# Power Triangle and Power factor

• Engineering
• user12323567

#### user12323567

Homework Statement
A group of induction motors with a total power of 500kW and power factor 0.8 lagging are to be partially replaced with a synchronous motors of the same efficiency but leading power factor of 0.707. As the replacement proceeds the overall power factor improves. What percentage of the load will have been replaced when the overall power reaches 0.9 lagging?(Answer is 15.2%)
Relevant Equations
Power factor(pf) = cosθ
Average power(Pave) = VIcosθ
Apparent power(Pa) = VI
Reactive power(Pr) = VIsinθ
Induction motors:

pf = 0.8
triangle:
θ = arccos(0.8) = 36.86 degrees
Pa = 500/0.8 = 625 VA
Pr = sqrt(625^2-500^2) = 375 VAR

Synchronous motors:

pf = 0.707
triangle:
θ = arccos(0.707) = 45 degrees
Pa = 500/0.707 = 707 VA
Pr = sqrt(707^2-500^2) = 499.85 VAR

I am uncertain of how I can relate this information with the load percentage, any hints/tips will be much appreciated.

x is the kW of remaining induction motors
y is the kW of new synchronous motor
x+y=500 kW
the final reactive power will be 500*[-tan(acos(0.9))]
the x part kVAr=-x*tan(acos(0.8))
the y part kVAr=+y*tan(acos(0.707))

• user12323567
According to reactive power sign convention, the inductive power has to be positive and capacitive negative and then the apparent power [sum of active and reactive power] it is the product of supply voltage with the conjugate of the current. So, I have to change the sign of the final reactive power [to be positive as inductive] and x part kVAr it is also positive but y part will be negative. The absolute values will the same as above.

But how do I use that information to obtain what they are asking for?

You need to solve the [equation] system:
x+y=500 [kW]
x*tan(acos(0.8))-y*tan(acos(0.707))= 500*[tan(acos(0.9))]
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