Would you believe 150,000 panels!

You asked a very open ended question. To provide a good estimate we need to make a few assumptions:

- 1 MW continuous ( 24 MWh / day)
- no other sources of energy are available
- operation for 3 days with heavy cloud cover
- the storage system will be recharges in 3 days
- system installed in New England (my home)
- able to provide 1 MW continuous even in the winter months

Based on these assumptions I’ll walk you though the process. Please adjust the numbers to fit your particular situation.

**Step 1: Estimate solar resources**

We start by looking at the solar resources available at the location of interest. Based on this map the solar insolation for my home on the Southeastern Connecticut is about 2.5 kWh/m^2/day.

**Step 2: Estimate size of array**

This simple equation is all we need:

**Energy =**

**AILS**

Where:

- (Greek letter eta) = Efficiency of the solar panel. Modern panels are 10 to 15 % efficient.
- A = Active area of the array in square meters.
- I = Insolation a.k.a solar irradiance in Watt-hours per square meter per day. This was taken from the NREL map in Step 1.
- L = System losses including heating of solar panel, dust, snow, atmospheric haze, battery charger, and inverter. A conservative coefficient is 0.75.
- S = Storage losses. Storage is not 100% efficient. Even batteries lose energy when they charge and discharge. A conservative coefficient is 0.75.

For this example we will estimate the daily energy produced as:

**Step 3: Adjust for storage and maintenance**

Solar is like a factory with a single production line. Storage is like a warehouse. On cloudy days the energy is pulled from the warehouse. The panels must be adjusted in size to power the load and restock the warehouse.

The opening assumptions were chosen to make this size adjustment easy. Let’s assume we have a completely empty storage. Every day we must power the load and fill 1/3 of the warehouse. Consequently, the size of the array must be doubled.

Again, this is the size of the active solar array. The actual amount of land would be perhaps doubled to accommodate the solar panel tilt and maintenance. Know that the power plant requires a field nearly 0.5 km^2 (about 120 acres).

**Step 4: Calculate number of panels:**

Earlier we assumed a panel with an efficiency of 15%. This is a good number for modern panels. A representative panel may be found here: https://www.gogreensolar.com/pro... This panel has a nameplate capacity of 250W, an efficiency of 15%, and an active surface area of 1.5 .

For reference, this solar farm will have a nameplate capacity of 38 MW.

Folks, I believe we need to step back and look at solar realistically. Solar is a very diffuse energy source. That 1 MW continuous solar power plant requires an extraordinary number of panels and nearly a 0.5 upon which to mount them. Land is expensive. Also, we need to add storage which will add considerable cost.

We must keep things in context. The diesel generator is the competition for plants of this size. A machine such as this could be installed in my basement along with the fuel to run it for days.

**Where do we go from here:**

We could drop the storage requirement and add the diesel generator as the solar backup. The cost and size of the array would drop considerably. IMO this is a very bad decision as the resulting system is high cost and emits considerable .

A much better solution is to combine nuclear, wind, solar, and a small amount of storage. Low cost reliable energy can be obtained by selecting the optimal proportion of each technology on a region by region basis. Add in hydo and geothermal if it is prudent to do so.

I encourage you to have this conversation with a colleague.

FUN FACT: There are 746 Watts per horsepower. This 1 MW solar power plant has a capacity of about 1300 hp.

Source and credits: quoran

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