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

Answers

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:

[tex]p.f'=0.960[/tex]

Explanation:

First motor Power [tex]P=1000W[/tex]

First motor Power factor [tex]P.f=0.80[/tex]

Second motor Power [tex]P=600W[/tex]

Second motor Power factor p.f=0.60

Voltage [tex]V=120Vrms[/tex]

Frequency [tex]F=60Hz[/tex]

Capacitor [tex]C=200\mu F[/tex]

Generally power in Var is given as

For First Motor

[tex]Q=\frac{1000}{0.8}\sqrt{1-0.8^2}[/tex]

[tex]Q=750Var[/tex]

For Second Motor

[tex]Q=\frac{600}{0.6}\sqrt{1-0.6^2}[/tex]

[tex]Q=800Var[/tex]

Generally the equation for The Reactive Power is mathematically given by

[tex]Q_c=\frac{V^2}{X_c}[/tex]

Where

[tex]X_c=\frac{1}{2 \pi fc}[/tex]

[tex]X_c=\frac{1}{2 \pi 60*200*10^{-6} }[/tex]

[tex]X_c=13.3[/tex]

Therefore

[tex]Q_c=-\frac{120^2}{13.3}\\\\Q_c=-1085.97j[/tex]

Giving

Total Power Drawn by Supply

[tex]P_t=(1000+j750)+(600+800)-j1085.97[/tex]

[tex]P_t=1600+464.03j[/tex]

Therefore

[tex]p.f'=\frac{1600}{\sqrt{1600^2+464.03^2}}[/tex]

[tex]p.f'=0.960[/tex]


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