PV power plant efficiency math problem?

  • #1

Homework Statement



The controversial Cheviot open-pit coal mine in Alberta, Canada occupies 7,455 hectares (one hectare= 10^4 m^2) and produces 1.4 million tonnes of coal per year.

a) use the energy content of coal of 29 MJ/kg to find the power in watts corresponding to this rate of coal production., assuming it is burned with 35% efficiency.

b) using the assumption of 254 W/m^2 and 15% PV efficiency, find the area of a PV plant with the power output, and compare that number with the area of the Cheviot mine.


Homework Equations





The Attempt at a Solution



For part A I did

1.4E9kg/yr * 1 yr/365 days * 1 day/24 hrs * 1 hr/3600 sec * 2.9E6 J/kg *.35 and got 450 MW

I have no idea how to start part B though

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
SteamKing
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In part a, 29 MJ/kg is NOT equal to 2.9*10^6 J/kg. Check again.
 
  • #3
In part a, 29 MJ/kg is NOT equal to 2.9*10^6 J/kg. Check again.

I am not sure what I am doing wrong
 
  • #4
SteamKing
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Is 29 million the same as 2.9 million? 10^6 = 1 million.
 
  • #5
CWatters
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Slightly bad question I think. Parts a and b aren't really comparing like with like. Part b is talking about the electrical power produced by PV where as part a is talking about the power embodied in the coal (which still has to be burnt to turn it into electrical power and that's not 100% efficient). Oh well lets ignore that discrepancy and just answer it as posed...

Part b tells you how many watts per square meter hit the solar panels (254 W/m^2). It tells you what percentage of that is converted to electrical power (15%). So you can calculate the electrical power produced per square meter of PV.

Part a gives you the total power output of the coal plant so you can work out how many square meters of PV you need to produce the same power output. (The word "same" is missing from the question).

Compare the area of PV with that of the coal plant. Perhaps express as a percentage?
 
  • #6
Slightly bad question I think. Parts a and b aren't really comparing like with like. Part b is talking about the electrical power produced by PV where as part a is talking about the power embodied in the coal (which still has to be burnt to turn it into electrical power and that's not 100% efficient). Oh well lets ignore that discrepancy and just answer it as posed...

Part b tells you how many watts per square meter hit the solar panels (254 W/m^2). It tells you what percentage of that is converted to electrical power (15%). So you can calculate the electrical power produced per square meter of PV.

Part a gives you the total power output of the coal plant so you can work out how many square meters of PV you need to produce the same power output. (The word "same" is missing from the question).

Compare the area of PV with that of the coal plant. Perhaps express as a percentage?


I am not sure how to calculate the electrical power produced per square meter? How would I do this?
 
  • #7
Would I be finding 15% of 254?


The answer in the back of the book is in hectares. I am not sure how to get to that
 
  • #8
CWatters
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Would I be finding 15% of 254?

Yes.

So each square meter produces 15% of 254W = 38.1W.

How many square meters of PV would you need to produce the same power output as the coal plant?

Convert the square meters to hectatares (the conversion factor is in the problem statement!)
 
  • #9
Yes.

So each square meter produces 15% of 254W = 38.1W.

How many square meters of PV would you need to produce the same power output as the coal plant?

Convert the square meters to hectatares (the conversion factor is in the problem statement!)

How would I go about finding the square meters?
 
  • #10
haruspex
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Slightly bad question I think. Parts a and b aren't really comparing like with like. Part b is talking about the electrical power produced by PV where as part a is talking about the power embodied in the coal (which still has to be burnt to turn it into electrical power and that's not 100% efficient).
As I read it, it is asking for the electrical power output of the mine, assuming 35% efficiency.
Of course, it overlooks that the PV can go on producing for millions of years, whereas the mine would eventually be worked out, leaving a large area of degraded land and spoil.
 
  • #11
As I read it, it is asking for the electrical power output of the mine, assuming 35% efficiency.
Of course, it overlooks that the PV can go on producing for millions of years, whereas the mine would eventually be worked out, leaving a large area of degraded land and spoil.

So how am I supposed to calculate part B?
 
  • #12
haruspex
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So how am I supposed to calculate part B?
You want the area of PV that would produce the electrical power as the coal mine, right? Suppose the area is A and the power output P. what equation can you write connecting them?
 
  • #13
CWatters
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How would I go about finding the square meters?

Example..

Suppose you needed to build a solar plant capable of producing 10 MW (=10 * 10^6 W). If each square meter produces just 38W then you will need 10 * 10^6 / 38 = 263,157 sqm of panels.
 
  • #14
Example..

Suppose you needed to build a solar plant capable of producing 10 MW (=10 * 10^6 W). If each square meter produces just 38W then you will need 10 * 10^6 / 38 = 263,157 sqm of panels.



So would I then do 254/38.1?
 
  • #15
haruspex
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So would I then do 254/38.1?
It often helps to keep track of the units. The 254 is a number of Watts per sq m. The 38.1 is a number of Watts. If you divide Wm-2 by W what units will you have?
 
  • #16
CWatters
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So would I then do 254/38.1?

Nope. Read my example again.

Replace the figure I used (10MW) with the figure you calculated for the power output of the coal mine.
 
  • #17
Nope. Read my example again.

Replace the figure I used (10MW) with the figure you calculated for the power output of the coal mine.

This would then be the from what I calculated in part A?

So then 4.5 x 10^6/ 38.1?
 
  • #18
If what I did above was right i got 118,110 m^2.

The answer in the back of the book is 2360 hectares. How was this gotten?
 
  • #19
haruspex
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This would then be the from what I calculated in part A?

So then 4.5 x 10^6/ 38.1?
You seem to have replaced 450E6 with 4.5E6. Correcting that will get you to one half of the book answer. Can't explain the remaining discrepancy yet.
 
  • #20
CWatters
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I also get 450MW as the answer for part a.

For part b I get..

450 * 10^6 / 38.1 = 11.8 million square meters = 1180 hectares.

Same as haruspex, I can't explain the factor of two difference either.
 
  • #21

Homework Statement



The controversial Cheviot open-pit coal mine in Alberta, Canada occupies 7,455 hectares (one hectare= 10^4 m^2) and produces 1.4 million tonnes of coal per year.

a) use the energy content of coal of 29 MJ/kg to find the power in watts corresponding to this rate of coal production., assuming it is burned with 35% efficiency.

b) using the assumption of 254 W/m^2 and 15% solar panel efficiency, find the area of a PV plant with the power output, and compare that number with the area of the Cheviot mine.


Homework Equations





The Attempt at a Solution



For part A I did

1.4E9kg/yr * 1 yr/365 days * 1 day/24 hrs * 1 hr/3600 sec * 2.9E6 J/kg *.35 and got 450 MW

I have no idea how to start part B though

Homework Statement





Homework Equations





The Attempt at a Solution


Were you able to find the exact equation? I am facing similar problem so please share some information..Waiting for reply thanks in advance:)
 
Last edited:
  • #22
gneill
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If what I did above was right i got 118,110 m^2.

The answer in the back of the book is 2360 hectares. How was this gotten?

Hint: You can burn coal at night.
 
  • #23
CWatters
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Hint: You can burn coal at night.

:redface: Now why didn't I think of that. :redface:
 
  • #24
haruspex
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Hint: You can burn coal at night.

I've a nasty suspicion that's the answer, but in that case I reject the question as flawed. The figure it gave for available sun power was a constant, not a function of time of day. On that basis (as is common practice) it should have represented a 24 hour average.
 

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