Assume that a feeder supplies an industrial consumer with a cumulative load of:

Question

Assume that a feeder supplies an industrial consumer with a cumulative load of: (i) induction motors totaling 200hp which run at an average efficiency of 90% and a lagging average power factor of 0.80; (ii) synchronous motors totaling 200 hp with an average efficiency of 80%, and a lagging average power factor of 0.80; and (iii) a heating load of 50 KW. The industrial consumer plans to use the synchronous motors to correct its overall power factor. Determine the required power factor of synchronous motors to correct the overall power factor at peak load to unity power factor.

Explanation / Answer

Eff=90% so after dividing the above KW value with efficiency

Induction motor= 165.89 KW

Eff=80% so after dividing the above KW value with efficiency

Synchronous motor =186.62 KW

Now KVA1 (Heat load) = 50 KW/ CosCos

=50 KVA

Now KVA2 (Induction motor)= 165.89 KW/ CosCos

= 187.65 KVA

Now KVA3 (Synchronous Motor)=186.62/ CosCos

= 233.27 KVA

KVAR1 = KVA1 x Sin Ø1= 50x0 =0

KVAR2 = KVA2 x Sin Ø2= 187.65x0.436 = 81.79 KVAR

KVAR3 = KVA3 x Sin Ø3= 233.27x0.6 = 139.96 KVAR

Total KW = 50+165.89+186.62= 402.51 KW

Total KVAR= 0+81.79+139.96 = 221.752 KVAR

Total KVA= Sq root of ((Total KW)^2 + (Total KVAR)^2))= 459.55 KVA

Power Factor = Total KW/ Total KVA

=402.51/459.55 = 0.876

So to synchronous motor must run at Power Factor of 0.876 to bring the overall power factor to unity.

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