# NCEES (version 2018) #122



## supra33202

I read the solution but I don't understand how to get the "Max Demand at Start of Interval (kW).

Please advise.

Thanks,


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## rg1

supra33202 said:


> I read the solution but I don't understand how to get the "Max Demand at Start of Interval (kW).
> 
> Please advise.
> 
> Thanks,


Can u please show up the question


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## supra33202




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## supra33202




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## JohnMdd

I second this question. I don't think I'm an idiot, but this doesn't make any sense to me. Perhaps there are terms or definitions that I don't understand.


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## supra33202

I understand that the "average power consumption" is just the area under the curve (graph).

1 square = 25 kW

Time interval 1: 25kW

Time interval 2: 25kW + 25kW (0.5) = 37.5 kW

Time interval 3: 25kW + 25kW + 25kW (0.5) = 62.5 kW

Time interval 4: 25kW + 25kW = 50kW

But for " "Max Demand at Start of Interval (kW)", I need help.

Why the max demand (35kW) is the same for both Time interval 1 and 2?

Thanks!


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## BirdGrave

Imagine there is a graph like the one in this problem but for the 60 seconds immediately preceding it.  The process you outlined of finding the maximum of the averages of the 15-second intervals showed that answer to be 35kW, as opposed to the answer for the graph in the problem, which is 62.5kW.  That is where that 35kW comes from.  I agree the wording is vague.


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## rg1

supra33202 said:


> View attachment 10941


In this question the Max demand given at t=0 is 35KW. Understand it like this. Every 15 minute period meter will calculate average KW of this 15minute period and then compare it with earlier data it has preserved- reject the new data if it is less than earlier data or select the new data if it is more than old data. So e.g. at t=15, the Max demand will remain 35KW because average KW between t=0 to t=15 is 25KW, similarly at t=30, the new value is 37.5KW, so it will pick up this value(37.5&gt;35), at t=45, the new value is 62.5KW, again it will pick up new value of Max demand as 62.5KW(62.5&gt;37.5), next at t=60, new data is 50KW so it will reject it and show the max demand as 62.5KW only. Is that the answer?

Did it make sense?


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## rg1

rg1 said:


> In this question the Max demand given at t=0 is 35KW. Understand it like this. Every 15 minute period meter will calculate average KW of this 15minute period and then compare it with earlier data it has preserved- reject the new data if it is less than earlier data or select the new data if it is more than old data. So e.g. at t=15, the Max demand will remain 35KW because average KW between t=0 to t=15 is 25KW, similarly at t=30, the new value is 37.5KW, so it will pick up this value(37.5&gt;35), at t=45, the new value is 62.5KW, again it will pick up new value of Max demand as 62.5KW(62.5&gt;37.5), next at t=60, new data is 50KW so it will reject it and show the max demand as 62.5KW only. Is that the answer?
> 
> Did it make sense?


This is general procedure Max demand is calculated. Generally it is 15 minute period in business, it can be something else say 30minutes- More is the time better it is for the consumer, lesser is the time better it is for the supply company. On the meter reading day, the Max demand reading is reset to zero after picking up the the reading for the monthly billing. You guys can refer Wildi ( Some chapters on use of power towards end of the book) or surf net you will get further idea.


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## rg1

supra33202 said:


> I understand that the "average power consumption" is just the area under the curve (graph). You are wrong. The area under graph is (Power X Time)= Energy. Average power will dividing this energy again by the time. This itself can be one question. To calculate energy or average energy from the graph.
> 
> 1 square = 25 kW-----For example 25X one 15minute period/one 15 minute period= 25KW . In this particular question, the graph is such that you will get the answer correct even if you are doing as you have done. This may not always be true.
> 
> Time interval 1: 25kW
> 
> Time interval 2: 25kW + 25kW (0.5) = 37.5 kW
> 
> Time interval 3: 25kW + 25kW + 25kW (0.5) = 62.5 kW
> 
> Time interval 4: 25kW + 25kW = 50kW
> 
> But for " "Max Demand at Start of Interval (kW)", I need help.
> 
> Why the max demand (35kW) is the same for both Time interval 1 and 2?
> 
> Thanks!


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## kswan

The values listed under "Average Power Consumption (kW)" refer to the END of the respective time interval.  However, the "Max Demand" values refer to the START of the respective time interval.  And remembering that in this context, "max demand" refers to maximum average demand measured in any single time interval, NOT the instantaneous maximum value measured.

Max demand at the start of interval #1 = 35 kW (given in the problem statement; we don't necessarily need to know how this number was obtained)

Max demand at the start of interval #2 = 35 kW (during interval #1, the average demand was only 25 kW, so the max demand registered did not change)

Max demand at the start of interval #3 = 37.5 kW (during interval #2, the average demand was 37.5 kW, so the new max demand registered is now 37.5 kW)

Max demand at the start of interval #4 = 62.5 kW (during interval #3, the average demand was 62.5 kW, so the new max demand registered is now 62.5 kW)

Max demand at the start of interval #5 = 62.5 kW (during interval #4, the average demand was 50 kW, so the max demand registered is still 62.5 kW)

Finally, the problem statement asks for the maximum demand registered at t=60 min, which is at the END of interval #4 and the START of interval #5.


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## rg1

kswan said:


> The values listed under "Average Power Consumption (kW)" refer to the END of the respective time interval.  However, the "Max Demand" values refer to the START of the respective time interval.  And remembering that in this context, "max demand" refers to maximum average demand measured in any single time interval, NOT the instantaneous maximum value measured.
> 
> Max demand at the start of interval #1 = 35 kW (given in the problem statement; we don't necessarily need to know how this number was obtained)
> 
> Max demand at the start of interval #2 = 35 kW (during interval #1, the average demand was only 25 kW, so the max demand registered did not change)
> 
> Max demand at the start of interval #3 = 37.5 kW (during interval #2, the average demand was 37.5 kW, so the new max demand registered is now 37.5 kW)
> 
> Max demand at the start of interval #4 = 62.5 kW (during interval #3, the average demand was 62.5 kW, so the new max demand registered is now 62.5 kW)
> 
> Max demand at the start of interval #5 = 62.5 kW (during interval #4, the average demand was 50 kW, so the max demand registered is still 62.5 kW)
> 
> Finally, the problem statement asks for the maximum demand registered at t=60 min, which is at the END of interval #4 and the START of interval #5.


Perfect. The instantaneous max demand is peak demand, different from max demand. Generally the supply companies will try to even/average out the KW supplied by putting a tariff based on Max demand averaged over 15/30 minutes period. The individual instantaneous peaks are rarely considered in tariff structures.  Do you know why is it done?. Why the supply companies not charge the customers only on the units (KWH) of electricity consumed as they do for house hold customers? This is called  two part tariff!!!! meant for commercial and Industrial consumers!!


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## SSG

rg1 said:


> In this question the Max demand given at t=0 is 35KW. Understand it like this. Every 15 minute period meter will calculate average KW of this 15minute period and then compare it with earlier data it has preserved- reject the new data if it is less than earlier data or select the new data if it is more than old data. So e.g. at t=15, the Max demand will remain 35KW because average KW between t=0 to t=15 is 25KW, similarly at t=30, the new value is 37.5KW, so it will pick up this value(37.5&gt;35), at t=45, the new value is 62.5KW, again it will pick up new value of Max demand as 62.5KW(62.5&gt;37.5), next at t=60, new data is 50KW so it will reject it and show the max demand as 62.5KW only. Is that the answer?
> 
> Did it make sense?


Thanks a lot, this made sense.


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## SSG

Even I assumed the area under the curve. Thanks for correcting.


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## SSG

Hi @rg1..

Average power made sense when I read it, but I got stuck while solving it. As you had mentioned, The first one will be (25 * 15mins)/ 15mins = 25 ; ((25 +50)*15mins)/30mins =37.5 ; ((50+75-50)*15mins)/ 45mins = ? so i am not getting the desired 62.5;  neither 50 ((50 *15)/60). What is it that I am doing wrong?

Thanks in advance.


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## rg1

SSG said:


> Hi @rg1..
> 
> Average power made sense when I read it, but I got stuck while solving it. As you had mentioned, The first one will be (25 * 15mins)/ 15mins = 25 ; ((25 +50)*15mins)/30mins =37.5 ; ((50+75-50)*15mins)/ 45mins = ? so i am not getting the desired 62.5;  neither 50 ((50 *15)/60). What is it that I am doing wrong? A lot of wrong here,  for the time being take average power as (max power+min power)/2 . please see below.
> 
> Thanks in advance.


Excuse me if I am being too heavy, actually I did not know a simpler way to explain it.

If you do not know the maths behind how to find area under the graph by integration then it will be good to understand average power as (Max+Min)/2. I think for PE that is sufficient. But you should know the concept that area under the graph gives you (power X time) energy.  For Straight line graphs  that energy can be found by Average power in KW multiplied by Time in Hrs to give you KWH.

Averaging is not done like you did, in graphs. It can be done dividing the the graph into small bars of time intervals say Delta t , multiplying the time interval with the corresponding power, summing up all such energies and then again dividing it by total time. I am sorry if I have confused you. I wanted to introduce only the concept. I will not ask you to waste time on this if you are not good at maths. PE exam will not ask tough questions on maths.


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## SSG

rg1 said:


> Excuse me if I am being too heavy, actually I did not know a simpler way to explain it.
> 
> If you do not know the maths behind how to find area under the graph by integration then it will be good to understand average power as (Max+Min)/2. I think for PE that is sufficient. But you should know the concept that area under the graph gives you (power X time) energy.  For Straight line graphs  that energy can be found by Average power in KW multiplied by Time in Hrs to give you KWH.
> 
> Averaging is not done like you did, in graphs. It can be done dividing the the graph into small bars of time intervals say Delta t , multiplying the time interval with the corresponding power, summing up all such energies and then again dividing it by total time. I am sorry if I have confused you. I wanted to introduce only the concept. I will not ask you to waste time on this if you are not good at maths. PE exam will not ask tough questions on maths.


Yes, your answer did confuse me. I did solve average using  (Max + Min)/2, but when you mentioned 

"1 square = 25 kW-----For example 25X one 15minute period/one 15 minute period= 25KW" 

I was trying to solve using your concept , but looks like I did not quiet follow what you intended. 

Sorry about that and thank you for clarifying.


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## rg1

SSG said:


> Yes, your answer did confuse me. I did solve average using  (Max + Min)/2, but when you mentioned
> 
> "1 square = 25 kW-----For example 25X one 15minute period/one 15 minute period= 25KW"
> 
> I was trying to solve using your concept , but looks like I did not quiet follow what you intended.
> 
> Sorry about that and thank you for clarifying.


Please carryon. Wish you all the best for the exam. Please express here if there is any doubt in understanding any question.


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## ARS

rg1 said:


> Perfect. The instantaneous max demand is peak demand, different from max demand. Generally the supply companies will try to even/average out the KW supplied by putting a tariff based on Max demand averaged over 15/30 minutes period. The individual instantaneous peaks are rarely considered in tariff structures.  Do you know why is it done?. Why the supply companies not charge the customers only on the units (KWH) of electricity consumed as they do for house hold customers? This is called  two part tariff!!!! meant for commercial and Industrial consumers!!


:thumbs:


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## Nashi

I truly despise this question. I understand up until interval 3. I took the average of the power at the start of interval 3 and to the peak of the triangle halfway through interval 3 and then the average of the second half of the interval 3 and both give me 62.5 which is great so I guess because the curve is flat and stays at that last amount the maximum demand registered at 60minutes is still 62.5.

I guess because they are showing a power curve and we are looking for the maximum Power demand its messing with my head looking at the graph.


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## iahi_2005

Hi Guys,

I am looking at the responses,but, none of them make sense, can you all please explain how the average power consumption in interval 2 is 37.5??


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## Sthabik PE

iahi_2005 said:


> Hi Guys,
> 
> I am looking at the responses,but, none of them make sense, can you all please explain how the average power consumption in interval 2 is 37.5??




Interval #2 Area; A2 = (25kW x 15min)+ (1/2 x 25kW  x 15min) = 562.5 kW min

Interval #2  time interval; t2: 15 min.

Avg power in interval 2 = A2/t2 = 562.5 kW min / 15 min = 37.5 kW




Does it make sense now?


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## iahi_2005

Thanks! it makes sense now!! so you are adding the area of square + area of triangle!


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## Ahmed S.

Hi Everyone , I was confused as well but after reading your comments it become easy to understand

Average power ( Pmax+Pmin)/2 

At T=0 , (25+25)/2 = 25

At T=15 , (50+25)/2 = 37.5

At T=30, (75+50)/2 = 62.5

At T=45 , (50+50)/2 = 50 

For the max power , I will quote the best comment above from SSG

In this question the Max demand given at t=0 is 35KW. Understand it like this. Every 15 minute period meter will calculate average KW of this 15minute period and then compare it with earlier data it has preserved- reject the new data if it is less than earlier data or select the new data if it is more than old data. So e.g. at t=15, the Max demand will remain 35KW because average KW between t=0 to t=15 is 25KW, similarly at t=30, the new value is 37.5KW, so it will pick up this value(37.5&gt;35), at t=45, the new value is 62.5KW, again it will pick up new value of Max demand as 62.5KW(62.5&gt;37.5), next at t=60, new data is 50KW so it will reject it and show the max demand as 62.5KW only. Is that the answer?


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## ahaq

Ahmed S. said:


> Hi Everyone , I was confused as well but after reading your comments it become easy to understand
> 
> Average power ( Pmax+Pmin)/2
> 
> At T=0 , (25+25)/2 = 25
> 
> At T=15 , (50+25)/2 = 37.5
> 
> At T=30, (75+50)/2 = 62.5
> 
> At T=45 , (50+50)/2 = 50
> 
> For the max power , I will quote the best comment above from SSG
> 
> In this question the Max demand given at t=0 is 35KW. Understand it like this. Every 15 minute period meter will calculate average KW of this 15minute period and then compare it with earlier data it has preserved- reject the new data if it is less than earlier data or select the new data if it is more than old data. So e.g. at t=15, the Max demand will remain 35KW because average KW between t=0 to t=15 is 25KW, similarly at t=30, the new value is 37.5KW, so it will pick up this value(37.5&gt;35), at t=45, the new value is 62.5KW, again it will pick up new value of Max demand as 62.5KW(62.5&gt;37.5), next at t=60, new data is 50KW so it will reject it and show the max demand as 62.5KW only. Is that the answer?





Ahmed S. said:


> Hi Everyone , I was confused as well but after reading your comments it become easy to understand
> 
> Average power ( Pmax+Pmin)/2
> 
> At T=0 , (25+25)/2 = 25
> 
> At T=15 , (50+25)/2 = 37.5
> 
> At T=30, (75+50)/2 = 62.5
> 
> At T=45 , (50+50)/2 = 50
> 
> For the max power , I will quote the best comment above from SSG
> 
> In this question the Max demand given at t=0 is 35KW. Understand it like this. Every 15 minute period meter will calculate average KW of this 15minute period and then compare it with earlier data it has preserved- reject the new data if it is less than earlier data or select the new data if it is more than old data. So e.g. at t=15, the Max demand will remain 35KW because average KW between t=0 to t=15 is 25KW, similarly at t=30, the new value is 37.5KW, so it will pick up this value(37.5&gt;35), at t=45, the new value is 62.5KW, again it will pick up new value of Max demand as 62.5KW(62.5&gt;37.5), next at t=60, new data is 50KW so it will reject it and show the max demand as 62.5KW only. Is that the answer?


How would the maximum power in interval 2 be 37.5 and not 50? Isn't the height of the triangle the maximum power?


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## ahaq

Ahmed S. said:


> Hi Everyone , I was confused as well but after reading your comments it become easy to understand
> 
> Average power ( Pmax+Pmin)/2
> 
> At T=0 , (25+25)/2 = 25
> 
> At T=15 , (50+25)/2 = 37.5
> 
> At T=30, (75+50)/2 = 62.5
> 
> At T=45 , (50+50)/2 = 50
> 
> For the max power , I will quote the best comment above from SSG
> 
> In this question the Max demand given at t=0 is 35KW. Understand it like this. Every 15 minute period meter will calculate average KW of this 15minute period and then compare it with earlier data it has preserved- reject the new data if it is less than earlier data or select the new data if it is more than old data. So e.g. at t=15, the Max demand will remain 35KW because average KW between t=0 to t=15 is 25KW, similarly at t=30, the new value is 37.5KW, so it will pick up this value(37.5&gt;35), at t=45, the new value is 62.5KW, again it will pick up new value of Max demand as 62.5KW(62.5&gt;37.5), next at t=60, new data is 50KW so it will reject it and show the max demand as 62.5KW only. Is that the answer?





kswan said:


> The values listed under "Average Power Consumption (kW)" refer to the END of the respective time interval. However, the "Max Demand" values refer to the START of the respective time interval. And remembering that in this context, "max demand" refers to maximum average demand measured in any single time interval, NOT the instantaneous maximum value measured.
> 
> Max demand at the start of interval #1 = 35 kW (given in the problem statement; we don't necessarily need to know how this number was obtained)
> 
> Max demand at the start of interval #2 = 35 kW (during interval #1, the average demand was only 25 kW, so the max demand registered did not change)
> 
> Max demand at the start of interval #3 = 37.5 kW (during interval #2, the average demand was 37.5 kW, so the new max demand registered is now 37.5 kW)
> 
> Max demand at the start of interval #4 = 62.5 kW (during interval #3, the average demand was 62.5 kW, so the new max demand registered is now 62.5 kW)
> 
> Max demand at the start of interval #5 = 62.5 kW (during interval #4, the average demand was 50 kW, so the max demand registered is still 62.5 kW)
> 
> Finally, the problem statement asks for the maximum demand registered at t=60 min, which is at the END of interval #4 and the START of interval #5.


*"And remembering that in this context, "max demand" refers to maximum average demand measured in any single time interval, NOT the instantaneous maximum value measured."* Most important piece of information that had no mention in the question.


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## akyip

Hey all. As a friendly FYI, attached is my work on problem 122.

I broke this problem down by each interval (each 15-minute interval):


For each interval, I listed the maximum demand and the average demand.
In each interval, the average demand is the area under the graph, within each 15-minute interval. This involved breaking down the shape of the area within each area into separate rectangles and triangles where necessary (this is especially important for the 15-30 min interval and the 30-45 min interval).
Whenever the average demand in one interval is greater than the maximum demand of the previous interval, that average demand now becomes the new maximum interval. You can see this occurring at the 15-30 min interval and the 30-45 min interval.
Hope this helps!


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