94 CIVIL WORKS GUIDELINES FOR MICRO-HYDROPOWER IN NEPAL
Example 6.1 Sizing of Galkot penstock pipe
The required data for the design of the Galkot penstock are as
follows:
Q = 42 l/s (calculated in Example 4.2)
hgross = 22m
two vertical bends, Þ = 20° & 42°, both mitred.
Penstock material: uncoated mild steel, 35 m long and
flange connected.
Turbine type: crossflow
Refer to Drawing 420/04/3A01 in Appendix C for the ground
profile and bend angles.
Pipe diameter calculation
Set V = 3.5 m/s
Calculate the internal pipe diameter:
d = 4Q / ΠV
d = (4 x 0.421)/( Π x 3.5)
= 0.391 m
Calculate wall loss:
From Table 4.3 choose k = 0.06 mm for uncoated mild steel.
k /d = 0.06 /391
= 0.00015
1.2Q/d = (1.2 x 0.421) / 0.391
= 1.29 m
f = 0.013 (Moody Chart, Chapter 4)
hwall loss
hwall loss
= f ( LV2 / d x 2g) = 0.013 ( 35 x 3.52 / 0.391 x 2g )
= 0.73 m
Inlet loss:
Kentrance
hInlet loss
= 0.5 for this case (Table 4.4 )
= Kentrance x V2 / 2g
= 0.5 x 3.52 /2g
= 0.31 m
Note that Exit loss = 0 since the flow at the end of the penstock
is converted into mechanical power by rotating the turbine run-
ner.
For mitred bends, from Table 4.4
è = 22°, Kbend = 0.11
è = 42°, Kbend = 0.21
Bend losses
Total head loss
% head loss
= (0.11 + 0.21) x (3.52 / 2g)= 0.20m
= 0.73 m + 0.31 m + 0.20 m = 1.24m
= 1.24 / 22 x 100 % = 5.6% < 10% OK.
Although the calculated pipe diameter was 391 mm internal,
Drawing 420/04/3C02 in Appendix C shows the outside pipe
diameter to be 388 mm, i.e. internal diameter 388 - 2 x 3 = 382
mm. This was chosen to correspond to a 1200 mm standard
plate size (1200/Π = 382). In practice, however, the manufac-
turer proposed 400 mm internal diameter pipes for the same
cost, because this corresponded to the plate size that he had..
Pipe wall thickness calculations
Calcutate the pressure wave velocity ‘a’
a = 1440 / 1 + [(2150 xd) / (E x t)]
E = 2.0 x 105 N/mm2 for mild steel (Table 6.2)
d = 400 mm
t = 3 mm
or, a = 1440 / 1 + [ (2150 x 400) /(2.0 x 105 x 3) ]
or, a = 923 m/s
Now calculate the critical time:
Tc = 2L / a
= 2 x 35 / 923
= 0.08 s .
Note that it would be impossible to close the valve in the power-
house in 0.08 seconds!
Choose closure time T = 10 s > 2 Tc = 0.16 s
K = [ LV / ghgrossT ] 2
= [35 x 3.5 / g x 22 x 10] 2
or, K = 0.003. Since K is less than 0.01
or, hsurge = hgross x ök = 22 x ö 0.003
or, hsurge = 1.20 m
or, htotal = hsurge + hgross = 23.2 m
The pipes were manufactured by welding (1.1) rolled flat steel
plates (1.2). 1.5 mm has been subtracted to allow for at least 15
years of life.
teffective
3 / (1.1 x 1.2) -1.5 =0.77 mm
Now check the safety factor:
SF = (200 x teffective x S) / (htotal x d)
S = 320 N/mm2 for ungraded mild steel (Table 6.2)
SF = (200 x 0.77 x 320) / (23.2 x 400)
SF = 5.3 > 3.5 OK .
Note that the safety factor is higher than required but the mini-
mum recommended thickness for flat rolled mild steel pipe is 3
mm.
The Galkot penstock alignment before and after pipe installation
can be seen in Photographs 6.7 and 6.8 respectively.