Chapter 9:
Hydraulic
turbines
ENEE 575 /ENEN 619
Lecture 2: Fundamentals of
Hydro Turbines
https://www.hydropower.
org/publications/2021hydropower-status-report
Possible
control
volume
fig_02_01
Suggested control volume
Simplified centrifugal pump for Tutorial
Assignment Q 2.
Ht
V −V
PA − PB
= Z A − ZB +
+
− losses A→ B
2g
g
2
A
PE change
2
B
KE change
Pressure change
Ht is the “head” available to the turbine
Losses due to friction in the pipes, valves, bends etc
Ht
V −V
PA − PB
= Z A − ZB +
+
− losses A→ B
2g
g
2
A
2
B
Apply equation between A and B as shown: VA = VB = 0 and PA = PB = 0
(gauge pressure). Also assume all losses are due to friction in the pipe
losses A→ B
V2 L
=
f
2g D
for a single pipe
V is average velocity in pipe f is the friction factor
L is length of pipe
D is pipe diameter
Use the relation between V and Q:
2
8Q 2
L
8
Q
fL
 
= Z A − ZB −
f   = Z A − ZB −
2 4
2 5
g D
g D
D
Ht
Ht
= H (1 − F ) where H
= Z A − ZB
F is the losses as a fraction
There are many expressions for f. One of the
best for fully developed turbulent pipe flow is
f
=
0.3086
log10  / (3.7 D)

1.11


2
 is the roughness length measuring the
roughness of the inside of the pipe
Example Q = 0.25 m3/s. Turbine h = 0.85
f
=
0.3086
 log10 0.046 / (3.7  300)

= 0.0130
1.11


2
2
Ht
8Q f L
= H−
g 2 D 5
8  0.252  0.0130  350
= 915 − 892 −
2
5
9.81    0.3
= 13.3m
Max. potential energy change through
turbine = gHt
Mass flow rate through turbine = Q
Power, W&is given by
W& = h g  QH
Q is the volume flow rate
h is the efficiency
t
= 0.85  9.81  10  0.25  13.3
3
= 27.72 kW
YoosefDoost, A., &
Lubitz, W. D.
(2020).. Sustainability,1
2(18), 7352.
The turbines on this chart are all “head-driven” hydroturbines relying on an elevation
difference to provide the potential energy to be converted into power. The second class
of hydroturbines are “run-of-the-river” turbines convertingthe kinetic energy of the
water. These turbines are similar to wind turbines.
Pelton Wheel Schematic
R is the radius assumed constant – the direction of R is into the page
Direction of rotation
U = WR
W is the angular velocity
R is the radius assumed
constant
V1 is the exit
velocity of
the jet which
is labelled Vj
on the
previous
slide
Torque = (mass flow rate) × radius × (tangential velocity entering CV –
tangential velocity leaving CV)
T
(
)
& V j − WR (1 − cos 2 )
= mR
Thus the power, W, is given by
(
)
W& = m&WR V j − WR (1 − cos 2 )
Maximum power, W, occurs when dW/dW = 0, i.e. when
W = Vj
W&max
( 2R )
So the maximum power is
2
2
&
&
= mV j (1 − cos  2 ) / 4  mV j / 2
So the wheel converts most of the jet’s KE into power
(
T
= mR V j − V2
V2
= 2W R − V j
)
V2 is the velocity of the water exiting the
stationary control volume
Idealized blade
Vj
2 = 180
V j − WR
WR − V j
 = 0.046 mm
0 =
losses A→ B
VA2 − VB2 PA − PB
H+
+
− losses A→ B
2g

V2
=
2g

f


L

 D +C+ K



C is the “minor pipe losses” = 1500 in this case
K is the remaining minor losses in bends etc = 5.25 in this case
If the generator efficiency = 0.85, find the maximum power output
and the r.p.m. of the wheel for that power.
VA = 0 and PA = PB
We want an equation for VB = Vj in the Pelton wheel example
8Q 2
8Q 2
0 = H−
−
2 4
g DB g 2 D 4

f


L

 D +C + K



which gives the volume flow rate Q.
f
=
0.3086
 log10  / (3.7 D) 


0.3086
1.11
=
2
 log10 0.046 / (3.7  1000)

= 0.0104
1.11


2
Q =  DB2
gH
  L

 DB4

8   f  + C  + K  4 + 1

D
  D

=   0.182 
9.81  670
8  0.0104  ( 6000 + 1500 ) + 5.25  0.184 + 1
= 2.80 m 3 /s
Thus Vj = 2.80×4/(×0.182) = 109.9 m/s
W&max = 0.5×103×2.80×109.92 = 16.92 MW
W = 109.9/(2×1.5) = 36.65 rad/s = 350 r.p.m.