The form factor of an AC waveform is the ratio of its RMS value to its average value. For a perfect sine wave, the form factor is approximately 1.11 (RMS value/Average Value = 1.11). This factor indicates how closely a waveform resembles a pure sine wave; a form factor closer to 1.11 suggests a waveform that is more sinusoidal.

$$kW_{t}=8000kW$$

$$\cos\phi_{t}=0.8$$

$$kW_1=6000KW$$

$$\cos\phi_1=0.9$$

To Find (a) kVA2 and cosϕ2

For alternator 1

$$\cos\phi_1=\frac{kW_1}{kVA_1}$$

$$0.9=\frac{6000}{kVA_1}$$

$$kVA_1=6666.667kVA$$

$$\sin\phi_1=\frac{kVAr_1}{kVA_1}$$

$$as\:\cos\phi=0.9;\:\phi=25.84\degree$$

$$so,\:\sin\phi=0.435$$

$$0.435=\frac{kVAr_1}{6666.667}$$

$$kVAr_1=-2905.932\:kVAr$$

$$Now,\:\cos\phi_{t}=0.8$$

$$\cos\phi_{t}=\frac{kW_{t}}{kVA_{t}}$$

$$0.8=\frac{8000}{kVA_{t}}$$

$$kVA_{t}=10000kVA$$

$$as\:\cos\phi_{t}=0.9\:\Rightarrow\:\phi_{t}=36.86\degree$$

$$so,\:\sin\phi_{t}=0.6$$

$$\sin\phi_{t}=\:\frac{kVAr_{t}}{kVA_{t}}$$

$$0.6=\frac{kVAr_{t}}{10000}$$

$$kVAr_{t}=-6000kVAr$$

For alternator 2:

$$kW_2=kW_{t}-kW_1$$

$$kW_2=8000-6000=2000kW$$

$$kVAr_2=kVAr_{t}-kVAr_1$$

$$kVAr_2=-6000-\left(-2905.932\right)$$

$$kVAr_2=3094.068kVAr$$

$$kVA_2=\sqrt{\left(kW_2\right)^2+\left(kVAR_2\right)^2}$$

$$kVA_2=\sqrt{\left(2000\right)^2+\left(-3094.068\right)^2}$$

$$kVA_2=3684.190kVA$$

$$as,\:\cos\phi_2=\frac{kW_2}{kVA_2}$$

$$\cos\phi_2=\frac{2000}{3684.190}$$

$$\cos\phi_2=0.542$$

$$thus,\:kVA\:rating=3684.19kVA$$

$$power\:fact\lor=0.542$$