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`

Step-by-step 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=(1000)/(0.8)√(1-0.8^2)`

`Q=750Var`

For Second Motor

`Q=(600)/(0.6)√(1-0.6^2)`

`Q=800Var`

Generally the equation for The Reactive Power is mathematically given by

`Q_c=(V^2)/(X_c)`

Where

`X_c=(1)/(2 \pi fc)`

`X_c=(1)/(2 \pi 60*200*10^(-6) )`

`X_c=13.3`

Therefore

`Q_c=-(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'=(1600)/(√(1600^2+464.03^2))`

`p.f'=0.960`