How to Plan a Mini Hydro Power Project - Revision history

Ranisha: /* Water into Watts */

2018-01-02T12:03:14Z

Water into Watts

Ranisha: /* Water into Watts */

2018-01-02T12:02:05Z

Water into Watts

Ranisha: /* Water into Watts */

2018-01-02T11:55:41Z

Water into Watts

Ranisha: /* Water into Watts */

2018-01-02T11:25:06Z

Water into Watts

Ranisha: /* Water into Watts */

2018-01-02T10:35:17Z

Water into Watts

Ranisha: /* Water into Watts */

2018-01-02T10:34:16Z

Water into Watts

Ranisha at 09:56, 2 January 2018

2018-01-02T09:56:09Z

Michel andre: /* Complexity of Micro-Hydropower (MHP) - Barriers to Success */

2015-06-18T14:57:25Z

Complexity of Micro-Hydropower (MHP) - Barriers to Success

Katharina Wiedemann: /* Scheme Components */

2014-11-18T06:41:45Z

Scheme Components

Wiki Gardener: /* Equip a Certain Place with Hydropower (A) */

2014-11-18T06:13:09Z

Equip a Certain Place with Hydropower (A)

@@ Line 56: / Line 56: @@
 [[How to Plan a Mini Hydro Power Project#toc|►Go to Top]]
@@ Line 64: / Line 65: @@
 To determine the power potential of the water flowing in a river or stream it is necessary to determine both the flow rate of the water and the head through which the water can be made to fall. The ''flow rate'' is the quantity of water flowing past a point in a given time. Typical flow rate units are litres per second or cubic meters per second. The ''head'' is the vertical height, in meters, from the turbine up to the point where the water enters the intake pipe or penstock.
 <blockquote>
-The potential power can be calculated as: '''P = g * Q * H * f<sub>eff</sub>'''<br/>'''''Example:'''''<i>A location with a head of 10 metres, flow of 300 liter / sec (= 0.3 m<sup>3</sup>/s) will have a potential power of 15 kW electricity:</i><br/>10m/s<sup>2&nbsp;</sup>* 0.3m<sup>3</sup>/s * 10m * 0.5 =&nbsp; &nbsp;<span style="font-size: 13.6px; background-color: rgb(255, 255, 255);">15m</span><sup style="background-color: rgb(255, 255, 255);">5</sup><span style="font-size: 13.6px; background-color: rgb(255, 255, 255);">/s</span><sup style="background-color: rgb(255, 255, 255);">3</sup>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 15m<sup>5</sup>/s<sup>3</sup>&nbsp;* 1000 kg/m<sup>3&nbsp;</sup>(density of water)<br/>
+The potential power can be calculated as: '''P = g * Q * H * f<sub>eff</sub>'''<br/>'''''Example:'''''<i>A location with a head of 10 metres, flow of 300 liter / sec (= 0.3 m<sup>3</sup>/s) will have a potential power of 15 kW electricity:</i><br/>10m/s<sup>2&nbsp;</sup>* 0.3m<sup>3</sup>/s * 10m * 0.5&nbsp; =&nbsp; <span style="font-size: 13.6px; background-color: rgb(255, 255, 255);">15m</span><sup style="background-color: rgb(255, 255, 255);">5</sup><span style="font-size: 13.6px; background-color: rgb(255, 255, 255);">/s</span><sup style="background-color: rgb(255, 255, 255);">3</sup> <br/>
-&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 15kW
+&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 15m<sup>5</sup>/s<sup>3</sup>&nbsp;* 1000 kg/m<sup>3&nbsp;</sup>(density of water)<br/>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 15000 J/s<br/>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 15000 W&nbsp;<br/>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 15kW
 </blockquote>

@@ Line 56: / Line 56: @@
 [[How to Plan a Mini Hydro Power Project#toc|►Go to Top]]
@@ Line 63: / Line 64: @@
 To determine the power potential of the water flowing in a river or stream it is necessary to determine both the flow rate of the water and the head through which the water can be made to fall. The ''flow rate'' is the quantity of water flowing past a point in a given time. Typical flow rate units are litres per second or cubic meters per second. The ''head'' is the vertical height, in meters, from the turbine up to the point where the water enters the intake pipe or penstock.
 <blockquote>
-The potential power can be calculated as: '''P = g * Q * H * f<sub>eff</sub>'''<br/>'''''Example:'''''<i>A location with a head of 10 metres, flow of 300 liter / sec (= 0.3 m<sup>3</sup>/s) will have a potential power of 15 kW electricity:<br/></i>
+The potential power can be calculated as: '''P = g * Q * H * f<sub>eff</sub>'''<br/>'''''Example:'''''<i>A location with a head of 10 metres, flow of 300 liter / sec (= 0.3 m<sup>3</sup>/s) will have a potential power of 15 kW electricity:</i><br/>10m/s<sup>2&nbsp;</sup>* 0.3m<sup>3</sup>/s * 10m * 0.5 =&nbsp; &nbsp;<span style="font-size: 13.6px; background-color: rgb(255, 255, 255);">15m</span><sup style="background-color: rgb(255, 255, 255);">5</sup><span style="font-size: 13.6px; background-color: rgb(255, 255, 255);">/s</span><sup style="background-color: rgb(255, 255, 255);">3</sup>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 15m<sup>5</sup>/s<sup>3</sup>&nbsp;* 1000 kg/m<sup>3&nbsp;</sup>(density of water)<br/>
-m/s<sup>2&nbsp;</sup>* 0.3m<sup>3</sup>/s * 10m * 0.5 = 15kW
+&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 15kW
 </blockquote>

@@ Line 63: / Line 63: @@
 To determine the power potential of the water flowing in a river or stream it is necessary to determine both the flow rate of the water and the head through which the water can be made to fall. The ''flow rate'' is the quantity of water flowing past a point in a given time. Typical flow rate units are litres per second or cubic meters per second. The ''head'' is the vertical height, in meters, from the turbine up to the point where the water enters the intake pipe or penstock.
 <blockquote>
-The potential power can be calculated as: '''P = g * Q * H * f<sub>eff</sub>'''<br/>'''''Example:'''''<i>A location with a head of 10 metres, flow of 300 liter / sec (= 0.3 m<sup>3</sup>/s) will have a potential power of 18 kW electricity:<br/></i>
+The potential power can be calculated as: '''P = g * Q * H * f<sub>eff</sub>'''<br/>'''''Example:'''''<i>A location with a head of 10 metres, flow of 300 liter / sec (= 0.3 m<sup>3</sup>/s) will have a potential power of 15 kW electricity:<br/></i>
-m/s<sup>2&nbsp;</sup>* 0.3m<sup>3</sup>/s * 10m * 0.5 = 18kW
+m/s<sup>2&nbsp;</sup>* 0.3m<sup>3</sup>/s * 10m * 0.5 = 15kW
 </blockquote>

@@ Line 68: / Line 68: @@
 </blockquote>
-Power in kW (P); Flow rate in m<sup>3 </sup>/s(Q); Head in m (H); Gravity constant = 9.81 m/s<sup>2</sup> (g); Efficiency factor (f<sub>eff</sub>) => 0.4 - 0.7 *
+Power in kW (P); Flow rate in m<sup>3 </sup>/s (Q); Head in m (H); Gravity constant = 9.81 m/s<sup>2</sup> (g); Efficiency factor (f<sub>eff</sub>) => 0.4 - 0.7 *
 <sup>*</sup>Small water turbines rarely have efficiency better than 80%. Generators' efficiency of ~ 90% and power will also be lost in the pipe carrying the water to the turbine, due to frictional losses. A rough guide used for small systems of a few kW rating is to take the overall efficiency as approximately 50%. Thus, the theoretical power must be multiplied by 0.50 for a more realistic figure

@@ Line 63: / Line 63: @@
 To determine the power potential of the water flowing in a river or stream it is necessary to determine both the flow rate of the water and the head through which the water can be made to fall. The ''flow rate'' is the quantity of water flowing past a point in a given time. Typical flow rate units are litres per second or cubic meters per second. The ''head'' is the vertical height, in meters, from the turbine up to the point where the water enters the intake pipe or penstock.
 <blockquote>
-The potential power can be calculated as: '''P = g x Q x H x f<sub>eff</sub>'''<br/>'''''Example:'''''<i>A location with a head of 10 metres, flow of 300 liter / sec (= 0.3 m<sup>3</sup>/s) will have a potential power of 18 kW electricity:<br/></i>
+The potential power can be calculated as: '''P = g * Q * H * f<sub>eff</sub>'''<br/>'''''Example:'''''<i>A location with a head of 10 metres, flow of 300 liter / sec (= 0.3 m<sup>3</sup>/s) will have a potential power of 18 kW electricity:<br/></i>
 m/s<sup>2&nbsp;</sup>* 0.3m<sup>3</sup>/s * 10m * 0.5 = 18kW

@@ Line 56: / Line 56: @@
 [[How to Plan a Mini Hydro Power Project#toc|►Go to Top]]
@@ Line 61: / Line 62: @@
 To determine the power potential of the water flowing in a river or stream it is necessary to determine both the flow rate of the water and the head through which the water can be made to fall. The ''flow rate'' is the quantity of water flowing past a point in a given time. Typical flow rate units are litres per second or cubic meters per second. The ''head'' is the vertical height, in meters, from the turbine up to the point where the water enters the intake pipe or penstock.
-<blockquote>The potential power can be calculated as: '''P = g x Q x H x f<sub>eff</sub>'''<br/>'''''Example:'''''<i>A location with a head of 10 metres, flow of 300 liter / sec (= 0.3 m<sup>3</sup>/sec) will have a potential power of 18 kW electricity:<br/>1</i>''0 m * 0.3 m<sup>3</sup> * 10m/s<sup>2</sup> * 0.5 = ''<span style="font-style: italic">18 kW</span></blockquote>
+<blockquote>
 Power in kW (P); Flow rate in m<sup>3 </sup>/s(Q); Head in m (H); Gravity constant = 9.81 m/s<sup>2</sup> (g); Efficiency factor (f<sub>eff</sub>) => 0.4 - 0.7 *
@@ Line 73: / Line 79: @@
 <u>This can be provided either:</u><br/>
 *directly to the home via a small electrical distribution system or,<br/>
 *can be supplied by means of batteries which are returned periodically to the power house for recharging - this system is common where the cost of direct electrification is prohibitive due to scattered housing (and hence an expensive distribution system),<br/>
@@ Line 79: / Line 86: @@
 [[How to Plan a Mini Hydro Power Project#toc|►Go to Top]]

@@ Line 1: / Line 1: @@
 [[Portal:Hydro|► Back to Hydro Portal]]
@@ Line 63: / Line 62: @@
 To determine the power potential of the water flowing in a river or stream it is necessary to determine both the flow rate of the water and the head through which the water can be made to fall. The ''flow rate'' is the quantity of water flowing past a point in a given time. Typical flow rate units are litres per second or cubic meters per second. The ''head'' is the vertical height, in meters, from the turbine up to the point where the water enters the intake pipe or penstock.
 <blockquote>The potential power can be calculated as: '''P = g x Q x H x f<sub>eff</sub>'''<br/>'''''Example:'''''<i>A location with a head of 10 metres, flow of 300 liter / sec (= 0.3 m<sup>3</sup>/sec) will have a potential power of 18 kW electricity:<br/>1</i>''0 m * 0.3 m<sup>3</sup> * 10m/s<sup>2</sup> * 0.5 = ''<span style="font-style: italic">18 kW</span></blockquote>
-Power in kW (P); Flow rate in m<sup>3 </sup>(Q); Head in m (H); Gravity constant = 9.81 m/s<sup>2</sup> (g); Efficiency factor (f<sub>eff</sub>) => 0.4 - 0.7 *
+Power in kW (P); Flow rate in m<sup>3 </sup>/s(Q); Head in m (H); Gravity constant = 9.81 m/s<sup>2</sup> (g); Efficiency factor (f<sub>eff</sub>) => 0.4 - 0.7 *
 <sup>*</sup>Small water turbines rarely have efficiency better than 80%. Generators' efficiency of ~ 90% and power will also be lost in the pipe carrying the water to the turbine, due to frictional losses. A rough guide used for small systems of a few kW rating is to take the overall efficiency as approximately 50%. Thus, the theoretical power must be multiplied by 0.50 for a more realistic figure
@@ Line 413: / Line 412: @@
 </div>
 [[Category:Small-Hydro/Mini-Hydro]]

@@ Line 230: / Line 230: @@
 </div>
 = Complexity of Micro-Hydropower (MHP) - Barriers to Success<br/> =
-A wide range of topics needs to be address to implement MHP projects successfully.
+A wide range of topics needs to be address to implement MHP projects successfully. The Thumbnail Diagramm ( click to enlarge) below provides an overview of ascpects to be considered when planning or implementing a MHP scheme. There is another article which provides an insight into [[Complexity of Micro-hydro Power (MHP) - Barriers to Success|Complexity of MHP and the barriers to success]].
-[[How to Plan a Mini Hydro Power Project#toc|►Go to Top]]
+{| cellspacing="1" cellpadding="1" border="0" style="width: 100%"

@@ Line 47: / Line 47: @@
 = Technical<br/> =
 <div class="booktext">
 == Scheme Components ==
 Figure 1 shows the main components of a run-of-the-river micro-hydro scheme. This type of scheme requires no water storage but instead diverts some of the water from the river which is channelled along the side of a valley before being 'dropped' into the turbine via a penstock. In figure 1, the turbine drives a generator that provides electricity for a workshop. The transmission line can be extended to a local village to supply domestic power for lighting and other uses.
 <center>
-[[File:MicroHydropowerplant01.gif|left|430px|Figure 1: Layout of a typical micro hydro scheme|alt=Micro-Hydropowerplant]]<br/><br/>
+[[File:MicroHydropowerplant01.gif|thumb|left|430px|Figure 1: Layout of a typical micro hydro scheme|alt=Micro-Hydropowerplant]]<br/><br/>
 </center>
 There are various other configurations which can be used depending on the topographical and hydrological conditions, but all adopt the same general principle.<br/>
 [[How to Plan a Mini Hydro Power Project#toc|►Go to Top]]

@@ Line 251: / Line 251: @@
 The information below provides a short summary for constructing a small hydro power.
 === Equip a Certain Place with Hydropower (A)<br/> ===
 '''1.''' '''Check the sites feasibility'''.
 <u>This means, gather the following Data:</u><br/>head, flow, number of consumers, location data (distance from grid...).
-► Go there and fill out this form => [[:File:Brief_site_assessment.pdf|a two page condensed site assessment]]<br/>► If you need further quick information go to [[Hydro Power Basics#Principle of Hydropower|Hydropower Basics]]
+► Go there and fill out this form => [[:File:Brief site assessment.pdf|a two page condensed site assessment]]<br/>► If you need further quick information go to [[Hydro Power Basics#Principle of Hydropower|Hydropower Basics]]
 '''2. Calculate the sites hydropower potential''':<br/>Multiply the (minimal) flow in liter/second with the available <span style="text-decoration: underline">height</span> difference'''. '''The result times 5 gives you an estimation on how many Watts can be produced.
 ► see also [[Hydro Power Basics#Principle of Hydropower|Hydropower Basics]]
 '''3. Estimate the <u>required power</u> at the site'''.
 Separate consumer power and productive use power. Keep in mind that transmission cables are costly.<br/>Consumers basics are light and TV. Light requires min. 5 -20 Watt per bulb (5 W = energy saving lamp, if available). A TV requires 40 - 150 W (depending on size and type). Machinery needs much more power. Motors may require from 1000 Watt (1kW = small) upward. Heating needs also a lot power (min. 500 W for a small water heater).<br/>
-'''4.''' <span style="font-weight: bold">C</span>'''ompare the local hydropower potential with the de''''''mand of electric power.'''<br/>If the potential is to be gained, check if the Data from point 1. is really valid.<br/>Is the measured flow available all year round?<br/>Is the head available on a short horizontal distance - means "how long have [[Hydro Power - Civil Engineering#Canal|canal]] and [[Hydro Power - Civil Engineering#Pipeline .28Penstock.29|penstock]] to be?"
+<br/>
 '''5. Cost estimation:'''
 <u>There is no general cost estimation possible as any site differs from the other. Biggest cost blocks are usually:</u><br/><u>Civil structure</u>: Ask a company to find out the local cost for building (material).<br/>mansion cost per m<sup>3</sup> weir - 1.5 m height x width of river; cost per m canal - size depends on water volume; cost per m<sup>2</sup> powerhouse ~ 10 - 25 m <sup>2</sup><br/><u>Penstock</u>: ask the local prices for pipes - size depends on water volume
@@ Line 295: / Line 296: @@
 [[How to Plan a Mini Hydro Power Project#toc|►Go to Top]]