Friday, April 16, 2010

Calculation of penstock diameter


By Ahmad Suhendra
Penstock serves to drain the water into the turbine, because water power is a combination of head (H) and flow (Q). Water will flow down and create pressure on the end of the pipe that provides power to rotate the turbine,






Example

  • Penstock length (Lp) = 15 m
  • Flow (Qp) = 2.0 m3/sec
  • H gross = 9 m
  • Manning Coef (n) = 0.012 (value of the Manning roughness coefficient for Mild steel )
What is the diameter and head loss of the penstock ?
1. Penstock diameter (Dp)

  • Dp = [C Qp / V ]0.5 
  • C (constants) = 1.273 = 4 / (phi)
  • V (water velocity ) = 1 - 2.8 m / sec.
  • Taken, V = 1.6 m / sec.)*
  • Dp = [1.273 (2.0 / 1.6)] 0.5 = 1.261 m ( penstock diameter)

2. Head loss
Assumed there are only head loss due to friction in penstock (applying the manning formulae), then :

  • Head loss = [(10.29 n 2 Qp2 )/ Dp5.333] L
  • = 0.114 m
  • Percent of Head loss = (0.114 m / 9 m ) * 100 % = 1.27 %)*
Note )* :
  • Generally, for economic reasons the percent of head loss between 5 % to 10 %
  • [Permissible velocity in Penstocks,V(m/s) = 0.125 (2 g H)^0.5, Ref : USBR (1961) (P J Bier)]
  • If using Sarkaria's eq ==>; Dp = 3.55 ((Qp^2/(2 g H ))^0.25 = 1.377 m
Ref : ESHA (European Small Hydropower Association),”Layman’s Handbook on How To Develop a Small Hydro Site,”2nd ed, 1998

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