Thermodynamics oil pump efficiency question

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Hi guys, could someone help me on this? I'm stuck :(

*A motor draws 35kW electricity, it operates at 90% efficiency. The motor is connected to an oil pump which has an inlet diameter of 8cm & an outlet diameter of 12cm. pressure rise in the oil pump is 400kPa. oil enters the pump at 0.1m^3/s. what is the mechanical efficiency of the oil pump?*


I understand the question, I know the mechanical input power is 31.5kW and that i need to figure out the output power from the info given, then the mech efficiency is that number/31.5kW ... but I've got no clue how to go about finding the output power. sooo... could I get some pointers pretty please? any help would be greatly appreciated.

 

[nilesh_pat]

Thanks! To calculate the output power of the oil pump, you need to calculate the kinetic energy of the oil exiting the pump. You can do this by using the Bernoulli equation. The Bernoulli equation states that the sum of the potential and kinetic energies of a fluid is constant. So, if you know the velocity of the oil entering the pump, the pressure drop across the pump, and the density of the oil, then you can calculate the velocity of the oil exiting the pump. This velocity can then be used to calculate the kinetic energy of the oil, which is the output power of the pump.

 

[piercas]

Hello there,

To calculate the output power of the oil pump, we need to use the equation for pump power: P = (Q*ΔP)/η. Q is the flow rate of oil, ΔP is the pressure rise, and η is the pump efficiency.

In this case, Q = 0.1 m^3/s and ΔP = 400 kPa. We can convert the flow rate to liters per second by multiplying it by 1000, giving us 100 L/s. We can also convert the pressure rise to Pa by multiplying it by 1000, giving us 400,000 Pa.

Now, we need to find the pump efficiency. This can be done by using the inlet and outlet diameters of the pump. The cross-sectional area of the inlet is A = π*(0.08/2)^2 = 0.00502 m^2, and the cross-sectional area of the outlet is A = π*(0.12/2)^2 = 0.01131 m^2. The pump efficiency can then be calculated as η = A_outlet/A_inlet = 0.01131/0.00502 = 2.25.

Plugging in these values into the pump power equation, we get P = (100*400,000)/2.25 = 177,778 W. This is the output power of the oil pump. To find the mechanical efficiency, we can divide this by the input power of 31.5 kW, giving us a mechanical efficiency of approximately 5.64%.

I hope this helps and clarifies the process for finding the output power and mechanical efficiency of the oil pump. Keep in mind that this is a simplified calculation and there may be other factors that could affect the efficiency of the pump in a real-world scenario.