The Installation of Home Water Powered Electrical Generator DVE(Practical, Simplified procedure)
Our Goal: Provide a sufficient quantity of electrical power for average operation of "Our" cabin, using a nearby water source.
Before considering the installation of DVE it is necessary to do some calculations on energy balance:
A) Source of Electrical Energy
B) Anticipated usage of Electrical Energy
A) Energy balance for water powered machine, operation of Home Water Powered Electrical Generator:Before we begin with the installation of the DVE, it will be necessary to research all ownerships of land including local land and water rights, building permits, and other applicable law and legal considerations if those issues have not been worked out.
In the country site where "Our" cabin is built, we will give an approximate estimate of the potential capacity to produce electrical energy from the water which will power the generator DVE. For our calculations we need two measurements:
1. Water down slope: H [m]
We will obtain "H" by simply measuring the differences between: upper water level, (the point where we will install the water intake pipe) and lower water level, (the point or location where the DVE will be installed).
In our example it was measured: H = 5m
2. Water volume: Q [l/s]
"Q" will be established at the location where the DVE will be installed by doing a few repeated measurements. This can be done by filling a larger container (barrel) of a known volume while timing the period needed to fill the container with a stopwatch.
In our example it was measured:
Barrel with a volume of 200 liters; average time to fill it: 25sec.
Water flow: Q=200/25=8 l/s
Q = 8 l/s
3. Calculating the sustained output of the electrical generator
With the assumption that the entire water volume will pass through the turbine and will produce usable energy, we will calculate the expected energy yield on the output contacts of the generator this way:
GIVEN: Mechanical Efficiency of Turbine: hM = 0,7 (70%)
GIVEN: Efficiency of Generator: hG = 0,5 (50%)
GIVEN: Coefficient of Hydraulic Loses in water supply pipes: 0,765
Electrical output PEL:
PEL= g * Q * H * hM * hG * 0,765 =
PEL= 9,81 * 8 * 5 * 0,7 * 0,5 * 0,765 = 105W
PEL= 105 W
Our water power source (in the example above) is capable of converting its energy potential to electrical power output of 105 W. The turbine SETUR will be equipped with a 3-phased synchronies generator with parameters as follows:
Output: 120 Watt
Voltage: 3x24 Volt AC (Alternating Current)
B) Energy Balance of electrical energy consumption in "Our" cabinThe basis for this calculation is "Our" useful layout of electrical devices throughout the dwelling and their rated consumption in Watts, as it can be seen on the next cabin floor drawing. At the same time we will establish the expected duration of their daily operation. All findings will be laid out in tables and from there we'll calculate electrical energy consumption.
For your orientation, table of used schematic symbols is provided below.
with permanent magnets.
installed regulator of
direct and alternating currents
(Compact Florescent Light)
(24V Direct current to
230 V, 50 Hz Alternating)
for energy conservation and
charging of the Batteries
|Basement - Cellar||Light bulb||9||*||1||=||9|
|Living room||Light bulb||9||*||1,5||=||13,5|
|Water Well||Water Pump||180||*||1,5||=||270|
|Losses in Regulation,|
Converters and Reserve
|10% (Chosen For Example)||=||173|
|Total Expected Daily Consumption Ad= 1899 Wh|
After calculating daily consumption (see Table I.) we will select battery voltage (12 V or 24 V DC). With regards to loses in power supply wiring it is better to choose higher voltage: USYST=24 V|
- Further we'll calculate basic capacity of Battery CA [Ah]:
Battery capacity calculated this way is for uninterrupted working regime of DVE, without regard to its reserve capacity and without consideration to the depth of battery cycling. It is safer to consider daily reduction of operation of DVE max. 0,5 hour. Coefficient for battery charging is thus:
kA= 24 / (24 - 0,5) = 1,021
In order to maximize life expectancy of the batteries, the depth of discharge should not exceed 50% of its capacity.
For those reasons the depth of discharge will be set to:
hV = 0,5
- Next we'll calculate the optimum capacity of the batteries for operation of our electrical devices:
C = CA* kA / hV = 79,125 * 1,021 / 0,5 = 161,617 Ah
From available selection of optimal battery capacities we have selected two 12Volt batteries with the rating of 80 Ah. They will be connected in series for a total of 24 Volts:
Total available energy output of the selected batteries equals: C=160 Ahand output Voltage of both batteries connected in series: U = 24 V (this coincides with USYST)
In order to operate the most energy demanding electrical devices simultaneously, it will be important to select appliances with a minimum power consumption designed for 230 V, 50 Hz AC power system in "Our" cabin.
These appliances are selected as an example:
From the above example it is self evident that the minimum setting of the DC to AC converter operating at system Voltage of 24 V, has to be:
Pstr. = 300 W
The required current consumption will be partly covered by the batteries and partly by the output of the DVE. The current on the DC side of the DC to AC converter will be:
Istr. = PM / USYST = 300 / 24 = 12,5 A
- Since there is a sizable current required by the DC to AC converter it is of paramount importance to properly dimension the electrical wiring. In our case with consideration to loses of 3% in the connective cabling to both sides (that is the + & - connections) of the DC to AC converter, wiring with a minimum of 6mm2cross sectional dimension and a maximum length of 8m (with wiring from batteries to the DC - AC converter). It is assumed that the current carrying capability is 2,5 A/ mm2.
This concludes the calculations of balance between consumption, suggested battery capacity, and output of DC to AC converter. For additional assurance we will confirm the capabilities of the power source (DVE) on reduced daily output in the time span from 6a.m. to 12p.m. or approximately 18 hours.
Ad = 1899 Wh / day = 24 hours
Adr = 1899 Wh / 18 hours -> will determine the continuous reduced output during the operation "Our" live-in accommodation (cabin):
P18 = Ad / 18 = 1899 / 18 = 105,5 W
Based on our above calculations, DVE is capable of a continuous output of PEL= 105 W
Comparing calculations and data from sections "A" and "B" we can conclude that the output of the electrical source is sufficient for our operation. In such case that the calculated data in P18 > PEL (greater then) is more than 5%, it will be necessary to reconsider the operation of "Our" cabin by readjusting consumption requirements in the appropriate table (Table I.) used above.
Should you require additional or more detailed information for the above example and also the possibility of energy power grid setup, we will be glad to supply it.