The pump of a water distribution system is powered by a 15-kW electric motor whose efficiency is 90 percent. The water flow rate through the pump is 50 L/s. The diameters of the inlet and outlet pipes are the same, and the elevation difference across the pump is negligible. If the pressures at the inlet and outlet of the pump are measured to be 100 kPa and 300 kPa (absolute), respectively, determine...
a.) The mechanical efficiency of the pump
b.) The temperature rise of water as it flows through the pump due to the mechanical inefficiency
6 comments:
Is the equation we use:
-Wdotpump = mdot * deltaHspecific
And with this work rate, we divide by (.9*15kW, the rate at which energy is provided by the motor to the pump) to find the mechanical efficiency?
If this is right, to find mdot, do we use the density of water in the process of converting the 50L/sec into a mass flow rate. Also, how do we go about finding the enthalpy at 1 and 2?
Audrey 7:28 PM
Your simplified 1st Law is correct.
mech. eff = (Wsh that actually enters the fluid) / (Wsh provided to the pump by the motor).
You can determine the Wsh that actually enters the fluid using the 1st law as you stated.
You can assume Tin=Tout=25 degC and use the density of sat liq water at 25 degC. The key is to assume that the water is incompressible. This means V^ depends only on T, not P. Therefore V^(subcooled liq@T&P) = V^(sat liq @ T and P*). This same relationship applies for U^, but NOT H. But deltaH^ = deltaU^ + delta(PV^). We assume the fluid does not change T in the pump...it is isothermal. Therefore, deltaU = 0. That means deltaH^ = V^ deltaP because the fluid is incompressible. Again, use V^ sat liq at 25 degC. That out to get you through this problem.
I did not realize all of the little weaknesses of this problem when I assigned it. I hope these comments help.
NEWS FLASH...
Do NOT do part (b) !
Assume Tin = Tout for part (a).
Do you mean do not do part b at all? Or just when solving part A?
Katie 1:27 PM
Do not do part (b) AT ALL.
Hi,
I have a similar problem. It is the blower in a home heating furnace. Can the efficiency of the furnace be partly determined by the heat rise across the heat exchanger. Assume the heat exchanger temperature is constant and the only thing that varies is air speed.
Gerry
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