Anybody good at engineerIng calculations?
30 October 2008, 02:56 PM
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Anybody good at engineerIng calculations?
I want to get peoples opinions on how they would work out the following sample calculation. I got the answer right, but my working out / solution was completely different from the one in the back of the answer book. The figures in brackets are powers as Scoobynet can't handle equations.
PRESENT POWER DEMANDS FROM ALL SOURCES ARE APPROX 10(12)W. WE NEED 10 TIMES THIS AMOUNT. SOLAR POWER PLANTS ARE ONLY 10% EFFICIENT. EARTH PRESENTS AN AREA OF ABOUT 10(15)M(2) (METRES SQUARED) TO THE SUN AND RECEIVES POWER AS SUNLIGHT AT ABOUT 1kWM(-2). WHAT FRACTION OF THE EARTHS SURFACE WOULD NEED TO BE DEDICATED TO POWER GENERTION IF ONLY SOLAR POWER WAS USED AS A SOURCE OF ENERGY?
The answer is 1 / 10,000. I'm interested to see how people arrive at the answer as its the route to the anwer I'd like to see.
PRESENT POWER DEMANDS FROM ALL SOURCES ARE APPROX 10(12)W. WE NEED 10 TIMES THIS AMOUNT. SOLAR POWER PLANTS ARE ONLY 10% EFFICIENT. EARTH PRESENTS AN AREA OF ABOUT 10(15)M(2) (METRES SQUARED) TO THE SUN AND RECEIVES POWER AS SUNLIGHT AT ABOUT 1kWM(-2). WHAT FRACTION OF THE EARTHS SURFACE WOULD NEED TO BE DEDICATED TO POWER GENERTION IF ONLY SOLAR POWER WAS USED AS A SOURCE OF ENERGY?
The answer is 1 / 10,000. I'm interested to see how people arrive at the answer as its the route to the anwer I'd like to see.
30 October 2008, 03:19 PM
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Total power demand = 10^12 W
Solar panel efficiency = 10% = 10/100 = 0.1, therefore total incident power required on solar panels to achieve present demand = (10^12) / 0.1 = 10^13 W.
The question then says that we need 10x that amount, for some reason. Assuming it's not just allowing for the inefficiency of the solar cells, that would imply that the total incident power required is 10x the above, which is 10^14 W.
Actual received power from sun = 1000 W/m2, therefore number of m2 required = 10^14 / 1000 = 10^11 m2.
Therefore, fraction of earth required to be covered = 10^11 / 10^15 = 1/10000.
Seems a bit odd that they'd suggest needing 10x enough power to meet present demand though, I think that's been included in the efficiency of the solar cells - coincidentally 10%. I think the real answer would be closer to 1/100,000.
Solar panel efficiency = 10% = 10/100 = 0.1, therefore total incident power required on solar panels to achieve present demand = (10^12) / 0.1 = 10^13 W.
The question then says that we need 10x that amount, for some reason. Assuming it's not just allowing for the inefficiency of the solar cells, that would imply that the total incident power required is 10x the above, which is 10^14 W.
Actual received power from sun = 1000 W/m2, therefore number of m2 required = 10^14 / 1000 = 10^11 m2.
Therefore, fraction of earth required to be covered = 10^11 / 10^15 = 1/10000.
Seems a bit odd that they'd suggest needing 10x enough power to meet present demand though, I think that's been included in the efficiency of the solar cells - coincidentally 10%. I think the real answer would be closer to 1/100,000.
30 October 2008, 04:30 PM
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Total power demand = 10^12 W
Solar panel efficiency = 10% = 10/100 = 0.1, therefore total incident power required on solar panels to achieve present demand = (10^12) / 0.1 = 10^13 W.
The question then says that we need 10x that amount, for some reason. Assuming it's not just allowing for the inefficiency of the solar cells, that would imply that the total incident power required is 10x the above, which is 10^14 W.
Actual received power from sun = 1000 W/m2, therefore number of m2 required = 10^14 / 1000 = 10^11 m2.
Therefore, fraction of earth required to be covered = 10^11 / 10^15 = 1/10000.
Seems a bit odd that they'd suggest needing 10x enough power to meet present demand though, I think that's been included in the efficiency of the solar cells - coincidentally 10%. I think the real answer would be closer to 1/100,000.
Solar panel efficiency = 10% = 10/100 = 0.1, therefore total incident power required on solar panels to achieve present demand = (10^12) / 0.1 = 10^13 W.
The question then says that we need 10x that amount, for some reason. Assuming it's not just allowing for the inefficiency of the solar cells, that would imply that the total incident power required is 10x the above, which is 10^14 W.
Actual received power from sun = 1000 W/m2, therefore number of m2 required = 10^14 / 1000 = 10^11 m2.
Therefore, fraction of earth required to be covered = 10^11 / 10^15 = 1/10000.
Seems a bit odd that they'd suggest needing 10x enough power to meet present demand though, I think that's been included in the efficiency of the solar cells - coincidentally 10%. I think the real answer would be closer to 1/100,000.
Thats an interesting answer. You actually start out solving it like the book does but finishing solving it the way I did it. It's like a combination of the two routes. I guess there is more than one way to skin a cat!!*
* NOTE : No actual cats were harmed during the solution to this calculation.
30 October 2008, 04:33 PM
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