Question

In the infinitesimal neighborhood surrounding a point in an inviscid flow, the small change in pressure, dP, that corresponds to a small change in velocity, dV, is given by the differential relation: dP=−rhoVdV. (a) Using this relation, derive a differential relation for the fractional change in density, drho/rho, as a function of the fractional change in velocity, dV/V, with the compressibility τ as a coefficient. (b) The velocity at a point in an isentropic flow of air is 10 m/s, and the density and pressure are 1.23 kg/m3 and 1.01 x 105 N/m2, respectively. The fractional change in velocity at the point is 0.01. Calculate the fractional change in density. (c) Repeat part (b), except for a local velocity at the point of 1000 m/s. Compare this result with that from part (b), and comment on the differences.

Answer 1

Answer:

(a). differential relation becomes dρ/ρ = -τρV2 dV/V

(b). fraction change in density; dρ/ρ = -8.7 ˣ 10⁻⁶

(c).  dρ/ρ = -8.7 ˣ 10⁻²

Explanation:

Let us begin,

(a).  given from the question we have that dp = -ρVdV

where dρ = ρ τ dp, i.e.

dp = dρ/ρτ ...............(1)

replacing value of dp we have,

-ρVdV = dρ/ρτ

so that dρ = -τp2 VdV

finally, dρ/ρ = -τp V2 dV/V

(b). from the question here, we were given Velocity to be = 10 m/s

density (ρ) =  1.23 kg/m3

pressure (p) =  1.01 x 10⁵ N/m2

from formula,

dρ/ρ = τs ρ V2 dV/V .............(2)

but τs = 1/γp = 1/(1.4× 1.01×10⁵) = 7.07 ˣ 10⁻⁶ m²/N

substituting value of τs  into equation (2) we have

dρ/ρ = τs ρ V2 dV/V =  (7.07 ˣ 10⁻⁶) ˣ (1.23) ˣ (10ˣ2) (0.01) = -8.7 ˣ 10⁻⁶

dρ/ρ = -8.7 ˣ 10⁻⁶

(c). from we question we know that dρ/ρ has a large ratio of (1000/10)²

so dρ/ρ = -8.7 ˣ 10⁻⁶ × (1000/10)² = -8.7 ˣ 10⁻²

dρ/ρ = -8.7 ˣ 10⁻².

comparing this result with part (b). we can see that when we increase the velocity of a factor 100, there is an increased factorial change in the density by a factor 104.