How to calculate the diameter and number of nozzles for Pelton turbine

By Ahmad Suhendra

Example :

A Pelton turbine is planned to be working on the head netto 80 m and a flow 20 liters / s , what is the number of nozzles installed on these turbines?

Pelton turbine requirements are :

• ns (specific speed , rpm) = 8 – 72 rpm
• Nominal speed = 500 – 1500 rpm

1.Specific speed of turbine

• Taken 1500 rpm as nominal turbine speed (N_turbine)
• Flow (Q) = 0.020 m3/s
• H_netto = 80 m

ns = 3.65 *N_turbine (rpm) * Q(m^3/s) 0.5 / H_netto(m)^ 0.75

= 3.65 * 1500 rpm * (0.020 m3/s)^0.5 / (80 m)^0.75

= 774.28 / 26.75 = 28.95 rpm

Based on the practical calculation,

• If ns >= (8) and < (29.5) then 1 nozzles
• If ns >= (29.5) and < (42) then 2 nozzles
• If ns >= (42) and < (54) then 3 nozzles
• If ns >= (54) and < (58) then 4 nozzles
• If ns >= (58) and <=(72) then 6 nozzles

ns = 28.95 rpm -------> Ok, Pelton turbine with one nozzles

2. Absolute velocity of water

kc ( nozzles coef.) = 0.96 to 0.98, Taken kc = 0.98

C = kc (2 * 9.8 m/s^2 * H_netto(m) )^ 0.5

= 0.98 (2 x 9.8 x 80 ) 0.5 = 38.81 m/s

3. Diameter of nozzles

Dn = [ (4 x Q/(number of nozzles)) / (phi x C) ] ^0.5

= [ 4 x (0.020/1) / (3.14 x 38.81)] ^0.5

= [ 0.08 / 121.86 ] 0.5 = 0.0256 m or 26 mm

Ref : Edy Sunarto dan Markus Einsering”,Turbin Pelton Mikro”, MHPG Andi Offset Yogyakarta 1992

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

Examples of open-channel calculation

1. What are the types of open channel flow?
2. Water is flowing at a velocity of 12 ft/s and depth of 10 ft in a channel of rectangular section. Find the change in depth and absolute water level produced by (a) the smooth upward step of 0.5 ft, (b) the smooth downward step of 1 ft in the channel bed. Also (c) find the maximum allowable size of upward step for the upstream to be possible as specified.
3. etc

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