How can I convert that Watt peak rating to Megawatt?

[t0mm02]

Homework Statement

E= r x GB x A x L

Relevant Equations

E= r x GB x A x L

Hello everyone. I need to calculate solar energy output (E) from a solar farm. The calculations are based on values from total module surface area in m² (A), panel efficiency (r), solar irradiance (GB) and losses (L) due to dust, cold temperature and ageing, which is estimated to be 0.75. Since the energy required from the PV modules has already been calculated, it is possible to analyze if the energy output is enough to satisfy the energy demand. The following formula is used to calculate the solar energy output (E):

• E= r x GB x A x L
• E= 19.08 x 1.30 x 10193.70 x 0.75 = 328697 Wp, which is the total watt-peak rating

How can I convert that Wp to Megawatt?

 

[PeroK]

Homework Statement:: E= r x GB x A x L
Relevant Equations:: E= r x GB x A x L

Hello everyone. I need to calculate solar energy output (E) from a solar farm. The calculations are based on values from total module surface area in m² (A), panel efficiency (r), annual average solar irradiance (GB) and losses (L) due to dust, cold temperature and ageing, which is estimated to be 0.75. Since the energy required from the PV modules has already been calculated, it is possible to analyze if the energy output is enough to satisfy the energy demand. The following formula is used to calculate the solar energy output (E):

• E= r x GB x A x L
• E= 19.08 x 1.30 x 10193.70 x 1.3 = 328697 Wp, which is the total watt-peak rating

How can I convert that Wp to Megawatt?

A Megawatt is a million watts, right?

 

[phyzguy]

Without units, your numbers don't mean much. You should add the units for each of the numbers. What does the efficiency r of 19.08 mean? 19.08%? If so, shouldn't you divide by 100? Also, you said L=0.75, but then you multiplied by 1.3. How can the loss be greater than 1? Also, since you have multiplied by the "annual average solar irradiance", why would you call this watt-peak? Isn't it average watts?

 

[t0mm02]

Without units, your numbers don't mean much. You should add the units for each of the numbers. What does the efficiency r of 19.08 mean? 19.08%? If so, shouldn't you divide by 100? Also, you said L=0.75, but then you multiplied by 1.3. How can the loss be greater than 1? Also, since you have multiplied by the "annual average solar irradiance", why would you call this watt-peak? Isn't it average watts?

19. 08 is the efficiency of the solar panel, therefore 19.08 %, L= 0,75 but I made a mistake so I will change it. Solar irradiance is 1.30 kW/ m². Now that everything is corrected, do you have any advice or help to offer? thank you

 

[phyzguy]

19. 08 is the efficiency of the solar panel, therefore 19.08 %, L= 0,75 but I made a mistake so I will change it. Solar irradiance is 1.30 kW/ m². Now that everything is corrected, do you have any advice or help to offer? thank you

Can you show me the new calculation, with the corrections, and with units added? Why do you call the 1.3 kW/m^2 the "annual average solar irradiance"? Isn't it the peak irradiance? Do you still need help on converting to MWatts, or is @PeroK 's comment enough? Are you trying to calculate the peak megawatts, or the annual megawatt-hours, or both?

 

[t0mm02]

Can you show me the new calculation, with the corrections, and with units added? Why do you call the 1.3 kW/m^2 the "annual average solar irradiance"? Isn't it the peak irradiance? Do you still need help on converting to MWatts, or is @PeroK 's comment enough? Are you trying to calculate the peak megawatts, or the annual megawatt-hours, or both?

I get a really low number that doesn't make sense as the solar farm has 5254 solar panels, I get less than a MW and doesn't matter what numbers I use my calculations do not make sense.

However, I have seen this calculations on a thesis that I have found and when I do the calculations for my solar farm it makes a lot more sense:
As described earlier, a solar panel in sunshine periods can normally generate 260 W of power. By assuming 30 solar panels mounted on the rooftop, the power generation for a tram operating on line 8 will be: Power generated by 30 panels = 30 × 260 = 7800 W = 7.8 kW

I see these calculations too simple though, he doesn't take into account sunshine hours or losses and I don't know how to introduce sunshine hours or losses into his calculations or even if it is needed to introducde them.

 

[phyzguy]

Please post the calculation again, like you did in post #1, with the corrections and with units. We can't help you if you don't show us your calculations.

 

[t0mm02]

Please post the calculation again, like you did in post #1, with the corrections and with units. We can't help you if you don't show us your calculations.

Power generated by 5254 panels = 5254 × 370 = 1943980 Wp = 1.944 MWp

These are my calculations but once again, I see these calculations too simple though, he doesn't take into account sunshine hours or losses and I don't know how to introduce sunshine hours or losses into his calculations or even if it is needed to introduce them.

 

[phyzguy]

Power generated by 5254 panels = 5254 × 370 = 1943980 Wp = 1.944 MWp

These are my calculations but once again, I see these calculations too simple though, he doesn't take into account sunshine hours or losses and I don't know how to introduce sunshine hours or losses into his calculations or even if it is needed to introduce them.

OK, two questions:

(1) You said in post # 6 "a solar panel in sunshine periods can normally generate 260 W of power". So why did you use 370W here?

(2) What are you trying to calculate? Are you trying to calculate the peak megawatts, or the annual megawatt-hours, or both?

 

[gmax137]

Power generated by 5254 panels = 5254 × 370 = 1943980 Wp = 1.944 MWp

These are my calculations but once again, I see these calculations too simple though, he doesn't take into account sunshine hours or losses and I don't know how to introduce sunshine hours or losses into his calculations or even if it is needed to introduce them.

here try this
$E = (r) (GB) (A) (L)$

That doesn't help much does it? Include in the units:
$E (MW) = (r) (GB \frac {kW} {m^2}) (A m^2) (L) ( \frac{MW} {1000 kW})$

You seem to know the number of panels; call that "N" and think of "A" as the area per panel:
$E (MW) = (r) (GB \frac {kW} {m^2}) (A \frac{m^2}{panel})(N(panel)) (L) \frac{MW} {1000 kW}$