Calculating a 3 phase motors mains input power

[Martin0951]

I have a question that is posing a problem regarding calculating the mains input power being consumed by a 3 phase induction motor that has the following parameters:
20HP
400Vac 50Hz
PF 0.9
Efficiency 0.85

I was led to believe that is was simply 20HP x 746W = 14.92kW but I think this is wrong?

Should you first calculate the motors line current i.e.
IL = Power (kW) / VL x 1.73 x PF x n
14,920 /400 x 1.73 x 0.9 x 0.85 = 28.18A

then

Pin = VL x IL x 1.73 x PF (does efficiency need to be used as well?)
Pin = 400 x 28.18 x 1.73 x 0.9 = 17.56kW

I have searched everywhere but I cannot seem to find correct solution to this the simple problem.

Could someone advise on how to do this correctly?
Also when you find the input power, is it this the power consumed by each phase or the TOTAL power consumed by the machine?

Many Thanks in advance,
Regards
Martin

 

[itsmoked]

Your first statement answers your specific question sans the efficiency.

You need to add more "mains power consumed" to cover just the inefficiency.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[jraef]

When you see a motor rated as HP (or kW), that is the maximum rated MECHANICAL power of the motor, i.e. (Tq x rpm)/5250. The electrical CONSUMED power is the kW / rated efficiency, but that ALSO depends on the actual LOAD on the motor, not the rated power capability.


" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden

 

[che12345]

Dear Mr. Martin0951 (Electrical)(OP)13 Jan 22 16:25
"... problem regarding calculating the mains input power being consumed by a 3 phase induction motor that has the following parameters:
20HP, 400Vac 50Hz, PF 0.9, Efficiency 0.85
I was led to believe that is was simply 20HP x 746W = 14.92kW but I think this is wrong? ..."

1. It is correct to say that the motor output (at full load) is capable to drive the mechanical load of Po=14.92kW.
2. But, the " the mains input power being consumed by a 3 phase induction motor " is the Pi=Po/efficiency i.e. 14.92kW/0.85 =17.55kW. This is the consumption 17.55kW " the electricity bill that you are charged ". The charge is in kWh , depending on the duration on hour that the load is loaded.
2.1 In most cases, for small consumers; PF is NOT taken into billing charge.
3. However, if the load is halved, then Pi=( Po/2 x eff2). Note: eff2 may differs from 0.85 , which is at full load.
Che Kuan Yau (Singapore)

 

[LionelHutz]

HP is power output. Try re-arranging your basic efficiency = Pin/Pout formula to solve for Pin.

 

[EngRepair]

At full load kVA = 20*0,746/(0.9*0.85)=19,5 kVA

 

[electricpete]

Good answers.

> Should you first calculate the motors line current i.e.
> IL = Power (kW) / VL x 1.73 x PF x n

Considering you didn't measure the current, and you have us a nice round number 20hp without any supporting calculations, I'm guessing you might have deduced 20hp from the nameplate. That would not be a correct approach in general. The motor outputs whatever power the driven load demands at the operating speed (provided the motor doesn't stall or trip). The 20hp rating is just a description of the maximum steady state power the motor can deliver while still maintaining a margin to thermal damage.

=====================================
(2B)+(2B)' ?