Solution Electrical Numerical 6

Numerical 6: A total load of 8000 kW at 0.8 power factor is supplied by two alternators in parallel. One alternator supplies 6000 kW at 0.9 power factor. Find the KVA rating of the other alternator and the power factor. 

Solution: In such questions, it is to be noted that KW is active component of KVA, while KVAr is reactive component.

The active components of all machines connected to a source can be added to know the total active load, and so the reactive load. But KVA can't be added or subtracted directly.

Alternator No.1:

$\displaystyle \mathrm{KW_1\ =\ 6000}$ 

$\displaystyle \mathrm{Cos\phi_1 \ =\ 0.9}$

$\displaystyle \mathrm{KVA_1\times Cos\phi_1 = KW_1}$

$\displaystyle \mathrm{KVA_1\times 0.9 = 6000}$ 

$\displaystyle \mathrm{KVA_1\ =\ 6666.667}$

$\displaystyle \mathrm{Cos\phi_1 = 0.9\ \Rightarrow\ Sin\phi_1 = 0.436\ }$

and, $\displaystyle \mathrm{KVA_1\times Sin\phi_1 = KVAr_1}$ 

$\displaystyle \mathrm{6666.667\times 0.436 = KVAr_1}$ 

thus, $\displaystyle \mathrm{KVAr_1 = 2906.67}$ 

Total:

$\displaystyle \mathrm{KW\ =\ 8000}$ 

$\displaystyle \mathrm{Cos\phi \ =\ 0.8}$

$\displaystyle \mathrm{KVA\times Cos\phi = KW}$

$\displaystyle \mathrm{KVA\times 0.8 = 8000}$ 

$\displaystyle \mathrm{KVA\ =\ 10,000}$

$\displaystyle \mathrm{Cos\phi = 0.8\ \Rightarrow\ Sin\phi = 0.6\ }$

and, $\displaystyle \mathrm{KVA\times Sin\phi = KVAr}$ 

$\displaystyle \mathrm{10,000\times 0.6 = KVAr}$ 

thus, $\displaystyle \mathrm{KVAr = 6000}$ 

Alternator No.2: 

$\displaystyle \mathrm{KW_2\ =\ KW-KW_1\ =\ 8000-6000\ =\ 2000}$

$\displaystyle \mathrm{KVAr_2\ =\ KVAr-KVAr_1\ =\ 6000-2906.667\ =\ 3093.333}$

$\displaystyle \mathrm{KVA_2\ =\ \sqrt{KW_2^2 + KVAr_2^2 }}$ 

$\displaystyle \mathrm{KVA_2\ =\ \sqrt{2000^2 + 3093.333^2 }=\ 3683.573}$ 

and, $\displaystyle \mathrm{Cos\phi _2 =\ \frac{KW_2}{KVA_2}}$

$\displaystyle \mathrm{Cos\phi _2 =\ \frac{2000}{3683.573}= 0.543}$  

Thus KVA rating of other alternator = 3683.573 KVA

and Power factor = 0.543