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 Q_p² )/ D_p^5.333] L_p 
• = 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