Question
Water is being pumped from a lake located at sea level to a storage reservoir
located on the side of a hill. For this arrangement, we are interested in
determining the amount of pumping power (in watts) that will be required to
move the water for different reservoir elevations. In all cases, the rate of water
transfer is to be 24,000 gallons per hour.
Create an excel spreadsheet that allows you to calculate the required pumping
power (Ws; watts) for different reservoir elevations. In completing the
calculations, estimate the total length (L) of the pipe to be equal to 2.5 times the
total elevation change in meters, + 20 meters:
L=20+2.5 Ah = 20 +2.5 (Zout - Zin)
.
The diameter of the pipe is constant everywhere, and equal to 12 centimeters.
The frictional losses across the pipe segments can be estimated per the
equation below, with f = 0.034.
h₁ = f ( t ) ·
212
(
pipe
2g
Also, since both the lake and the upper reservoir are open to the atmosphere,
PINLET POUTLET = PATM-
Take the water levels as constant at both locations, with Vinlet = Voutlet = 0.
What is the Pumping Power, in Watts, required to pump the water for an outlet
elevation of 125 meters? Report your answer to the nearest whole number.
Type your answer...
located on the side of a hill. For this arrangement, we are interested in
determining the amount of pumping power (in watts) that will be required to
move the water for different reservoir elevations. In all cases, the rate of water
transfer is to be 24,000 gallons per hour.
Create an excel spreadsheet that allows you to calculate the required pumping
power (Ws; watts) for different reservoir elevations. In completing the
calculations, estimate the total length (L) of the pipe to be equal to 2.5 times the
total elevation change in meters, + 20 meters:
L=20+2.5 Ah = 20 +2.5 (Zout - Zin)
.
The diameter of the pipe is constant everywhere, and equal to 12 centimeters.
The frictional losses across the pipe segments can be estimated per the
equation below, with f = 0.034.
h₁ = f ( t ) ·
212
(
pipe
2g
Also, since both the lake and the upper reservoir are open to the atmosphere,
PINLET POUTLET = PATM-
Take the water levels as constant at both locations, with Vinlet = Voutlet = 0.
What is the Pumping Power, in Watts, required to pump the water for an outlet
elevation of 125 meters? Report your answer to the nearest whole number.
Type your answer...
Answers
GPT 3.5
To calculate the required pumping power (Watts), we need to consider the frictional losses across the pipe segments and the elevation change.
First, let's calculate the total elevation change:
Elevation change (Zout - Zin) = 125 meters
Next, we can use the equation to find the total length of the pipe:
L = 20 + 2.5 * (Zout - Zin)
L = 20 + 2.5 * (125 - 0)
L = 20 + 312.5
L = 332.5 meters
Now, we can calculate the frictional losses across the pipe segments using the equation provided:
h₁ = f * (2 * L)/(pipe diameter^2 * g)
h₁ = 0.034 * (2 * 332.5)/(0.12^2 * 9.81)
h₁ = 6.14 meters
Finally, we can calculate the required pumping power (Ws; Watts) using the formula:
Ws = (Poutlet - Pinlet) * (flow rate / 3600) + (h₁ * flow rate * water density * 9.81)
Since both the lake and the upper reservoir are open to the atmosphere, PINLET - POUTLET = PATM = 101325 Pa
Flow rate = 24000 gallons per hour = 90.84 m^3/h
Ws = (101325) * (24000 / 3600) + (6.14 * 90.84 * 1000 * 9.81)
Ws = 675500 + 534392.1776
Ws = 1,209,892.1776 Watts
Rounded to the nearest whole number, the required pumping power for an outlet elevation of 125 meters is approximately 1,209,892 Watts.
First, let's calculate the total elevation change:
Elevation change (Zout - Zin) = 125 meters
Next, we can use the equation to find the total length of the pipe:
L = 20 + 2.5 * (Zout - Zin)
L = 20 + 2.5 * (125 - 0)
L = 20 + 312.5
L = 332.5 meters
Now, we can calculate the frictional losses across the pipe segments using the equation provided:
h₁ = f * (2 * L)/(pipe diameter^2 * g)
h₁ = 0.034 * (2 * 332.5)/(0.12^2 * 9.81)
h₁ = 6.14 meters
Finally, we can calculate the required pumping power (Ws; Watts) using the formula:
Ws = (Poutlet - Pinlet) * (flow rate / 3600) + (h₁ * flow rate * water density * 9.81)
Since both the lake and the upper reservoir are open to the atmosphere, PINLET - POUTLET = PATM = 101325 Pa
Flow rate = 24000 gallons per hour = 90.84 m^3/h
Ws = (101325) * (24000 / 3600) + (6.14 * 90.84 * 1000 * 9.81)
Ws = 675500 + 534392.1776
Ws = 1,209,892.1776 Watts
Rounded to the nearest whole number, the required pumping power for an outlet elevation of 125 meters is approximately 1,209,892 Watts.