What does this mean...anything?

[7Eggyjla4]

I'm somewhat unfamiliar with the process below and am not sure what it really means or if it is even correct.

3Hp motor; 3Ph; 415V; 84%Eff; 0.8pf

The problem started by:

(3Hp * 0.7457 * 0.8pf)/(415V * sqrt(3) * 0.84) = 3.0A [Active]

(3Hp * 0.7457 * sqrt(1-0.8^2))/(415V * sqrt(3) * 0.84) = 2.2A [Reactive]

(3Hp * 0.7457)/(415V * sqrt(3) * 0.84) = sqrt(Active^2 + Reactive^2) = 3.7A [Apparent]

Is there any validity to the above?

My understanding would be: (3Hp * 0.7457)/(415V * 0.8pf * sqrt(3) * 0.84) = 4.6A

 

[rcw retired EE]

Your first two equations are wrong. Do not multiply by power factor 0.8. (Equation is also missing a 1000).

O.8 = power factor = KW/KVA, and KW = HP x 0.746. Multiplying HP by power factor gives units of (KW^2)/KVA. You already had KW since HP = k x KW.

Your last equation is correct.

Note that most motors have power factors in the 0.8-0.9 range and efficiency around 0.85 - 0.9. Assume 0.86 for both, 0.86^2 = 0.747, which cancels with 0.747 KW/HP in the numerator. That is why for most motors we can simplify and say 1 HP = 1 KVA and skip the power factor and efficiency except when very accurate detail is needed.
This assumption is close except for large and very small motors.

Amps = HP *1000/ (V x 1.732)

 

[racookpe1978]

Thank you for showing us that simplification.

The 1 HP / 1KW means 1 HP (output) or "useable developed motor load", and the 1 kW is "input" or "actual required power" right?

 

[waross]

When the motor power is given in HP I assume that kW is input power.
When the motor power is given in kW then I assume that kW is useable developed motor load.


Bill
--------------------
"Why not the best?"
Jimmy Carter