Hi,

I have these 2 question which i cant solve, can anyone teach me how to solve these 2 questions? Sorry for the bad image.. have to edit them to cut the file size. Thanks

 

I'm not going to work it out for you, but it seems like you're given the power and phase angle from G1, you should be able to calculate the current from G1. Since you know the impedance of the line between G1 and the load, you should be able to find the voltage and phase angle at the load. Given that information, you should be able to find the reactive powers supplied by G2. Don't forget about the impedance of the lines, which are complex impedances.

For the second part, isn't it just a common ABCD parameter problem?

 

I'm not going to work it out for you, but it seems like you're given the power and phase angle from G1, you should be able to calculate the current from G1. Since you know the impedance of the line between G1 and the load, you should be able to find the voltage and phase angle at the load. Given that information, you should be able to find the reactive powers supplied by G2. Don't forget about the impedance of the lines, which are complex impedances.

For the second part, isn't it just a common ABCD parameter problem?

Hi,

I am abit lost... am i right to say that the power 15kw needs to be divided by √3VLcosϕ to get the IL?

Since it is Y connection, the IL = Ip = IG1??

Also may i know what is terminal voltage?

 

Hi,

I am abit lost... am i right to say that the power 15kw needs to be divided by √3VLcosϕ to get the IL?

Since it is Y connection, the IL = Ip = IG1??

Also may i know what is terminal voltage?

The total power is 15 kw, so the power per phase is 5 kw. Your voltage is given in line-line, so you need to convert it to phase or line voltage. So you divide the line-line voltage by √3. Your cos θ is already known. It's the power factor of 0.8. If you must know the angle, you can take the arcos of 0.8, which is 36.9°.

Yes, you're correct about the Y connection details.

There's another snag here. The line impedances appear to be given in per unit values. Have you studied those? You need to select base values and convert your generator and load values to a common base. In this case, I would probably use 800 VLL and 15 kW as my base. You have to decide if you're going to work in single phase or 3-phase and choose your base on that. For example, instead of 800 VLL, you might want to use 462 VL and 5 kW as your base. Whatever you choose, you can then derive the other bases from them. For example on a phase basis, base kVA = base kW, and base current = base KVA / base VL, and so on. From there you can convert your generator and load parameters (voltages, currents, powers, etc.) to per unit (p.u.) values. If everything is in per unit values, then it is easy to work with them. At the end of the problem you'll convert back by multiplying times the base.

As for the terminal voltage, it would be the generator terminal voltage of G2. It's the unknown generator voltage.

 

Hello mlog,

I think you are incorrect with your interpretation of the line impedance values being given in per unit values.

Take the case of the G1 source. The G1 line current would be of the order 13.5A rms. Using (one of) your proposed basis of VA base = 5000W and voltage base as 462V would give a per unit impedance base of 42.3 ohms. So the G1 line impedance would then be 42.3(1.2+j1.8)=91.6Ω @ angle 56.3°. With the line current of 13.5A the rms line voltage drop would be 1236V which is an impractical result.

I think the indicated line impedances should be considered as actual rather than per-unit values.

 

Hello mlog,

I think you are incorrect with your interpretation of the line impedance values being given in per unit values.

Take the case of the G1 source. The G1 line current would be of the order 13.5A rms. Using (one of) your proposed basis of VA base = 5000W and voltage base as 462V would give a per unit impedance base of 42.3 ohms. So the G1 line impedance would then be 42.3(1.2+j1.8)=91.6Ω @ angle 56.3°. With the line current of 13.5A the rms line voltage drop would be 1236V which is an impractical result.

I think the indicated line impedances should be considered as actual rather than per-unit values.

You're absolutely right. I'm glad you ran the numbers, because I wondered if they were per unit. One ohm seemed small for actual numbers, so I assumed they were per unit. The lines must be short. Thanks for not letting me steer him down the wrong path.

 

You're absolutely right. I'm glad you ran the numbers, because I wondered if they were per unit. One ohm seemed small for actual numbers, so I assumed they were per unit. The lines must be short. Thanks for not letting me steer him down the wrong path.

just to clarify one more issue. should i take the impedence given as a line impedence or should i take it as a phase impedence?

 

just to clarify one more issue. should i take the impedence given as a line impedence or should i take it as a phase impedence?

I'm not sure I understand the question. The impedances given between the terminals of the generators and the load are between the terminals. The current in the lines through the line impedance doesn't care how you treat the generator or the load, as long as you use the correct values. For example, the generator and load can be a black box with lines between them. No matter what is in the black boxes, the line impedance remains the same.

It's the generator and the load where you have to make a decision of line vs. phase, and that depends on what makes the calculations easier. I think it's usually easier to work in phase where I would treat the generators and loads (after conversion if necessary) as a wye.