Calculating Energy Usage and Demand Charges for a Peak Day: A Bill Analysis

[Adam Bourque]

Homework Statement

EXCEL TABLE AS ATTACHMENT[/B]
a.) What is the kWh used for this day?
b.) Assuming that this is the peak day for the previous year and TOD rates shown below apply, what is the energy usage and demand charges for this day?
c) What would be the change in charges if the second occurrence of process 2 was moved to 45 minutes later?

Homework Equations

None provided.

The Attempt at a Solution

For part a. I simply added up all the kwH used throughout the day from the excel table
kwH used for the day: 9497.28
For part b. I calculated the energy usage by multiplying the cost per kwH by the total kwH.
Energy Usage I got: $.03773 * 9497.28 = $358.33
I found the demand charge by adding up the total kwH from 1pm to 7pm (peak demand), and replacing the cost of kwH with the maximum load charge since the question stated "assuming peak day".
Demand Charge I got: $4.55 * 3341.61 = $15,204.33
For part c. I believe we go from 1:00pm to 1:45pm for process 2 (since this is the second occurrence period); however, I don't know what the change in charges would be since it seems it is during the intermediate period regardless. So no change?

Appreciate any help, if you are familiar with this subject I do have a few other relevant questions so I am in search of an online tutor, thank you.

 

[mfb]

For (b), the maximum load charge is per kW. You multiply a "charge per kW" with a number of kWh, the result cannot be correct. You'll have to find the peak demand (in kW) that happens at some point during the time window, for all three time windows.

 

[Adam Bourque]

For (b), the maximum load charge is per kW. You multiply a "charge per kW" with a number of kWh, the result cannot be correct. You'll have to find the peak demand (in kW) that happens at some point during the time window, for all three time windows.

You raise a good point here, thank you. Firstly, did I do part (a) correctly or do I have to do something for the conversion of kW to kWh (divide by hours?). I found the peak demand in kW to be 264.34, so what do I do from here? Thank you.

 

[mfb]

Didn't look at the file before. You have to fix (a) as well. Dividing kW by hours won't give kW*h, however.

I found the peak demand in kW to be 264.34, so what do I do from here?

In which period does it occur? What do you have to pay per kW peak load in this period?

What about the other periods?

 

[Adam Bourque]

Didn't look at the file before. You have to fix (a) as well. Dividing kW by hours won't give kW*h, however.
In which period does it occur? What do you have to pay per kW peak load in this period?

What about the other periods?

Okay so for part a, I would multiply 9497.28 (total kW) * 24 hours = 227934.72 kWh right?

It occurs during peak period for summer and intermediate period for other months.
So
Summer = $4.55 * 264.34 = 1202.747
Winter = $2.95 * 264.34 = 779.803

Is this correct? & If so how do I go about part c.

Thank you for all your help

 

[mfb]

"Total kW" doesn't make sense, and the multiplication with 24 hours doesn't make sense either.
You can add kWh, but then you have to find kWh values for every 15 minute step (!) first.

Imagine a similar situation in classical mechanics: At 0 am, you walk at a speed of 3 kilometers per hour. At 0:15 am, you still walk at this speed. At 0:30, you still walk at this speed. And so on the whole day. If you walk at 3 kilometers per hour for 24 hours, how far do you get?

What you tried to do here: Add all (24*4) 3km/h values to get 288 km/h walking speed, and then multiply that by 24 hours to get 6912 km distance walked in a single day.

So
Summer = $4.55 * 264.34 = 1202.747
Winter = $2.95 * 264.34 = 779.803

Looks fine. The other two periods are still missing.

(c) works like (b), but you have to shift the time where process 2 works and then repeat everything done in (b). Get (a) and (b) right before you come back to (c).

 

[Adam Bourque]

"Total kW" doesn't make sense, and the multiplication with 24 hours doesn't make sense either.
You can add kWh, but then you have to find kWh values for every 15 minute step (!) first.

Imagine a similar situation in classical mechanics: At 0 am, you walk at a speed of 3 kilometers per hour. At 0:15 am, you still walk at this speed. At 0:30, you still walk at this speed. And so on the whole day. If you walk at 3 kilometers per hour for 24 hours, how far do you get?

What you tried to do here: Add all (24*4) 3km/h values to get 288 km/h walking speed, and then multiply that by 24 hours to get 6912 km distance walked in a single day.
Looks fine. The other two periods are still missing.

(c) works like (b), but you have to shift the time where process 2 works and then repeat everything done in (b). Get (a) and (b) right before you come back to (c).

You made me decide to get a gold membership on this forums with all the useful help you are giving me.

Okay so for part (a) I understand what you are saying now, but am still confused. It is a 15 minute interval, so since kWh is in hours I am assuming I have to multiply the average of consecutive times by 0.25?

For part (b) [Can you verify that I did the correct times]
Also this solves demand charge, but what about energy usage? Is that .03773 multiplying by everything not in demand?

Period 1:
Summer = $2.95 * 263.28 = $776.68 (Intermediate Time Charge) -263.28 comes before intermediate, but is in that period so still charged for it right?
Winter = $2.95 * 263.28 = $776.68 (Intermediate Time Charge)

Period 2:
Summer = $4.55 * 264.36 = $1,202.84 (Peak Time Charge)
Winter = $2.95 * 264.36 = $779.86 (Intermediate Time Charge)

Period 3:
Summer = $4.55 * 174.21 = $792.66 (Peak Time Charge)
Winter = $2.95 * 174.21 = $513.92 (Intermediate Time Charge)

Thanks for all the help, sorry for being difficult (have had no engineering/physics background, am actually more biology/chemistry oriented)

 

[mfb]

Okay so for part (a) I understand what you are saying now, but am still confused. It is a 15 minute interval, so since kWh is in hours I am assuming I have to multiply the average of consecutive times by 0.25?

Multiply the power by its duration: 0.25h. That way you get the correct energy and the correct units at the same time. Units are always part of the quantities, and if you treat them correctly they will always work out. If they don't fit, you did something wrong.

The $.03773/kWh should apply everywhere in the same way.

Period 1:
Summer = $2.95 * 263.28 = $776.68 (Intermediate Time Charge) -263.28 comes before intermediate, but is in that period so still charged for it right?
Winter = $2.95 * 263.28 = $776.68 (Intermediate Time Charge)

Period 2:
Summer = $4.55 * 264.36 = $1,202.84 (Peak Time Charge)
Winter = $2.95 * 264.36 = $779.86 (Intermediate Time Charge)

Period 3:
Summer = $4.55 * 174.21 = $792.66 (Peak Time Charge)
Winter = $2.95 * 174.21 = $513.92 (Intermediate Time Charge)

I don't understand what you did here, and I would have expected the "base demand period" charge somewhere.

As far as I understand the problem, the final bill will have 5 different parts:
- the base service charge of $200
- $.03773/kWh that applies to the total energy delivered
- $4.55 times the maximal power delivered at some point during the "peak demand period"
- $2.95 times the maximal power delivered at some point during the "intermediate demand period"
- $3.62 times the maximal power delivered at some point during the "base demand period"

 

[Adam Bourque]

Multiply the power by its duration: 0.25h. That way you get the correct energy and the correct units at the same time. Units are always part of the quantities, and if you treat them correctly they will always work out. If they don't fit, you did something wrong.

The $.03773/kWh should apply everywhere in the same way.I don't understand what you did here, and I would have expected the "base demand period" charge somewhere.

As far as I understand the problem, the final bill will have 5 different parts:
- the base service charge of $200
- $.03773/kWh that applies to the total energy delivered
- $4.55 times the maximal power delivered at some point during the "peak demand period"
- $2.95 times the maximal power delivered at some point during the "intermediate demand period"
- $3.62 times the maximal power delivered at some point during the "base demand period"

(a) So since it is asking for the kWh for the entire day, do I add up all the kW and multiply it by .25 ? (9497.28 * .25) = 2,374.32

(b) I read your response here wrong, I thought you meant during the periods when process 1 & 2 were active for some reason. This makes way more sense, I will solve this in a minute, just want to make sure I did part a correctly.

 

[mfb]

(a) So since it is asking for the kWh for the entire day, do I add up all the kW and multiply it by .25 ? (9497.28 * .25) = 2,374.32

Work with units.

You multiply every kW value by its duration (0.25 h) and then sum all. You'll get the same number, but that way the units are correct.

 

[Adam Bourque]

Work with units.

You multiply every kW value by its duration (0.25 h) and then sum all. You'll get the same number, but that way the units are correct.

Okay I did this using sum function in excel, does put it in better perspective. For b, I found the maximal powers during each period, all are the same before calculating the cost except for peak maximal during winter (does this seem right?) Or am I suppose to exclude the overlapping of the periods? (As in since intermediate is 10am-10pm but peak is 1pm-7pm would I only consider the maximum to be from 10am-12:45 and 7pm-10pm for intermediate).

Then I multiplied by the cost for each period like you said to do.

Summer
Peak Maximal: 264.36 * 4.55
Intermediate Maximal: 264.36 * 2.95
Base Demand Maximal: 264.36 * 3.62

Winter
Peak Maximal: 263.28 * 4.55
Intermediate Maximal: 264.36 * 2.95
Base Demand Maximal: 264.36 *3.62

 

[mfb]

Or am I suppose to exclude the overlapping of the periods? (As in since intermediate is 10am-10pm but peak is 1pm-7pm would I only consider the maximum to be from 10am-12:45 and 7pm-10pm for intermediate).

I would expect that the periods overlap (otherwise the time frames would have been written differently), but I'm not sure.

Sounds right. For (c), you'll see that at least one of those values changes.

 

[Adam Bourque]

I would expect that the periods overlap (otherwise the time frames would have been written differently), but I'm not sure.

Sounds right. For (c), you'll see that at least one of those values changes.

See that's what I am confused about because the second occurrence of process 2 starts at 1pm-2:30pm so that would just move the occurrence to 2:30pm through 3:15pm which is the same time period for peak/base/intermediate?

 

[mfb]

It changes the power values, and you'll get a different peak power value.

 

[Adam Bourque]

It changes the power values, and you'll get a different peak power value.

I realized this after posting, thanks for the verification though, I wasn't considering the other operations. Do you mind me asking a related question to this

It changes the power values, and you'll get a different peak power value.

Definitely makes a difference getting a different peak power value (maximum), which is now 185.19 versus before at 264.36. So change in charges is calculated like:
4.55 * 185.19 = 842.61 - previous charge
2.95 * 185.19
3.62 * 185.19

 

[mfb]

Didn't check the file again now, but there is no value exceeding 185.19 in other periods? Then it is fine.