AGUS HARYANTO
The objective of a hydropower scheme is to
convert the potential energy of a mass of
water, flowing in a stream with a certain fall
to the turbine (termed the "head"), into
electric energy at the lower end of the
scheme, where the powerhouse is located.
The power output from the scheme is
proportional to the flow and to the head.
POWER HOUSE
RESEVOIR
PENSTOCK
DAM
TURBINE
INTAKE
TRANSFORMER
GENERATOR
POWER LINE
 The
movement of water can be used to
make electricity. Energy from water is
created by the force of water moving
from a higher elevation to a lower
elevation through a large pipe penstock).
When the water reaches the end of the
pipe, it hits and spins a water wheel or
turbine. The turbine rotates the
connected shaft, which then turns the
generator, producing electricity.
A
water intake must be able to divert the
required amount of water into a power
canal or into a penstock without
producing a negative impact on the local
environment.
 “Conveying
water from the intake to the
power house”.
The water in the reservoir is considered
stored energy (potential energy).
When the gate opens the water flowing
through the penstock becomes kinetic
energy because it is in motion.

The water strikes and
turns the large blades
of a turbine, which is
attached to a generator
above it by way of a
shaft. The most
common type of
turbine for hydropower
plants is the Kaplan
Turbine, Francis
Turbine, and Pelton
Turbine.
 Low
head (from 70
meter and down to
5 meter)
 Large flow rates
 The runner vanes
can be governed
 Good efficiency
over a vide range
 Heads
between
15 and 700 m.
 Medium Flow
Rates
 Good efficiency
 = 0.96 for
modern
machines




Large heads (from
100 meter to 1800
meter)
Relatively small
flow rate
Maximum of 6
nozzles
Good efficiency
over a vide range
*Q = 28,5 m3/s
*H = 1130 m
*P = 288 MW
Jostedal, Norway
 After
passing through the turbine the
water returns to the river trough a short
canal called a tailrace.
 As
the turbine blades turn, so do a series
of magnets inside the generator. Giant
magnets rotate past copper coils,
producing alternating current (AC) by
moving electrons.
 Basic parts of generator :
1. Shaft
2. Excitor
3. Rotor
4. Stator
The heart of the
hydroelectric power plant
is the generator.
The basic process of
generating electricity in this
manner is to rotate a
series of magnets inside
coils of wire. This process
moves electrons, which
produces electrical current.
 As
the turbine turns, the excitor sends an
electrical current to the rotor. The rotor is
a series of large electromagnets that
spins inside a tightly-wound coil of
copper wire, called the stator. The
magnetic field between the coil and the
magnets creates an electric current.
is a device that transfers electrical energy from
one circuit to another through a shared magnetic
field. A changing current IP in the first circuit (the
primary) creates a changing magnetic field; in
turn, this magnetic field induces a voltage VS in
the second circuit (the secondary). The secondary
circuit mimics the primary circuit, but it need not
carry the same current and voltage as the primary
circuit. Instead, an ideal transformer keeps the
product of the current and the voltage the same in
the primary and secondary circuits.
 Used
water is carried through pipelines,
called tailraces, and re-enters the river
downstream.


In the scheme of hydropower the role of power house is
to protect the electromechanical equipment that convert
the potential energy of water into electricity.
Following are the equipments of power plant:
1.Valve
2.Turbine
3.Generator
4.Control System
5.Condensor
6.Protection System
7.DC emergency Supply
8.Power and current transformer
Flowing water creates energy that can be captured and
turned into electricity. This is called hydropower.
Hydropower is currently the largest source of renewable
power, generating nearly 10% of the electricity used in
the United States.
The most common type of hydropower plant uses a dam
on a river to store water in a reservoir. Water released
from the reservoir flows through a turbine, spinning it,
which, in turn, activates a generator to produce
electricity.
But hydropower doesn't necessarily require a large dam.
Some hydropower plants just use a small canal to channel
the river water through a turbine.
Sumber air pada ketinggian h = 14 ft memberikan debit
Q = 9 ft3/dt untuk menggerakkan kincir air berdiameter
D = 12 ft. Kincir air berputar pada kecepatan N = 5 RPM
dan memiliki 36 buah mangkuk yang terisi air tidak lebih
dari separohnya.
 Berapakah volume tiap mangkuk?
 Bila luas penampang mangkuk 1.2 ft2, berapa lebar
kincir?
 Berapa kecepatan spesifik? (Asumsi  = 60%)
 Volume
mangkuk:
N = 5 RPM  Tiap putaran = 12 dt
Q = 9 ft3/dt  Vol air jatuh tiap putaran
= (9 ft3/dt)*12 dt = 108 ft3
3
Vol tiap mangkuk = 108 ft  6 ft 3
36  0.5
 Lebar kincir = 6 ft3/1.2 ft2 = 5 ft
 Daya
yang dihasilkan:
P=Qgh
= (60%)(9 ft3/dt)(62.4 lb/ft3)(14 ft)
= 4717.4 ft.lb/dt
= 8.58 Hp
 Kecepatan spesifik
Ns = (5)(8.58)0.5(14)-1.25 = 0.54
 Kecepatan
jet: V  C 2 gh
C = koefisien transmisi
 Kecepatan titik di luar roda Pelton
U = a.V  a = konstanta
U = DN
 Daya
=Qgh
Sebuah roda Pelton dipakai untuk menghasilkan
daya dari air berketinggian h = 14 ft. Jika U =
0.45 V dengan V adalah kecepatan jet air pada
nozel. N = 300 RPM. Pertanyaan:
 Berapa diameter roda Pelton? (Asumsi C =
0.95)
 Bila diameter nozel 2 in, berapa debit air?
 Berapa daya (HP) jika efisiensi 88%?
 Berapa kecepatan spesifik roda Pelton?
 Diameter
roda Pelton

2(32.2)(75)
V  C 2 gh V = 0.95
= 66 ft/dt
U = 0.45 V = 0.45(66) = 29.7 ft/dt
D = U/N = 29.7/(0.88*300) = 1.89’ = 22.7”
 Debit
air Q = A.V
Anozel = d2/4 = (2/12)2/4 = (/144) ft2
Q = A.V = (/144) ft2 (66 ft/dt)
Q = 1.44 ft3/dt
P =  Q  g h
= (0.88)(62.24)(1.44)(75)/550
P = 10.78 HP
 Ns
= N(Daya)0.5 (h)-1.25
Ns = 300(10.78)0.5 (75)-1.25
Ns = 4.46
Note: Ns optimum untuk turbin roda Pelton
adalah 5