Load Analysis for PV Panel

[Prototype1]

Can you please help and review this? I am the only EE at my job and not sure If I am doing it right. This is my first time sizing a PV panels.

Load Analysis for PV Panels:

Load analysis for the public latrine in the park, calculate the total electrical load of the building. Calculate the load of each device:

• GFCI Two duplex receptacles (men's and women's restrooms): 2 x 2 x 120V x 20A = 960W (assuming 20A per receptacle)
• LED lights: assume 10-15 LED lights, each consuming 10W = 100-150W (let's use 125W)
• Blow hair dryers: assume 2 hair dryers, each consuming 1500W = 3000W (let's use 3000W)
• Exhaust fans (2 x 50 CFM): assume 2 fans, each consuming 50W = 100W (let's use 100W)
• Small boiler: assume a small boiler consuming 2000W = 2000W (let's use 2000W)

1. Calculate the total load:
   • Total load = 960W (receptacles) + 125W (LED lights) + 3000W (hair dryers) + 100W (exhaust fans) + 2000W (boiler) = 6185W
2. Determine the load factor:
   • Load factor is the ratio of the average load to the peak load. For a public latrine, a load factor of 0.5-0.7 is reasonable. Let's use 0.6.
   • Adjusted load = Total load x Load factor = 6185W x 0.6 = 3711W
3. Determine the required PV array size:
   • To determine the required PV array size, we'll need to calculate the total energy consumption per day.
   • Assume the public latrine operates for 12 hours a day (8am-8pm).
   • Total energy consumption per day = Adjusted load x Operating hours = 3711W x 12h = 44532 Wh or 44.53 kWh
4. Determine the PV array size:
   • To calculate the required PV array size, we'll need to know the peak sun hours (PSH) available at the site. Assume an average PSH of 5 hours per day.
   • Required PV array size = Total energy consumption per day / PSH = 44.53 kWh / 5h = 8.91 kW
5. Select the PV panel size:
   • Let's assume we'll use a standard PV panel size of 300W.
   • Number of PV panels required = Required PV array size / PV panel size = 8.91 kW / 0.3 kW = 29.7 panels
   • To account for efficiency losses and other factors, let's round up to the nearest whole number. We'll need approximately 30 PV panels.

PV Panel Configuration:

• 30 PV panels x 300W each = 9 kW total capacity
• Array configuration: 5 rows x 6 columns (30 panels total)
• Inverter size: 9 kW or larger to accommodate the total PV array capacity

 

[jaggedben]

Is this for a grid tied system?
Is the goal to offset the energy on an annual basis?

 

[solarken]

Can you please help and review this? I am the only EE at my job and not sure If I am doing it right. This is my first time sizing a PV panels.

Load Analysis for PV Panels:

Load analysis for the public latrine in the park, calculate the total electrical load of the building. Calculate the load of each device:

• GFCI Two duplex receptacles (men's and women's restrooms): 2 x 2 x 120V x 20A = 960W (assuming 20A per receptacle)
• LED lights: assume 10-15 LED lights, each consuming 10W = 100-150W (let's use 125W)
• Blow hair dryers: assume 2 hair dryers, each consuming 1500W = 3000W (let's use 3000W)
• Exhaust fans (2 x 50 CFM): assume 2 fans, each consuming 50W = 100W (let's use 100W)
• Small boiler: assume a small boiler consuming 2000W = 2000W (let's use 2000W)

1. Calculate the total load:
   • Total load = 960W (receptacles) + 125W (LED lights) + 3000W (hair dryers) + 100W (exhaust fans) + 2000W (boiler) = 6185W
2. Determine the load factor:
   • Load factor is the ratio of the average load to the peak load. For a public latrine, a load factor of 0.5-0.7 is reasonable. Let's use 0.6.
   • Adjusted load = Total load x Load factor = 6185W x 0.6 = 3711W
3. Determine the required PV array size:
   • To determine the required PV array size, we'll need to calculate the total energy consumption per day.
   • Assume the public latrine operates for 12 hours a day (8am-8pm).
   • Total energy consumption per day = Adjusted load x Operating hours = 3711W x 12h = 44532 Wh or 44.53 kWh
4. Determine the PV array size:
   • To calculate the required PV array size, we'll need to know the peak sun hours (PSH) available at the site. Assume an average PSH of 5 hours per day.
   • Required PV array size = Total energy consumption per day / PSH = 44.53 kWh / 5h = 8.91 kW
5. Select the PV panel size:
   • Let's assume we'll use a standard PV panel size of 300W.
   • Number of PV panels required = Required PV array size / PV panel size = 8.91 kW / 0.3 kW = 29.7 panels
   • To account for efficiency losses and other factors, let's round up to the nearest whole number. We'll need approximately 30 PV panels.

PV Panel Configuration:

• 30 PV panels x 300W each = 9 kW total capacity
• Array configuration: 5 rows x 6 columns (30 panels total)
• Inverter size: 9 kW or larger to accommodate the total PV array capacity

This is not really how it is typically done. Ideally energy history is available, like from an electric bill. If that data is not available or if it is not yet built, then an estimate would be made like you have done. However, I suspect your numbers are high for a public latrine. For many households in the US, 30 modules will offset all their energy use.
If the goal is to just offset the annual energy use within the latrine with a grid-tied solar PV system, a modeling/simulation tool is used after looking at the available roof space or ground space for modules, to size a system based on the desired energy production and with the array orientation and position as it would be on the site, and with insolation data available at that location. If you have not used modeling tools before, you can try PVWatts, or Solaredge Designer. Both are free. And you really won't find 300W modules anymore, more like 400W or higher. And you don't match the AC capacity of the inverter to be the same as the array DC capacity, you can over size the array by 20% to 50% or so, depending on orientation and any shading.

If the goal is to design an off-grid PV system to operate those loads, that is a completely different animal.

 

[Prototype1]

And you don't match the AC capacity of the inverter to be the same as the array DC capacity, you can over size the array by 20% to 50% or so, depending on orientation and any shading.

So, the ac capacity of the inverter has to be same size as DC capacity?

Ideally energy history is available, like from an electric bill. If that data is not available or if it is not yet built, then an estimate would be made like you have done.

The thing is this building is still not build. I don't have a energy history. My boss want do a load analysis for this future build.

If the goal is to design an off-grid PV system to operate those loads, that is a completely different animal.

Luckily sir, it not off-grid, It will be hook up to utility power. I am still getting battery storage.

I suspect your numbers are high for a public latrine

Yes, it is too high, I change my calculations. If you notice in my new calculation below, I went extreme, and assume that equipment will be use the entire time as long as the latherin is open, like from 8 hrs a day. We know that it is not the case.

So, I refined my calculations after looking at google and other sites and took in account of solar sun hours average for summer and winter months through the year. How is it looking now? I got 400W panels that are 5ft long and 3ft wide.

Load Analysis for PV Panels:

To perform a load analysis for the public latrine in the park, calculate the total electrical load of the building. Calculate the load of each device:

Device	Quantity	Power (W)	Usage (Hours)	Daily Energy (Wh)
GFCI Receptacles	2	120	8	960
LED Lights (Interior)	10	10	8	800
LED Lights (Exterior)	5	15	10	750
Hand Dryers	2	1500	0.5	1500
Exhaust Fans	2	150	4	1200
Small Boiler	1	1500	2	3000
Other( Safety Margin)				1090
Total Daily				9300 Wh/day

960+800+750+1500+1200+3000 = 8210 Wh/day

Adding safety margin is 10 to 20% of the calculate load.

I chose 13% 8210 * 0.13 = 1066.3 Wh/day ~ 1090 Wh/day.

8210+1090 = 9300 Wh/day

Number of PV panels.

# of panels: Daily Energy Requirement (Wh) / (Panel Power (W) * Average Sun hours/day)

Daily Energy Requirement: 9300 Wh/day

Panel Power: 400 W (0.4 kW)

PVWatts​

Estimates the energy production and cost of energy of grid-connected photovoltaic (PV) energy systems throughout the world. It allows homeowners, small building owners, installers and manufacturers to easily develop estimates of the performance of potential PV installations

pvwatts.nrel.gov

Average Sun Hours/Day: 5 (annual average in Leavenworth)

# of Panels: 9300/400*5 = 9300/2000 = 4.65

Included safety factor of 1.5 to account for inefficiencies like inverter losses, shading and heat and other factors.

4.65 * 1.5= 6.975 ~ 7 panels

Adjustments for Winter (Lower Solar)

During winter months, the solar drops tot 3.1 sun hours/day.

Same formula: 9300/400*3.1 = 7.5

Adjust winter panels (safety margin) = 7.5 *1.2 ~ 9 panels

Annual average is 7 panels (400W) for 5 sun hours /day

Winter: 9 panel of (400 W) for 3 sun hours/day.

So, 9 panels throughout the year.

 

[pv_n00b]

You are sizing this like it is an off grid system. If this is a grid connected system you should be more concerned with annual energy usage and annual PV production. If you have TOU rate adjustments and a split between how much the utility pays for exported energy and how much the energy will cost from the utility then you need to do a more detailed hourly energy import/export analysis. You want to zero out the annual cost of energy, not the annual PV production and site energy usage. If you want to add a battery you need to know why you want to add a battery.

 

[ggunn]

So, the ac capacity of the inverter has to be same size as DC capacity?

Not at all. The DC:AC wattage ratio is typically 1.2 or more, and it could be significantly more depending on several factors, including array tilt, array azimuth, shading, and array location (mainly latitude). Most inverters have a published maximum DC:AC ratio, and it is usually 1.5.

 

[ggunn]

Not at all. The DC:AC wattage ratio is typically 1.2 or more...

This is because PV modules rarely if ever produce their full power rating, and that is mostly because ST conditions rarely if ever exist in the real world.

 

[PWDickerson]

I can't think of the last time I saw a receptacle in use in a public restroom.

 

[solarken]

So, the ac capacity of the inverter has to be same size as DC capacity?

No, your initial post sized a 9kW or larger inverter for a 9kW array. My reply stated an array can generally be 120% to 150% of the kW inverter capacity, depending on the orientation and shading.

Luckily sir, it not off-grid, It will be hook up to utility power. I am still getting battery storage.

Why are you adding battery storage if this is a grid-tied system? You really should define what your reason for battery storage is.

If your goal is just to offset the annual energy usage, and you have net metering in your location, just forget the batteries and calculate the annual kWh usage estimate, then run a simulation for a few PV array sizes in PVWatts or in SolarEdge designer, and see how much array you will need to install at that location. You seem to keep going back to Sun-hrs/day for different months instead of using the tools as they are meant to be used.

You don't design for worst case, winter hrs, if your goal is to offset annual energy use with a grid-tied system. You design for the total annual energy needed to be equal to the estimated annual kWh usage.

 

[Prototype1]

I can't think of the last time I saw a receptacle in use in a public restroom.

We are not sure, if we will add a receptacle but I just added to my load analysis if my boss decided to or the PM.

 

[Prototype1]

Why are you adding battery storage if this is a grid-tied system? You really should define what your reason for battery storage is.

Thank you for mentioning, I was following the calculation from different websites and did size the batteries too. Now, I look at this post it make sense as we have utility side power. I need to use my common sense