Power factor and frequency change

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
7 replies · 8K views
DT71
Messages
7
Reaction score
0

Homework Statement


240V,50 Hz electrical appliance is rated at 2KW and has a lagging power factor of0.7.
a)Determine the appliance's power factor when it is used on a 60Hz supply?
b)Calculate the supply voltage required to maintain the appliance at its rated power when operated off a 60Hz supply?


Homework Equations


p.f.=P/S
S^2=P^2+Q^2 to obtain Q
S=VsI to obtain I
Z=Vs/I to obtain Z
p.f. = R/Z to obtain R
Z^2=R^2+Xl^2 to obtain Xl
Xl=2pifL to obtain L at 50Hz



The Attempt at a Solution



I have done the above then used the value for L in Xl=2pifL at 60 Hz but not sure were to go from here.
 
Physics news on Phys.org
As you have already found Xl at 60 Hz you can find power factor at 60 Hz from value of new Z and R (Note R is unchanged)

Since you know Z and P (rated power) you can find V
 
Thanks, I get the first section now but I don't see how I can work out the supply voltage using the new Z and the rated power.
 
should i be using P=V^2/Z
 
Note V^2/Z gives the apparent power S.
 
I'm confused now, how do I calculate the supply voltage required to maintain the appliance at its rated power at 60Hz.

I have the rated power and the impedence at 60Hz.
 
Hi, you know P and Z, from Z you can find p.f.
From above values you can find S. Once you know S you can find V.
 
I have worked out the pf to be 0.63
P(rated power) = 2000 w
Z=22.31 ohms

Would it be correct to use P(rated power)= V^2/R as R remains the same to obtain V(supply voltage)