HYDROELECTRIC POWER PLANT
PROBLEMS AND SOLUTION
1. A Pelton wheel is required to develop 4500 kW at 400 rpm operating under an available
head of360 m. There are two equal jets and the bucket angle is 170 0. The bucket pitch
circle diameter is 1.82 m. Taking k for the buckets as 0.85, determine (a) the efficiency of
the runner and (b) the diameter of each jet.
Given:
Required:
P = 4500 kW
N = 400 rpm
H = 300 m
N=2
ϴ = 1700
D = 1.82 m
k 0.8
(a) ηD = ?
(b) d = ?
Solution:
v 1= √ 2 gH
√(
¿ 2 9.81
¿ √ 2 gH k v
m
( 360 m )( 0.85 )
s2
)
¿ 77.4836757
m
s
vb =
¿ 38.11799086
(a)
η D=
πDN
=
60
m
s
1−kcosθ
2 2g
v 1 v b −v b ) 2
(
g
v1
(
π ( 1.82 m ) 40
60
m
min
)
0
¿
[
η D=0 . 9183
(b)
Power develop,
P=
4500 kW
=4900.36 kW
0.9183
Power develop per jet,
P=
4900.36 kW
=2450.18 kW
2
P=γQH =γA V 1 H =γ
π d2
V1H
4
( )
2
4 P=γπ d V 1 H
2
d=
d=
4P
γπ V 1 H
√
( )]
1−0.85 cos 170
m
(77.48 )( 78.12 )−( 38.12 )2
m
s
9.81 2
s
4 ( 2950.18 kw )
m
( 9.81 )( π ) 77.4836757
(360 m )
s
(
d=0.106773422 m
)
2
[ ]
m
2
s
2
m
77.48
s
2 9.81
(
)
(
)
2. A Pelton wheel develops 8 MW under a net head of 130 m at a speed of 200 rpm.
Assuming cv = 0.98, hydraulic efficiency = 87 percent, speed ratio = 0.46 and the ratio of
jet-to-wheel diameter = 1/9, determine (a) the flow required (b) the diameter of the wheel
(c) the diameter and number of jets of jets needed.
Given: P = 8 MW
H = 130 m
N = 200 rpm
Cv
= 0.98
η D=87
ϕ=0.46
m=
1
4
a.) Q = ?
b.) D = ?
c.) d = ? ; n = ?
Solution:
a.) Q = ?
P=γQH η D
Q=
P
=
γH η D
(
8000 kw
=7.21 m3 /s
kN
9.81 3 ( 130 m ) (0.87)
m
)
b.) D = ?
V b=
πDN
60
√(
V b=ϕ √ 2 gH =( 0.46 ) 2 9.81
m
m
( 130 m ) =23.231
s
s
)
( ms ) =2.22m
rev
π (200
min )
( 60 ) 23.231
D=
c.) d=?
m=
D
d
D 1
=
d 9
d=
2.22 m
=0.2467 m
9
n=?
n=
7.21
π ( 0.2467 )2
0.98 √ 2 ( 9.81 )( 130 ) (
)
4
=3
3. A Pelton wheel has to develop 12 MW under a head of 300 m at a speed of 500 rpm. If
the diameter of the jets is not to exceed 1/9 of the wheel diameter, estimate the number
and diameter of the bucket wheel and the quantity of flow. Assume overall efficiency =
88 percent, cv = 0.97 and Ф = 0.45.
Given: P = 12 MW
H = 300 rpm
N = 500 rpm
d 1
=
D 9
ηo =88
C v =0.97
ϕ=0.45
Solution:
Q=?
V 1=C v √ 2 gH =0.97 √ 2 ( 9.81 ) (300 m)=74.42 m/s
Q= A V 1
P=QγH η0
√(
V b=ϕ √ 2 gH =0.45 2 9.81
Q=
P
=
γH η o
(
D=
)
πDN
60
60(34.52 mls)
=1.32 m
π ( 500 rpm )
d=?
1 d
=
9 D
d=
)
12000 kw
=4.63 m3 / s
kN
9.81 3 ( 300 m) (0.88)
m
D=?
V b=
m
m
( 300 m )=34.52
2
s
s
1.32
=0.147 m
9
n=?
Q= A V 1 n
3
n=
[
4.63 m /s
=3.67∨4
2
π (0.147)
( 74.42 m/s )
4
]
4. A Pelton wheel to be designed is to run at 300 m under an effective head of 150 m. The
ratio of the nozzle diameter to the pitch circle diameter is 1/12. Assuming efficiency =
84%, cv = 0.98 and speed ratio = 0.45, determine (a) diameter of the wheel, (b) diameter
of the jet (c) the quantity of water flow (d) the minimum number of buckets required; and
(e) the power developed.
Given: N = 300 rpm
H = 150 m
d 1
=
D 12
η1=84
C v =0.98
ϕ=0.45
Solution:
√(
V 1=C v √2 gH =0.98 2 9.81
√(
V b=0.45 2 9.81
a.) D = ?
m
m
( 150 m )=53.16
2
s
s
)
m
(150 m)=24.41mls
s2
)
V b=
D=
πDN
60
60(24.41 mls)
=1.55 m
π (300 rpm)
b.) d =?
d 1
=
D 12
d=
1.55
=0.129 m
12
c.) Q = ?
[
2
]
π (0.129)
Q=
( 53.16 mls )=0.695 m3 /s
4
d.) z = ?
z=
m
12
+15= + 15=21
2
2
e.) P = ?
(
P= 9.81
kN
m3
0.695
( 150 m ) ( 0.84 )=859.0617 k
3
s
m
)(
)
5.
A francis turbine with an overall efficiency of 76 percent is required to produce 180kW. It
is working under a head of 8 m. The peripheral velocity is 0.25 (2gH) 1/2 and the radial velocity of
flow is 0.95 (2gH)1/2. The wheel runs at 150 rpm and the hydraulic losses in the turbine are 20
percent of the available energy. Assuming radial discharge, determine (a) the guide vane angle
(b) the wheel vane angle at inlet (c) the diameter of the wheel at inlet.
Given: ηo = 76%
Peripheral velocity = 0.25 (2gH)1/2 = Vf = 11.902 m/s
P = 180 kW
Radial velocity = 0.95 (2gH)1/2 = Vb = 3.1321m/s
H=8m
N = 150 rpm
Solution:
o 
shaft power 180 kW

 0.76
water power QgH
o 
180 kW
 0.76
kg
m
1 kN
Q (1000 3 ) (9.81 2 ) (
) (8 m)
m
s 1000 N
180 kW
kg
m
1 kN
 o (1000 3 ) (9.81 2 ) (
) (8 m)
m
s 1000 N
180 kW

kg
m
1 kN
(0.76) (1000 3 ) (9.81 2 ) (
) (8 m)
m
s 1000 N
Q
Q  3.018
m3
s
P  QgH h
h 
P

QgH
 h  0.76
180 kW
m
kg
m
1 kN
(3.018 ) (1000 3 ) (9.81 2 ) (
) (8 m)
s
m
s 1000 N
3
Vw Vb
gH
gH h
(9.81)(8)(. 76)
Vw 

Vb
3.1321
h 
Vw  19.043
tan  
m
s
Vf
Vw
 Vf 
11.902 
  tan  

 19.043 
 Vw 
  tan  
  32.00565
  32 0' 20.35"
ans (a)
Vb  DN
D
Vb
3.1321

N
 150 


 60 
D  0.399 m
ans (c)
Vw = Vb + X
X = Vw + Vb
= 19.043 m/s – 3.1321 m/s
X = 15.911 m/s
tan  
Vf
X
 Vf
  tan  
 X



 11.902 
 tan  

 15.911 
  36.7978
  36 47 ' 52.08 "
ans (b)
6.
An inward flow reaction turbine having an overall efficiency of 75 percent delivers 132
kW. The head H is 9 m, velocity of the periphery of the wheel is 54 m/s and the radial velocity is
18 m/s. The wheel makes 120 rpm. The hydraulic losses in the turbine are 20 percent of the
available energy. Determine the discharge at the inlet, the guide blade angle, the wheel blade
angle and the diameter and width of the wheel. Assume radial discharge.
Given: Vf = 18 m/s
ηo = 25 %
N = 120 rpm
P = 132 kW
Vb = 54 m/s
H=9m
Solution:
a)
Q=?
ηl = 20 %
P  QH o
Q
P
H o
Q
132 kW
9.81 m s 2  9 m  0.75
Q  1.99 m
s
∝1=?
b)
∝1=tan−1
ηH=
3
vf1
vw 1
vw 1 vb1
gH
o
0.8=
( v w1 ) ( 54 m/s )
( 9.81m/s 2 ) ( 9 m )
v w 1=1.308 m/ s
α 1=tan −1
vf 1
vw 1
α 1=tan −1
18 m/s
1.31 m/s
α 1=850 50'
c) β1 = ?
η H =1−η L =1−0.2=0.8
β 1=tan −1
β 1=tan
−1
0
vf 1
v b 1−v w 1
18 m/ s
(54−18 ) m/ s
β 1=26 33
'
d) D = ?
v b 1=πDN
D=
54 m/ s
π ( 120 rev /min ) ( 1min / sec )
D=8.60 m
e) B1 = ?
Q= AV
Q=πD B1 v f 1
B 1=
1.99 m3 /s
π ( 8.6 m/s ) (18 m/ s )
B 1=0.0041 m
7.
From the performance curves of turbines it is seen that a turbine of one meter diameter
under a head of one meter develops a speed of 25 rpm. What diameter should the prototype have
it is to develop 1000 kW, working under a head of 200 m with a specific speed of 150?
Given:
Nm = 400 m
Q = 4.5 m3/s
ηt = 82%
Ns = 60 rpm
Solution:
Dp = diameter for prototype = ?
(√ )
Dm
=
DP
D p=1 m
H p Nm
Hm Np
25 rpm
(√ 2001 mm ) 150
rpm
D p=2.36 m
8)
The following data relates to a hydroelectric plant, Head = 400 m, Discharge = 4.5 m3/s.
Turbine efficiency = 82%, Specific speed = 60, Determine (a) the power develop (b) the type of
turbine (c) the speed of the turbine.
Given:
H = 400 m
Q = 4.5 m3/s
ηt = 82%
Ns = 60 rpm
Solution:
a) P = ?
P=QγH ηt
P=( 4.5 m3 /s ) ( 9.81m/ s2 ) ( 400 m )( 0.82 )
P=14479.56 kW
b) The Pelton turbine which has a capacity from 300 m up to 2000 m of heads.
c) N = ?
N√P
NS=
H
60=
5
4
N √ 14479.56 kW
5
400 4
N=891.97 rpm
9)
A runner blade is to be designed for a Francis turbine, 1.5 m outer diameter and 0.75 m
inner diameter, to operate under a head of 120 m with a specific speed of 150 and generate 14000
kW. Assume hydraulic efficiency is 92 percent. What should be the inlet and outlet blade angles
if the water has to enter a wheel an angle of 11020’ and leave the blade with no velocity of whirl?
Given:
Outer diameter = 1.5 m
Inner diameter = 0.75 m
H = 120 m
Ns = 150 rpm
P = 14000 kW
ηH = 92%
α1 = 11020’
Solution:
NS=
N √P
H
N=
5
4
150 ( 120 m )
√ 14000 kW
N=503.505 rpm
v w =π Do N
v w =π ( 1.5 m )( 503.505 rev /min )
v w 1=39.545
tan α 1=
( 601 minsec )
m
s
vf1
vw 1
v f 1=v w1 tanα 1
v f 1=( 39.545 m/s )( tan 11.333 )
v f 1=7.926 m/s
η D=
v w1 v b 1
gH
v b 1=
( 9.81 m/s 2 ) ( 120m ) ( 0.92 )
39.545 m/s
v b 1=27.387 m/s
x=v w1 −v b
x=12.16 m/s
β 1=tan
−1
vf
9.23 m/ s
=
x ❑ 12.16 m/ s
0
0
β 1=33.10 =33 6
β 2=tan −1
[
'
vf 1
v
7.23 m/s
= f1 =
v b 1 πDN π ( 0.75 m )( 503.51 rev /min )( 1 min/60 sec )
β 2 = 21.840 =210 50'
]