KWh calculation for utility power supply

Well, In general, for a balanced 3ph system- (which is not going to be entirely correct)

Power=(1.732)*(line voltage)*(line current)*(power factor)

But this will not be too accurate

Call your utility company and get the power consumption history for your address, this might be somewhat more realistic, and most likely a bit easier.

hehe, it seems less answer to this thread. I like to help those who come from africa.

its this,

measure your phase voltage at same time. then you will get.

phase boltage x 50 x 1.73 x 3600/1000 = * kwh.

* kwh x 50 cents x 30 (31) = * moth charge.

haha, forget a 24 hours a day.

so * kwh x 24 x 50 x 30 = * monthcharge

Power = sqrt(3) *V * I * Power factor

Take PF = approx 0.8 ( usaul standard unless You have specific reactive loads)

You will land up in a

Power = 1.4 * V * I

Energy Consumed = Power (KW) * Time (Hours)

= 1.4 * V * I * Time(Hours) / 1000

All these are known to you and not provided in the solution

Assuming V=400, I=50, Per day 16 Hrs of load, 30 working days / month it is

= 1.4* 400 * 50 * 16 * 30 /1000

= 13440 Units

which will translate into approx $6720

This is approximate and with lot of assumption, feed your data and get approximate value.

Does Cnpower not help people from other regions or he is just being racist ?

hehe, neednt shout. why not show your calculation? you are very active at motor thread ...

wngito said pinguin dont live in europe. when did you arrive in northern hemisphere?

holeinthesnow, why dont act as active as you do in the motor threads?

is it you think the paster comes from africa so neednt to put out your suggestion?

ah, do you think which one is the best in the pasts?

give out your opinion. I'm too impression on your action in the motor part. ha-ha.

also noticed that the current is given as 20/40. I would have expected something like 23/40.

400/230 ==2V?

ha ha, current is 0.5 ? I'm just starting with my elementary math.

you appear to be very enthusiasm. and like to find very tiny mistake from others. except yourself.

ohm law is very difficult, you may explain more clear than others.

of cause as well as delta/star connection. i hvnt read your continuous to the motor issues. its not shame by other checked up and found error. hehe

I shall learn dc ac from others out there. I really dont know what they are!!!! hehe. need endeavor.

hehe, your math is really more better than mine.

doesnt matter, because I dont know ohm law. and ddc aac.

eh, I think its better for me to show all who visit here. in order to leave a good funy material for them laughing at and talking about when work off, during a cup of tea,

" look that lovely chinese, he didnt do such easy math which even a pupil can do it."

hehe. continue, continue, Im waiting for your more suggestion, and more answer about <| / Y.

no no, not chinese joke but this cr4 joke.

you lost nothing as you are very careful, and know ohm law very well, you can calculate 230/40, you can work out this 10/20 or 400/230 as well.

dont use tipic off function, Im learning abstruse da2c subject from you now. others may be benifit from you , too.

hehe, no shame. Im always curious why current can rotate?

why someone often acts so strangely like a.c.?

A power factor of 0.8 seems to be slightly on the high side.

A power factor of approx 0.6 would be a better place to start with.

while i am not against the calculations ,why dont you just install a KWH meter and build up an accurate history of your consumption ,you may as well get your KWH readings from the utility meter and record.

This will help you in predicting your future consumptions and estimate your monthly bills also.

Yes, though it will cost money to pay for instrument C.Ts and the meters. But, installing a KWH Meter or an Energy Meter will make life easy. Where by, from your accumulated energy consumption history, you may some time in the future wish to reduce your consumption through energy reduction methods.

From which you could save money from, say instead of using 36W light fittings, you decide to replace them with their equivalent of eco-lights which will work at 25W with an improved power factor of 0.98. These lights are inbuilt with very good power factor correction devices, and I've seen them work. Or if your from an industrial/manufacturing plant, you may want to improve your power factor correction from say 0.70 to 0.98 and save on money.

Having a good history is important when it comes to improving and reducing energy consumption.

You did not mention the voltage but i am assuming 400 volts and 0.85 pf.

so your hourly consumption = 1.732 x 400 x 50 x 0.85 = 29.4 kw

daily= 29.4 x 24= 705.6 kwh

Daily costs= 705.6 x 0.5= 352 dollars and 80 cents.

cheers.

Where is original author ?

sb, The original poster probably got pissed off reading the nonsense going on between myself and cnpower.

Sorry about that.

Regards

Mr. W.A Snow

Greetings,

Your question was to estimate your utility bill, you can use history (if available) or calculate based upon the load you have and the utility rates in effect,

there are several calcuclations here to arrive at the instantaneous KWH (kilowatthour, one kilowatt demand/used for one hour) consumption, of which getting the hourly history from your utility (which they may have a smart meter to record or if a dumb meter would need to install a meter to record),

but no reference to the rate per hour calculation, electricity may be priced at multiple rates depending upon time-of-day use or it may just be one flat charge irregardless of time of day or day of the week, or month of the year, you can get this information from your utility, the rates may vary by over 100% so they are an important part of the estimation, you may be able to get the utility customer service rep to plug your info into his equations/spreadsheet and give you a quick answer as he will know all the rates for your connection.

there is also probably a minimum service charge just to be connected to the utility.

Hello All. Iam gratefull for the response. Thanks

Hi,

It is simply power multiply by the time use. Power has a unit of Killowatt and time has it unit of hour from there Killowatt-hour was derived. To calculate basic formula as follows:

P = (1.732)*V*I*P.F.

where:

P - real power

1.732 is sqrt of 3 and this will only be present when your system is 3 phase

V is the volatge system

I is the phase current

P.F. is power factor (basically in a design and consulting firm we always assume power factor to be 0.8 or 80%, for the reason in most country, electric utilly company have fine for a system that's having less that 0.8 or 80% power factor. With this more than 0.8 or 80% power factor system has a discount that's why most larger facilities put capacitor banks to their system for them to increase the power factor.)

KWH will be P*Hour

KWH consumption will be KWH as presented above time the KWH rate of your Electric Power Utility company.

Hope this somehow helps.

You can also, email me to my addresses. CC copy both emails.

bennegregsiman@yahoo.com

bsiman@meinhardt.com.ph

Regards,

Benn Siman

1.good answer

U need to calculate p = 1.732 X VI cos(phy);

p x no of hrs... u get kwh

kwh x 50 cents.... u get cost to pay..