The Number of Blades for Crossflow Turbine

By Ahmad Suhendra

Example :

A crossflow turbine is designed with blade angle β1=30 0 (angle between the relative velocity and tangential velocity) and α1=16 0 (angle between the absolute velocity and tangential velocity). The turbine have a rotation speed of 750 rpm on the net head 10 m, flow 100 lt / s, what is the specific speed, absolute and tangential velocity , outside diameter of the disc, blades spacing and number of baldes?

Crossflow turbine requirements are :

• ns = 42 – 170 (classical based on metric HP)
• Ns = 0.86* ns (True metric specific speed)
  
• Nt = 100 -1000 rpm (nominal speed)

1.Specific Speed

ns (rpm) = 3.65 x Nt (rpm) x Q(m^3/s)^ 0.5 / H_netto(m)^ 0.75

= 3.65 * 750 * 0.01^ 0.5 / 10^ 0.75

= 48.7 rpm ----> Ok, Crossflow

2.Absolute Velocity of water

kc = 0.967 Coefficient dependent upon the nozzle

C = kc * (2 * g * Head)^ 0.5

= 0.967 (2 * 9.8 * 10 )^0.5

= 13.54 m/s

3.Tangential velocity

U1 = ( C * Cos α1) / 2

= ( 13.54 * Cos 16 )/ 2

= 6.51 m/s

4. Outside diameter of disc

D1 = 60 * U1 / (phi * Nt)

= 60 * 6.51 / 3.14 * 750

= 0.166 m

Then :

S1 = k * D1----> k = 0.075 – 0.1 (respectively)

Taken k = 0.087

S1 = 0.087 * D1

= 0.087 * 0.166

= 0.0144 m

5. Spacing of blades in disc

t = S1 / Sin β1

= 0.0144 / Sin 30

= 0.0288 m

6.Number of Blades

Z = phi * D1 / t

= 3.14 * 0.166 / 0.0288

= 18 blades (20 were used for design)

Ref : Mockmore C.A and Merryfield Fred " The Banki Water Turbine " , Bulletin Series No:25 Oregon State College Corvallis

What is the blade length (bo) limits on crossflow turbine ?

By Ahmad Suhendra

Crossflow turbine blade has a length limit that must be met in the design, for the turbine at an angle α1=16 o, nozzles coef. kc = 0.98 and k = 0.075 to 0.10 are as follows:

1). C (m/sec) = kc.(2.g.H)^0.5
2). Phi.D1.Nt / 60 = (C. cos α1)/2

Then:
The diameter of blade is :
3). D1 (m) = 39.85 H ^ 0.5 / Nt

Blade length (bo) is obtained from the following equation(4)
4). Q(m^3/sec ) =(So.bo)[ kc ( 2.g.H )^0.5 ]
5). So(m) = k D1

By entering the equation (5) and (3) into equation (4) is obtained :
The blade length (bo) limits is :

6). bo = ( 0.058 Q.Nt / H ) to ( 0.077 Q.Nt / H )

Where:

• g is the constant of gravity 9.8 m/s^2
• C is the absolute velocity of water (m/sec)
• Phi is 3.14
• Nt is the nominal turbine speed (rpm)
• D1 is the diameter of Disc (m)
• So is the thickness of the water jet (m)
• H is the net head (m)

Ref : Edy Sunarto , dkk,” Pedoman Rekayasa Tenaga Air (Hydropower Engineering Guidelines) ”, UPT Hidro Elektris BPPT, Jakarta 1991

What is the maximum performance of crossflow turbine?

By Amad Suhendra

Sonnek (1923) modified the banki theory assuming a constant angle of the blade equal to 30 , so that the expression resulted in maximum performance:

ηmax = 0.863-0.264 *(D / H)

D is the diameter of turbine (disc) and H is the net head

Ref : Ariel R.Marchegiani , "Turbina de Flujo Transversal O Michell-Banki", Universidad Nacional Del Comahue

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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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