GATE Mechanical Engineering (ME) 2018 Shift 2 Solved Paper

Question : 51 of 65

Marks: +1, -0

Air flows at the rate of

1.5 {m}^{3} {/} {s}

through a horizontal pipe with a gradually reducing cross-section as shown in the figure. The two cross-sections of the pipe have diameters of

400 {mm}

and

200 {mm}

. Take the air density as

1.2 {kg} {/} {m}^{3}

and assume inviscid incompressible flow. The change in pressure

(p_{2}-p_{1})

(in kPa) between sections 1 and 2 is

Solution:

Q=a_{1} v_{1}=a_{2} v_{2}=1.5 {m}^{3} {/} {sec}

d_{1}^{2} v_{1}=d_{2}^{2} v_{2}

v_{2}=4 v_{1}

a l s o, v_{1}=\;{Q}/{a_{1}}=11.93 {m}{/} {sec}

\;{P_{1}}/{ρ g}+\;{v_{1}^{2}}/{2 g}+z_{1}=\;{P_{2}}/{ρ g}+\;{v_{2}^{2}}/{2 g}+z_{2}

\;{P_{1}-P_{2}}/{ρ g}=15 \;{v_{1}^{2}}/{2 g}

P_{1}-P_{2}=15 \;{v_{1}^{2} × ρ}/{2}=1282.346 P a=1.28 k P a

P_{2}-P_{1}=-1.28 {kPa}

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