Mathematical Explanation on converting Watt to Ampere 3

[Station 2]

Please I need mathematical explanation on how we arrive at 60ampere is equal to 1 Mega-watt (1 MW). This is the conversation value will use on our load analysis here at my distribution Station. For every 1 Mega-watt, will divide it by 60Amperes. 1 MW= Ampere/60

I am confused on how the calculation is being arrived. I have make used of P=sqrt.3 X IV X PF.

 

[sushilksk]

There is no constant relation between Power and current. As mentioned by you P=sqrt.3 x I x V x PF. P will depend on V & PF also. However V and PF are aproximately constant for a perticular type of load. Resultant value may be such that for every 60A power is 1W. But it is a crude method.

 

[crshears]

What sushilksk said.

My utility uses the following conversion factors:

To find MW @...

500 kV: amps/1.1

230 kV: amps/2.4

115 kV: amps/4.8

44 kV: amps/13

28 kV: amps/20

14 kV: amps/41

8.3 kV: amps/120

4.8 kV: amps/208


In line with what sushilksk has indicated, these figures are based on approximations only, and are used as "rule of thumb" values; I have often found them to be meaningless when performing load transfers between different substations, i.e. when there are [too many] capacitors in service at one station but none in service at the other, and [ after parallel has been made ] the operator is attempting to derive vectorial reactive flows based on transformer secondary telemetry since the feeder breakers are only equipped with ampere telemetry, the goal being to achieve minimal reactive flow at [ by "moving the null point" to ] the intended parallel breaking point [ typically by the manually initiated raising and/or lowering of ULTCs of the supplying transformers on either side of the parallel ] prior to the actual opening of the device.

Load transfers must always be performed carefully, not only to provide minimum voltage disturbances to supplied customers but to prevent equipment damage, and even more crucially for safety reasons; if the switch being opened is being manually operated by a field agent, and is an older horn-gap air break switch with reduced arc break capabilities, and the control room operator miscalculates, molten copper may well rain down upon the head of the field agent on the switch handle below, resulting in curses being rained down upon the progeny of the control room operator - thankfully I've never had that happen to me, neither as the agent nor the controller .

Incidentally, for sixty amperes of current to flow producing one watt of real power would involve extremely low voltages; perhaps the OP meant 60A = 1 MW?


CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[Station 2]

@ sushilksk, thanks for your response. From my observation on our own MV AB 11KV Switch Board Panel, for every value of ampere is divided by 60,it gives us the actual value of watt (power) displayed on the panel. You admitted that its a crude method of calculation, can you enlighten me on the latest method of conversion.

 

[Station 2]

@crshears, please how do your company arrive at all the conversion value?, what will then be 60A to power on a 11KV distribution breaker? Thanks

 

[Station 2]

@sushilksk
WIth your experience, can you explain how these conversions were derived?

To find MW @...

500 kV: amps/1.1

230 kV: amps/2.4

115 kV: amps/4.8

44 kV: amps/13

28 kV: amps/20

14 kV: amps/41

8.3 kV: amps/120

4.8 kV: amps/208

 

[crshears]

Hello Station 2,

The Rule of Thumb my utility always used in the past was that every two megawatts of real load will have a reactive draw of one megavar; I believe the conversion factors provided were derived using that rule.

You could try drawing yourself a power triangle using those values to derive a rough conversion factor for 11 kV.

Note however that there is no "latest method of conversion" providing a direct correlation between megawatts and amperes, as that is considered too simplistic nowadays. The preferred modern method as alluded to by sushilksk, calculates from MVA to amps, meaning both in-phase and lagging currents are automatically factored into the equation as computed by the SCADA system using the telemetered values and, sometimes, also by calculating parallel state estimated quantities.

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[Station 2]

Ok. Thanks

 

[che12345]

Dear Mr. Station 2

Q. " Please I need mathematical explanation on how we arrive at 60ampere is equal to 1 Mega-watt (1 MW). I have make used of P=sqrt.3 X IV X PF. "
A. It is not a simple relation between Power (MW) and Current (A). There are other factors to be taken into consideration. The relationship is what you had known " P=sqrt.3 X IV X PF" , but I would add an efficiency factor. So, a more complete formula would be P= 1.73 x VI x 0.8 x 0.95 assuming pf=0.8 and efficiency = 0.95.
Then P= 1.73 x VI x 0.8 x 0.95 = VI x (1.73 x 0.8 x 0.95) = VI (1.31)
Where Power P in watt, Voltage V in volt, I current in ampere.
Solving for V, where V=P/ I(1.31)
Take I=60A, P in W, V=P/ 60 x (1.31)= P/78.6
Let P=1 MW=1000,000 W,
Then V=1000,000/78.6 = 12,722 Volt = 12.7kV
Conclusion: If the system voltage is 12.7kV, pf=0.8 and Efficient =0.95; it would draw 60A for 1MW load.
then " 60ampere is equal to 1 Mega-watt (1 MW)" would be a good "indication".
Che Kuan Yau (Singapore)

 

[sushilksk]

Why to take efficiency in this calculation for deriving True Power (Watt) for a distribution substation?

You can measure true power at input and output both sides of a distribution station. Therefore, efficiency not to be used in calculation.

 

[Station 2]

@che12345, thanks so much for your explanation. but our distribution network is 11KV, are you saying using 60ampere to make 1 Mega-Watt is wrong? I have an instinct that our conversion may be wrong, this was what actually prompted me to seek for help here. Is it compulsory for me to consider efficiency in this situation?

We have a 33KV distribution network also and we convert 20Amperes to Make 1 Mega-Watt, no mathematical explanation is given by the manufacturer. I need someone to actually assist. The explanation given by crshears on the Rule of Thumb I have applied it, I am still VERY Lost.

 

[sushilksk]

If you need calculation, it is like: 1.732 x 33kV x 20A x 0.9PF = 1.028MW

But voltage may vary between 31kV to 35kV, and PF between 0.7 to 0.95 (or even more in worst conditions), therefore 20A = 1 MW is just a crude calculation based on assumptions. Measuring power through MW meter/ Transducer/ IED is better way to do this.

 

[Station 2]

Thanks @all

 

[che12345]

Dear Mr. sushilksk

Q. " Why to take efficiency in this calculation for deriving True Power (Watt) for a distribution substation? You can measure true power at input and output both sides of a distribution station. Therefore, efficiency not to be used in calculation.
A1. The formula:
a) for Power P= (1.73 x VI x pf x efficiency) unit in [watt].
b) Q = (1.73 x VI x pf) unit in [VA].
A2. Mr. Station 2 post states the Power in ( MW ). Therefore, about a) shall be used and NOT b).

Dear Mr. Station 2
Q. ".... but our distribution network is 11KV, are you saying using 60ampere to make 1 Mega-Watt is
wrong? "
A. See my post : * ... a (more complete formula) would be [P= 1.73 x VI x 0.8 x 0.95]; (assuming pf=0.8 and efficiency = 0.95)... *.
The factors that I had assumed are pf=0.8 and efficiency=0.95. If the system is running at pf and efficiency deffer from (0.8 and 0.95 respectively), the [Power value will change}.
Conclusion: 1. The general formula [P = 1.7 x VI x pf x efficiency] in watt, is valid.
2. The system that you are referring to is NOT running at pf=0.8 and efficiency is NOT =0.95.
BTW: This does NOT mean that the system MUST run at pf=0.8 and efficiency=0.95. I had chosen these values just to explain the mathematical derivation.
Che Kuan Yau (Singapore)

 

[Station 2]

Mr che12345 thank you so much for your time.