DoItYourself's Blog
Frequently Asked Questions:
Hydro or Solar, which is best for me?
You either have access to flowing water or you don't. If you do, hydro is without doubt the way to go. There is far more energy available in a mass of water falling some height than in 'the soft rain of photons' striking silicon. Even if the water source is half a mile away, it may be less expensive to string a long power line than invest in solar modules. Hydro will produce 24H / 365D, and when the demand is greatest, typically in winter, the water source will generally be high.
If your situation is more typical, you will not have access to flowing water. However, the sun does (occasionally) shine on all of us. There are thousands of modules supplying electricity to remote sites in BC. Unfortunately, the climate in BC, especially at the coast, is not always conducive for high output. It would take a large array of photo-voltaic modules to come close to supplying the energy obtainable from even a modest hydro plant. You really need to do an analysis of your loads, and determine the solar influx available at your site to see if it is practical to proceed with a home sized system. For stand alone telemetry sites, data logging or other low current demands, solar will win hands down.
What is the life span of the generating equipment?
Solar panels are rated as retaining 90% of their output for 25 years or more. Since there are no moving parts, a well installed system will have a life span which far exceeds the current level of technology.
Turbines are fitted with bearings which are replaceable. Stream Engines will run for several years continually under heavy load before the bearings may need attention. They are an inexpensive automotive type, and can be replaced easily. There are no brushes in these units, and nothing else to replace.
Old auto alternators require brush replacement on a more frequent schedule. This depends on the load and rpm the machine is operating at. They should be checked monthly, as every system is unique. Six months to two years is an approximate range.
The life span of larger AC hydro plants is measured in decades with very little maintenance. There is no reason they can not be kept operational indefinitely, with periodic bearing replacement.
What is the life span of inverters and control equipment?
I have not yet seen any of this equipment fail in normal use. Modern electronics is so reliable that unless there is an external fault or a lightning strike, there is no reason the equipment can't me kept in service for many years. That is not to say there can never be a failure. It is important to correctly fuse all current carrying wires, and in some situations, you may want to carry some spare components.
What is the noise level of a hydro plant?
The small 'Stream Engines' pictured here are extremely quiet. There is a soft whirring heard when standing next to the units, but in the house, or standing away from the unit, the sound is barely if at all noticeable. Noise can be virtually eliminated by careful mounting on non resonating surfaces, then enclosing in an insulated box.
The tail race, or the water that has passed thru the turbine, will flow directly back to the creek. This is either discharged under water or the pipe end covered with rocks to dampen any sound carried thru the pipe.
How should the building be wired?
If you are planning new construction and anticipate having an AC home power system, you should wire the cabin as you would for a grid tied home. If using one inverter, you will likely be producing 120 volts. Most distribution panels are designed for 120 / 240 volt use. It is important to wire the circuits so that at no time can a neutral conductor carry more than its rated current. Problems could occur if you had two hot wires and one neutral feeding a kitchen outlet in a so called 'Edison Circuit' where normally that output would be fed with a three wire circuit from a 120 / 240 volt source typical to most grid connected homes.
Use the provincial standards, and return all circuits to a main distribution panel. Keep in mind the inverter capacity you might use, usually from 1500 watts to 5000 watts. It doesn't hurt to have many circuits, but the main breaker must be sized to match the inverter used. Obviously you can't have all the loads possible turned on at once if you only have a 1500 watt inverter. Here is where you must practice 'load management' manually. In all cases, you should have a copy of the electrical code handy, and follow all the rules for your area. Don't forget the system grounding and ground rod, this is mandatory.
For DC only systems, wiring must be sized to keep voltage drops to a practical level. See the wiring chart in the section on tables. Often you will be installing wires in an existing building, and won't be opening up walls. In this case, you should protect the conductors from damage and be sure to fuse all circuits. Armored cable and conduit is available from many suppliers, and makes a neat job.
What voltage should I produce?
This depends on the type of loads you intend to use, and also on the distance from the generating source to the loads, and the available prime energy source, either solar, water or wind. Most small systems for cabin use produce low voltage as in the DC systems described here. All of these produce from 12 to 24 volts, with some of the power being used directly, and most being transformed to AC with an inverter. Transmit ion distances are short enough that line loss is minimal. If an out building requires power, it is usually distributed at the AC level.
The AC systems described all produce 117 volts directly. Battery storage is not used as the total peak loads are comparatively large in comparison to those of DC systems. For plants in the 5 to 10 kW range, single phase is standard. For larger generators, it is better to go with 3 phase wiring.
How much will it save in other fuels?
If you are currently using a gas generator, you will know the fuel consumption and cost per hour. Expect this to increase with time. Of the hydro systems described here, the gas generators have not been used since they went on line. The savings are not limited to fuel cost and generator repairs. You will benefit from the instant availability of AC power, the quiet, lack of fumes, and not having to pack gas in the middle of a rainy night.
Propane savings can be realized if you connect the load controller to a water tank. See the section on hot water below.
Can I get hot water from the energy produced?
Yes. This is only really practical on hydro systems. There are several things you can do. On DC plants, the excess energy must be diverted from the battery bank once full. This can be done by connecting special low voltage water elements to the load controller. In the case of full batteries and no other base loads on, all the power will be shunted to the heating element. Over the night, this can heat a good sized tank of water. See the section on tables and formulas to figure the Joules and BTU produced.
One fellow connected a 2000 watt 240 volt element to his trace DR3624 inverter. At 117 volts, a 2kW element will draw one quarter the power as it would at 240 volts, that is, 500 watts. His hydro would easily produce this, so at night he kept the heater on. This is a pretty slick way to go, as the losses in inverter efficiency don't matter as he had the available water.
What are the risks?
For hydro systems, annual precipitation may vary so they can't run at full capacity all the time. Tree falls and land movement can have a potential impact on pipelines. Land slides are generally restricted to narrow gullies on steep hill sides. Risk from dealing with electricity need be no greater than what you are use to every day. Pipe lines can freeze or burst, but again, this need not pose any great danger, and can usually be prevented by thoughtful design and installation.
With solar modules, breakage from vandalism or theft are potential hazards. Natural local hazards can largely be avoided. Variations in sunshine due to local weather patterns can vary the total output from year to year, but this is usually anticipated.
Are these systems upgradeable?
Generally once a penstock is installed, the upper limit on power is fixed. The only way to increase it is to go higher up the hill, bearing in mind the pressure rating of the lower sections pipe used. Batteries banks can be expanded within reason, but the total energy from the generator must be considered when doing this. Don't over size the storage and consumption and keep the supply fixed or you will have chronically weak batteries.
Solar systems can always be expanded, provided that you also increase the wiring and protection devices. t takes a lot of solar to match even a small hydro plant, so battery storage capacity is not as great an issue, but you still have to watch the total maximum charge rate.
What is load management?
A means where by there can be a greater load connected to the power distribution system than the installed capacity of the generator. By automatically selectively shutting off lower priority loads, available power will be delivered to higher priority loads. Low priority loads that are controlled are typically water and space heating. This is common on AC direct systems, but seldom if ever used on small DC hydro plants with an inverter. I have not seen load management on a solar installation.
Any Questions?
Use the e-mail link on the bottom of any page. All questions will be responded to.
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Power Calculations :
In SI units, (metric) The energy available in falling water is expressed as, E = mgh where...
e = energy = ( m g h ) Joules (a Joule is one watt for one second)
m = mass, the product of density (p) x volume (V)
g = gravitational constant of acceleration, 9.8 meters per second per second
h = head in meters
Note: time is not a factor in this relationship for energy.
Power is defined as energy released over a specific time, so substituting in the values above,
Power = ( p V g h ) Joules / second (same as watts) The density of water is 1000 kg/m^3, (or 1 kg/liter), and the gravitational constant of acceleration on earth is 9.8 m/s^2, (the force that gives 'mass' its weight). In hydro applications, the letter 'Q' is used to represent the volume of water combined with its density. Substituting in gives gross power equal to,
Power = ( 1000 x Q x 9.8 x h ) watts, or in kW Power = (1000 x Q x 9.8 x h ) / 1000 kilowatts, and futher reduced to...
Power = (Q x 9.8 x h ) kilowatts, ( with Q in cubic meters per second )
Note: as mentioned above, see how the "V" for volume has been replaced by "Q" for quantity representing 1 cubic meter of water.
Note: See how this is similar to the energy equation above, energy = m(Q) x g(9.8) X h
Overall system efficiency is often around 55 - 65 % in systems under a kW, so you can round off the above to the following for rough working.
Power = ( Q x h x 10 x 0.6 ) kW ... for small systems where Q is volume in cubic meters per second, and h is head in meters.
For plants in the multi kW range, efficiencies reach 70 percent or higher. Actual efficiency depends on the water flow characteristics, nozzle design, turbine design, coupling, and generator. Power line losses, battery charge / discharge losses and inverter effiency must also be considered. Example:
You have a stream with a flow of 3 liters per second and a drop of 50 meters available.
What is the gross power potential in watts ?
3 liters per second is 0.003 cubic meters per second)
Power = ( 1000 x Q x h x 9.8 x efficiency) watts
Power = ( 1000 x 0.003 x 50 x 10 ) watts
Now multiply this result by the expected overall efficiency. (say 60%)
Power = ( 3 x 50 x 10 ) x 0.6 = 900 watts (rounded off)
In imperial units:
Total energy available is expressed as,
energy = Q x 62.4 x h
Q = cubic feet of water per minute
62.4 = density of water, @ 62.4 lbs. per cubic foot
h = net head in feet
This is really foot pounds of potential energy per minute, (so many pounds at so many feet head.)
One horsepower is equivalent to 33,000 foot pounds of work done in 1 minute, so to figure in HP,
Q X 62.4 X h / 33,000 or Q x h / 529 Since 1 Hp is equivalent to 745 electrical watts, convert to kW by Power in kW =(Q x 62.4 x h / 33,000 ) X 0.745
or,
Power in kW = (Q x H / 709) where Q is in cubic feet per minute.
In all cases it is important to multiply the gross power result by the system efficiency, again use 55- 65 % for small systems, and perhaps 70 % for larger system. On larger sites, it is often easier to think of cubic feet in one second, so convert the above to...
Power in kW = ( Q x h / 11.8 ) where Q is now in cubic feet per second.
For general work on small systems, the following formula is easy to remember and works well. It assumes an overall efficiency of 55 %, so the result is close to the power you will get.
Power in watts = (net head in feet x flow in USGM) / 10 (again, at an overall efficiency of 55 %)
Volumes and conversions. 1.0 litre (1 kg)
1.0 cubic ft (62.4 pounds)
1.0 us gallon (8.3 pounds)
1.0 imp gallon (10 pounds)
1.0 cubic foot (62.4 pounds)
1.0 cubic meter (1000 kg)
| = US gallons * 0.26442
= litres * 28.313
= litres * 3.785
= litres * 4.5459
= US gallons * 7.49
= 1000 litres |
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Some practical fuel cost examples:
Natural gas
is sold by its heating value in 'Giga Joules', not by its volume because the heating ability per unit volume can vary. Since it is a commodity, its price can vary.
Electricity
is sold by the 'kilowatt hour'. In BC, the cost for 1 kwh is currently $0.0615. That is 6.15 cents for 1,000 watts for one hour - or 100 watts for ten hours. So to run a typical 1500 watt electric heater for one hour it costs (1.5kW X 6.15 cents, or about 10 cents an hour)
Some definitions you have to understand for this to make sense:
' / ' means per or divided by. * or X mean to multiply.
1 Joule = 1 watt for 1 second ( a tiny amount of energy )
1 Giga Joule = 1 billion Joules ( a fixed amount of energy used for billing purposes)
1 Giga Joule is the same amount of energy as 277 kilowatt hours. That is about the amount of electricity used in 15 days in an average home.
The cost of 277 kW hours of electricity (1 GJ worth) in November = 277 X 6.15 cents/kwh = $ 17.03
The cost of 1 GJ of natural gas (277 kW worth) in November 2007 is hovers about $ 7.00 per GJ
So you can see that heating with natural gas is less than half the cost as heating with electricity.
Facts about Oil:
The cost of 1 liter of heating oil at the end of Nov 2007 is $ 0.92. It was $0.77 at end of Sept 2007, the price difference a result of a barrel of oil costing nearly $ 100. Oil, like natural gas, is traded as a commodity on the stock markets.
A typical home oil tank holds 300 US gallons, or 1130 liters.
The heating value of 1 liter of # 2 heating oil is 37,000 BTU, or British Thermal Units. (An old, but useful measuring system - based on 140,000 BTU per USA gallon)
At 37,000 BTU per liter / 3,413 BTU per kW hour = 10.8 kW hour per liter of heating oil. This means that there is the same amount of energy in one liter of oil as that used by a 1000 watt electric heater running for 10.8 hours.
10.8 kW of electricity at $ 0.0615 per kW = (10.8 X 6.15) = 66.4 cents
And as said above, 1 liter of heating oil costs 92 cents. Add to that the lower efficiency of some gas or oil furnaces and the real cost is somewhat more. If the furnace is 75 % efficient, then 25 % of the cost is going to waste. Electric heat on the other hand, although also quite expensive, is nearly 100 % efficient, so all the power used is turned to heat at the electric heater.
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Units of measure:
Any time we deal with the subject of energy, power, pressure or heat, various definitions are required to make meaningful calculations and comparisons. This section explains the meaning of the 'SI' or metric units, and compares them to the imperial or US units.
Every attempt has been made to ensure accuracy on these figures. In all your calculations, be sure the result makes sense, and that you are not getting a result that seems out of proportion to the input numbers. In dealing with energy transfer, be sure to factor in losses due to in-efficiencies. This is very important when comparing costs from the BTU's produced from burning fossil fuels to KWh / BTU's achieved from electrical heat sources.
Definitions Ampere
| -Unit of electrical current, a quantity of electron flow equal to 6 * 1018 electrons / second. Analogous to water volume or quantity flowing in a pipe.
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Volt
| -Unit of potential difference or electromotive force. Analogous to water pressure in a pipe. In a house, normal wall voltage is 117 volts. Stoves and dryers use 240 volts.
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Watt
| -Unit of power, the product of amps times volts. The rate of work done in a set unit of time. Equal to the energy spent by one amp flowing through one ohm for one second.
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Ohm
| -The measure of electrical resistance equal to that resistance which dissipates energy at the rate of one watt from a current of one ampere.
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Joules - kilowatts - BTU's 1 Joule
1 Joule
1 Joule
1 GJ
1 BTU
1 kW
1 kWh
1 kWh
| 1 watt flowing for one second
1 Newton / metre
0.000948 BTU
948,000 BTU
1054.6 Joule
3412 BTU / hr
3412 BTU
3.6 Mega Joule
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Definition of a BTU
Energy required to heat one pound of water by 1 degree F.
10 BTU = 10 pounds (1 imperial gallon ) heated by 1 degree F.
Energy 1 Watt hour
1 kWh
1 Therm
| 3.413 BTU
3413 BTU
100000 BTU
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Power and Heat flow 1 kW
1 kW
1 Ton
1 HP
1 hp
| 0.948 BTU/sec = 3413 BTU/hr
1.3415 HP = 738 ft lb/sec = 44,268 ft lb/min
12000 BTU/hr
0.764 kW = 2546 BTU/hr
0.7455 kW = 550 ft lb/sec = 33000 ft lb/min
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Volumes and conversions. litre
cubic ft
us gallon
imp gallon
cubic foot
cubic meter
| = US gal * 0.26442
= litres * 28.313
= litre * 3.785
= litre * 4.5459
= US gallons * 7.49
= 1000 litres |
Length
meters
feet
km
miles
| = feet * 0.3048
= meters * 3.281
= miles * 1.609
= km * 0.621
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Pressure
1 PSI = 6894.76 Newtons / sq. m = 6894.76 Pascal
Absolute Pressure = psia = Gauge pressure (psig) + 14.7
Chemical content of fuels
1 ft^3 of natural gas = 1020 BTU (chemical energy constant)
1 gallon (US) # 2 fuel oil = 140,000 BTU
1 gallon (US) propane = 91,200 BTU
Calorie (c)
Heat required to raise 1 gm. water by 1 degree C.
Kilocalorie (C) = 1000 calories. C = food calorie
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Pipe sizing example: Email to info@homepower.ca for a printable copy of this example. A site has 50 meters of head and a flow of 5 liters per second, a very typical situation here on the coast. The pipeline will have to be 150 meters long, and is going to be polyethylene plastic, (poly pipe). We will accept a head loss of 10% at maximum flow, that is, 5 meters of height lost to pipe friction. The power line length is 50 meters to the battery.
Refer to the section on "Power Calculations" for more detailed explanations of these formula.
Step 1.
Determine the available power.
5 liters per second is 0.005 cubic meters per second )
Power = ( 1000 x Q x h x 9.8 ) x efficiency) watts
Power = ( 1000 x 0.005 x 50 x 9.8 ) x 0.5 ) watts
Power = ( 5 x 50 x 10 ) x 0.5 = 1250 watts
Note: the acceleration due to gravity value of "9.8" was rounded up to "10"
The final output will be somewhat less as the net head will be 45 meters at full flow, or around 1125 watts, 90 % of the last calculation. This is close enough, and we decide to proceed with the project. Step 2.
Plot a flow of 300 liters per minute on a friction chart applicable to the pipe material you are using, so the flow velocity is less than 5 feet per second. Note the pipe size indicated, and draw lines for several sizes that are available. This will likely be 2 inch, 2.5 inch, 3 inch and 4 inch. Pipe friction charts are available from pipe suppliers. Each type of pipe will have a different coefficient of friction, so a chart used for steel will not be accurate for plastic pipe.
From the chart ...
2 inch gives a velocity of 8.2 feet per second which is too fast.
2.5 inch results in 5.2 feet per second.
3 inch gives 3.7 feet per second.
Pipe inside diameters vary with pipe schedule, (wall thickness), so you will seldom get exactly the diameters listed. Adjust the line you drew on the graph to match the pipe diameter you will be using.
Our allowable head loss is 5 meters in 150 meters of pipe length, or 3.33 meters per 100 meters of pipe. Charts are usually normalized to meters per 100 meters or feet per 100 feet, so you may want to convert at this point. See that 2.5 inch poly pipe will deliver the required maximum flow with 3.3 meters of head loss per 100 meters of pipe, and the flow velocity is right on 5 feet per second.
It is possible to run more water through the 2.5 inch pipe by increasing the nozzle size. This will increase the head loss beyond our limits, but it will produce more power. In fact we could drop to 70% of the gross head before the power curve would start to fall. However, wide ranges of pressure result in variable jet velocities which in turn determine turbine RPM. In DC systems this is often acceptable, but in AC systems, RPM must be maintained, and a low jet velocity results is a significant loss of efficiency. This is why hydro equipment is so site specific, what worked well at one site may perform badly at another. .Step 3.
Determine the correct nozzle diameter to accept a maximum of 5 liters per second. Often, several nozzles are cut and installed to match flow conditions. 'Stream Emgine' turbines used on some of the example sites featured here are equipped with 2, 3 or 4 nozzles. Each nozzle can have a control valve permitting a wide range of flow to suit conditions. In larger AC systems, spear nozzles are often used which permit continuous adjustments of flow by varying the position of a movable spear within the nozzle.
Refer to a nozzle flow table to determine the correct size to use, or start small and gradually increase the size while measuring the flow in a bucket or through a weir.
Since water is non compressible, the flow velocity and nozzle size can be calculated from scratch. One only needs these formulas. The flow will typically be 5% less than the jet area suggests, so cut it a fraction big. Also, jet velocities are often about 5% less than calculated due to inefficiencies in the nozzle and other complex factors.
Area = 3.1428 x R^2
Circumference = 3.1428 x D
Jet velocity = 8 x H ^ 1/2 ( 8 times the root of the net height in feet)
Imagine a solid bar of water traveling at the jet velocity. One only needs to determine the bars volume per unit time as it passes by. The diameter and hence cross sectional area are found first, then simply multiply this by the 'bars' length passing per second to arrive at the volume. This is easy to do in cubic inches and linear inches, then convert to gallons or litres. |
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