2 more questions

[Rei]

1. Several references that I read stated that the starting current of a motor is 6 times the rated current. However, I have seen several sample question asked to find the starting current of the motor and they just calculate the current, I=S/(1.73*V). Don't you also need to multiply by 6 after calculated I=S/(1.73*V)?

2. Is it true that you can't add apparent power? A secondary of a transformer has the following loads: 1 induction motor given kVA & pf, 1 synchronous motor given kVA & pf, and a lighting load in kW with 100% pf. They ask for the total power factor at the secondary of the transformer. I total up the S and P value and divide: pf=P/S and got the wrong answer. The solution total up P and Q to find S=sqrt(P^2+Q^2) and then divide: pf=P/S.

 

[Flyer_PE]

1. Several references that I read stated that the starting current of a motor is 6 times the rated current. However, I have seen several sample question asked to find the starting current of the motor and they just calculate the current, I=S/(1.73*V). Don't you also need to multiply by 6 after calculated I=S/(1.73*V)?

That's the equation to find current given voltage and apparent power. If the apparent power given is under running conditions, then you are correct to multiply the answer by six to estimate inrush current for the motor.

2. Is it true that you can't add apparent power? A secondary of a transformer has the following loads: 1 induction motor given kVA & pf, 1 synchronous motor given kVA & pf, and a lighting load in kW with 100% pf. They ask for the total power factor at the secondary of the transformer. I total up the S and P value and divide: pf=P/S and got the wrong answer. The solution total up P and Q to find S=sqrt(P^2+Q^2) and then divide: pf=P/S.

Given P and Q, S2 = P2 + Q2

 

[Rei]

That's the equation to find current given voltage and apparent power. If the apparent power given is under running conditions, then you are correct to multiply the answer by six to estimate inrush current for the motor.

starting current is the inrush current, right?

Given P and Q, S2 = P2 + Q2

So do you say that for giving different type of load, you can't add kVA together?

 

[Flyer_PE]

starting current is the inrush current, right?

Correct.

So do you say that for giving different type of load, you can't add kVA together?

You have to remember that S is not a scalar quantity, it's a vector. You have to account for the angle for each of the loads. You can add all the P's together to get total real power and all the Q's together to get the total reactive power.