Electrical energy calculation

As I recall, a HP is about 750 W. Varying conversion factors in generators and motors makes the exact # moot. Of course, 1 hp into a generator is much less W, and much more W has to go into a 1 hp motor.

1 HP = 746W so you can do the math to convert the load from HP to W.

Now look at efficiency. If your system is 75% efficient that means you only get 75W of desired output for every 100W of input, the other 25% is output in undesired ways - typically heat. Divide the desired output power by the efficiency factor to determine the absolute minimum power input. Then add margin to the capacity of the source for de-rating to ensure long term power capacity. example...

I need 100W output from a 75% efficient system and I de-rating my input source by 30% to ensure it can meet the demands, surges etc and last a long time.

100W / 0.75 = 133.4W input power requirement to support the load

Use 0.70 for source calculation, 100% - 30% = 70%.

133.4W / 0.70 = 190.5W capable power source to provide de-rating.

Question; What if you find a way to use the heat that is dissipated or was undesired.

Answer; It is now a desired output and your efficiency is now higher

Right, and for a generator, multiply not divide: HP x efficiency x 746 W

For small organisation that used low power consumption does not required Maximum Demand calculation. Maximun demand calculation, is for heavy load, where, sometime the total connected load can go as high as 30MW or 65MW load.

I think what you mean is to "How to calculate the power consumption" for that particular organisation.

If that so, Just convert the HP to KW and multiply it with per hr, multiply again with the electricity cost per unit and you will get the power consumption cost per hr (KWh). If you want to know the power consumption for a day, just multiple with 24hr or for a month will be 24hr x 30 days.

if a organisation required 254 hp power load,

how to calculate maximum demand or unit consumption..??

Ans:- Above question has 3 things..

1> Power factor (efficiency of the System)

2> MD Sanction required.

3> Units (Actual power consumption)

1> First I suggest you to get a power factor corrected up to 0.99 (99%). you may need around 200-300Kvar (Depends upon How much Inductive & resistive load is connected in the total plant) capacitors connected paralell to the mains Input. For this you can use Automatic Power Factor Correction Relay (APFCR) Panel.

Then, Consider your Plant efficiency is now around 95%. It means, If you are using 100kw(130Hp) continuously for 1 Hour, you pay for 105 units (100/0.95=105kw).

2> Now, The Max Demand Comes in to picture..

Max Demand is the Average kw per minute (calculated & stored(locked) by the Energy Meter) consumed by you in 30 mnts slot (15mnts*2slots).

If your consumption structure is; 100kw for 10mnts, 20kw for 20mnts, 10kw for 30mnts, MD will lock on 28.33kw(Average per minute) at this hour...This data will remain same till the end of the month, until you cross it...If on some other day & hour the MD locked at 34.6kw, your plant MD is considered 34.6 (for Billing). And this indicates that you need Around 40kw MD sanctioned by Electricity supply authorities.

In Some Cases, The MD shoots beyond the Authorised limit (40kw). You need to pay heavy penalty charges.

To get rid of this overshoot problem, you can get MD Controller (Conzerv make) in the market. This device Monitors your plant MD simultaneously with the Energy Meter & Alerts (or trips the load) if there is a possibility of overshoot.

3> Unit Consumption Calculation is simple; How much Kilowatts per hour you consume, that much units are considered for billing.

I hope this data will be helpfull,

Regards,

-Veeresh Kotagi

Hi kveeresh2,

A good answer and I've given you a plaudit! Your MD tracking device and what happens next raises some interesting options for Demand Side Load Management systems. Clearly these must be applied to non-critical loads which are also selected in rotation; otherwise you might end up with the same piece of kit being clobbered incessantly.

I've heard recently that even in major cities such as London that site demands are rising at such a rate that the frequency with which authorised capacity is being exceeded is now providing a significant revenue stream for the utility Coys!

Clearly automation is no longer an 'addy-on' in the management of this aspect of the electricity account management.

Regards.

1 H.P=746 W

254 H.P=746*254

=189484 W or 190 KW

WE CAN USE A POWER SOURCE OF EFFINCIANCY 80%

SO REQUIRED KVA POWER SOURCE IS =190/.80

=236 KW

SO AROUND 250 KVA DEMAND YOU REQUIRED FOR YOUR COMPONEY

tHIS IS ONLY WHEN YOU SUPPOSE YOUR FACTOR IS UNITY WHICH IS NOT POSSIBLE

IF WE SUPPOSE POWER FACTOR .90THEN LOAD IN KW IS 236*0.90=213 KW

It's been a while since I did power stuff, but if the device draws extra current on the lag (usual with motor windings, which are inductive loads) while generating max HP, how dows that reduce the KW? I'd think the KVA would be something like KW/(Power Factor)!

Maybe the devices that control MD can also correct power factor.

Hi, DGPicket,

The MD Controller doesn't make any corrections (like PF) in system parameters.

It's not at all linked to the System PF. It just notesdown the KW or KVA (user configerable) from the input supply cables, and adds up 30 readings & then devides it by 30 to get an average kw or kva per minute (i.e MD).

Also gives alarm output before 1-2mnts.

I suppose the MD fines might be on either KW or KVA, that you can configure it either way. While KVA is about local wires melting, and power factor is about stressing the supplier systems with currnet that is not doing work, in the aggregate load of a supplier the lagging current is diluted by the more normal phase of the average load, so the KW load stressing the generation capacity might be their target.

Does it actually "control," or just "alarm"? This sounds like the quality culture thing -- the utility monitors so it can fine, and the user monitors so it won't. This is so like Demming: QA/QC at the suppplier is redundant to QA/QC at the consumer, and both may be superceded by a quality process throughout.

If it controls MD by lowering voltage delievered or switching off circuits of secondary importance, it might have the necessary power control circuits to also fix power factor. I have heard of systems that reconvert the power to take in power at power factor 1.0 and deliver it at any power factor needed. While fixing power factor, they can also control voltage fluctuations and even integrate battery/generator backup. Sometimes they just buck/boost each incoming line with a series transformer to make the input current and output voltage bahave.

Sometimes, it is more efficient to fix power factor at each big motor, as many motors can run more efficiently and as near 1.0 power factor and higher efficiency if the voltage is lowered. This green though is pretty old, but likly still applicable. The leading current is a way of shedding the excess AC motor capacity -- the current surfs forward on the voltage wave to achieve equilibrium with the load. However, some motors experience a loss of RPM with lower voltage that might not be tolerable, an some motors have loads that vary suddenly with every cycle of their machine.

Hi, Wasbright,

Very Good answer....!

Yes, but I believe that the original post asked what would the MAX Demand be. Would you not have to figure the starting current that the motor would draw? I would guess that the starting current would have to be higher than the current drawn to sustain the motor operation once running.

Please don't flame me as it has been MANY MANY years since I studied electric motors... 8^)

I always wondered if one could contrive a controller that limited the starting current to the max running current, and just let it gain speed more gradually? If the motor load is less than 100%, maybe not that much more gradually, and with less stress on the windings, mounting, load and coupling. While the circuitry would be expensive, the supply could be cheaper. Since many problems require such circuitry, this could be an added benefit of good control.

I am guessing that start up current is more of a return-from-power-failure concern to the supplier, which they may also have as a concern, whereas MD is about continuous capacity.

You are absolutely right. The start current can easily be 5 - 7 times or more than the running current. However, most motors with normal loads get over that (in seconds or less) so quickly that the Peak Demand over 15 to 30 minutes is not affected that much. If you have alot of starts in a normal day and/or a lot of motors starting at the same time it can definetly affect the Peak Demand.

hi friend,

1 HP = 735 W

235 HP = 172.725 kw

Assume your plant PF is 0.8 lag then KVA load will be 215.9 KVA

the maximum demand may be fixed 30% extra so you con select 280 KVA

The unit consumption per month will be at full load =172*24*30 unit

You can calculate the maximum demand by plotting a graph with time in the x-axis and load in the y-axis for a period of time of operation,say from 8am to 5pm for each machine.For this you should discuss with the production staff regarding the starting time,stopping time,partial or full loading etc of each machine.Simply add the current/kw/kva at 15min intervals of time vertically and record the total under the total column at the bottom row.Then you will know the magnitude and time of maximum demand.Similarly you can calculate the approximate kwh.