Question

Water is being pumped through a horizontal pipe that is 36.0 m long and 13.5 cm in diameter. A pump maintains a gauge pressure of 742 Pa at a large open tank at one end of the pipe. The other end of the pipe is open to the air. If this pipe is replaced by a second one of length 9 m and diameter 5.0 cm, what gauge pressure must the pump provide to give the same volume flow rate as for the first pipe?

Answer 1

The energy required to pump the fluid through Pipe #1 = the energy required to pump the fluid through Pipe #2

The energy is used to do the work of forcing the fluid to move the length of the pipe.

Work = Force * distance
Pressure = Force ÷ area
Force = Pressure * area

Work = Pressure * area * distance
The work required to pump the fluid through Pipe #1 = the work required to pump the fluid through Pipe #2

Pressure1 * area1 * distance1 = Pressure2 * area2 * distance2

Volume of Pipe #1 = Cross sectional area * length
Radius = 13.5/2 = 6.75 cm = 0.0675 m
Area = π * 0.0675^2 = 0.0143066 meter^2

Volume of Pipe #1 = 0.0143066 * 36 m^3
Volume of Pipe #1 = 0.5150376 m^3

Volume of Pipe #2 = Cross sectional area * length
Pipe #2
Radius = 2.5 cm = 0.025 m
Area = π * 0.025^2 = 0.0019625meter^2

742 * 0.0143066 * 36 = Pressure2 * 0.0019625 * 9


867 * 5.25625 * 21.6 = Pressure2 * 0.0019625 * 9
Pressure 2 = 5.57 * 10^6 Pa

Calculations maybe faulty, logic is correct.

Hope this helps :)