The horizontal pipe shown in the figure (Figure 1) has a cross-sectional area A

Question

The horizontal pipe shown in the figure (Figure 1) has a cross-sectional area A1 = 40.5 cm2 at the wider portions and A2 = 10.2 cm2 at the constriction. Water is flowing in the pipe, and the discharge from the pipe is 6.05×103 m3/s (6.05 L/s ).

Part A

Find the flow speed at the wide portion.

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Part B

Find the flow speed at the narrow portion.

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Part C

Find the pressure difference between these portions.

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Part D

Find the difference in height between the mercury columns in the U-shaped tube.

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The horizontal pipe shown in the figure (Figure 1) has a cross-sectional area A1 = 40.5 cm2 at the wider portions and A2 = 10.2 cm2 at the constriction. Water is flowing in the pipe, and the discharge from the pipe is 6.05×103 m3/s (6.05 L/s ).

Part A

Find the flow speed at the wide portion.

v =   m/s  

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Part B

Find the flow speed at the narrow portion.

v =   m/s  

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Part C

Find the pressure difference between these portions.

p =   Pa  

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Part D

Find the difference in height between the mercury columns in the U-shaped tube.

Explanation / Answer

volume flow rate is dV/dt = A1*v1 = 6.05*10^-3

40.5*10^-4*v1 = 6.05*10^-3

v1 = 1.5 m.s

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b) A2*v2 = 6.05*10^-3

10.2*10^-4*v2 = 6.05*10^-3

v2 = 5.93 m/s

c) using Bernoulies principle

P1+ 0.5*rho*v1^2 = P2+0.5*rho*v2^2

P1-P2 = 0.5*(rho)*(v2^2-v1^2)

P1-P2 = 0.5*1000*(5.93^2-1.5^2) = 1.645*10^4 Pa

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P1-P2 = rho*g*h

P1-P2 = 13593*9.81*h = 1.645*10^4

h = 0.123 m = 12.3 cm

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