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I reside in Kathmandu where the current power supply is so weak one has to face between 8 to 18 h power cut per day. In order to work, I need to have continuous supply of energy for my laptop and a bulb a night.

I have looked into this Q&A already and found some answers, some of which are too broad or assume I know something about electronics. In spite of my interest, I have always been very bad with electronics: I need please a case-specific and for-dummy answer.

I own a Dell Vostro 1510 laptop, which will run all day long (let's assume 24 hours as I need to download data from a slow Internet connection), and I will use an energy saving bulb to light up my desk at night.

From what I understand, I need a solar panel, connecting into a power controller, connecting into a battery, connecting into an inverter.

How do I find out what solar panel I need? (A local store which I don't really trust, proposed I buy their most expensive model, a 60W panel) I am concerned about the battery too, in term of efficiency and environmental impact. What are my options?

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The suggested 60 Watt panel will not be too small for you.
You will also need battery storage.

If at all possible run the laptop or laptop charger directly off the battery. Use of an inverter adds extra inefficincies.

For (almost) best possible efficiency you may be able to supply battery power to the Vostro's actual battery connections. You need somebody who know what they are doing to do this as laptop destruction is made easier by playing in this way.


To work out PV (photovoltaic) panel Wattage needed you need to know how much solar insolation is available.

The marvellous Gaisma site for Kathmandhu says that the average daily sunshine hours by month, January to December are

3.42 January
3.93 February ... 4.69
5.37
5.96
5.64
5.02
4.51
4.52
4.73
3.96
3.44 December.

I use the term SSH or Sunshine-Hours for what they call kWh/m2/day.
One standard sun = 1000 W/m^2 or 1 kW/m^2 so in 1 hour you get 1 kWh/m^2.
So in a day which has 4.5 hours of EQUIVALENT full sun you get 4.5 kWh/m^2/day.
So in September in Kathmandu you get 4.52 kWh/m^2/day./

Solar panels are rated in output in one standard sun at 25 degree C with light spectrum of AM1.5.

AM1.5 is the spectrum that you get from sunlight when it has passed through 1.5 x as much air mass as it would have if the sun shone directly down (eg equator at midday). You don't need to worry much about that except to note that at high lattitudes you'll get more losses at some spectral points (such as deep blue, near UV) due to longer average air paths. And at high altitudes (such as Kathmandu) you'll get less losses for eg blue.

SO to work out what energy per day you'll get from a panel in good condition and which is optimally pointed relative to the sun you take the max power wattage rating = Wmp and multiply it by the sunshine hours (or kWh/m^2) from gaisma for your location and month. If the panel gets hot (as happens) output will drop somewhat. (Maybe 10%).

A panel fixed in one position but optimised for lattitude and time of year will get perhaps 70% - 90% of absolute peak achievable output. In Summer you will probably get less of max possible as the sun swings through more than 180 degrees during the day (240 degrees in Kathmandu in June) and a fixed panel doesn't work well when the sun is behind it :-) - or even at +/- 90 degrees to the direction it's facing. Most of the energy occurs during the middle of the day so it's probably not worth chasing every last bit.

SO How much energy can you get? If you have N SSH/day and a panel is rate at Wmp = W Watts then energy per day max =

Watts x hours = W x N Watt hours.

If you want to equipment for H hours per day then the energy provided by the panel will be spread over H hours so the load you can operate for H hours from a panel of Wmp = W and SSh = N = energy from panel / H

= W x N / H Watts load

Finally, some of the energy from the panel will be lost when stored in a battery and some of the battery energy will be lost in converting it to drive the load.
If you specify the overall ratio.
k = (Energy into load)/(Energy from panel) = end to end energy efficiency (0 - 1)

Energy into load = Energy from panel - losses into battery - losses out of battery.

then you get Load wattage able to be operated
for H hours per day
from panel of Wattage Wmp
with N SSH = N hours/day of full sun equivalent

Load Watts = Wmp x SSH x k /H

or by rearranging

Required panel Watts = Wmp = Load_Watts x H /SSH /k

Example:

(1) Given 60 Wmp panel and 5 SSH and k = 0.66, what load can I run for 8 hours / day?
Load Watts = Wmp x SSH x k /H x = 60 x 5 x 0.66 /8 = 25 Watts continuous for 8 hours from a 60 Watt panel.

(2) Given a 20 Watt load and 10 hours/day operation, in mid winter with SSH = 3.4 hours/day, what size panel is needed.

Required panel Watts = Wmp = Load_Watts x H /SSH /k
= 20 W x 10h /3.4 / 0.66 = 88 Watt panel

These are best case figures. Non tracking of the sun adds 10% - 20%. Hot panels, dirty panels, worse than assumed matching of panel to battery or battery losses etc lead to larger panels or smaller loads or less hours of operation.


Calculating k

I used "k" as a measure of panel rated power to actual delivered power. (Actually energy_out/energy_in but power and energy are somewhat interchangeable here to make what is being said clearer. Power is NOT equal to energy and usually using one term in place of the other is both "plain wrong" and also misleading as a bonus.

Details follow but -

PV panel Wmp is rated at optimum V & I.. When loaded below this point I will increase slightly as V drops.
A nominal 12V panel is usually rate at Vmp = 18V !. For lead acid you need about 14.4V max on battery for equalisation. Most output is taken at 12 - 12.5 Volt.
Even if your panel was optimised to work at 14.4V then if battery output is at 12.5V average the efficiency = 12.5/14.4 =~ 87%. Add a diode with 0.6V drop (low) and you get 83%.

*Panel to battery *

The battery will not store all input energy - some will be lost in secondary cheimcal reactions, resistive losses etc. Losses will be a greater percentage of input energy at higher charge rates as voltage drop from R_internal will be a greater % or V_vbattery. and as capacity approaches 100% (as Vin and Vo/c diverge due to chemical and other processes. A lead acid

A Lithium Ferro Phosphate battery has a 99%+ current charge efficiency (no side reactions - what you input is what you get out - and this **improves with use.

A Lithium Ion cell will probably be similar but tbd.

a lead acid battery is good by many standards - over 90% at slow rates of charge over much of range.

NimH is worse than lead acid. Stated and achieved rates both vary widely but as good as 85% may be achieved and much worse can be managed under heavy charging and worse again close to full capacity.

NiCd may be similar to NimH.

Overall panel to battery with voltage mismatch (say 80%-85%) and battery acceptance losses (say 85% overall for lead acid leads to say 0.85 x 0.85 = 0.66 - 0.75 range.

Battery in to battery out.

A say 12V lead acid battery will charge at from 12V or less (0 capacity) to 13.8V terminal voltage (float) or 14.4V on boost for equalisation. Say mean charge voltage = 13V. Vout to load depends on C rate but is say 12.5 to 12V. Not much capacity below 12V and best unused if longer cycle life wanted. Say Vout = 12.25V average. So Energy out/ Energy stored = ~~ 12.25/13 = 0.94. Probably 90%-95% in practice.

Battery to load:

Energy actually delivered to load wrt energy out of battery depends greatly on what is in the path. Run a light bulb and what you see is what you get (literally) + heat. Run an LED or a number of them and it deep-ends muchly on your driver. A good buck converter may give 99-95% efficiency input to load. A not unusual one may give 80%-90% and less is known. A boost converter will usually be slightly less efficient than a buck converter with equal effort but say 85% mean with 80%-90% reasonably typical. With great care and choice of LEDs and battery technology you can get 90-95% for most of LED drive and very close to 100% across part of the range. Fine details of that may be available from me as a paid consultancy :-) - or buy one of my lights in due course :-).

If you are running eg a Dell Vostro [tm]
- from a 12V battery

  • via a 12VDC to 110 VAC or 230 VAC inverter and then using

  • a Dell switching power supply to provide typically 110/230 VAC to 19VDC to the Dell

  • which then charges the internal battery via a buck converter

  • and then uses a switching regulator to provide internal voltages.

then you shouldn't be !!! :-)

If you figures even say 85% D C-AC, 85% AC-DC, 85% DC19V-> 85% say LiIon charge then you get 0.85 x 0.85 x 0.85 = 61% efficiency. The internal battery to internal voltages follows this.

Far far better is to store the panel energy at a voltage that falls in the Dell battery voltage range and apply this voltage at the battery terminals (NOT the charge input terminals) of thr Dell. You save about 40% energy losses or get about 100/60 = 1.66x as long run time. Thats as effective as getting a 100W panel for the price of a 60W panel AQND not having to pay for any extra batt5ery capacity.

Obviously you want the supply you feed to the Dell internals to be as stable as the battery would have been (preferably better in some cases). If you allow eg the Dell Vbattery input to rise to the PV panel full sun unloaded output of 20V+ then you may be needing a new PC. Simple means can protect against this.

Overall:

If you accept the above figures then overall you get

PV Wmp x 0.7 Panel out to battery out x 0.7 say if you go DC-AC-DC as you shouldn't.

So k = say 0.5 with a mains inverter and say 0.7 if you use battery DC direct into laptop battery terminals. The 2nd is possibly a bit generous - say use 0.66 for starters.

In summary - with a mains inverter you need twice as much PV Watts as you may expect from load calculations. With direct feed to laptop battery you need 50% more PV watts than you may expect.

PV to LiIon direct:

With due care you may be able to feed the PV panel voltage directly to a suitable LiIon charger that charges the laptop battery directly. End results could be reasonably good.

Laptop internal charger use? : Your laptop MAY have such charger built in. Or not. If so this could be a very efficient way to operate.

Some modern laptops with a 19V charger spec will not run at much under 18V and some demand nearly the full rated current to operate. eg a 19V 4A power pack may not work if under 3A is available.

However - others may work from 19V down to near 12V and MAY be tolerant if current is less than may be desired. This probably works with laptops with 3 cell batteries of about 13.8V full voltage (14.5V up needed) or 3ScP arrangements (3/6/9) or 2 cell laptops (rare) with Vbattery = 8.4V max or 2SxP arrangements (2 4 6 8 ...). 4 cells in series (Vmax = 16.8V) and 4SxP arrangements 4 8 12) probably want 18V or more to charge.

I have a HP netbook that originally has a 4 cell 4S battery. I bought an "extra capacity" battery and was surprised to see it had 6 cells in a 3S2P arrangement so that terminal voltage was lower. The internal charger and power supply was equally happy with either arrangement. Whether a D ell will allow 4S or 3S arrangements and lower Vin is TBD.

I am told of laptops that will charge from 12V. Many won't. Such laptops will have an internal buck-boost chargers and so allow 3SxP or even 4SxP batteries with 12 Volts in. It's good luck if you have one.


Rushing - hopefully will come back and tide up later.

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  • \$\begingroup\$ Thanks for this amazing answer Russell. I need to process all this information, to which I am not used. I may come back for further details. Meanwhile, is there some simple literature on the subject you can recommend? Thanks again. \$\endgroup\$ – Benjamin Sep 13 '12 at 13:40
  • \$\begingroup\$ OK here I come with my questions: 1. how does one figure out k? 2. I don't understand then how a 60 Wmp solar panel is supposed to light 8 bulbs (according to vendor) nor how a laptop (app.70 W) and one bulb could run on a 60 Wmp panel if it produces merely 25 W / hour? Or is there something I am missing out? \$\endgroup\$ – Benjamin Sep 13 '12 at 15:35
  • \$\begingroup\$ when I plan my power organising module, should I start from battery up to solar panel or the other way around? One thing I am not sure of (and which I should probably ask in another question), is what battery to choose? Especially, I am not sure about the Amperes I should pick up. \$\endgroup\$ – Benjamin Oct 19 '12 at 16:22

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