- Francis turbines are radial flow turbines used in hydroelectric power generation. They have adjustable guide vanes and runners that control water flow. - Key dimensions include the diameters and widths at the inlet and outlet, as well as the rotational speed. These dimensions are calculated based on design criteria like flow rate, head, and hydraulic efficiency. - Dimensions are chosen to satisfy requirements for net positive suction head (NPSH) and minimize hydraulic losses while maintaining high hydraulic efficiency.

1. Francis turbines
•
•
•
•
Examples
Losses in Francis turbines
NPSH
Main dimensions

4. Traditional runner
X blade runner

5. SVARTISEN
P = 350 MW
H = 543 m
Q* = 71,5 m3/S
D0 = 4,86 m
D1 = 4,31m
D2 = 2,35 m
B0 = 0,28 m
n = 333 rpm

6. La Grande, Canada
P = 169 MW
H = 72 m
Q = 265 m3/s
D0 = 6,68 m
D1e = 5,71m
D1i = 2,35 m
B0 = 1,4 m
n = 112,5 rpm

8. Outlet
draft tube
Outlet
runner
Inlet
runner
Outlet
guide vane
Inlet
guide vane

9. Hydraulic efficiency
ηh =
c1u ⋅ u1 − c2u ⋅ u 2
=
g ⋅ Hn
1
2
2


 
c1
c3
 g ⋅ h1 + + z1  −  g ⋅ h 3 + + z 3  − losses

 

2
2

 

2
2


 
c1
c3
 g ⋅ h 1 + + z1  −  g ⋅ h 3 + + z 3 

 

2
2

 


10. Losses in Francis Turbines
Hydraulic Efficiency [%]
Draft tube
Output Energy
Head [m]

11. Hydraulic Efficiency [%]
Losses in Francis Turbines
Output Energy
Output [%]

12. Friction losses between runner
and covers
Friction losses

13. Gap losses
Gap losses

14. Friction losses
Friction losses in
the spiral casing
and stay vanes
Guide vane losses
Gap losses
Runner losses
Draft tube losses

16. Relative path
Absolute path without
the runner installed
Absolute path with the runner installed

17. Velocity triangles

18. Net Positive Suction Head, NPSH
2
2
2
c3
c3
c2
h2 +
+ z 2 = hb + h3 +
+ z3 + ζ ⋅
2⋅g
2⋅g
2⋅g
2
c3
− H s = h3 +
+ z3 − z 2
2⋅g
c2
h 2 + 2 + z2
2⋅g
c2
h2 + 2
2⋅g
2
c3
= h b − Hs + z2 + ζ ⋅
2⋅g
J
= h b − Hs + J
1
'
h 2 = hb + h 2

19. NPSH
h2
2
 c2


= hb − H s − 
 2 ⋅ g − J  > hva


NPSH
NPSH = h b − h 2 − H s
NB:
HS has a negative
value in this figure.

20. NPSH required
NPSH R
< h b − h va − H s
2
m2
= NPSH A
2
2
c
u
NPSH R = a ⋅
+b⋅
2⋅ g
2⋅ g
Turbines
Pumps
a 1.05 < a < 1.15 1.6 < a < 2.0
b 0.05 < b < 0.15 0.2 < b < 0.25

21. Main dimensions
• Dimensions of the outlet
• Speed
• Dimensions of the inlet
D1
D2

22. Dimensions of the outlet
We assume cu2= 0 and choose β2 and u2 from NPSHR:
2
c
u
a ⋅ ( u 2 ⋅ tan β 2 ) + b ⋅ u 2
NPSH R = a ⋅
+b⋅
=
2⋅ g
2⋅ g
2⋅ g
2
m2
13o <
35 <
1,05 <
0,05 <
β2
u2
a
b
<
<
<
<
2
2
2
22o
(Lowest value for highest head)
43 m/s (Highest value for highest head)
1,15
0,15

23. Diameter at the outlet
c m 2 = u 2 tan β 2
Connection between cm2and choose D2 :
D2
Q = π ⋅ 2 ⋅ cm2
4
⇒ cm2
Q⋅4
=
π ⋅ D2
2
D1
4⋅Q
⇒ D2 =
π ⋅ cm 2
D2

24. Speed
Connection between n and choose u2 :
π ⋅ D2 ⋅ n
u2 =
60
u 2 ⋅ 60
⇒ n=
π ⋅ D2

25. Correction of the speed
The speed of the generator is given from the
number of poles and the net frequency
3000
n=
zp
for
f = 50Hz

26. Example
Given data:
Flow rate
Head
Q = 71.5 m3/s
H = 543 m
We choose:
a = 1,10
b = 0,10
β2 = 22o
u2 = 40 m/s
(
)
 2
1,1⋅ 40 ⋅ tan 22 + 0,1 ⋅ 40 2
NPSH R =
= 22,8 m
2⋅g

27. Find D2 from:
4⋅Q
4⋅Q
D2 =
=
π ⋅ cm 2
π ⋅ u2 ⋅ tan β 2
4 ⋅ 71,5
D2 =
= 2.37 m

π ⋅ 40 ⋅ tan 22

28. Find speed from:
u2 ⋅ 60 40 ⋅ 60
n=
=
= 322 rpm
π ⋅ D2 π ⋅ 2.37

29. Correct the speed with synchronic speed:
3000
zp =
= 9.3
n
3000
choose z p = 9 ⇒ n K =
= 333 rpm
9

30. We keep the velocity triangle at the outlet:
u2
u2K
β2
c2
c2K
w2
cm 2 cmK
tan β 2 =
=
u2 u2 K
4⋅Q
4⋅Q
π ⋅ D22 K
π ⋅ D22
=
=
π
π
⋅ n ⋅ D2
⋅ nK ⋅ D2 K
60
60
⇓
3
3
nK ⋅ D2 K = n ⋅ D2
⇓
D2 K
3
n ⋅ D2
322
=3
= 2.373 ⋅ 3
= 2.35m
nK
333

31. Dimensions of the inlet
u 1 ⋅ c u1 − u 2 ⋅ c u 2
ηh =
= 2 ⋅ ( u 1 ⋅ c u1 − u 2 ⋅ c u 2 )
g⋅H
At best efficiency point, cu2= 0
u 1 ⋅ c u1
ηh ≈ 0,96 =
= 2 ⋅ u 1 ⋅ c u1
g⋅H
ηh
0,96
c u1 =
=
2 ⋅ u1 2 ⋅ u1

32. Diameter at the inlet
We choose:
0,7 < u1 < 0,75
D1 n ⋅ 2 ⋅ π D1
u1 = ω ⋅
=
⋅
2
60
2
⇓
u1 ⋅ 60
D1 =
n⋅π
D1
D2

33. Height of the inlet
Continuity gives:
c m1 ⋅ A1 = c m 2 ⋅ A 2
We choose:
cm2 = 1,1 · cm1
1.1 ⋅ π ⋅ D
B1 ⋅ D1 ⋅ π =
4
2
2
B1 = B0

34. Inlet angle
cu1
cm1
c1
c m1
tan β 1 =
u1 − c u1
u1
β1
w1

35. Example continues
Given data:
Flow rate
Head
Q = 71.5 m3/s
H = 543 m
We choose:
ηh = 0,96
u1 = 0,728
cu2 = 0
η h = 2 ⋅ u 1 ⋅ c u1
ηh
0.96
⇒ c u1 =
=
= 0,66
2 ⋅ u1 2 ⋅ 0,728

36. Diameter at the inlet
u1 = u1 ⋅ 2 ⋅ g ⋅ h = 0,728 ⋅ 2 ⋅ 9,81⋅ 543 = 75,15 m s
u1 ⋅ 60 75,15 ⋅ 60
D1 =
=
= 4,31 m
n ⋅π
333 ⋅ π
D1

37. Height of the inlet
1,1⋅ π ⋅ D
1,1⋅ 2.35
B1 =
=
= 0,35 m
4 ⋅ D1 ⋅ π
4 ⋅ 4,31
2
2
B1
2

38. Inlet angle
cu1
cm1
c1
u1
β1
w1
cm 2 sin β 2 ⋅ u2 sin 22 ⋅ 40,9 m s
cm1 =
=
=
= 13,9 m s
1,1
1,1
1,1
c m1 =
c m1
13,9
=
= 0,135
2⋅g ⋅h
2 ⋅ 9,81 ⋅ 543
 c m1 
0,135


 = arctan
β1 = arctan
 = 62.9
u −c 
 0,728 − 0,659 
 1 u1 

39. SVARTISEN
P = 350 MW
H = 543 m
Q* = 71,5 m3/S
D0 = 4,86 m
D1 = 4,31m
D2 = 2,35 m
B0 = 0,28 m
n = 333 rpm