GATE Mechanical Engineering (ME) 2020 Shift 1 Solved Paper

Question : 49 of 65

Marks: +1, -0

Air discharges steadily through a horizontal nozzle and impinges on a stationary vertical plate as shown in figure.

The inlet and outlet areas of the nozzle are

0.1 {m}^{2}

and

0.02 {m}^{2}

, respectively. Take air density as constant and equal to

1.2 {kg} {/} {m}^{2}

. If the inlet gauge pressure of air is

0.36 {kPa}

, the gauge pressure at point

{O}

on the plate is

{kPa}

(round off to two decimal places).

Solution:

Given

{P}_{1}=0.36 {KPa} \;\text " (At Inlet) "\;

{P}_{2}=0 \text " (At outlet) "\;

{A}_{1}=0.1 {m} 2 \;\; ρ \;\text " air "\; {P}_{ {o}}=\;{ {f}_{ {x}}}{/}{ {A}}=ρ {v}^{2}= {A} .2 {kg} {/} {m}^{3}

{A}_{2}=0.02 {m}2[\text "Constant"]

{P}_{0}=?

Apply Bernoulli's equation at Inlet and outlet section.

\;{ {P}_{1}}/{ρ_{ {g}}}+\;{ {V}_{1}^{2}}/{2_{9}}+ {z}_{1}=\;{ {P}_{2}}/{ρ_{ {g}}}+\;{ {V}_{2}^{2}}/{2_{ {g}}}+ {z}_{2}

{Z}_{1}+ {Z}_{2}, \;\; {P}_{2}=0

\;{P_{1}}/{ρ_{g}}+\;{V_{1}^{2}}/{2_{g}}=\;{v_{2}^{2}}/{2_{g}}

...............(i)
By Continuity equation.

ρ_{1} {A}_{1} {V}_{1}=ρ_{2} {A}_{2} {V}_{2}

ρ 1=ρ 2

{V}_{1}=\;{ {A}_{2}}/{ {A}_{1}} {V}_{2}

{V}_{1}=\;{0.02}/{0.1} {V}_{2}

{V} 1=0.2 {V}_{2}

Putting in equation

\;{360}/{1.2}=\;{ {V}_{2}^{2}-(0.2 {V}_{2})^{2}}/{2}

300=\;{0.96 {V}_{2} 2}/{2}

V_2 = 25 m{/}s

Apply Bernoulli's equation between 2 and

0 .

\;{P_{2}}/{ρ g}+\;{V_{2}^{2}}/{2 g}+Z_{1}=\;{P_{0}}/{ρ g}+\;{V_{0}^{2}}/{2 g}+τ_{0}

Z_{1}=z_{0}

P_{2}=0

V_{0}=0

\;{V_{2}^{2}}/{2 g}=\;{P_{0}}/{ρ g}

P_{0}=\;{ρ V_{2}^{2}}/{2}

P_{0}=\;{1.2 ×(25)^{2}}/{2}

P_{0}=375 P_{a}

P_{0}=0.375 {KPa}

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