GATE Papers >> Mechanical >> 2016 >> Question No 223

Question No. 223 Mechanical | GATE 2016
A channel of width 450 mm branches into two sub-channels having width 300 mm and 200 mm as shown in figure. If the volumetric flow rate (taking unit depth) of an incompressible flow through the main channel is $ 0.9\;\mathrm m^3/\mathrm s $ and the velocity in the sub-channel of width 200 mm is 3 m/s, the velocity in the sub-channel of width 300 mm is _____________ m/s.
Assume both inlet and outlet to be at the same elevation.


Answer : 0.99 : 1.01

Solution of Question No 223 of GATE 2016 Mechanical Paper

$ \left(\mathrm Q\right)=0.9\;\frac{\mathrm m^3}\sec\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm{velocity}=3\;\frac{\mathrm N}{\mathrm m^2} $

As fluid is incompressible

From equation of continuity

$ \mathrm Q={\mathrm Q}_1+{\mathrm Q}_2 $

For unit depth

Flow rate $ ({\mathrm Q}_2)=\mathrm{velocity}\times\mathrm{Area} $


$ \begin{array}{l}\mathrm{Area}=\mathrm{width}\times\mathrm{depth}=3\times\left(0.2\times1\right)=0.6\;\frac{\mathrm m^3}\sec\\\therefore{\mathrm Q}_1=\mathrm Q-{\mathrm Q}_2=\left(09-0.6\right)\;\frac{\mathrm m^3}\sec=0.3\;\frac{\mathrm m^3}\sec\\\therefore\mathrm{Velocity}\;\mathrm{in}\;\mathrm{channel}\;1\;\left({\mathrm v}_1\right)\\{\mathrm v}_1=\frac{\mathrm{Volume}\;\mathrm{flow}\;\mathrm{rate}\;\mathrm{in}\;1\left({\mathrm Q}_1\right)}{\mathrm{Area}\;\mathrm{of}\;\mathrm{channel}\;1\left(\mathrm{width}\times\mathrm{Height}\right)}=\frac{0.3}{0.3\times1}\\{\mathrm v}_1=1\;\frac{\mathrm m}\sec\end{array} $

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