Fluid Mechanics Viscosity Question 292
Question: A horizontal tube has different cross sections at points A and B. The area of cross section are $ a_{1} $ and$ a_{2} $ , respectively, and pressures at these points are $ p_{1}=\rho gh_{1} $ and$ p_{2}=\rho gh_{2} $ , where p is the density of liquid flowing in the tube and $ h_{1} $ and $ h_{2} $ are heights of liquid columns in vertical tubes connected at A and B. If$ h_{1}-h_{2}=h $ , then the flow rate of the liquid in the horizontal tube is
Options:
A) $ a_{1}a_{2}\sqrt{\frac{2gh}{a_{1}^{2}-a_{2}^{2}}} $
B) $ a_{1}a_{2}\sqrt{\frac{2g}{h\left( a_{1}^{2}-a_{2}^{2} \right)}} $
C) $ a_{1}a_{2}\sqrt{\frac{(a_{1}^{2}+a_{2}^{2})h}{2g\left( a_{1}^{2}-a_{2}^{2} \right)}} $
D) $ \frac{2a_{1}a_{2}gh}{\sqrt{a_{1}^{2}-a_{2}^{2}}} $
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Answer:
Correct Answer: A
Solution:
[a] Let $ a_{1} $ and $ a_{2} $ be cross sectional areas and$ v_{1} $ and $ v_{2} $ the velocities of liquid flow at A and B. If $ p_{1} $ and $ p_{2} $ are pressures of liquid recorded by manometer, then $ P_{1}+\frac{1}{2}pv_{1}^{2}=p_{2}+\frac{1}{2}\rho v_{2}^{2} $ $ \Rightarrow p_{1}-p_{2}=\frac{1}{2}\rho v_{1}^{2}\left( \frac{v_{2}^{2}}{v_{1}^{2}}-1 \right) $ By equation of continuity, $ a_{1}v_{1}=a_{2}v_{2} $ Also, $ p_{1}-p_{2}=h\rho g $ Substituting these values, we have $ v_{1}=\sqrt{\frac{2gh}{\frac{a_{1}^{2}}{a_{2}^{2}}-1}} $ Rate of flow of liquid is$ a_{1}v_{1}=a_{1}a_{2}\sqrt{\frac{2gh}{a_{1}^{2}-a_{2}^{2}}} $
BETA
