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Pressure drop at power station pumping power law fluid
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[QUOTE="Chestermiller, post: 4659678, member: 345636"] What you do is, you first choose the two points between two of the pumps, and get the pressure drop in the section of pipe. Next you choose the two points right before and right after one of the pumps. The pressure change across each pump is equal to the pressure drop in the section of pipe between pumps. From the first law of thermo, the change in enthalpy per unit mass of fluid passing through the pump is equal to the shaft work per unit mass (since the pump is assumed adiabatic). Also, since the temperature rise across the pump is going to be close to zero (assuming negligible viscous heating in the pump), the change in enthalpy per unit mass is just equal to the Δ(PV)=VΔP. So the shaft work per unit mass is just VΔP. Chet [/QUOTE]
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Pressure drop at power station pumping power law fluid
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