Maximum and average demand (PE exam question)

[Andres Arias]

Good morning.

I want to ask you all about how to understand this equation, from the PE exam: Power, pool of questions provided by NCEES. I do not how the maximum demand is obtained for the following intervals, given the maximum demand at the beginning of the first interval. See attached the question and solution. Your answers are very appreciated. Thank you.

 

[d0nut]

The maximum demand is a ratcheting register. Once a higher value is recorded, it won't reset until the end of the metering period. With the values given in the above question, you don't actually have to do any math. If you look at the graph, you will see that the demand for the 15 minute period between 30 and 45 minutes will be the highest demand. It starts at 50kW and goes up to 75kW, so the value must be between 50 and 75kW. There is only one value in that range given in the answers.

 

[Elect117]

The highest recorded demand was between 30-45 min. Since nothing peaked over that number, that is the last recorded peak.

 

[Andres Arias]

Notice that the book shows 62.5 kW as the correct answer. Why ?

 

[Elect117]

The utility meter records the highest demand.

You could extend this graph to 30 days and as long as 62.5kW is still the highest recorded value, that is the value you will see on your electric bill.

Is your question on how the average for the 15min cycle was calculated?

75-50 / 2 = 12.5, 12.5 + 50 = 62.5.

 

[Elect117]

For the table, the average power consumption is less than or greater than the max demand at the start of the interval. If yes, then that value becomes the max demand at the start of the interval.

 

[Andres Arias]

My question is, why is the maximum at the start of fifth interval (t = 60 minutes) 62.5kW?. I know how the average is computed at each interval, bu my confusion is on how to compute the maximum demand (3rd column on table in the solution).

 

[wwhitney]

My question is, why is the maximum at the start of fifth interval (t = 60 minutes) 62.5kW?. I know how the average is computed at each interval, bu my confusion is on how to compute the maximum demand (3rd column on table in the solution).

The entry in each row of the 3rd column is the maximum of the entries in the 2nd column of all the earlier rows.

Except for the first two rows, as this table is really an excerpt of a larger, unspecified table, for which the largest of the entries in the 2nd column appearing in the earlier, unspecifed rows is 35, as per the problem statement.

Cheers, Wayne

 

[Elect117]

The 3rd column is the saved highest average demand.

Was there any demand higher than 62.5kW?

 

[Elect117]

If you think the answer is 75kW then you did the same thing I did when I did the practice test.

The devil is in the details of the question. The highest peak demand is calculated as an average of the 15min cycle.

If you think the answers are 35 or 50kw then you are misunderstanding the function of "recording the highest demand".

 

[Andres Arias]

If you think the answer is 75kW then you did the same thing I did when I did the practice test.

The devil is in the details of the question. The highest peak demand is calculated as an average of the 15min cycle.

If you think the answers are 35 or 50kw then you are misunderstanding the function of "recording the highest demand".

It's not so clear for me why we have to record the highest demand from the historic average?

 

[wwhitney]

It's not so clear for me why we have to record the highest demand from the historic average?

Because that's part of how the customer is billed. Basically the POCO has to pay for enough infrastructure to deliver that level of power, even if the customer only uses it once a month, so the customer bears some of that cost.

Cheers, Wayne

 

[Elect117]

And that is also how the question is worded. "that registers maximum demand for each billing period".

 

[gadfly56]

Because that's part of how the customer is billed. Basically the POCO has to pay for enough infrastructure to deliver that level of power, even if the customer only uses it once a month, so the customer bears some of that cost.

Cheers, Wayne

...and for 15 minutes. It's a pretty sharp stick to poke the customer into minimizing his peak demand. If you have some process that requires starting up huge loads, like an aluminum smelter, you'd be wise to bring them on line 20 minutes apart or so.

 

[Elect117]

...and for 15 minutes. It's a pretty sharp stick to poke the customer into minimizing his peak demand. If you have some process that requires starting up huge loads, like an aluminum smelter, you'd be wise to bring them on line 20 minutes apart or so.

Peak demand pricing is used to offset the cost of the transformer and wire they have to install for you. If you are 1000kva for one minute, once a month, and 100kva for the average over that month, they still need a 1000kva transformer and secondary... It is not like they can install a 100kva transformer for you.

 

[gadfly56]

Peak demand pricing is used to offset the cost of the transformer and wire they have to install for you. If you are 1000kva for one minute, once a month, and 100kva for the average over that month, they still need a 1000kva transformer and secondary... It is not like they can install a 100kva transformer for you.

Kinda the point I was trying to make. If you have 10 loads that are 10kVA when operating, but they pull 100kVA on startup, you can get away with a 200kVA transformer instead of the 1,000kVA.

 

[ron]

One item to keep in mind is that it is not cost effective for the utility to measure peak every milli-second to determine demand. Demand metering / charges as have been pointed out is for the utility to recover cost for their infrastructure.
They utilize the average over 15 minute periods (in reality, it is easy to measure kWh, so they measure the kWh over 15 minutes, which is a continuous counter, then multiply the usage it by 4 (since there are 4 instances of 15 minute periods in an hour) to determine the average demand for that 15 minute period) from a practical standpoint.

 

[ggunn]

The plot on the chart is instantaneous demand. The period between 30 and 45 minutes is where the peak demand of 75kW occurred. What is the average demand over that period? You have to estimate it, but by inspection it's less than 75kW and more than 50 kW, and 62.5kW is the only answer in that range. As Sherlock Holmes once said, once you have eliminated the impossible...

 

[wwhitney]

The plot on the chart is instantaneous demand. The period between 30 and 45 minutes is where the peak demand of 75kW occurred. What is the average demand over that period?

If, as appears clear, the graph is 2-piece linear over that time period, peaks at 75 kW, and passes through (30,50) and (45,50), the average value is exactly 1/2*(50+75). The area of a triangle of a given altitude is 1/2 base * height, regardless of where that top vertex is located over the base.

Cheers, Wayne

 

[ggunn]

If, as appears clear, the graph is 2-piece linear over that time period, peaks at 75 kW, and passes through (30,50) and (45,50), the average value is exactly 1/2*(50+75). The area of a triangle of a given altitude is 1/2 base * height, regardless of where that top vertex is located over the base.

Cheers, Wayne

62.5kW is the only possible answer; the precise number is irrelevant. Testing is often designed to give the testee the opportunity to waste time doing calculations where applying a bit of logic gets you on to the next question faster. Test taking 101: Check to see what choices you can eliminate before picking up the pencil or touching the calculator.