Power Triangle and Power factor

In summary, induction motors have a power factor of 0.8, resulting in an angle of 36.86 degrees and an apparent power of 625 VA. Synchronous motors have a power factor of 0.707, resulting in an angle of 45 degrees and an apparent power of 707 VA. To find the load percentage, solve the equation system x+y=500 and x*tan(acos(0.8))-y*tan(acos(0.707))= 500*tan(acos(0.9)). The answer will be y/500[%].
  • #1
user12323567
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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.
 
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  • #2
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))
 
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  • #3
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.
 
  • #4
But how do I use that information to obtain what they are asking for?
 
  • #5
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))]
The answer will be y/500[%]
 
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  • #6
Babadag said:
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))]
The answer will be y/500[%]
Thank you, that helped!
 

Related to Power Triangle and Power factor

What is a power triangle?

A power triangle is a graphical representation of the relationship between real power, reactive power, and apparent power in an electrical circuit. It is used to calculate the power factor of a circuit.

What is power factor?

Power factor is the ratio of real power to apparent power in an electrical circuit. It is a measure of how efficiently a circuit uses the power it receives.

Why is power factor important?

Power factor is important because it affects the efficiency and cost of an electrical system. A low power factor means that more energy is being wasted, resulting in higher electricity bills and potential damage to equipment.

How is power factor calculated?

Power factor is calculated by dividing the real power by the apparent power. It is typically expressed as a decimal between 0 and 1, or as a percentage between 0% and 100%.

How can power factor be improved?

Power factor can be improved by using power factor correction techniques, such as adding capacitors to the circuit, to reduce the amount of reactive power and increase the power factor. Properly sizing and maintaining equipment can also help improve power factor.

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