Power Factor Improvement

To understand power factor improvement, it's required that we have to understand about power triangle. This concept is very important to support our sharing today. let's look at figure below.

From the figure above, we can know that power factor or cos pi is comparison of active power (P) to apperent power (S). To improve cos pi of electric power system is required reactive power (Q) supplied to the electrical system. How much is Q required? It depends on the cos pi to be achieved. So, how to determine Q ? look at figure below.

Q or Q (C) can be calculated by knowing active power (P) and initial cos pi and final cos pi to be achieved.
Let's learn an example below.

Suppose we wish to increase power factor from 0,8 to 0,9 in a tree phase system (V= 400 Volt) absorbing an active power of 100 kW.

• The absorbed current shall be :
• I   = P/(1,732*V*Cos pi) = 100*1000/(1,732*400*0,8) = 180 A
• Q = P*(tan pi0-tan pi1) kVar = 100*(0,75-0,48) = 26 kVar

Due to power factor improvement, the absorbed current decreases from 180 A to 160 A.

•  I   = P/(1,732*V*Cos pi) = 100*1000/(1,732*400*0,9) = 160 A

From the example above, there are several adventages by improving power factor :

•  Reduction of losses
•  Reduction of voltage drops
•  Increase the performance of electrical equipment and line

Please comment after reading my posting.
Thank you

Regards,
HI

Voltage Drop Calculation with MVA Method

Voltage Drop (VD) is useful to determine cable sizing of electric power system. By selecting the appropriate cable  will  improve stability. To calculate voltage drop at bus we can use MVA Method. In this session we will use the same one line diagram on article before.

Look at the one line diagram below.

In this case study, we want to determine how is the voltage drop at bus 12 kV?. On article before, we got that MVAsc series is 153 MVAsc and MVAsc motor is 75 MVAsc. So, to calculate we use formula :

• VD at bus 12 kV = MVAsc series/(MVAsc motor + MVAsc series) = 153/(75+153) = 67%
• It means that the voltage at bus 12 kV decrease 33% when motor starting to run.
  

Now, we try to calculate VD at bus 69 kV.

• MVAsc for motor, transformer and line are connected series to the bus 69 kV. So, we have to sum up like a parallel network.
• 1/MVAsc series = 1/MVAsc motor + 1/MVAsc transformer + 1/MVAsc line = 1/(75)+1/(198)+1/(1230) = 52 MVAsc
• VD at bus 69 kV = MVAsc generator/(MVAsc generator + MVAsc series) = 1500/(1500+52) = 96%
• It means that the voltage at bus 69 kV decrease 4% when motor starting to run.

By using MVA method, we can solve the voltage drop esier than other method. To check whether the result is correct, you can use ETAP or similiar software.

Thank you.
Please comment after reading my posting.

Regards,
HI

Short Circuit Current

SCC calculation is important to determine rating of breaker, cable feeder, busbar, protection relay  setting, etc. How to calculate SCC (Short Circuit Current) ?There are three methods to solve it i.e. Ohmic Method, PU Method and MVA Method. In this session I want to share about one power full method, i.e. MVA Method. Why MVA method? There are several reasons :

•  This method doesn't require MVA base to calculate
•  No need to convert impedance based on the voltage
•  Easy and not complex
•  Suitable for the whole system one line diagram (OLD)
•  The equation is simple and easy to remember.
•  MVA is the most accurate method.

Let's look at the one line diagram below.

Following are steps to convert the OLD  with MVA Method:

• Generator : MVA/Xd" = 1500/1 = 1500 MVAsc
• Line : kV^2/x = (60 kV)^2/(3,87) = 1230 MVAsc
• Transformer : MVA/x = 15/0,078 = 198 MVAsc
• Motor : MVA/Xd" = 15/0,2 = 75 MVAsc

Generator and motor are a source of SCC.  Now, how to calculate SCC 3 phase at the system above if the fault at bus 12 kV?

The generator, transformer and line are connected in series so that we have to sum up it like a parallel network  and if those connected in parallel we have to sum up it like a series network.

• 1/(MVAsc series) = 1/(MVAsc Generator)+1/(MVAsc Line)+1/(MVAsc Transformer) = 1/(1500)+1/(1230)+1/(198) = 153 MVAsc (MVAsc series)
• MVAsc series is connected in parallel to the motor at bus 12 kV. So, here is the result :
• MVAsc total = MVAsc series + MVAsc motor = 153 + 75 = 228 MVAsc
• To get SCC (Isc 3ph) at bus 12 kV, please use this formula :
• Isc = MVAsc total/(1,732*kV) = 228/(1,732*12) = 11 kA

You can try to solve LL, LG, LLG, and LLLG. To check whether the result is correct, you can use ETAP or similiar software.

If any question or suggestion please deliver to me. I am glad to hear your comment. Before closing my first posting at least there are two questions that we can to discuss futher :

• Is lighting panel could be a source of SCC when the fault happened ?
• How about the line and transformer? are they a source of SCC too?

Thank you.
Please comment after reading my posting.

Regards,
HI