The discussion focuses on calculating the power consumption of a pump operating at 500 kPa and a flow rate of 60 m³/h with an efficiency of 80%. The relevant formula for power is established as P = q_v * ρ * g * (p / (ρ * g)), where q_v is the volumetric flow rate, ρ is the fluid density (1,200 kg/m³), and g is the acceleration due to gravity. The participants clarify that the pressure can be used to derive the energy head, allowing for the simplification of the formula to P = p / (ρ * g). This leads to a definitive understanding of how to incorporate efficiency into the power calculation.
PREREQUISITESEngineers, fluid mechanics students, and professionals involved in pump design and performance optimization will benefit from this discussion.
The formula looks useful. What is qm?jderulo said:I believe it to be: P = q_m ρ g h_p
mfb said:The formula looks useful. What is qm?
Instead of g and a height, you are given the pressure. How are those concepts related?
Similar enough.jderulo said:g is gravity, h isn't height but energy head..
mfb said:Similar enough.
The main idea is that you can write down a similar formula where you replace ρ, g and hp 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.
Right. You can simplify that formula, ρ and g will drop out and you are left with a simple product.jderulo said:h_p = p / ρ g
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Am I right in thinking the final formula would be
P = q_v ρ g (p / ρ g)