## Introduction: Hybrid Design: Photovoltaic/Eolic for Rural Communities

This is an adaptation to a formal Paper published by the Universidad del Nordeste, Corrientes, Argentina.

It starts with this:**Diseño de un sistema híbrido solar-eólico para proveer de energía a una comunidad.
(Design of an hybrid solar-eolic system to provide energy to a community)
Sogari, Noemí
Depto. de Física - Facultad de Cs. Exactas y Naturales y Agrimensura - UNNE.
Av. Libertad 5450 - (3400) Corrientes - Argentina.
Tel./Fax: +54 (03783) 473931 int. 134
E-mail: nsogari@exa.unne.edu.ar**

She is the true autor.

So, I am not the autor of the paper, but I have modified his procedure a little bit, so I could use this kind of method in my course.

It uses 2 sources, wind and solar, so we are going to be working with both at the end.

The method itself is called, or to be true is like I like to callit:

**Energetic balance.**

Since you use a balance, in watts, to acchieve your goal: have energy stored to be used later. Some of the parts of this method are similar to the other I have described in my earlier Instructable (see here: https://www.instructables.com/id/Current-method-for-Photovoltaic-Calculations/)

We will size:

**1) The PV Generator (Photovoltaics)**

As I said before, this is hybrid, so we use both, solar AND wind. In Argentina we use this scale: cell → Module → PanelThe PANEL is an array of several modules, plugged in series or parallels to achieve the right configuration. In this example we will have everything in 12 V systems, so 32-cells modules are going to be OK.

The MODULE is composed of several cells in series and parallel to achieve the right current and voltage.The CELL is the smaller component. Is made of silicon, with the electric contacts in front and rear to pick up some electrons, running around the silicon when some photons hits them.

**2) The charge regulator**

This is a critic component, since it keeps your batteries fresh and healthier. A bad chosed charge regulator will end up in an early death of the batteries. Here we could chose two regulators, one for each area, or one hybrid regulator.

**3) The battery bank**

Deep cycle solar batteries aren't cheap, nor light weight. So choosing the right size of that battery bank, suitable for your budget and energy needs is going to be criticall.

**4) The wind generator**

The machine here will be fixed, but with some searches you could adapt this to our own machine.

Well, lets start, and our first step is always the same:

**the basic knowledge**

## Step 1: Basic Knowledge (added to the One in Our First Instructable)

OK, we have already seen some basic knowledge on my previus instructable, but now we add a lot more of variables, since wind is more unstable than solar.

We are going to use **WIND SPEED**. And you will see later, that wind speed goes cubed on the power over the shaft, so is the variable we whant to increase.**Roughtness of the ground**, wich means how much wind is drawn by the terrain, or the trees. General rule: sea or lake = the best; city of forest = bad.

With wind power location is everything. So we will take a look at some HomePower issues, that you could download for free from it's web site, and read on, to catch up with some not-that-basic knowledge.

In this methos we use watts in everything. So get familiar with this. When you size your battery bank, you will have to choose 12 Vcc, 24 Vcc or 48 Vcc. Here we will get different currents that will travel across the conductors (cable), and since A x V = W; more power means more amps at the same voltage.

We will use here the "Peak Sun Hours" too, so lets see it again:

W = A x V

MJ / 3,6 = Peak Hours (HSP) in further formulas referred as "H"

The full formula looks like this: 1 HSP = [(1000 W x 1 h ) / m2] x [3600 s / 1 h] x [1 J/s / 1 W] = 3,6 MJ/m2

I assume in all the process that you have already take a look at my previus Instructable, so I don't have to put everything back again here.

Let's go to on the **Energy Needs **of this case.

## Step 2: Energy Needs

Ok, let's move on.

I am makin for most of this a copy of the step in the previus instructable, but adding some considerations about this new method.

I have to say, this method is really good for small scale installations, but if you have to make a whole house installation... well you better talk to a local spetialist.

To make this easier, we will use a table.You will have to note like this:

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Room or ambient | Appliance | Power [W] | Quantity | Hours of Use [h] | [W . h / day]

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Bedroom | Light | 15 W | 2 | 1 h | 30 Wh/day

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Bedroom | Light | 30 W | 1 | 3 h | 90 Wh/day

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Kitchen | Light | 15 W | 1 | 3 h | 45 Wh/day

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Kitchen | Light | 15 W | 1 | 1 h | 15 Wh/day

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Kitchen | Radio | 10 W | 1 | 7 h | 70 Wh/day

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Workshop | Light | 30 W | 2 | 3 h | 180 Wh/day

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| MSD | 160 W | TOTAL | 430 Wh/day

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So we have here some subtotals, on how much energy in watts are we going to need on each room for light.

Notice that I am using small appliances and a few hours for each, just to make things easy to follow. Once you have some practice with this, you could add as many appliances as you want to.

So here we are going to take this numbers and make 2 more calculations on this subject.

1) The “maximum simultaneous demand” or MSD (in spanish is DPMS).

2) The total daily energy demand, which we will cal ED.

For the MSD you have to take the “Power [W]” column and multiply this by the “Quantity” column, and make a sum on each result.

So here you have the maximum energy you will have to manage on you cables, and with your regulator.

Since we use Amps on each case to run some calculations, you will have to divide this power (54 W) by the system voltage (12 V).

A = W/V

A = 54 W / 12 V = > A = 4,5 A

The total daily energy demand is the sum of each Wh/day on the last column

OK, in this method, we use Watt.

So, you have here some interesting values.

You need 430Wh each day, tu use your appliances. We are considering here an inverter, to use 220Vac, and his efficiency has to be teked into account. Take a look at the specifications of your equipement, we are using a 93% of efficiency (0,93).

Also, you know that your MSD is 160 W, so an inverter of 200 W could be used. Inverters are becoming cheapper and cheapper, and having more control over the load or the battery usage, so as I always put here, choose a good one.

In recent years, instant inverters has hit the market. With those inverters, you have only a small power use during most of the time, with a lower-energy phase on the inverter, and when you start, lets say, the kitchen light, the inverter detects this energy demand, wich is a "wake up call" for it, and starts a power phase, with full 200 W capabilitty. I hope this explanation was clear enought.

We take our ENERGY NEED (430Wh/d) and apply the lost by the inverter: 430Wh/d / 0,93.

This is adding to the energy need (430Wh/d) the lost that occurs in the inverter, so now we have: 462,35Wh/d

It means, we are going to need a generator of 462,35 Wh/d to use our real 430 Wh/d.

See where we are going? We add all the losses.

And also, considering a loss in all the installation, with cable runs, and connections. We take here an acceptable 3%. This is 462,35Wh/d x 1,03. With this, we add the lost too, and we have a new energy need: 476,71Wh/d. **We are calling this ENERGY NEED [EN]**

This is our first value to calculate the ammount of energy we need to capture from the sun.

To put in pla our generator, we will need again to go for the Resource. Here, to make it quick, I recommend to read the other post:

https://www.instructables.com/id/Current-method-for-Photovoltaic-Calculations/step3/Resource/https://www.instructables.com/id/Current-method-for-Photovoltaic-Calculations/step3/Resource/

And when you are done with it, let's go to **The Generator.**

## Step 3: The Generator and Battery Bank

Well, you have to be confident in your use of Watts, since we use all in watts here.

To determinate the number of modules we are going to need for our calculation, we run this numbers:

Nm = (Sf x EN) / (H x PFP)

Where:

Nm: Is the NUMBER OF MODULES, we are going to need.

Sf: Security factor, wich considers all losses by dust, agging, etc. = 1,2

EN: Is our COMPLETE energy need, with the losses in there = 476 W h / día

H: Those are the equivalent hours, or Sun Peak Hours.

PFP: Is the peak power of our choosed module. We choose here 60 Wp. This value is presented by the mannufacturer.

Nm = (1,2 x 476Wh/d) / (3,3 h/d x 60 Wp) = 2,88

And here we take 3 modules.

Notice we are not talking about Volts or Amps, just the watts, so is up to you choose your system voltage, and pair everything to that voltage: modules, batteries, regulators, inverters.

Pay spetial attention to the wind generator, since they sometimes work at higher voltages.

Ok, let's move on to **Battery bank.**

Well here we use a very similar formula to get our Number of Batteries [NB]:

NB = (EN x TA) / (ALM x CM)

Where:

NB: Is the NUMBER OF BATTERIES, we are going to need.

EN: Is our COMPLETE energy need, with the losses in there = 476 W h / día

TA: Is TIME OF AUTONOMY, in days.

ALM: Is the BATTERY CAPACITY IN WATTS HOUR. How do you get this? Well 12V and 165 Ah battery is a 1980 Wh battery.

CM: is like our previus instructable DoD, ot DEPTH OF DISCHARGE. We keep 0,7.

NB = (476 [Wh / día] x 3 [días]) / (1980 [Wh] x 0,7) = 1

This is 1 battery.

If we use 5 days of autonomy, and a 250 Ah and 6V battery (1500 Wh)

NB = 476 [Wh / día] x 5 [días] = 2,26

(1500 [Wh] x 0,7)

## Step 4: Wind Power

To calculate the power and energy incorporated by the wind power generator, and known the speed of the wind on a particular place (at rotor hight) we calculate the average power at certain speed and the daily energy.

The generator choosen for this is KURIANT, with this specs:

• Nominal Power = 15 Kw

• Cut speed = 25 m/s

• Power controll : Stall

• Axis hight : 18 m

• Speed at site: V = 5,05 m/s

Taking in count the hight at the rotor axis, and knowing that the annemometer is located at 10m and estimating a rough terraing between class 1 and 2, we assignated a value of á= 0,15 A, knowing this, we calculate the average speed at rotor hight using this ecuation:

V18 = V10 (L18 / L10 ) á

V18 : wind speed at 18 m

V10 : wind speed at 10 m

L18 : hight of the generator, in this case 18 m

L10 : hight at the annemometer postion, 10 m

αα : terrain roughness coeficient: á = 0,15, due to the bushes around.

V18 = 5,51 m/s

Now we use the power curve of the generator for this speed (Pm):

Pm = 1,67 Kw

Now, the average energy given this machine is = 40 Kwh/día

(NOTE: I am not an expert, and this is just a translation of the original article.

could you please put the original article i speak spanish and i think that im gonna understand better in spanish