Sorry. that's a fundamentally incorrect assumption. You can equate kWh and kW.
Using KwH to add load
- Thread starter mannyb
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kwired
Electron manager
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- NE Nebraska
Which is what most of of us have been trying to tell OP. We still don't know if he has told us the max instantaneous load as he has changed just exactly what his numbers mean a couple times already, but each time has also assumed whatever figures have been presented are at a constant value 24/7, which may or may not be accurate we don't really know enough about the load(s) measured to say otherwise.
Charlies analogy of how fast can you drive a certain distance can have some comparison issues, but overall is a pretty fair analogy. The root issue is very similar. Assuming for the moment you have constant speed the entire trip you can drive the 25 miles in 1 hour at a rate of 25mph or you can drive it in half an hour at 50 mph - same work is done either way but at double the rate for half of the time in the second example. That vehicle may have it's own on board power source making it somewhat different analogy, but regardless it still must be capable of delivering twice the power (disregarding any inefficiencies that may alter the results) for the second example or it will not be able to do the task.
Now if you have stop and go or other varying conditions in that 25 mile trip it gets even more complicated to calculate the peak energy demand vs the average demand over any specific time period.
He mentioned doing it without getting a speeding ticket, that sort of is something that isn't part of the analogy but it does indicate that one way is being done at a different rate than the other way.
- Location
- Lockport, IL
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- Retired Electrical Engineer
I am sure that you meant that you agree with me. :happyyes:
My analogy is perfectly correlated with the OP's question. Of course you can't use total miles driven to determine what your rate of speed at any given moment might have been. That is exactly my point. The OP wants to use a unit of measure that does not have a time element (the analogy being miles) to determine a unit of measure that does include a time element (the analogy being miles per hour). Since KW is a unit of energy per unit time, it is the one that has a time element. KWH does not have a time element, since it equates to (energy per unit time) multiplied by time, which leaves you only with energy.
RumRunner
Senior Member
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- SCV Ca, USA
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- Retired EE
The point of discussion here is OP's exigent need to know whether any response that-- he could glean from this forum-- could help him decide if he could add additional load based on the information available . We are not supposed to be going through the semantics that defines the differences between kWh and kW.
You haven't really contributed in finding a solution to OP's concern other than your usual sardonic remarks.
You are an engineer-- OP is an electrician. . . try and be helpful.
This is what this forum is all about.
RumRunner
Senior Member
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- SCV Ca, USA
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- Retired EE
Energy as in power can only can be put to work with the necessary “vehicle” and deliver it to a point where we can harness it. It is clear that OP has only a limited information to work on—this is where engineers like you come to the rescue.
If you were commissioned to figure out how much water is flowing in the Mississippi River, for example, you are not equipped with a giant bucket to measure how many gallons you are able to scoop in an hour.
A vessel of pressurized steam for example or radioactive energy source sitting idle is not doing us any good if we don't provide a way of transferring them. We need heat exchangers for them to be usable as an example.
In essence, this is what OP is trying to get.
Vast oil reserves in the earth mean nothing if we don't extract them and build refineries to make our cars run.
Now, that's another allegorical parallelism.
Adamjamma
Senior Member
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- Jamaica and london
If you have a unit like KWh, and I know I may be asking the obvious here, but don’t you then just multiply that by the hours of use to get your need? If continuous for 24 hours don’t you just multiply by 24? Or is it just too easy and we use another formula??
I mean, it is holiday and formula one time... not too much thinking right now..
I mean, it is holiday and formula one time... not too much thinking right now..
I made a typo.
"Sorry. that's a fundamentally incorrect assumption. You can equate kWh and kW."
Should be "Sorry. that's a fundamentally incorrect assumption. You can't equate kWh and kW."
Posting stuff that is simply incorrect is not helpful.
kwired
Electron manager
- Location
- NE Nebraska
Only if you know that the net kWhr recorded was at a continuous kw during the duration the reading was taken.
If measurement was 10 kWhr in 24 hours, it could have been steady 416 watts for 24 hours, or it could have been 9 kWhr in 10 hours at a steady rate and remainder varied over the next 14 hours, or all sorts of combinations. for the purpose of what OP is trying to do though, one needs to know the peak demand in order to properly select conductors, switchgear, transformers, etc. that must be able to carry that peak demand when it is called for.
If you are bailing water out of a vessel one bucket at a time and it takes a million buckets to completely empty it, you use same energy to empty it, but if you do it ten buckets a minute vs 1000 buckets a minute it will be at different energy rates and of course the same work is completed but in different time as well. Calories consumed per minute is also going to vary greatly, even though same work was accomplished, this all assumes you don't tire out and can maintain steady work output at either rate.
winnie
Senior Member
- Location
- Springfield, MA, USA
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- Electric motor research
I am going to try to answer mannyb's question directly. I will be repeating what other people have already said and implied in this thread.
The question: can you use kWh information to determine if a feeder has excess capacity. We are given that it is a 1200A 480V feeder, and that the usage is 3500 kWh per day.
The answer is that we can make a _guess_ but only by making certain assumptions.
At the full 1200A and 480V running continuously,24000 kWh would be used every day. So as a rough approximation only 15% of the feeder capacity is being used.
However this requires the _assumption_ that the existing load is continuous and steady. It is quite possible that the existing load is using the _full_ capacity of the feeder but only for 15% of the time during the day. In this case, there is no room to add any more load unless you have some provision to make sure that the new load doesn't run at the same time as the old load.
To get a better answer you will need to watch the existing load energy consumption over a shorter time period than a day. Perhaps your existing meter has a tool for recording maximum demand or instantaneous power consumption.
Best regards,
Jon Edelson
The question: can you use kWh information to determine if a feeder has excess capacity. We are given that it is a 1200A 480V feeder, and that the usage is 3500 kWh per day.
The answer is that we can make a _guess_ but only by making certain assumptions.
At the full 1200A and 480V running continuously,24000 kWh would be used every day. So as a rough approximation only 15% of the feeder capacity is being used.
However this requires the _assumption_ that the existing load is continuous and steady. It is quite possible that the existing load is using the _full_ capacity of the feeder but only for 15% of the time during the day. In this case, there is no room to add any more load unless you have some provision to make sure that the new load doesn't run at the same time as the old load.
To get a better answer you will need to watch the existing load energy consumption over a shorter time period than a day. Perhaps your existing meter has a tool for recording maximum demand or instantaneous power consumption.
Best regards,
Jon Edelson
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