# Power System Problem - Schaun's outline



## patelpe (Sep 16, 2010)

Que. A three phase, three wire, 500 V, 60 HZ sourse supplies a three phase induction motor, a wye connected capacitor bank that draws 2 Kvar per phase, and a balanced three phase heater that draws a total of 10 KW. The induction motor is operating at its rated 75 hp and has an efficiency and PF of 90.5 and 89.5 percent, respectively. Draw a one line diagram for the system and determine

(a) the system KW

(b) sthe system Kvar

© the system KVA

answer (a) 71.82 KW (b) 24.81 kvar © 75.98 KVA

I got an answer for (a) as below. Isn't this should be answer for © total KVA rather then KW

IM KW = 75 hp * 746

=55950/efficiency

=61823 + 10000

= 71.823 KW

Can anyone take a look and post the answer with explanation please for a,b,c ?

Thanks,


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## VectrenEng (Sep 16, 2010)

This is my first post, so here we go.

OK - I was able to get the answers. Your work for Part A is correct (at least it matches mine). Break this problem into 3 loads: motor, cap bank, and heater (each with its own power triangle) and then bring them back together at the end.

Motor:

P = 75hp x 0.746kW/hp / 0.905 = 61.82kW

pf = acos(0.895) = 26.49º

Q = P tan(angle) = 61.82kW tan(26.49º) = 30.81kVAR

Cap Bank

P = 0kW (no real power)

Q = 6kVAR (3-phase)

Heater

P = 10kW

Q = 0kVAR (pure resistance load)

No add this for your total load.

Pt = 61.82 + 10 = 71.82kW

Qt = 30.81 - 6 = 24.81kVAR (cap bank offsets inductive load of motor)

St = SQRT(71.82^2 + 24.81^2) = 75.96kVA

I hope this helps.


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## patelpe (Sep 16, 2010)

Thanks, got it.



VectrenEng said:


> This is my first post, so here we go.
> OK - I was able to get the answers. Your work for Part A is correct (at least it matches mine). Break this problem into 3 loads: motor, cap bank, and heater (each with its own power triangle) and then bring them back together at the end.
> 
> Motor:
> ...


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## Gnana (Sep 17, 2010)

Motor:

P = 75hp x 0.746kW/hp / 0.905 = 61.82kW

pf = acos(0.895) = 26.49º

Q = P tan(angle) = 61.82kW tan(26.49º) = 30.81kVAR

Okay not sure if this is an obvious answer or not

Why did you use the input power for the motor and not the output power to calculate the sytem KW?


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## HornTootinEE (Sep 17, 2010)

Gnana said:


> Motor:
> P = 75hp x 0.746kW/hp / 0.905 = 61.82kW
> 
> pf = acos(0.895) = 26.49º
> ...


Anyone else find that sometimes the answer key for Schaum's may be suspect?


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## VectrenEng (Sep 17, 2010)

Gnana said:


> Motor:
> P = 75hp x 0.746kW/hp / 0.905 = 61.82kW
> 
> pf = acos(0.895) = 26.49º
> ...


The input power of the motor is the load "seen" by the source. The motor's output is 15hp, but requires more given the efficiency of the motor.


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## knight1fox3 (Sep 17, 2010)

VectrenEng said:


> The motor's output is 15hp,


Err....75HP


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