Can a chainsaw be ever powered by a built-in solar-panel?

Miroslav Kolář, February 7, 2016

 

In the Jan. 13, 2016 issue of The Bird's Eye (www.thebirdseye.ca) there was an interesting article "The coming of the Solar Age" by Greg Ross. He expects that photovoltaic (PV) solar panels will in the future become not only much cheaper and ubiquitous, but also so efficient that "we each will have just one solar panel the size of a tablet to power our home, car, and all electrical devices we own," and that "a small solar panel built into your chainsaw and other tools will allow you to use the tool nonstop all day, every day, just like today's solar calculators."

 

PV panels will no doubt continue to become more efficient and affordable, and I can't wait till the solar age comes in earnest. However, I was sceptical about the expectation that a small solar panel will sometime be able to power a chainsaw. The limitation is not so much the efficiency of the PV panels as the fact that the solar radiation is not that much concentrated. One has to harvest solar radiation from an area proportional to the power requirements of whatever is to be powered. The power requirements of a simple calculator are almost zero. I have not found specifications for any more recent calculator, but already in 1979 there was a Sharp calculator powered by only 0.0002 W (Watts). Today's calculators most probably require even less than that. But let us stay with 0.0002 W.  An electric chainsaw needs at least 1,000 W to do any useful work. That is 5 million times more than a simple calculator. Therefore, to power a chainsaw  directly by sun, one has to collect solar radiation from an area 5 million times larger than the one needed for a solar calculator. I have a calculator with a solar panel with an area of about 3 cm². Multiplying this by 5 millions gives 1,500 m². My calculator's little solar panel may still have the efficiency of only 5%. The maximum possible theoretical efficiency of a solar panel is 95%. For such a panel 1,500 m² would decrease to 79 m². This rough estimate of the solar panel area needed to produce a power of 1kW cannot be considered an authoritative figure because my solar calculator's panel may be over-dimensioned.

 

It is not difficult to calculate a much more exact size of a solar panel needed for any power requirement from the data on the amount of sun radiation reaching the surface of the Earth that has been measured all over the Globe for decades. I have found, and will demonstrate in some detail below, that there will never be a chainsaw powered fully by a small built-in solar panel. On the other hand, slower light-weight vehicles powered directly (and only) by solar panels are possible, although with a surface area of quite a few square meters.

 

First a brief physics refresher: Watt is a unit of power, or rate at which energy is delivered (or work done). Thus Power = Energy/time, and one of the units of energy is Ws (watt-second) or Wh (watt-hour; 1 Wh = 3600 Ws; 1 kWh = 1,000 Wh). Another name of Ws is Joule (J). And 1 Ws = 1 J = 0.239005736 Cal (Calories, familiar from the packaged food nutrition facts tables). And for the units of  power, 1 kW = 1.35963 hp (metric or German horsepower, or PS DIN), and 1 kW = 1.34102 bhp (British/American or break horsepower).

 

The solar energy influx at the top of the Earth atmosphere is 1,370 W/m² (Watt per square meter – measured on a surface perpendicular to the Sun rays). Some of it is absorbed in the atmosphere, and so at the sea level one only gets about 1,000 W/m² (or 1kW/m²), and that is the maximum amount when the Sun is overhead at noon and there are no clouds. (This maximum solar input is assumed in the rating of PV panels. Thus if you buy a 100 W panel, you will get a 100 W output only when the panel is irradiated with1kW/m². From the panel rating and its surface area, one can thus calculate its efficiency.) The actual amount of available sun energy varies with time of the day, panel orientation, weather, and the season of the year. Because of this variability, it only makes sense to measure for a given place on the Globe, the total energy absorbed by a unit area over a longer period of time, given in units of kWh/m². Daily doses are usually averaged over each month or over the whole year from data collected over many decades.

 

Natural Resources Canada (http://pv.nrcan.gc.ca) has a map of this "mean daily global insolation" for different orientation of the absorbing surface, averaged annually or monthly. The largest insolation is for a panel that tracks the sun all day (2 axis sun-tracking), the smallest for a horizontally positioned panel. It is a colour coded map, where each colour corresponds to a range of values. NASA published exact values for various cities around the globe, but the ones I found are only for the horizontal position. NASA's values for Nanaimo and Vancouver (http://goo.gl/zn3UQV) are identical, and almost the same as those for Victoria. And they very well correspond to the upper bound of the NRCAN ranges for the horizontal panel position for the southern Vancouver Island. Thus we can also use the corresponding NRCAN upper bound values for the 2-axis tracking. The largest daily insolation occurs in July, the smallest in December.  In Table 1, the average daily insolation values (and the sunrise to sunset period) from the above sources are copied for these two months and for the whole year:

 

Table 1. Daily insolation for southern Vancouver Island (kWh/m²), and sunlight duration (hours)

Period:

All year average

July average

December average

Horizontal panel

3.31 *

5.75

0.86

2-axis sun-tracking panel

5.8

9.0

2.5

 

 

 

 

Sunrise to sunset (hours)

12

16

8

* The 'All year average' for the horizontal panel was obtained by the averaging the NASA values over all 12 months, assuming the average number of days in February to be 28.25.

 

In Canada, only the Prairies and Southern Ontario have higher annual insolation than V.I. The closer one gets to the equator, the larger the insolation. Deserts and oceans in the tropical region can get on average daily insolation of more than 7.5 kWh/m² on the horizontal surface.

 

The last thing we need is the maximum theoretical efficiency of solar panels. As has been mentioned above, this is 95% - the efficiency of the Carnot heat engine operating between the temperature of the surface of the Sun and that of the Earth. This means that at most only 95% of the energy arriving at a panel can be converted into useful output in the form of electricity, the remaining 5% just turns into heat.  Currently the most efficient PV panels produced have an efficiency of 44.7%, but they are too expensive. Most of the affordable panels in the marketplace have an efficiency of only 10-20%, with 20% slowly becoming the norm. All existing PV panels are of the p-n cell semiconductor type, which have without the light concentration an inherent maximum efficiency of 30% for the single p-n junction, or up to 68% for the multilayer structure of stacked p-n cells. With the concentration of light, by mirrors or lenses, from an area larger than the active surface, one can increase the efficiency at most to 86%. Thus with 44.7% we are already somewhat more than half way to the maximum 86% achievable by the current PV p-n technology. To get over 86% up to the maximum 95%, one would first need to come up with a different way to convert solar radiation into electricity than by p-n semiconductor cells. Currently only the visible part of the solar spectrum, which constitutes only 35% of the total spectrum, is used. If sometime in the more distant future people learn to also convert at least some of the invisible components (infrared, microwaves, X-rays, ...), then the amount of convertible energy per unit area would increase, and all the minimum solar-panel areas derived below would decrease by up to 65% if all invisible components could be converted. Table 1 of course refers only to visible light.

 

To obtain the solar input power available in the Vancouver Island area for the conversion into electricity, arriving onto 1 m² of a solar panel, one just needs to multiply the average daily insolation values of Table 1 by the (average) duration of the sunlight (the last row of Table 1). Of this solar input, one can convert (at most) 95% into electricity. Thus we further need to multiply the result by 0.95, to get the maximum theoretical electrical power output of any solar panel. The results are in Table 2:

 

 Table 2. Maximum theoretical output of a solar panel on the southern Vancouver Island (W/m²)

Period:

Annual average

July average

December average

Horizontal panel

262

341

102

2-axis sun-tracking panel

458

534

297

 

Now we are ready to return to the CHAINSAWS. Their power consumption currently ranges from about 1.2 kW (light-duty) to 3.5 kW (heavy-duty). But a 1kW efficient electric motor may be enough for some firewood cutting. To get the minimum surface area of a PV panel needed to power a 1kW chainsaw (or any other 1 kW machine), let us simply divide 1,000 W by the panel output of Table 2. The results are in Table 3:

 

Table 3. Area (in m²) to harvest solar radiation needed to produce an average electrical output of 1kW over the given period

Period:

All year

July

December

Horizontal panel

3.82

2.93

9.80

Sun-tracking panel

2.18

1.87

3.37

 

The 'All year' areas are producing 1kW on average over the all year. That means much more than 1kW in July, and much less in December. Not really much useful for powering an autonomous tool that have no means of storing the summer surplus output till the winter. The 'All year' numbers are useful for roof-top panels when you want to make sure that you contribute to the grid a certain average output over a long period. The 'Sun-tracking panel' values are also applicable mainly to permanent installations mounted on 2-axis sun-tracking racks.

For the current best panels with 44.7%-efficiency, one has to multiply the surface areas of Table 3 by 95/44.7 = 2.13.

 

If you really wanted to operate a 1kW chainsaw “all day, every day,” and power it fully solely by a built-in PV panel, its area would have to be at least 9.8 m² (or 105 sq ft), the above value for December. (At least, because at exactly 9.8 m², there would still not be enough power on December mornings and afternoons and on more cloudy days and probably around the winter solstice, because the December values are averaged over all daylight hours of all December days.) If you wanted to use your chainsaw in the summer only, you could get by with a 3-4 m² panel (in July with 2.93 m², in June and August you already need more, because the respective daily insolation values are less than for July – see the NASA link above). Even in the Sahara and similar deserts getting 7.5 kWh/m², you would need a 1.7 m² panel (3.82*3.31/7.5 = 1.7 – using for this prorating the 'All year' value of Tables 1 and 3, because near the equator the daily insolation is about the same throughout the year).

 

Even if new types of solar panels will exist in the future that would work efficiently without cooling with highly concentrated light, although the panels itself could then be quite small, one would still need to collect the light from exactly the same area as given in Table 3. Whatever size the converting element itself might be, one always needs a structure with the collecting surface of several square meters to collect enough sun energy to power a 1 kW tool or vehicle in real time. Even if made of ultralight materials, if attached rigidly to the chainsaw, such a monster would be an obstacle to most of the cutting, not to mention the safety concerns. Thus we must conclude that no chainsaw powered only by a small built-in panel is possible. With more efficient  solar panels, a built-in panel could at best be used to replenish the chainsaw battery during work breaks, possibly with the help of a separate light concentrator.

For use near a house, a corded electric chainsaw powered by big roof-top panels or the electric grid would continue to be the best option. For heavy-duty work on the forest floor, direct use of solar power would be problematic in any case because of limited sunshine there. Corded electric chainsaws would be a safety hazard (and a mobile solar-panel source for the cords would be usually positioned too far away, in a sunny spot). Thus until something new is invented, the choice is between a gasoline powered chainsaw or a battery powered electric one. And the clear winner will be the gasoline chainsaw until a breakthrough in the battery development is achieved. Namely, rechargeable Li-ion batteries have the energy density of only 0.1-0.265 kWh/kg, whilst gasoline of the same weight contains up to hundred times more energy, it has an energy density of 12.2 kWh/kg. Because a gasoline engine has large losses, only 1.7 kWh/kg of gasoline is for example delivered to the wheels of a car. But that still means that gasoline of the same weight as a Li-ion battery delivers 10 times more energy than the battery (or that to replace just one gas tank fill, one needs to bring with them into the forest 5-10 charged Li-ion packs – electric motor is lighter that the gasoline one, thus battery packs could be somewhat heavier than a full gas tank for the same chainsaw weight).

 

There is a still experimental Li-sulphur battery that achieved 0.5 kWh/kg (still only less than 1/3 of deliverable gasoline density). And since 1990s researchers work on promising Li-air batteries that are theoretically expected to achieve 5-15 times higher energy density than the Li-ion ones. If this is ever achieved, they would finally made battery-powered electric chainsaws fully competitive in convenience with the gasoline ones. For now, the periodic reports of breakthroughs in Li-air research alternate with news of more hurdles (instability, small number of recharging cycles).

 

The high density of energy stored in gasoline helps to explain our society's addiction to gasoline, and why it is so difficult to substitute the convenience of using gasoline with something else. The energy in fossil fuels is nothing else than the stored solar energy accumulated over millions of years. If you realize that, it seems the more stupid to try to extract everything in a wasteful way as fast as possible for the profit of a small minority. We should instead concentrate on energy conservation by introducing more efficient technologies, and use fossil fuels at much slower pace than now and only for the purposes where they cannot be yet easily replaced by renewable energy sources, and keep the rest in the ground for emergency purposes. All that independently of the climate change issues (and your view on them).

 

In the end, lets turn our attention to cars and trains:

 

SOLAR VEHICLES: The world smallest cars Peel 50 (1 seat) and Peel Trident (2 seat) have both in their electric version only a 1.5  kW  motor which is powerful enough for top speed of 50 km/h! They have footprints of 1.43 m² and 1.95 m², respectively. Not enough large areas to harvest 1.5 kW, unless the panels would stick out a lot on all their sides (adding 0.5 m on all sides of the Trident would increase total area to almost 6 m²,  which would be enough to power the car in the summer with current solar panels – by the way, they are producing another limited batch right now again, starting price from US$22k apiece – somebody ask Peel Engineering to build a hybrid solar/battery version?). It gives an idea what kind of solar cars would be possible. One could probably decrease the motor power to 1 kW  if all modern power sawing technologies are used (regenerative braking, light-weight materials, high-quality bearings). Even then switching to 1kW motor may still lead to a decrease in the top speed.

 

There already exist solar “cars” without batteries powered by currently available panels. But they look more like four-wheel bicycles with a canopy, like this one in the sunny southern California, https://goo.gl/YFfrnn, running in full sun at16 km/h on a level road on a power of 720 W, produced by 920 W rated panels (thus they are getting there close to the maximally available insolation of 1 kW/m² at sea level).

 

Higher speeds require more powerful motors, and normal size car roofs are not large enough to accommodate panels to fully power for example a 10 kW motor even in the summer at this latitude (29.3 m² is the panel area needed for 10kW with the ideal 95% efficiency panel). One would definitely never be able to power directly by solar panels a Tesla S car that has a 290 kW motor (over-dimensioned mainly to achieve an extra high acceleration).

 

More practical solution is to use roof-top panels as an auxiliary energy source together with batteries that are being charged between trips from other sources. Many groups have been working on this concept for some time. Since 1987 there has been an annual 3000 km Darwin to Adelaide (route going through another sunny desert) solar car race “World Solar Challenge”, http://goo.gl/KOXgJi. Competitors can still fully charge their vehicles from the grid before the start, and in some categories also in the middle of the race.

Ford CMAX Solar car has a 1.5 m² 300 W panel, and Ford is working on an interesting solar light concentrator to charge this car through its roof panel between the trips: https://goo.gl/zn27ek.

Honda has been testing a solar charging station for its electric vehicles since 2010 (a six hour solar charge is needed for a160 km drive).

 

One of the first demonstrations of a solar-powered car was in 1960 when an 1912 Baker electric car was fitted with then available solar panels: https://goo.gl/BZadJT. In this experiment 10 hours of solar charging enabled one hour of driving at 32 km/h.

 

SOLAR TRAINS?: Train is my favourite means of transportation that I miss in Western Canada, and so I was also interested to find how much solar power can be harvested from the roof of passenger trains and how it compares with their power needs. I did it for three cases for which I could quickly find all relevant specifications. All panel areas below are again based on the ideal 95% panel efficiency.

Amtrak18 coach electrified train: The coaches are 85 feet long, roof area of one coach is about 78 m². This area would collect about 20 kW of solar energy on average throughout the year (78 divided by 3.82 of the last table; maybe a bit more than 20k as it operates south of us). The electric locomotive is somewhat shorter, and so its roof could harvest another about 15 kW. That would give a total of 375 kW  for the whole train, which is just a small fraction of its present locomotive's power input of 6,400 kWh. Such high power is needed to accelerate the train to 200 km/h in less than 8 minutes.

British Rail Class 466 Networker: Two coach electric multiple unit (EMU) with top speed of (only) 120 km/h.  The total roof area of both coaches is 117 m², collecting thus on average about 31 kW (assuming similar insolation in Britain as here). There is just one motor in the whole unit, rated 600 kW, which is again much bigger than what the panels could ever in the future provide.

Liverpool Overhead Railway (historic). Built in 1893, it was using the first ever built EMUs. Each lightweight coach had one motor with power input of 45 kW  (initially; over the about 60 years of operation of this railways, the motors were replaced by more powerful ones, first a 52 kW, then a 75 kW). Roof area of each coach was 42 m² and could thus collect on average about 11 kW, still only a 1/4 of the motor power. Even for such a simple, light-weight, slower train, solar panels would never be sufficient.

The conclusion: solar panels on train roof-tops could only be an auxiliary power source for trains. The main solar generating stations (best in combination with wind turbines for night travel, unless cheap storage of solar electricity is available) would have to be put on the ground along the tracks. For this, all the tracks need to be electrified, as already done in Europe, except for some local routes.