Flow energy usually represents the forces of gravity, pressure, heat, mechanical systems or some combination of these applied to a mass of fluid, vapor or gas over a period of time. This causes the flowing media to impart its energy or power to any containment or engine that may harness the flow energy to use the power to do useful work. Examples would be devices that impart flow energy such as pumps, boilers and compressors, as well as devices that convert flow energy back to mechanical energy such as turbines, engines and nozzles like rocket motors.
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Define the application. A water pump sends 6,000 gallons up a 150foot pipe to an elevated municipal water tank every hour. If you know that the pump has an efficiency of 68 percent and the pipe is 3 inches in diameter, you can calculate the flow energy imparted to the system by the pump and the flow energy that would be released by letting it flow down again.

Calculate the mechanical work in raising the water 150 feet. The 6,000 gallons of water weighs 8.33 lbs/gallon so the total weight of water pumped is 6,000 lbs X 8.33 = 49,980 lbs, which requires 49,980 X 150 feet = 7,497,000 foot lbs of work to raise it.

Calculate the energy lost as friction against the piping. Data for Schedule 40 Steel Pipe shows that the 6,000 gph/60 min/hr = 100 gpm flow sustains a 3.1 pounds per square inch (psi) pressure drop per 100 feet of pipe. The loss over 150 feet would be 150/100 feet X 3.1 psi = 4.65 psi. This represents a loss percentage: 4.65 psi/the 150 foot elevation/2.31 feet/psi = 4.65/64.94 psi = 0.0716 or 7.16 percent extra work the pump has to perform to push the water through the pipe.

Calculate the total work performed by the pump in lifting the 6,000 gallons of water by multiplying the 7,497,000 foot/lbs of work by a factor of 1.071 to account for the lifting work plus the frictional pipe losses, which yields 8,033,860 foot lbs of work to pump the water up.

Calculate pump horsepower. Total work/33,000 foot lbs/minute/hp/60 minutes/hr = 4.06 hp. Since the pump is only 68 percent efficient, 4.06/0.68 = 5.97 horsepower. A pump with a minimum of 6 horsepower would be required for this job.

Calculate the total flow energy (expressed in kilowatts) required by the pump to raise the liquid and the total flow energy that would be released by letting the water flow back out again. Converting the 5.97 horsepower required to pump it up to the top to kW yields 5.97 hp/0.746 kW/hp = 8 kW. Since the only difference between pumping it up and letting it back out again is the pump efficiency, 8 kW X 0.68 = 5.44 kW recaptured by draining the water. This might be further frustrated by a turbogenerator efficiency equivalent to pump efficiency.
Tips & Warnings
 The less manipulation done to a flow, the more of its energy that is conserved.
 Flow energy in the form of highly compressed gases can release with dangerous or fatal results if their containment piping ruptures.
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 Photo Credit Blue Water Tower image by Jim Mills from Fotolia.com motors image by Dusan Radivojevic from Fotolia.com rocket motor 1 image by Aaron Kohr from Fotolia.com pipes on line image by timur1970 from Fotolia.com