Power Consumption of a Pump: Solve at 500kPa, 60m3/h, 80% Efficiency

[jderulo]

Homework Statement

Determine the power consumed by a pump when it is pumping a fluid of density 1,200kg/m^-3 60m3/h at 500kPa with an 80% overall efficiency.

Homework Equations

I believe it to be: $$P = q_m ρ g h_p$$

The Attempt at a Solution

Stuck at the above part, I don't have enough info?

 

[mfb]

I believe it to be: $$P = q_m ρ g h_p$$

The formula looks useful. What is q_m?

Instead of g and a height, you are given the pressure. How are those concepts related?

 

[jderulo]

The formula looks useful. What is q_m?

Instead of g and a height, you are given the pressure. How are those concepts related?

q_m is mass flow-rate. g is gravity, h isn't height but energy head..

 

[mfb]

g is gravity, h isn't height but energy head..

Similar enough.

The main idea is that you can write down a similar formula where you replace ρ, g and h_p to include pressure instead of those two parameters.

I am surprised that it is the mass flow rate and not the volume flow rate.

Anyway, you need the delivered power output of the pump (that's what your formula can do).
Once you have that, you can see how you can include the 80% efficiency.

 

[jderulo]

Similar enough.

The main idea is that you can write down a similar formula where you replace ρ, g and h_p to include pressure instead of those two parameters.

I am surprised that it is the mass flow rate and not the volume flow rate.

Anyway, you need the delivered power output of the pump (that's what your formula can do).
Once you have that, you can see how you can include the 80% efficiency.

My mistake it's volumetric flow. - but unsure how to do the first point.

 

[jderulo]

$$h_p = p / ρ g$$

?

Am I right in thinking the final formula would be

$$P = q_v ρ g (p / ρ g)$$

 

[mfb]

$$h_p = p / ρ g$$

?

Am I right in thinking the final formula would be

$$P = q_v ρ g (p / ρ g)$$

Right. You can simplify that formula, ρ and g will drop out and you are left with a simple product.