GATE Mechanical Engineering (ME) 2020 Shift 1 Solved Paper

Question : 52 of 65

Marks: +1, -0

Consider steady, viscous, fully developed flow of a fluid through a circular pipe of internal diameter D. We know that the velocity profile forms a paraboloid about the pipe centre line, given by :

V=-C(r^{2}-\;{D^{2}}/{4}) m{/}s

, where

C

is constant. The rate of kinetic energy (in

{J} {/} {s}

) at the control surface

A-B

, as shown in the figure, is proportional to

D^{n}

, The value of

n

is ________________

Your Answer:

Validate

Solution:

Given

v=-c(r^{2}-\;{D^{2}}/{4})=-c(r^{2}-R^{2})

v=c(R^{2}-r^{2})

d(K E)=\;{1}/{2} d m(v^{2})=\;{1}/{2} v^{2} ρ d v=\;{1}/{2} ρ v^{2} × v d A

d(K E)=\;{1}/{2} ρ v^{3} d A

KE ∫ d(K E)=∫ \;{1}/{2} ρ c(R^{2}-r^{2})^{3} × 2 π r d r

\;\; =\;{1}/{2} ρ c^{3} ∫ R^{6}+r^{6}-3 R^{2} r^{2}-3 R^{2} r^{2}(R^{2}-r^{2}) 2 π r d r

\;{1}/{2} ∫↙{0}^{2 π p c^{3}} R^{R} R+r^{7}-3 R^{4} r^{3}+3 R^{2} r^{5} d r

K E=π p c^{3} R^{8} × 0.625

K E=0.625 × π p c^{3} × R^{8}

{KE} ∝ {R}^{8}

K E ∝ D^{8}

K E ∝ D^{n} ⇒ n=8

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