{2}– 1 value in the area – velocity relation to be positive. Thus for divergent sections where cross – sectional area increases with x, the value of velocity increases.

7. State true or false. Divergent – convergent nozzles are used to achieve supersonic flow.

a) True

b) False

View Answer

Explanation: In the case of subsonic flow, convergent nozzle leads of increased velocity using the area – velocity relation. Similarly, for supersonic flow the divergent nozzle leads to increased flow velocity. Thus, in order to obtain the supersonic flow we make use of convergent – divergent nozzle.

8. What is the differential form of energy equation for quasi one – dimensional flow?

a) dh – u^{2}du = 0

b) dh – udu = 0

c) dh + u^{2}du = 0

d) dh + udu = 0

View Answer

Explanation: The energy equation for quasi one – dimensional flow is given by:

h + \(\frac {u^2}{2}\) = constant

On differentiating the above equation we arrive at the differential energy equation for the quasi one – dimensional flow:

dh + udu = 0

9. The area – Mach number relation yields how many solutions for a given Mach number?

a) 2

b) 4

c) 6

d) 0

View Answer

Explanation: The area – Mach relation which is the ratio of local area to throat area as a function of Mach number yields two solutions for a given Mach number. One is a subsonic value and the other is the corresponding supersonic value.

10. In case of flow inside a nozzle, how should the exit pressure be in relation to the inlet pressure?

a) Exit pressure = Inlet pressure

b) Exit pressure > Inlet pressure

c) Exit pressure < Inlet pressure

d) There is no relation between the two pressures

View Answer

Explanation: The air in the nozzle will never flow by itself considering that the area ratio between the inlet and the throat is very high. There needs to be a pressure difference created for the air to move. This happens only when the exit pressure is less than the the inlet pressure (p

_{e}< p

_{0}).

11. Which of these conditions is not applicable for choked flow?

a) The Mach number at the throat is zero

b) Mass flow is constant

c) Exit pressure is lower than inlet pressure

d) Exit pressure is same as the inlet pressure

View Answer

Explanation: Choked flow is a condition where the exit pressure is lower than inlet pressure to a point where the flow after throat section becomes frozen. After this, despite lowering the value of exit pressure, inlet pressure has no effect. The mass flow remains constant despite reducing the exit pressure.

12. What is the relation between back pressure and exit pressure in case of subsonic flow at the nizzle’s exit?

a) p_{B} = p_{e}

b) p_{B} > p_{e}

c) p_{B} < p_{e}

d) p_{B} = 2p_{e}

View Answer

Explanation: Back pressure, which is the pressure downstream the nozzle’s exit is equal to the exit pressure when the flow at the exit is subsonic. This is because of a discrepancy between maintaining a steady subsonic flow in case of pressure difference.

**Sanfoundry Global Education & Learning Series – Aerodynamics.**

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