Question

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1)Suppose we have a steady fluid flow through a section of a cylindrical pipe with varying area, with control volume.  At the entrance the radius of the pipe r1 = 4m, the flow velocity is U1 = 5 i m/s and at the exit the radius of the pipe is r2 = 2.5 m and the velocity of the flow is U2 = 5 i m/s. Use the integral form of the continuity equation to show that the ratio of the density of the fluid entering the pipe to that exiting the pipe is

ρ1/ ρ2 ≈ 0.4

Explain the result?

Answer 1

e integral form of the continuity equation is given as a t Since, the fluid fow is steady through the sechon of Rthe lindrical pipe, therefore, Эр D. So, the integral form Ot educes to ที่. (pajdV =0 Use Ghauss Diver gence theorem to convert the volume integral tb the surpace integral, i.e., This implies 2 S. Rewrite the integral on the three surfaces, namely, nlet surface,si, outlet surface, So, and bounding surfae along the sides as, S

Consider surface Si, || Řul (ti. ).ti) dS pu.nds | = U) note here nalso Likewise, For surface So, u.nds note here n1 ond, For the Surfoce S since at the bounding surface the velocity is along the surface so alwaus nor mal to the surgace, thereforr, Vi , е, о, so,

Substitute the expressions from (2h (3), and (4) into (1) gives Since, p and U are uniformly distributed on both the inlet and outlet Surfoces So, 512ニ25,~ 0.4 64 2. Thus, _ ~ 0.4 As, the area increases, For Flow with constant velocity the density decreases, therefore,