Small scale Hydro power

Small scale Hydro power

The small hydroelectric system can produce power rating up to a few mage watt  there are two version of hydroelectric system diversion type and reservoir type. The reservoir type may require a small dam to store water at higher elevation. The diversion type does not require a dam, depend upon the speed of the current to generate electricity.

Diversion type Hydropower

This type of system does not require a dam so it consider more environmentally sensitive.the river for this small hydroelectric system must have stronger enough current for a realistic current generation  the KE entering the turbine KE is given by

KE = .5* mv²    = .5Aspv³t

Pt = KEt/t =      .5Aspv³t

Reservoir Type hydropower

it is either a natural lake at high elevation, then a downstream river or lake created by dam. The system consist mainly  a pen-stock, a turbine and generator. If the water is allowed to flow to lower elevation through the pen stock the potential energy of water converted into KE where some of it is captured by the turbine.
the PE of the water behind the reservoir PE can be found from
 Per WH = mgH

Where

W = the weight of the water, Kg

H = the elevation of the water with respect to the turbine, m;

m = the water mass of the reservoir, kg;

g = the gravitational speed, m/s

if m (kg) is mass of the water of the water entering the pen-stock can be applied to find the input PE of the pen-stock as given by

                          PE pin = mgh

The water flow fw inside the pen-stock is defined as the mass of the water m passing through the pen stock during a time travel t that,

Fw = m/t

PE pin = fw tghH

Which is converted into KE inside the pen-stock. The outpout of KE can be found

KE(out) = .5Apvt pv²

Where

T= the duration of water flow, S;

V = the speed of water exiting the pen-stock,m/s;

Ap = the cross section area of penstock m²

Due to losses in side the penstock such as the water friction, the KE existing the pen-stock KE(output) o\is less then the PE(p_in) at the entrance of the penstock. Thus the efficiency of the penstock is given by

Efficiency = KE/ PEp-in

 The blades of the turbine cannot capture all the kinitaic energy KE existing the penstock.

Cp = KEblade/ KEp-out

Turbine Efficiency

The energy captured by the blade KE is not all captured in to the mechanical energy entering the generator KE due to various losses in the turbine. The ratio of two energies is know as turbine efficiency

The output electrical energy of the generator Eg its equal to its KEm minus the loss of the generator. Thus the generator efficiency ng is defined as

Generator efficiency  = Eg/ KEm

The out put electrical energy as a function of water flow and head.

Eg = fGHt(Cp np nt ng)