WIRING AND LOAD CENTER 

Every system requires some wiring to connect the various components. Load centers are available as a complete package that easily facilitates the connection of loads and power source(s). All circuits in the system should use wire of adequate size and have fuses or breakers of sufficient capacity to carry the expected load current. Even the LH1000 must be fused since it can suffer from a short or similar fault just like anything else in the system.

Inside the "junction box", are two terminal lugs for the battery wiring. The negative terminal lug is bolted to the box and the positive terminal lug is bolted to the clear plastic terminal block. Transmission wire ends are inserted into these two connectors (after being stripped of insulation) and then tightened.

The precision shunt installed in the junction box will give a readout of the hydro output in amperes if the digital multimeter is plugged into the jacks (color coded in the shunt body), and turned to 200m (the 9 o'clock position). A voltmeter connected to the batteries will roughly indicate the charge level, as described in "Charge Level" above, and an ammeter will indicate the output of the machine.

PIPE FRICTION LOSS - PVC Class 160 PSI Plastic Pipe

Pressure Loss from Friction in Feet of Head 
per 100 Feet of Pipe

POWER OUTPUT IN WATTS (CONTINUOUS)
PUT IN OUTPUT CHART

DESIGN EXAMPLE 

This example shows how to proceed with a complete installation. The parameters of the example site are:

     - 6 feet (2m) of head over a distance of 50 feet (15m) 
     - a flow of at least1000 GPM (63l/s) 
     - 100 feet distance from the house to the hydro machine 
     - 12 volt system 

The first thing to do is determine the pipe size. Given that there is friction between water and the pipe in which it flows, this friction can be reduced by increasing the size of the pipe to minimize the friction to acceptable limits. Therefore, pipe size must be optimized based on economics and performance.

The pipe flow charts show us that eight-inch (approx. 20cm) diameter PVC pipe has a head loss of 0.97 feet of head per 100 feet (30m) of pipe at a flow rate of 800 GPM (50 l/s). This is about 0.5 feet (15cm) of loss for 50 feet (15m) of pipe. PVC comes in short lengths and is glued together or purchased with gaskets.

The maximum output occurs with a flow of about 800 GPM (50 l/s). Note that with this machine, the flow is determined by the head, as there are no nozzles that can be adjusted that would change the flow.

1 foot loss/100 feet pipe=x feet loss/50 feet pipe
x=0.5 feet (15cm) of head loss

Next, we subtract the head losses from the measured head (often referred to as static, or gross head. Abbreviated: Hg) in order to determine the actual, operating head (often referred to as dynamic, or net head. Abbreviated Hn):

6 feet head (Hg)-0.5 feet head losses =
5.5 feet (1.85m) actual head (Hn)

It is now known that the LH 1000 will be operating at an actual, or dynamic, head of 5.5 feet (1.85m) Hn. By referring back to the output chart, it can be determined that the LH1000 can, realistically, be expected to produce approximately 400w.

 
COPPER WIRE RESISTANCE

Since we require 12 volts and the transmission distance is short, we can generate and transmit 12 volts using the LH1000. This LH1000 could also be used for higher voltages like 24 and 48, and power could be transmitted longer distances. We need to go 100'(30m) with 400 watts at our site. The amperage can be determined using the formula: volts x amperage = watts. So, a 12v system usually operates at an actual voltage of about 15v, therefore: 400/15 = 26.7 amps. The machine will need to be wired parallel delta for this site.

This will be about 26.7 amps at 15 volts at the generator. Note that there will be some voltage drop in the line and 12-volt batteries require somewhat higher voltages than nominal to become charged. So the 26.7 amps must pass through 200'(60m) of wire for the distance to the batteries and back which completes the circuit. As there is friction between water and the pipe that carries it, causing losses, so there is resistance between electricity and the conductor that carries it, and is measured in units called ohms. Resistance losses should be kept as low as economics permit, just like the pipeline losses. Let's assume that a 5% loss is acceptable at this site, resulting in the loss of 25 watts. The formula to calculate resistance losses is I (amps) x I (amps) x R (resistance) = w (watts). We put our known figures into the formula to learn the resistance that we require in a copper conductor to achieve this.

26.7 x 26.7 x R = 25w 
711 x R = 25w 
R = 0.04 ohms

It has been calculated that a copper conductor with losses of 0.04 ohms over a total distance of 200 feet (60m) will result in an acceptable 5% loss. The Wire Loss Chart shows losses per 1000' (300m) of wire, so: 

1000'/200' x 0.04 ohms = 0.2 ohms per 1000'. 

The chart shows 2 ga. wire has a resistance of 0.16 ohms per 1000', so: 

200'/1000' x 0.16 ohms = 0.032 ohms. 

This is close enough to the desired level, that with a little more investigation we can determine whether this will result in acceptable power losses: 

26.7 amps x 26.7 amps x 0.032 ohms = 22.8watts of loss.

Increasing the wire size can further reduce the losses, but can also increase costs, as larger wire is usually more expensive. Resistance in a length of wire results in power loss that is seen as a voltage drop from one point in the line to another. For example, if your voltage, as measured at the generator, is 15vdc, then it could be assumed that if the voltage were measured along the line to the batteries, it would be lower as you got further from the generator: Voltage drop= I (amps) x R (ohms resistance in your circuit). So:

Voltage drop (v) = 26.7 amps x 0.032 ohms = 0.85 volts

Hence, if your generator voltage is 15vdc, your battery voltage will be 14.15vdc. Keep in mind that it is always the batteries that determine the system voltage, as they are the stabilizing force in your system. All voltages in the system will rise and fall corresponding to the battery voltage, or the battery's state of charge. At the site, we would be generating 26.7 amps continuously. Typically, a battery bank is sized to have two days storage capacity. If we choose lead acid batteries and wish to have two days of storage capacity, then we use the formula: amps x hours x days = amp/hrs capacity. So:

33 amps x 24 hrs x 2 days = 1584 amp. Hrs. Capacity 

The Trojan L-16 has a rating of 6vdc and 350 amp/hr. Using these you would require at least eight batteries; there would be four strings paralleled, with each string consisting of two batteries in series to give the 12vdc system voltage we have chosen. This would give 1400 amp/hrs at 12vdc capacity, which is about two days storage. An inverter and charge controller are usually used in the system. The diagram for such a system would look like this: