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Power – Energy Relationships

Power – Energy Relationships. Power = Energy / time. Energy units: BTU, calories, ergs, joules Power units: Hp, watts. Unit conversions: Energy Power 1 joule = 9.5 x 10 -4 BTU 1 watt = 1.34 x 10 -3 hp 0.239 cal 10 7 ergs. Pumps and Pipes.

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Power – Energy Relationships

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  1. Power – Energy Relationships Power = Energy / time Energy units: BTU, calories, ergs, joules Power units: Hp, watts

  2. Unit conversions: EnergyPower 1 joule = 9.5 x 10-4 BTU 1 watt = 1.34 x 10-3 hp 0.239 cal 107 ergs

  3. Pumps and Pipes Gravity flow will only work if there is enough elevation difference between the source and the delivery location. This elevation difference must be enough to overcome frictional head loss in pipes and to provide enough kinetic energy to provide adequate flow rate. If elevation difference is not adequate pumps need to supply this energy.

  4. Moving fluid from point 1 to point 2 requires energy. Bernoulli equation can be used to calculate energy balance. V = fluid velocity P = pressure g = acceleration of gravity g = specific weight of water z = elevation hL = frictional head loss

  5. Frictional head loss is a function of turbulence level and roughness of pipe. One method to predict frictional head loss is the Darcy- Weisbach equation: f = “friction factor” L = length of pipe D = diameter of pipe

  6. Q L h2 Q h1

  7. Example – A pipe with dia. = 0.15 ft, length = 200 ft and friction factor, f, = 0.1 is to carry a flow of 10 gal/min (0.022 ft3/min) from point 1 to point 2. Elevation difference between point 1 and 2 = 25 ft. Determine the “head” a pump must provide to deliver this flow. “Head” provided by the pump must be the sum of the elevation + friction head loss. (assuming that velocity and pressure in the pipe are constant)

  8. First calculate frictional head loss: Note: V = Q/A Total pump head = 3.3 ft + 25 ft = 28.5 ft. (or in terms of pressure = 12.13 psi (head x 62.4 lb/ft3)

  9. Head Pump selection: Pumps are classified by “pump curves”. Objective is to find a pump with a performance curve that meets your design. A typical pump curve is shown below. Q

  10. Objective is to find a pump that will deliver the required flow at the head required. Generally you will not be able to find a pump to meet exact design requirements so go to the next bigger size and throttle flow with a valve.

  11. Pump curves usually have more information such as efficiency at various flows. See example on next slide.

  12. To compute the power required to drive the pump use: • Where: • Q has units of ft3/sec • = 62.4 lbs/ft3 • H = hL + elevation difference • 550 = conversion from ft-lb/sec to hp Efficiency of motor and pump (overall) is typically 50 – 60%

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