Question

g A motor driving a 1000-W water pump has a power factor of 0.80 lagging; a second motor driving a 600-W water pump has a power factor of 0.60 lagging assuming the motors are working under 120- Vrms, 60-HZ AC. (a) When both motors working together what is combined power factor

Answer 1

complete Question

A motor driving a 1000-W water pump has a power factor of 0.80 lagging; a second motor driving a 600-W water pump has a power factor of 0.60 lagging assuming the motors are working under 120-Vrms, 60-HZAC. When both motors working together what is the combined power factor? If a 200-μF capacitor is connected to the above system (two motors) what is the new combined power factor?

Answer:

$p.f'=0.960$

Explanation:

First motor Power $P=1000W$

First motor Power factor $P.f=0.80$

Second motor Power $P=600W$

Second motor Power factor p.f=0.60

Voltage $V=120Vrms$

Frequency $F=60Hz$

Capacitor $C=200\mu F$

Generally power in Var is given as

For First Motor

$Q=\frac{1000}{0.8}\sqrt{1-0.8^2}$

$Q=750Var$

For Second Motor

$Q=\frac{600}{0.6}\sqrt{1-0.6^2}$

$Q=800Var$

Generally the equation for The Reactive Power is mathematically given by

$Q_c=\frac{V^2}{X_c}$

Where

$X_c=\frac{1}{2 \pi fc}$

$X_c=\frac{1}{2 \pi 60*200*10^{-6} }$

$X_c=13.3$

Therefore

$Q_c=-\frac{120^2}{13.3}\\\\Q_c=-1085.97j$

Giving

Total Power Drawn by Supply

$P_t=(1000+j750)+(600+800)-j1085.97$

$P_t=1600+464.03j$

Therefore

$p.f'=\frac{1600}{\sqrt{1600^2+464.03^2}}$

$p.f'=0.960$