# Motor Power Factor

#### Three Phase

Joined Jul 11, 2007
13
A 240 Volt, 60Hz supply has a heater and a single-phase motor connected in parallel to it. The heater draws a current of 5.5 amps and the total line current is 35 amps at a power factor of 0.60. The motor power factor is 0.484. By the way, the answer to this question is 0.484 (pf). I just can't figure on what to calculate to arrive at that answer.

#### recca02

Joined Apr 2, 2007
1,212
i dont get your question is .484 given or to be found out?
the data seems incomplete to me.

#### Three Phase

Joined Jul 11, 2007
13
i dont get your question is .484 given or to be found out?
the data seems incomplete to me.
The power factor .484 is to be found out.

#### JoeJester

Joined Apr 26, 2005
4,390
What is the question?

I see in input voltage [230V 60Hz], the Total I w/pf [35 A .. 0.6 pf], the heater's amps w/o a pf [5]. Then the next statement is the pf of the motor [0.454].

What I didn't see was a question?

#### Three Phase

Joined Jul 11, 2007
13
A 240 volt, 60 Hz supply has a heater and a single-phase motor connected in parallel to it. The heater draws a current of 5.5 amps and the total line current is 35 amps at a power factor of 0.60. What is the power factor of the motor?

#### recca02

Joined Apr 2, 2007
1,212
ok
i have tried it with the little data that was given,
my answer does not match ur's.
however i will give the explanation try to correct me if i m wrong.
we assume heater to be purely resistive since it must be a heat dissipating resistance.
.6 is total pf right?
so total active power supplied
.6*35*240
active power consumed by heater
240*35*1.
subtract and we get active power consumed by the motor.
since the current draw is 29.5 by motor (35-5.5)
apparent power consumed
29.5*240
pf= active/apparent.
but it comes to abt .525 something.
another approach wud be to actually find out the impedance using above idea but i think that wud yield the same answer and i m little lazy to find it out (maybe later).

#### Three Phase

Joined Jul 11, 2007
13
Thank you for your help, recca02.

#### recca02

Joined Apr 2, 2007
1,212
ok

active power consumed by heater
240*35*1.
.
oops sorry,
240*5.5*1
it was a typing error
but that wont change the answer.
someone with a correct answer?

#### spar59

Joined Aug 4, 2007
64
Here goes:-

The total power factor is 0.6 and the total current drawn is 35A.
Active current = Total current * Power factor = 35A * 0.6 = 21A.

The active and reactive current are at 90 degrees to each other,
therefore pythagoras' theorem applies.
i.e. Total current squared = Active current squared + Reactive current squared.

Hence:- Reactive current squared = Total current squared - Active current squared.
In this case, Reactive current squared = (35 * 35) - (21 * 21) = 784.
So the total Reactive current is 28A.

The current drawn by the heater (taken to be purely resistive and hence entirely active power) is 5.5A

The active current drawn by the motor is therefore equal to the total active current of 21A minus the heater current of 5.5A = 15.5A.
The reactive curent of the motor is the entire reactive current of 28A.
Total motor current squared = 15.5 squared + 28 squared = 1024.25
Hence total motor current = 32A

Power factor = Active current / Total current
Hence the motor Power factor = 15.5A / 32A = 0.484 !

Hope this helps, Steve.

#### recca02

Joined Apr 2, 2007
1,212
i wud like to be pointed where exactly i went wrong.
hope three phase sees yer reply.

#### spar59

Joined Aug 4, 2007
64
Hi there.

I reckon it started to go wrong here:-

"subtract and we get active power consumed by the motor.
since the current draw is 29.5 by motor (35-5.5)"

The 35A is at 0.6 power factor (presumably lagging since the motor will possess inductive reactance). So the total current of 35A is at an angle of 53 degrees. The 5.5A drawn by the purely resistive heater is at unity power factor i.e. 0 degrees. You then arithmetically subtracted the two figures rather than considering it as a vectorial operation.

Steve.

#### recca02

Joined Apr 2, 2007
1,212
got it.
and i was so lost in getting what did i actually do wrong thanks.