A pump adds 48000 ft*lb/s to a steady flow rate of 5 ft^3/s . the fluid is water

Question

A pump adds 48000 ft*lb/s to a steady flow rate of 5 ft^3/s . the fluid is water at 60 F. At the entrance to the pump , the diameter of the pipe is 1.5 ft and the elevation is 624.00 ft. At the exit of the pump, the diameter is 1.25 ft and the elevation is 626.00 ft. the temprature of the inflow to the pump is approximately the same as the temperature of the outflow. determine the pressure difference between the entrance and the exit of the pump .

only answer is your sure , and show all calculation steps and formulas to be rated.

Explanation / Answer

Flow rate [F] = 5 ft^3/s

Density of water at60 F (D)= 1.93 slugs/ft^3

mass flow rate (m) =F*D = 9.65 slugs/s

Initial Conditions : [Enterance]

Elevation [ h1 ] = 624ft

Diameter [D1]= 1.5 ft

Area [A1] = (pi*D1^2)/4 = 3.14 * 1.5*1.5/4 ft^2 = 1.766 ft^2

Velocity -> V1 = F/A1 = 2.83 ft/s

Intial pressure ->P1

Final Conditions : [Exit]

Elevation [ h2 ] = 626ft

Diameter [D2]= 1.25 ft

Area [A2] = (pi*D2^2)/4 = 3.14 * 1.25*1.25/4 ft^2 = 1.226 ft^2

Velocity -> V2

Final pressure ->P2

Using the continuity Equation :

A1*V1 = A2*V2

(or) V2 = A1*V1/A2 =1.766*2.83/1.226 ft/s =4.076 ft/s

V2 =4.076 ft/s

Rate of energy entering :{e1]

P1 * F + 0.5 *m *V1^2 + m*g*h1 = P1 *5 + 0.5 *9.65 *2.83^2 + 9.65*32.2*624 = 5P1 + 193934.163 ft*lb/s

Rate of energy Exiting: : [e2]

P2 * F + 0.5 *m *V2^2 + m*g*h2 = P2 *5 + 0.5 *9.65 *1.226^2 + 9.65*32.2*626 = 5P2 + 194524.23 ft*lb/s

There e2-e1 = Energy added = 48000 ft*lb/s

or 5 (P2-P1)+194524.23 ft*lb/s-193934.163 ft*lb/s = 48000 ft*lb/s

or 5 (P2-P1) + 590.06 =48000 ft*lb/s

Pressure difference = P2-P1 = 9481.98 lb/ft^2 [ANS]

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