Engineering Questions with Answers - Multiple Choice Questions
MCQs on Governing Equations for Quasi One Dimensional Flow
1 - Question
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
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. Quasi one – dimensional flow 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
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 rearraging 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. Find Euler’s equation for given quasi – one dimensional flow 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: AdP + Adρu2 + ρu2 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: ρu2 dA + ρuAdu + Au2 dρ = 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
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
Explanation: The area – velocity relation is given by: dAA = (M2 – 1)duu According to this formula, for subsonic flows the value of M2 – 1 becomes negative since 0 < M < 1. Thus, with decreasing cross – sectional area, the velocity increases. Decreasing area is achieved by convergent duct.
6 - Question
What happens to the velocity of the supersonic flow in the divergent duct?
c) Remains the same
d) Changes periodically
Explanation: Supersonic flows have M > 1 which results in M2 – 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 - Question
State true or false. Divergent – convergent nozzles are used to achieve supersonic flow.
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 - 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
9 - Question
The area – Mach number relation yields how many solutions for a given Mach number?
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 - Question
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
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
11 - Question
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
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 - Question
What is the relation between back pressure and exit pressure in case of subsonic flow at the nizzle’s exit?
a) pB = pe
b) pB > pe
c) pB < pe
d) pB = 2pe
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.