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
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
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
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
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