Question

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

(a) p1A1 + ρ1u\(_1^2\)A1 + \(\int_{A_1}^{A_2}\)pdA = p2A2 + ρ2u\(_2^2\)A2

(b) p1A1u1+ ρ1u\(_1^2\)A1 + \(\int_{A_1}^{A_2}\)pdA = p2A2u2+ ρ2u\(_2^2\)A2

(c) p1A1 + ρ1u\(_1^2\)A1 = p2A2 + ρ2u\(_2^2\)A2

(d) p1A1u1+ ρ1u\(_1^2\)A1 = p2A2u2+ ρ2u\(_2^2\)A2

This question was posed to me in unit test.

The question is from Governing Equations in section Quasi-One-Dimensional Flow of Aerodynamics

Answer 1

Right option is (a) p1A1 + ρ1u\(_1^2\)A1 + \(\int_{A_1}^{A_2}\)pdA = p2A2 + ρ2u\(_2^2\)A2

To explain: 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 – ρ1u\(_1^2\)A1 + ρ2u\(_2^2\)A2.

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:

 – \(\int_{A_1}^{A_2}\) – 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:

 – ρ1u\(_1^2\)A1 + ρ2u\(_2^2\)A2 = -(-p1A1 + p2A2 ) + \(\int_{A_1}^{A_2}\)pdA

On rearranging the terms we get:

p1A1 + ρ1u\(_1^2\)A1 + \(\int_{A_1}^{A_2}\)pdA = p2A2 + ρ2u\(_2^2\)A2