Engineering Questions with Answers - Multiple Choice Questions

# MCQs on 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

a
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.
Find the flow variation from the given diagram

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

a
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

c
Explanation: We consider a duct of variable cross sectional area with two stations 1 and 2, having properties as given in the figure.
Determine the equation in cross sectional area of given diagram
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

b
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

d
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?
Find the value of surface integral volume from the given diagram
a) V.dS
b) dS
c) Zero
d) V2.dS

c
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

a
Explanation: The energy equation stated for steady quasi – one – dimensional flow is given by h1 + u212 = h2 + u222. According to this formula which is derived for the adiabatic flow, the total enthalpy remains constant (h0 = 0).

8 - Question

The governing equations for quasi – one – dimensional flow is used to find properties of flow in a nozzle.
a) True
b) False