GATE Mechanical Engineering (ME) 2018 Shift 1 Solved Paper

Question : 51 of 65

Marks: +1, -0

A sprinkler shown in the figure rotates about its hinge point in a horizontal plane due to water flow discharged through its two exit nozzles.

The total flow rate

Q

through the sprinkler is 1 litre/sec and the cross-sectional area of each exit nozzle is

1 {cm}^{2}

. Assuming equal flow rate through both arms and a frictionless hinge, the steady state angular speed of rotation (rad/s) of the sprinkler is_________ (correct to two decimal places).

Your Answer:

Validate

Solution:

Relative velocities of water with sprinkler

V_{A}=\;{Q {/} 2}/{A}=\;{1 × 10^{-3}}/{2 × 10^{-4}}=5 {m}{/} {s}

V_{B}=5 {m} \/ {s}

Absolute velocity from B side

V_{A b s}^{'}-(+r_{B} ω)=V_{B}

V_{\;\text "Abs "\;}^{'}=V_{B}+r_{B} ω

=5+0.1 ω

Absolute velocity from A side

V_{A b s}-(-r_{A} ω)=V_{A}

V_{A b s}=V_{A}+r_{A} ω

V_{A b s}=5-0.2 ω

The external torque to the sprinkler is zero.
So,

\;\; Σ T=0

m_AV_{\Abs}r_A-m_BV_{\Abs}r_B=0

ρ(\;{Q}/{2})\{5-0.2 ω\} 0.2-ρ \;{Q}/{2}\{5+0.1 ω\} 0.1=0

1-0.04 ω-0.5-0.01 ω=0

0.05 ω=0.5

ω=10 {rad} {/} {s}

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