Alternative energy Question

Kippure

Work out an alternative energy system to run a refrigerator using solar photovoltaic panels and batteries. The average power consumption of the refrigerator is 150 watt, its compressor start up current is 5 amps. Given that average weekly hours of sunshine is 40 hours.

Calculate the size of the PV panels, batteries, voltage regulator, start current and voltage through the system.

So far ive done this

P=I * V
V = P/I
V =150/5
V= 30


then 150 watts by 24 hours a day = 3600 watts

then 40 hours rated sunshine divided 3600 watts = 90 watts per hour

now i have the watts per hour and the voltage.

so the running current is I = P/V

I = 90/30
I = 3amps.


From this a i understand i have the watts per hour 90w, current per hour 3amps and the voltage 30 volts


so

i need a two 17 volt 100 watt PV panels.

this works out like this

2 by 100 watt panels = 200 watts

17 by 2 = 34 volts ( batteries in series you add the voltage.)

therefore

P= I by V
I = 200/34
I = 5.8 amps

so im covered. Ive done this so far. Am I correct

So 200 watt PV panels, 34 volts and 5.8 amps.

Now i need a voltage regulator.

The nearest one i can buy is a 26 - 40 volt 10 amp voltage regulator.

backboiler

Few things awry here:

Surely your 'fridge needs a mains supply (~230 V AC @ 50 Hz)? You need to be able to supply a peak of 5 A but much less than this continuously. The compressor motor's coil is effectively a short circuit before it starts spinning so that's mainly why there's a start-up surge. That 5 A is only needed for a fraction of a second though.

Watts is a unit of power (the amount of energy per unit time, whether it be instantaneous or average) so quoting watts per hour makes no sense.
If you keep the units alongside your calculations you can help avoid these mistakes.

You mention sunshine per week but calculate energy needs for the appliance for just one day.

In summary, you're going about this all wrong. You need to have an energy reservoir (battery) that gets charged up by the PV panel. The battery then feeds an inverter or something to get the mains supply that powers the 'fridge. You'll need some kind of power management circuitry to handle all that.
It'll be less than 100% efficient - that you can be sure of - so if your 'fridge needs average 150 W the panels will need to supply more than that (e.g. if you get 80% efficiency then your panels need to supply 187.5 W. 50% would mean 300 W panel capacity).
17 V should be plenty to charge a 12-15 V battery so don't worry about that bit.

backboiler

backboiler wrote: »

It'll be less than 100% efficient - that you can be sure of - so if your 'fridge needs average 150 W the panels will need to supply more than that (e.g. if you get 80% efficiency then your panels need to supply 187.5 W. 50% would mean 300 W panel capacity).
17 V should be plenty to charge a 12-15 V battery so don't worry about that bit.

Actually that's wrong. If your 40 hours a week of sun is correct it means that you're only getting power from the PVs for a quarter of the time (168 hours in a week) so the PV panel capacities above will need to be multiplied by 4. Maybe I've a mistake there because that seems very high just to power a 'fridge. The 150 W you quoted probably only applies when the compressor is running. Since it switches on and off depending on your room temperature and temperature setting it'll probably actually use much less than 150 W on average.

Kippure

The set up will consist of the PV panels then the voltage regulator, then the batteries then the invertor then the frigde.

The average power consumption is only 150 watts. The start up current is 5 amps instantanous, so .83 amps it will pull during normal operation.

From what i have im just trying to work out a simply system.

backboiler

Simple is always good but unfortunately motors are not simple to control.
Say you go for a 200 VA inverter to power the 150 W load. Sounds safe enough. The thing is that the inverter will probably have some kind of overload protection that will kick in when the current draw gets above the 200 VA capacity. Probably around 1 amp. You've already said that the motor pulls 5 A on start so the inverter won't get over this bump and the motor won't start.
Then you say well I'll just get an inverter that can handle 5 A so you go and get a 1.2 kVA inverter and then you find that its eficiency is only 20% at the low levels so it's pulling 600 W from your battery (that's 50 A at 12 V so big fat wires needed), meaning that you need more than 2.4 kW from your PVs for 40 hours a week, even if you have no losses in your charging circuit and battery. I'm making up figures but you get the idea hopefully

All that said, unless it's a very big fridge I really doubt the average draw will be 150 W, that more likely only applies when the motor is running. The motor is switched on and off by a thermostat when the temperature in the fridge rises.

Kippure

backboiler wrote: »

Simple is always good but unfortunately motors are not simple to control.
Say you go for a 200 VA inverter to power the 150 W load. Sounds safe enough. The thing is that the inverter will probably have some kind of overload protection that will kick in when the current draw gets above the 200 VA capacity. Probably around 1 amp. You've already said that the motor pulls 5 A on start so the inverter won't get over this bump and the motor won't start.
Then you say well I'll just get an inverter that can handle 5 A so you go and get a 1.2 kVA inverter and then you find that its eficiency is only 20% at the low levels so it's pulling 600 W from your battery (that's 50 A at 12 V so big fat wires needed), meaning that you need more than 2.4 kW from your PVs for 40 hours a week, even if you have no losses in your charging circuit and battery. I'm making up figures but you get the idea hopefully

All that said, unless it's a very big fridge I really doubt the average draw will be 150 W, that more likely only applies when the motor is running. The motor is switched on and off by a thermostat when the temperature in the fridge rises.

Cheers mate, when i finally work it out, ill have something to compare it too. You will see what i,ve missed out and vice a versa

Heaven Net

How Much Energy is Returned for the Energy Invested (EROEI)? Have all energy costs been taken into account? This is where too many alternative energy sources fall flat after the simplest examination.
Commercial hydrogen offers one clear example of how it takes more energy to produce the fuel than can be obtained from burning it. The current feedstock from which hydrogen is produced is natural gas. The natural gas is then treated with steam. Steam is water that is boiled using more natural gas, oil, or coal, either in the form of direct fuel or to generate electricity which is used to boil the water. Common sense dictates that this cannot be a solution because it still relies on fossil fuels.
Converting water to hydrogen is done through electrolysis. Scientist David Pimentel has established that it takes 1.3 billion kWh (Kilowatt hours) of electricity to produce the equivalent of 1 billion kWh of hydrogen. (BioScience, Vol. 44, No. 8, September 1994.)
Even a small positive EROEI, if obtainable, is not a solution because fossil fuels on the whole return many times the energy invested, not just a fraction. That's why we use them.

Heaven Net

As oil and gas prices continue to rise, the sun has apparently set on the development of solar power and other forms of alternative energy, despite official claims that the United States is committed to making them a success. The explosion in oil and gas prices has been attributed to numerous causes, but little attention has been given to the lackadaisical effort to develop alternative fuel sources and the continuous quest by the oil industry to discover more oil. Big oil has both money and power, and it shouldn't be any surprise how much can be accomplished, or prevented, with such a potent combination.
The application of solar power is not a new idea. The ancient Greeks and Romans developed mirrors that would direct the sun's rays and cause a target to burst into flames within seconds. Nearly two centuries ago in 1839, Edmund Becqurel, a French experimental physicist, discovered that sunlight could produce electricity--almost fifty years before the first successful internal combustion engine was built.
During the late 1800s, harnessing the sun's rays to produce hot water was a booming business in the United States. Although the Industrial Revolution was in high gear and remarkable discoveries and inventions abounded, it took over 100 years for the first photovoltaic cell (a cell capable of producing wattage when exposed to radiant energy) to be developed by Bell Laboratories in 1954. Considering that photovoltaic cells have been the exclusive power source for satellites since the 1960s, and how rapidly television evolved during an era known as the Atomic Age, it is a wonder that solar technology hasn't advanced further.
The utilization of solar energy was briefly resurrected during the 1970s when the United States appeared to be committed to pursuing a technology that had the potential to reduce our dependency on fossil fuels. In April 1977, in the midst of an "energy crisis," President Jimmy Carter began a bold initiative to develop solar energy and other alternative fuels when he unveiled his National Energy Plan, which included setting an example by placing solar panels on the White House. Carter announced a "national goal of achieving 20 percent of the nation's energy from the sun and other renewable resources by the year 2000," and he introduced legislation that would provide homeowners with tax breaks for investing in this promising technology. In 1979, Congress followed Carter's lead and approved a $20 billion development fund for synthetic and alternative fuels. It appeared that the alternative energy industry was finally getting the financial backing it needed to have a profound impact on the nation's energy needs.
During the period in which financial support for solar energy was growing and a "windfall tax" on the profits of the oil industry was imposed, the proponents of big oil were gathering their own resources on Capitol Hill. Political action committees (PACs) that were affiliated with oil and gas interests began to sprout and, from 1977 to 1979, they contributed over $2.6 million to House and Senate candidates. A report by Alan Berlow and Laura Weiss in Congressional Quarterly concluded that most of the money went to candidates "with strong pro-industry voting." Support for alternative energy took a downward spiral when Ronald Reagan (a former spokesperson for General Electric) was elected U.S. president and became a staunch ally of corporate America.

allternative energy

Kippure

Disregard the calclations above.

Work out an alternative energy system to run a refrigerator using solar photovoltaic panels and batteries. The average power consumption of the refrigerator is 150 watt, its compressor start up current is 5 amps. Given that average weekly hours of sunshine is 40 hours.

Calculate the size of the PV panels, batteries, voltage regulator, start current and voltage through the system.





Heres this question again.

A fridge 150 watts. runs 24 hours a day . That equals 3.6 kwh per day.

With an averge of 40 hours of weekly sunshine, Thats 40 divided by the 7 days a week. Which gives me 5.7 hours of sunshine a day.

I need to size my photovoltaic panels.

So 3.6kw divided by 5.7 equals 631 watts . This is the power required by from my panels.

So I,ll choose A 700 to 800 watt PV panel.
So ill choose to run a 12 volt system as componets are cheap to buy.


From my PV panels they will deliver 700 -800 watts and 17 to 24 volts into my voltage regulator.

Note that 17 to 24 volts is the average output of a pv panel.

So i need to know the amps coming out of my VR to size the cables.


700watts power out of my panel

12 volts is the rating of my VR

I = p/v

I = 700/12

I = 58 amps so i need 16 square cable.

Now i need to size the battery bank.

I know how much power i require i need every day = 3.6kwh
I know the size of my VR = 12

So

I= p/v
I = 3.6/12
I = 300amphour battery system

So to get the voltage i need and the amp hour capacity.

Ill, need 10 12 volt 40 ah batteries in parallel

As you see i have gone over by a 100 amp hour to cover myself.

Now coming out of my batteries i need to know the current to size the cables to feed my inverter.

(I also choose a inverter rated at 1kw, that covers my 631 watts. And that it has a peak of 2kw.) Its the way you buy them.

I know the voltage going into the invertor 12dc volts and i want to change that to 230ac

I know from the start i have a peak current of 5 amps.

So I NEED TO FIND THE POWER PEAK LOAD .

V by I = P
230 by 5 = 1150 watts peak load watts


Therefore 1150 divided by 12volts = 95.8 amps

So i need 25 square cable to feed the inverter from the battery bank.

Ill do a drawing later to show you the whole system

backboiler

Remember your 25 sq calculation is based on a short-lived peak lasting a fraction of a second. Steady-state current draw is only 150 W.
The other thing is your voltage regulator, anything not passed through to the battery is absorbed by this regulator and turns into heat, needing a big wing of a heatsink to dissipate it. Something more sophisticated may be needed to manage swapping between charge and drain on the batteries.