Using KwH to add load

[mannyb]

The customer has a KwH sub meter. The average is 143 kwhs. The want to add equipment and want to know if the feeder has enough room for new load. Without a load analysis but given the information. CaN the kwhrs be used to make that decision? The feeder is a 1200a feeder 480 3pH 4wire.

 

[gar]

181120-2359 EST

mannyb:

You need to understand what kWH means, which I don't think you presently do. Additionally you need to know how current varies with time, how balanced it is, and possibly what normal peak current can occur.

.

 

[Jraef]

If this is a commercial or industrial customer, the utility may have a kW demand meter, which will give them the peak demand in kW, which can be used to determine worst case. But because kWh has a time component, it can’t tell you anything useful for this purpose. A kWH meter can’t tell the difference between 143 kW for 1 hour or 1 kW for 143 hours.

 

[mannyb]

If this is a commercial or industrial customer, the utility may have a kW demand meter, which will give them the peak demand in kW, which can be used to determine worst case. But because kWh has a time component, it can’t tell you anything useful for this purpose. A kWH meter can’t tell the difference between 143 kW for 1 hour or 1 kW for 143 hours.

Yes it's commercial. The customer sub meter is part of a larger system. ThE cusotcustomer has a dedicated 1200a feeder for their equipment. They juat have a sub meterr for their billing. I was asking was there a way to use KwH information for adding a new load. I could put a mwter to see the amperage as is but currently we don't have one that size to read the parallel feed into panel. I know that's not how you do it but I would have a better reading thaN using the kWh. Thanks for help.

 

[GoldDigger]

Yes it's commercial. The customer sub meter is part of a larger system. ThE cusotcustomer has a dedicated 1200a feeder for their equipment. They juat have a sub meterr for their billing. I was asking was there a way to use KwH information for adding a new load. I could put a mwter to see the amperage as is but currently we don't have one that size to read the parallel feed into panel.

If you want to use the kWh information from the sub meter, you need to total the kWh for a specific period of time and you must also know how the load varies with time to be able to get a kW value. It is the kW values (peak and continuous) that determine the wire and service size needed, not the kWh by itself.

You said the average is 143 kWh (kilowatt-hours) but you do not say whether that is per day, per month, or what.

 

[mannyb]

If you want to use the kWh information from the sub meter, you need to total the kWh for a specific period of time and you must also know how the load varies with time to be able to get a kW value. It is the kW values (peak and continuous) that determine the wire and service size needed, not the kWh by itself.

You said the average is 143 kWh (kilowatt-hours) but you do not say whether that is per day, per month, or what.

Yes Sir per day 24 hr period continuous use. Some time but not much they do have down town for maintenance and or repair. They it's running again.

 

[GoldDigger]

Yes Sir per day 24 hr period continuous use. Some time but not much they do have down town for maintenance and or repair. They it's running again.

OK. If your energy figure of 143kWh per day is correct and the load is more or less constant and unchanging over the 24 hour period, you could state that the power demand averages (143/24) kW or just under 6kW. That seems to be a very small power consumption to me, corresponding to just 25A at 240V single phase. In any case, motor starting/stopping or heater cycling could easily lead to a short term power demand up to twice that figure.
What sort of operation is this?

 

[mannyb]

OK. If your energy figure of 143kWh per day is correct and the load is more or less constant and unchanging over the 24 hour period, you could state that the power demand averages (143/24) kW or just under 6kW. That seems to be a very small power consumption to me, corresponding to just 25A at 240V single phase. In any case, motor starting/stopping or heater cycling could easily lead to a short term power demand up to twice that figure.
What sort of operation is this?

It's a wash plant for mining. The system is 277480v 3ph 4wire. I am taking all information under advisement. I am looking the right equipment but until it gets here I am getting iinput on the kWh information. Thanks for your help. Do you think I am missing information for your calculation?can you share your calculation using the kWh? I want to visit the site tomorrow and see if the customer reading are true or they are providin wrong info.

 

[kwired]

It's a wash plant for mining. The system is 277480v 3ph 4wire. I am taking all information under advisement. I am looking the right equipment but until it gets here I am getting iinput on the kWh information. Thanks for your help. Do you think I am missing information for your calculation?can you share your calculation using the kWh? I want to visit the site tomorrow and see if the customer reading are true or they are providin wrong info.

What you are missing is the time/demand element. As already mentioned you could have 143 kW for one hour or 1 kW for 143 hours and a watt hour meter gives same result. So first thing to ask is the load a constant load or does it vary? If it is constant this may be somewhat easy, but if it is fairly idle most of the time then has a much higher peak for say a couple hours each day, that peak demand is what you need to be most concerned with, especially if whatever you are adding will also run during that peak.

 

[Jraef]

I would be surprised if a 480V 3 phase service did not come with a Demand Meter. That would be very useful information because in order for them to charge a “peak demand” penalty, they now (with smart meters) keep track of a running average demand. That the info you need. Call their utility rep, they likely have that info for you.

 

[charlie b]

What you are trying to do is like asking this: Given that I drive an average of 25 miles during my daily commute to and from work, and based on that information alone, can I drive faster without violating the speed limits?

 

[Besoeker]

The customer has a KwH sub meter. The average is 143 kwhs. The want to add equipment and want to know if the feeder has enough room for new load. Without a load analysis but given the information. CaN the kwhrs be used to make that decision? The feeder is a 1200a feeder 480 3pH 4wire.

kWh is energy.
What you need is load in kW.

 

[gar]

181121-1040 EST

mannyb:

You need to understand the relationship of voltage, current, power, and energy to each other.

Instantaneous power to a two terminal load is the instantaneous voltage across the load times the instantaneous current thru the load. Assuming sine wave voltage and current waveforms, but not necessarily in phase, if you take a number of, like 40, equally spaced instantaneous power measurements over a one cycle period, and average these you will get a reasonably good estimate of of average (real) power. You can get the energy transferred over this time by multiplying that average power by the time period of measurement.

From your first post you mentioned 143 kWH as an average energy use. But you did not indicate a time period. Later you said 24 hours. So your average load power was 143/24 = 6 kW. This is only 2 to 6 times my use at home.

Then you mentioned a 1200 A feeder at 480 V. To a balanced resistive load this about 1000 kW capability. 6 kW is peanuts to a 1000 kW capability. Now suppose all those 143 kWH occurred in 1 hour, then we are up to about a 15% load relative to the source capability. Cut the time period to 1 minute, then we have an average power of 8580 kW over that 1 minute period. Which is way more than the system capability.

How much load do you want to add? Is it a constant invariant value? Is it single phase, or a balanced 3 phase load? How does your present load vary with time? Can you estimate a peak load presently, and a peak load for the added load?

It gets back to you understanding the relationship of power and energy to each other.

You mention other measuring equipment that is on order. What is it? How long to get it? What will it tell you?

Immediately you could put a time lapse camera in front of the meter, and record the reading every 1 minute, or some other time period. From the pictures you can determine how average power varies over time.

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[mopowr steve]

Check with your power company they may be able to assist you in finding what your max or average KW usage is if they would offer to come out and hook up a wattage recorder at your sub-panel.

 

[topgone]

Yes it's commercial. The customer sub meter is part of a larger system. ThE cusotcustomer has a dedicated 1200a feeder for their equipment. They juat have a sub meterr for their billing. I was asking was there a way to use KwH information for adding a new load. I could put a mwter to see the amperage as is but currently we don't have one that size to read the parallel feed into panel. I know that's not how you do it but I would have a better reading thaN using the kWh. Thanks for help.

We sometimes do estimate of things in the old days, when GE meters were rotating disc type. You take the Kh of the meter (WHr/rev or kWHr/rev, found in the nameplate) then count the number of revolutions of the disk during peak hours! We usually consider ten trials for each complete turn of the disc and record the time it took the meter to make the turn--> example 1 turn = 5 seconds, time consumed per turn = 5 seconds per turn. If the Kh is 7.2 WHr per rev, then the instantaneous demand is = 7.2 WHr/turn / (5/3600) Hr/turn = 5184 watts!

 

[mannyb]

What you are missing is the time/demand element. As already mentioned you could have 143 kW for one hour or 1 kW for 143 hours and a watt hour meter gives same result. So first thing to ask is the load a constant load or does it vary? If it is constant this may be somewhat easy, but if it is fairly idle most of the time then has a much higher peak for say a couple hours each day, that peak demand is what you need to be most concerned with, especially if whatever you are adding will also run during that peak.

Please don't give up on me just yet. The daily kWh usage is around 3,500kWh /24= 143 per hour.I am typing without.a calculator This is a constant every day. The machine operates 24/7 and the 3500kwh is average. I don't want to waist you guys valuable time I understand this is the long way of downing it and probably wais of time but I just figure some has had success with using kwh. Thanks

 

[gar]

1871121-2213 EST

mannyb:

Your information has changed again. It now appears that on average you have a 143 kW load. This corresponds to about 15% of capacity. Next question is how constant is this load.

Look at the device that is the load. Does it appear to have a constant mechanical load, or is it cyclic?

At a time when the machine is running have someone record the kWH meter reading every 5 minutes. If each 5 minute increment is about the same, then assume the load is constant at 143 kW.

Is the load a balanced 3 phase load? Is the new load a balanced 3 phase load? How big is the new load? Assuming your present load is in fact an essentially constant balanced 3 phase load, then unless the new load is very large and/or peaked you have a lot of capacity.

.

 

[RumRunner]

What you are trying to do is like asking this: Given that I drive an average of 25 miles during my daily commute to and from work, and based on that information alone, can I drive faster without violating the speed limits?

Your statement is “extraneous analogy” that I don't agree.


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Distance from your house has nothing to do with you getting a ticket for speeding. . .a metaphor perhaps and which is true but this is not correlated to OP's question.


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Given OP's info (143 kWh), this consumption is within a 24-hour period.

Whether this energy is consumed in one hour or 24 hours doesn't make a difference. We know that 143 kW is needed for the plant to operate or a particular load. If it is spread over the 24 hour period, the consumption with still stand.


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Since we know the watts (energy ) required for the plant's operation or load, we can-- by calculation-- determine the amount of current going through the system wiring.


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And here is where the trusty OHM's LAW gets a dusting.


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The plant has 480 volts and 143 kWh are given numbers to perform the job. So, if we derive amperes from this info we will get:


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P=E x I​

143 kW = 480 volts times X (where X is Amps)


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By Ohms Law, using the generic non- three phase for simplification
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we will have:


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X = 143kW divided by 480 volts


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143,000 (watts) divided by 480 volts equals 297.91 Amps (this would be the demand for the proposed load)

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So, 297.91 Amp hour would be the peak current going through the system which, according to OP is rated 1200 Amps. Peak consumption of the proposed load if everything is turned on.


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As further stated by OP, this 143kW is what they use in the 24-hour period. So they are not using all machines normally but still result in the total consumption of 143kW at the end of the day.


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Looking from a different perspective, look no further than a car battery.


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Ordinary family car has a battery rated 100 Ah (ampere Hour.) Through this info we can deduce that the battery will supply power at 100 Amps for a period of one hour at 12 volts. Keep in mind that wattage is not a given number when it comes to car batteries.


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Again by Ohms Law:​

P = E x I​

12 times 100 = 1200 watts will be supplied by the battery for a time frame of one hour. From here, we can figure out the needed appurtenances that will support to handle and distribute this energy that is available-- e.g. wires, fuses, breakers etc.

Note:

( The battery discharge rate is theoretically (and pragmatically) presented without having to deal with the vagaries of Peukert's Law for simplicity.)


Happy Thanksgiving​

 

[gar]

181122-2200 EST

myspark:

In post #1 mannyb told us that he had an average energy measurement of 143 kWh. But he never defined over what time period. Later he defined it as over 24 hrs. This turned out to be incorrect. If load is really constant, then most recently he said it was 3500 kWh in a 24 hour period. Which is 143 kWh in a 1 hour time, and thus an average load of 143 kW. He has said the load is constant, but we have no measurement data or other information that would verify this.

Assuming it is a constant and a balanced load, then his total load is 143 kW. And this is simply 143/(3*1200*277/1000) = about 15% of the source capability at 100% power factor. Pick some power factor and adjust for that assumption.

.

 

[topgone]

181122-2200 EST

myspark:

In post #1 mannyb told us that he had an average energy measurement of 143 kWh. But he never defined over what time period. Later he defined it as over 24 hrs. This turned out to be incorrect. If load is really constant, then most recently he said it was 3500 kWh in a 24 hour period. Which is 143 kWh in a 1 hour time, and thus an average load of 143 kW. He has said the load is constant, but we have no measurement data or other information that would verify this.

Assuming it is a constant and a balanced load, then his total load is 143 kW. And this is simply 143/(3*1200*277/1000) = about 15% of the source capability at 100% power factor. Pick some power factor and adjust for that assumption.

Instantaneous demand is more important for utilities, IMHO. It is the amount of generation that needs to be provided, if a service cannot deliver, the system downgrades and may become unstable. That is why utilities require the applicants to write down all the connected loads. Back to the OP, you have to measure your max instantaneous load, then decide.