Question

# An upward flow of oil (mass density 800 kg/m3, dynamic viscosity 0.8 kg/m-s) takes place under laminar conditions in an inclined pipe of 0.1 m diameter as shown in the figure. The pressures at sections 1 and 2 are measured as p1=435 kN/m2 and p2=200kN/m2. The discharge in the pipe is equal to

A
0.100 m3/s
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B
0.127 m3/s
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C
0.144 m3/s
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D
0.161 m3/s
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Solution

## The correct option is B 0.127 m3/s Apply energy equation between (1) and (2) P1ρg+V212g+Z1=P2ρg+V222g+Z2+hf Where hf=32μVLρgD2 435×103800×9.81+V22g+0=200×103800×9.81 +V22g+5√2+32(0.8)V(5)(800)(9.81)(0.1)2 ⇒ 435×103−200×103800×9.81−5√2=32×0.8×V×5800×9.81×(0.1)2 ⇒ V=16.19 m/s ∴ Q=AV =π×(0.1)24×16.19 =0.127 m3/s

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