Correct answer is (c) dp = -ρudu

To explain I would say: 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 + ρ1u\(_1^2\)A1 + \(\int_{A_1}^{A_2}\)pdA = p2A2 + ρ2u\(_2^2\)A2

We get,

pA + ρu^2A + pdA = (p + dp)(A + dA) + (ρ + dρ)(u + du)^2 (A + dA)

pA + ρu^2A + 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^2dA + 2ρuAdu = 0 ➔ eqn 1

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:

ρu^2dA + ρuAdu + Au^2dρ = 0 ➔ eqn 2

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

dp = -ρudu