Gec223 Fluid Mechanics Hydropower Impulse Turbine Pelton Wheel
Gec223 Fluid Mechanics Hydropower Impulse Turbine Pelton Wheel
Gec223 Fluid Mechanics Hydropower Impulse Turbine Pelton Wheel
AMASCE, GMICE]
GEC223: FLUID MECHANICS
MODULE 4: HYDROPOWER SYSTEMS
TOPIC: IMPULSE TURBINES-PELTON WHEEL
DEPARTMENT OF CIVIL ENGINEERING, LANDMARK UNIVERSITY, KWARA STATE,
NIGERIA
1650-1700.
u1
motion of vane/bucket.
V r1
, with
, Vf1
V1
V 2 = Velocity of jet leaving the vane or velocity of jet at the outlet of the
vane
V r 2 = Relative velocity of jet with respect to (w.r.t.) the outlet vane
V r2
with
V2
= V 1 - u2 = V 1 - u (Since u1 = u2 =u)
= V 1 and = 0 and = 0
= K V r1
Depending on the magnitude of the peripheral speed (u), the unit may
have a slow, medium or fast runner and the blade angle
and V w 2 will
vary as follows:
< 900; V w 2 is +ve
i.
Slow runner
ii.
iii.
Fast runner
aV 1
d2
4
aV 1
( V w 1+ V w 1 ) x u
Work done per second per unit weight of water striking the buckets =
aV 1(V w 1+V w1 ) x u
aV 1 x g
=
3
1
[ V +V ]
g w1 w1 u
Recall, m = Q and Q = a V 1 , m = a V 1
2
K.E. of jet per second = ( a V 1 ) x V 1
Hydraulic Efficiency,
Hydraulic Efficiency,
2( V w 1+V w1 )x u
V 12
_______Equation 1
= V 1 ; V r 1 = V 1 - u1 = V 1 - u
Vw2
= V r 2 cos - u2 = V r 2 cos u
Since V r 2 = k V r 1 and V r 1 = V 1 - u
Vw2
= kV r 1 cos u = k ( V 1 -u)cos -u
2 [ V 1 +k (V 1u)cos u ] u
V 12
Rearranging gives,
h
V
( 1u)(1+k cos )
u
2
_____________Equation 2
aV 1(V w 1+V w1 ) x u
( a V 1 ) x V 12
V
( 1u)(1+k cos )
2 [ u V 12 ]
=0
du
2(1+ k cos )u
V 12
Since
when
=0
This implies
Or
V1
2( 1+ k cos )u
V 12
d
V 1 uu2 )
(
du
d
du
(V 1 uu 2) = 0
0
=0
V1
= 2u and u =
V1
2 _______________Equation 3
( h ) max =
V
V1
( 1 )(1+k cos )
2
V1
2
2
V1
V
(1+ k cos ) 1
2
2
2
V1
V1
2
in
( h ) max =
1+ kcos
2
_________________Equation 4
( h ) max =
1+ cos
2
__________________Equation 5
4 fL V 2
D X 2g
=
=
w
Q (V V w 2) u
g a w1
w Qa H
( V w 1 V w 2)u
gH
( V w 1 V w 2)u
g
ii.
Shaft Power
Bucket Power
P
w Qa
Since Hr =
(V
w1
+V w2
u
g
( V w 1 V w 2)u
g
P
w Qa H r
iii.
= Water Power
Shaft Power
P
wQH
= h x m x v
Hr
Q
P
x
x a
H w Qa H r Q
P
wQH
P
wQH
x g
K u 2 gH
D=
60 U
N
5. Jet ratio (m) = Ratio of pitch diameter of pelton wheel to diameter of the
jet (d).
m =
D
d
practice, m = 12 is adopted.
6. Number of Jets: Practically,
D
2d
Example 1
9
= 15 + 0.5 d
i.
ii.
iii.
i.
4800
9.81 x 420 x 0.85
1.37m3/s
Velocity of jet , V1 = C v 2 gH
a = 2 x 88.05
ii.
iii.
= 7.779 x 10-3 m2
DN
60
D=
D x 480
60
= 1.66m
= 0.913 = 91.3%
Example 2
The water available for a pelton wheel is 4m 3/s and the total head from
the reservoir to the nozzle is 250m. The pipe is 3km long. The efficiency of
transmission through the pipeline and the nozzle is 91% and the efficiency
of each runner is 90%. The velocity coefficient of each nozzle is 0.975 and
coefficient of friction 4f for the pipe is 0.0045. Determine:
i. The power developed by the turbine
ii. Diameter of the jet
iii. Diameter of the pipeline
Solution
Rate of flow, Q = 4m3/s
Total or gross head, H g = 250m
11
H gh f
Hg
250h f
250
Q V 1
8486439W
Water power = 8486.44KW
Hydraulic Efficiency,
0.91 =
= x 1000 x 4 x
65.14
= 8486439Nm/s =
2) Diameter of jet, d
Discharge per jet, q =
Total discharge
No . of jets
12
4
4
= 1.0m3/s
q=
d2 x V 1
x 65.14
d=
1 /2
d 2 x 65.14
4
1x4
x 65.14
1 /2
= 0.14m
3) Diameter of pipeline, D
Head lost due to friction, hf =
4 fL V 2
D x2 g
Q
Area
Q
2
D
4
4Q
D2
hf =
0.0045 x 3000 x
Dx 2g
0.0045 x 3000 x 16 Q 2
2
4
D x 2 x 9.81 x x D
22.5 =
D5 =
( 4DQ )
=
0.0045 x 3000 x 16 x 4 2
5
2
D x 2 x 9.81 x
17.85
D5
17.85
22.5
17.85
22.5
1/ 5
= 0.955m
Example 3
A pelton wheel nozzle, for which C v is 0.97, is below the water surface of
a lake. The jet diameter is 80mm, the pipe diameter is 0.6m, its length is
4km, and f is 0.032 in the formula,
fL V 2
D x 2 g . The buckets deflect the
jet through 1650 and they run at o.48times the jet speed, bucket friction
reducing the velocity at outlet by 15% of the relative velocity at inlet.
Mechanical efficiency is 90%. Determine:
i. The flow rate
ii. Shaft power developed by the turbine
13
Solution
1) Flow rate, Q
V = Velocity of water in pipe
V 1 = Velocity of jet of water
Using equation of continuity,
AV = a V 1
Where A = Area of pipe and a = area of jet
D2
4
xV=
V=
d2
D2
d2
4
x V1
V1
0.082
0.6 2
x V 1 = 0.0177 V 1
Applying Bernoullis equation to free water surface in the reservoir and the
outlet of the nozzle,
Head at reservoir = Kinetic head of jet of water + head lost to friction in
pipe + head lost in nozzle
14
V 12
2g
fL V
D x2 g
+ hnozzle __________equation 1
Cv
1
or (V 1)th = C v
(V 1)th 2
2g
V1 1
1
2 g C v2
V 12
2g
V1
Cv
( )
1
2g
V 12
2g
V 12
2g
0.032 x 4000 V 2
0.6 X 2 X 9.81
fL V 2
D x2 g
H reservoir
400 =
V1
2g
400 =
0.032 x 4000(0.0177 V 1 )2
0.6 X 2 X 9.81
V1 1
1
2 g C v2
V1
2
2 x 9.81 x 0.97
V1
+ 2 x 9.81 x 0.972
2
2
2
400 = 0.0034 V 1 + 0.054 V 1 = 0.0574 V 1
V1
400
0.0574
1 /2
= 83.48m/s
x 83.48
Q = 0.419m3/s
2) Shaft power
Velocity of bucket, u1 = 0.48 V 1 = 0.48 X 83.48 = 40.07M/S
From the inlet velocity triangle,
V r 1 = V 1 - u1 = 83.48-40.07 = 43.4m/s
V w1 =
V1
= 83.48m/s
15
2
x 0.08
Mechanical efficiency,
Power gi ven
Shaft power
runner
( V w 1V w2 ) x u
Shaft power =
wQ
g
> 900.
( V w 1V w2 ) x u1
= 0.9 x
9.81
9.81 x0.419(83.48-
4.44) 40.07
Shaft power = 1194.3KW
Example 4
The following data relate to a pelton wheel:
Head______________72m
Speed of wheel____________240rpm
Shaft power of wheel_______115KW
Speed ratio_______________0.45
C v _______________________0.98
Overall efficiency___________0.85
Design the Pelton wheel
Solution
Effective head, H = 72m
Speed of wheel, N = 240rpm
Shaft power, P = 115KW
Speed ratio, K u = 0.45
Cv
= 0.98
u=
60 x 16.56
x 240
= 1.32m
2) Diameter of jet, d
Overall efficiency,
0
60 u
N
Shaft power
Water power
P
wQH
115
9.81 x Q x 72
115
0.85 x 9.81 x 72
Q=
= 0.1915m3/s
d2
4
0.1915 x 4
x 36.8
V1
d2
x 36.8
4
1 /2
= 0.0814m = 81.4mm
3) Size of buckets
Width of bucket, B = 3d to 4d
choosing B = 3.5d
D
2d
= 15 +
1.32 x 1000
2 x 81.4
= 23
Number of buckets, Z = 23
17
Example 5
A pelton wheel of 1.1 mean diameter works under a head of 500m. The
deflectionof jet is 1650 and its relative velocity is reduced over the
bucket by 15% due to friction. If the diameter of jet is 100mm and the
water is to leave the bucket without any whirl, determine:
i.
ii.
iii.
iv.
v.
vi.
Solution
Main bucket diameter, D = 1.1m
Net head, H = 500mm
Jet deflection = 1650
Reduction of relative velocity due to friction = 15%
Jet diameter, d = 100mm = 0.1m
Cv
= 0.97
i.
Bucket speed, u1 = u2 = u
Relative velocity at inlet, V r 1 = V 1u1 = 96.07-u
18
Relative
velocity
at
outlet,
V r2
0.85 V r 1
0.85(96.07-u)
_____________equation 1
Blade angle at exit, = 1850-1650 =150
Since jet leaves bucket without whirl,
V r 2 cos
=90
u = 1.821
Recall, u =
= 43.31m/s
DN
60
x 1.1 x N
60
43.31 x 60
x 1.1
u
V1
= 752rpm
iii.
43.31
96.07
ii.
V1
= 0.45
x 0.12
x 96.07
4
0.7545m3/s
Impulsive force on buckets, F =
( V w 1 V w 2 Since V w 2 = 0
19
V 12 = x 1000 x
2
0.7545 x 96.07 = 3481808W =3481.8KW
vi.
Efficiency of wheel,
wheel
Power input
Power developed by wheel
buckets
3139.3
3481.3
H nozzle
0.9016 = 90.16%
H nozzle
2
= (1- C v ) H =
H buckets
V r 12
2g
2
(1- K
20
HL