Saddles, supporting blocks and expansion joints

By Ahmad Suhendra

The saddles are designed to support the weight of the penstock full of water, but not to resist significant longitudinal forces. The vertical component of the weight to be supported, in kN, has a value of :

F1 = (Wp + Ww) * L Cos θ
where :
Wp = weight of pipe per meter (kN/m)
Ww = weight of water per meter of pipe (kN/m)
L = length of pipe between mid points of each span (m)
θ = Angle of pipe with horizontal

The design of support rings is based on the elastic theory of thin cylindrical shells. The pipe shell is subject to beam and hoop stresses, and the loads are transmitted to the support ring by shear. If penstocks are continuously supported at a number of points, the bending moment at any point of penstock may be calculated assuming that it is a continuous beam, and using the corresponding equation. The rings are welded to the pipe shell with two full length fillet welds and are tied together with diaphragm plates.

The span between supports L is determined by the value of the maximum permissible deflection L/65000. Therefore the maximum length between supports is given by the equation:

L= 182.61 * [(Dp + 0.0147)^4 - Dp^4)]^(1/3) / (Wp + Ww)

Example :
What is The vertical component of the weight to be supported if such data is given below:

1.Diameter of pipe (Dp) = 0.636 m
2.Pipe thickness (e) = 0.005 m (5 mm)
3.Density of pipe (ρ steel) =7.9 ton/m^3
4.Density of water (ρ water) = 1 ton/m^3
5.Angle of pipe with horizontal (θ )= 5 deg


I. Wp (weight of pipe per meter)
= phi * ( Dp + e) * e * ρ steel
= 3.14 (0.636 + 0.005)* 0.005 * 7.9
= 0.079 ton/m

II. Ww (weight of water per meter of pipe)
= [(phi x ( Dp^2) / 4] * ρ water
= [(3.14 x 0.636^2)/4] * 1
= 0.32 ton /m

III. Total weight (Wp + Ww)
= 0.399 ton/m (Wtotal)
= 0.399 * 9.81
= 3.914 kN/m

IV. The maximum length between supports.
L= 182.61 * [(Dp + 0.0147)^4 - Dp^4)]^(1/3) / (Wp + Ww)
= 182.61 * [(0.636 + 0.0147)^4 - 0.636^4]^(1/3) / (3.914)
= 182.61 * [0. 01566 ]^(1/3) / (3.914)
= 182.61 *[0.2502]/3.914
= 11.67 m

V. The vertical component of the weight to be supported.
F1 = (Wp + Ww) * L Cos θ
=3.914 kN/m * 11.67 m* Cos 5
= 45.503 kN

Ref :
ESHA (European Small Hydropower Association),”Layman’s Handbook on How To Develop a Small Hydro Site,”2nd ed, 1998 http://www.scribd.com/doc/8885765/Layman-Handbook-for-hydro-electric-power-plants

Photo
http://www.fr.aps-sales.com/documentos/downloads/HydroPower%20and%20penstock%20applications.pdf

Sebintang River (0 36 22.4N ; 113 02 41.7E) in West Borneo, Indonesia

The Choice of Materials For Water Turbines

Photo by Ahmad Suhendra

Determination of material for water turbine is very important, because turbines work on conditions such as high pressure, abrasion caused by sand carried by the flow, and also of pH levels and salinity that can cause corrosion, so that the appropriate materials needed to treat the condition. There are several types of water turbines and materials used in their manufacture.
Penentuan bahan untuk membuat turbin air sangatlah penting, karena turbin bekerja pada tekanan tinggi, adanya pengikisan karena pasir yang terbawa air , juga derajat keasaman (pH air) dan kadar garam dalam air yang dapat menyebabkan korosi, maka untuk itu diperlukan bahan yang tepat untuk mengatasi hal tersebut.
Disini diperlihatkan beberapa jenis turbin air dan bahan yang digunakan untuk membuatnya.

Read More

Ref:
http://www.ivt.ntnu.no/ept/fag/tep4200/innhold/The%20choice%20of%20materials.pdf

Shaft Diameter and Shaft Critical Speed

By.Ahmad Suhendra
The size of the diameter of the shaft is very influential at critical speed. If the shaft is rotating close to or exceed the critical speed, the shaft becomes very fast vibrate and cause the bearings become damaged, so the correct practice should be working below the shaft critical speed.

Example calculations.
A Pelton turbine disc has a weight of 30 kg , and planned to have a nominal rotational speed is 1500 rpm and Power = 100 kilowatts. Shaft material (solid cylinder shaft) to be used is a standard steel, with the number: AISI-SAE 1040, which has Ultimate tensile stress (UTS) 90.000 psi and Young's modulus (E) 29 10^6 psi .
What is the diameter (D,mm) of the shaft is used?
If the distance between the bearings and disc is L= 50 mm (see picture),
what is the critical rotational speed (Nc)?
Did diameter shaft that is used safely?

Given :
N = 1500 rpm
P =100 kW
m = 30 kg
UTS = 90,000 psi (620.5 N/mm^2)
E = 29*10^6 psi (199948 N/mm^2)

I.) Torsional or twisting moment (T, N.mm).
T = 9.55 10^6 P/N................................(1)
= 9.55 10^6 * 100 / 1500
= 636667 N.mm

II.) Ultimate shear stress (UST, N/mm^2).
UST = 0.75 *UTS........................................(2)
= 0.75 * 620.5
= 465 N/mm^2

The allowable stresses (Ss) that are generally used in practice are:
4000 psi ( 28 N/mm^2) for main power-transmitting shafts (Ss < UST).

III.) Polar Section Modulus (Zp, mm^3).
Zp =T / Ss............................................(3)
= 636667/28
=22738 mm^3
and,
for circular solid shaft is :
Zp = (phi D^3) / 16................................(4)
D^3 = Zp *16 /phi
= 22738 * 16 / 3.14
D = 115863^(1/3)
= 49 mm or 2 inchi. (Shaft diameter)

IV.) Polar moment of inertia (I ,mm^4) of a circular solid shaft.
I=1/4* R^4.............................................(5)
I = phi D^4 / 64
= 3.14 ( 49^4)/64
=282836 mm^4

V.) Critical speed.
shaft weight is negligible.
E =199948 N/mm^2
I=282836 mm^4
L=50 mm
m=30 kg

Nc =[ 3 * E * I / (m *L^3)]^0.5 / (2*phi)..............(6)
= [ 3 * 199948 * 282836 / (30 * 50^3)] ^0.5 / ( 6.28)
= 45242^0.5 / 6.28
=213 /6.28
=34 rev/s
=2040 rpm

Nc > N, 1500 rpm (OK, can be used safely).

N(1500) / Nc(2040) = 0.735
0.735 < 0.8 (subcritical operation)

Note:
lower than 0.8 * Critical speed - subcritical operation
higher than 1.25 * Critical speed - above critical operation


Ref:
http://www.tribology-abc.com/calculators/uts.htm
http://www.roymech.co.uk/Useful_Tables/Matter/shear_tensile.htm
http://www.engineeringtoolbox.com/torsion-shafts-d_947.html
http://www.roymech.co.uk/Useful_Tables/Drive/Shaft_Critical_Speed.html
http://jimnevins.net/Stuff/Machinery%27s%20%20handbook%2026th%20Edition/MH26/yc.pdf

An example design of the crossflow turbine, type BYS-T3

Picture : www.boutiquepower.com.au

You can use an example design of the crossflow turbine type BYS/T3 by Ueli Meier as a teaching or a comparison with the turbine you plan, if you look at and study the sample design,it will open your insights about how to design a crossflow turbine. For those just learning about the turbines may be found in many less obvious things but do not worry , a lot of reading on the internet resources that can help you to understand it. We hope you become part of people who care about the future of the planet by using renewable energy . Thanks

Read more
Ref : Ueli Meier , " Design of Crossflow Turbine BYS/T3 " , Swiss Center Appropriate Technology - Varnbuelstrasse 14 CH-9000 St.Gallen Switzerland

Resistance Coefficient ( Ke ) For Entrance

Inward projecting----Sharp edged---Slightly Rounded---Well Rounded

he = Ke * V₂ ² / (2*g)

he = Entrance losses (m)
Ke = Depends upon the shape of the intake opening
V2 = The average velocity (m/s) of water in penstock
g = Gravitational constant (9.8 m/s^2)

Ref :
ESHA (European Small Hydropower Association),”Layman’s Handbook on How To Develop a Small Hydro Site,”2nd ed, 1998 http://www.scribd.com/doc/8885765/Layman-Handbook-for-hydro-electric-power-plants

Resistance Coefficient K for Sudden Expansion-Contraction

The losses through these fitting are generally evaluated by first obtaining

ß = d2 / d1

Important Note:

the resulting K values as tabled below are based on the flow velocity in the larger pipe,

if the flow velocity in the small pipe is used to evaluate the head loss then the K values tabled below should be multiplied by 

( ß)^4 = (d2 / d1) ^4

Head loss (h):

h_expansion(he) = Ke*(v2)^2 / (2*g)

h_contraction(hc) = Kc*(v2)^2 / (2*g)

v2= Average velocity (m/s) of water in small pipe

g =  Constant of gravity 9.8 m/s^2

Table of Ke & Kc against β = d2 / d1

β	Ke	Kc
0.15	1887.42	965.43
0.2	576	300
0.25	225	120
0.3	102.23	56.17
0.35	51.31	29.24
0.4	27.56	16.41
0.45	15.51	9.72
0.5	9	6
0.55	5.32	3.81
0.6	3.16	2.47
0.65	1.87	1.62
0.7	1.08	1.06
0.75	0.6	0.69
0.8	0.32	0.44
0.85	0.15	0.27
0.9	0.06	0.14
0.95	0.01	0.06
1	0	0

Ref : http://www.roymech.co.uk/Related/Fluids/Fluids_Pipe.html

Trash rack (or screen) losses

By Ahmad Suhendra

A screen is always required at the entrance of a pressure pipe . The function of screen (trash rack) is :

• keep debris away from the entrance to the outlet works where the debris will not clog the critical portions of the structure;
  
• capture debris in such a way that relatively easy removal is possible;

The flow of water through the rack also gives rise to a head loss. Though usually small, it can be calculated by a formula due to Kirchmer .

ht = [Kt * (t / b)^(4 / 3) * (Vo ^2 * Sin α )] / (2 * g)

ht = Screen head loss (m)
Kt = Resistance coefficient
t = Bar thickness (mm)
b = Width between bars (mm)
Vo= Approach velocity (m/s)
g = Gravitational constant (9.8 m/s^2)
α = Angle of inclination from horizontal


Ref :
ESHA (European Small Hydropower Association),”Layman’s Handbook on How To Develop a Small Hydro Site,”2nd ed, 1998 http://www.scribd.com/doc/8885765/Layman-Handbook-for-hydro-electric-power-plants

http://www.knoxcounty.org/stormwater/pdfs/vol2/3-3-5%20Trash%20Racks%20and%20Safety%20Grates.pdf