DIY nano Hydro Power generator in a Jungle

[GreenPelton]

TL;DR Summary

Need some help how to calculate the needed kinetic energy and water speed to bring a 20w nano hydro power generator with a pelton wheel into motion!?

Hi

I am usually living off grid in simple huts in remote locations in Asia. A few months per year (rainy season) I don't have enough sun to produce solar power. I need very little electricity. My 20w solar panel produces enough electricity for my needs (lighting + charging a few small gadgets).
I thought to use the small streams with head close by to produce hydro power during that rainy and misty season.
For the nano hydro power project I got now a cheap small 20w hand crank (picture attached) and replaced the hand crank with a 3d printed pelton wheel.

These are the hand cranks specifications :
Stable Power: 3v/6.6A, 5v/4A, 6v/3.3A, 9v/2.2A, 12v/1.6A, 15v/1.3A for selection
■ Max Cranking power: 20W
■ Rotating speed: 2000rmp/min
Speed: 120 RPM (Recommended)

I need some help to calculate how much kinetic energy / water speed I need to start to turn the wheel in order to calculate how much head, pipe length and pipe diameter is required.
A friend tested that one needs to hang on the handle a bag with 500ml water to bring the handle into motion. The handle is 17cm long.

The diameter of the pelton wheel is 35cm. One spoon can hold about 50ml of water.

Can someone tell me how to calculate how much kinetic energy is needed to start the crank and how much is needed to reach 120rpm? Approximately (close by) would be enough.
As well how to calculate the necessary water speed to generate the necessary kinetic energy to bring the pelton wheel into motion?

I want to check before how much head, pipe length and diameter would be necessary for that kind of wheel. Maybe it isn't feasible (to much pipe length necessary since I have head but not very steep as well difficult to maintain and protect something which is out of sightfrom monkeys playing with it etc) and cost effective and I should better change the design and switch to a classic water wheel?

Hope someone knowledgeable can help me out!

Much appreciated!

 

[anorlunda]

Good for you. You already did the number 1 thing that many DIY fans forget --- you reduced your power demand to a very manageable number 20 w.

Here is a very comprehensive paper on pelton wheel design
http://www.ivt.ntnu.no/ept/fag/tep4...ger/forelesninger 2006/5 - Pelton Turbine.pdf

But i fear that the paper focuses on huge projects in the 20-200 MW range rather than 20 w range. There are many articles about micro hydro DIY projects, but even those consider 5 kW as micro. 20w would be nano size. That makes it hard to find precedents.

Hydro plants use head in m and volume flow rate $\frac{m^3}{s}$ as the primary parameters, not kinetic energy. Let's start there. What do you have available?

• How much flow? $\frac{m^3}{s}$
• How much head? m

 

[GreenPelton]

Good for you. You already did the number 1 thing that many DIY fans forget --- you reduced your power demand to a very manageable number 20 w.

Here is a very comprehensive paper on pelton wheel design
http://www.ivt.ntnu.no/ept/fag/tep4195/innhold/Forelesninger/forelesninger 2006/5 - Pelton Turbine.pdf

But i fear that the paper focuses on huge projects in the 20-200 MW range rather than 20 w range. There are many articles about micro hydro DIY projects, but even those consider 5 kW as micro. 20w would be nano size. That makes it hard to find precedents.

Hydro plants use head in m and volume flow rate $\frac{m^3}{s}$ as the primary parameters, not kinetic energy. Let's start there. What do you have available?

• How much flow? $\frac{m^3}{s}$
• How much head? m

Thanks a lot for your reply!

Well I don't have my nano hpg yet installed because I want to know first if it even makes sense or not.
I could get easily 7 to 8m head, the pipe would be maybe 20m long.
The stream has a flow of about more than 1l per sec. I didn't messure it yet but should be enough for this nano generator .
Right now we tried to get the generator moving with tap water and 4m head but it was motionless ;). Tab has 0.1l per sec of flow.

So I want to know formulas to calculate kinetic energy in my setting.
As well I want to know how much kinetic energy has 0.5kg on a 17cm long crank handle. Then I know how much joule the jet stream has to produce to get the turbine moving. Then I can calculate how much head and flow I need. I want to go the other way around.

You are right what you find online is for big systems, not a nano system like mine. I couldn't find the info I needed as well I lack the physics k knowledge to help myself. That's why I posted here in the hope a smart capable professional person reads it and gives me the tools and knowledge to figure out what I need. I could also do trial and error but I fear I would waste a lot of materials and time. As well I like to know how to calculate it, not just the answers so I can help others to do the same (I know others how have the same lifestyle and power issues).

Thanks a lot again!

 

[anorlunda]

Start with the basics, ignoring details of a pelton wheel. We are looking for an order of magnitude guess, and ignoring efficiency and losses. Use an online hydro power calculator such as this.
https://power-calculation.com/hydroelectricity-energy-calculator.php

7 m head, and 1 l/s flow.

If the pipe has 10 cm diameter, the velocity of the water is .13 m/s

Acceleration of gravity 9.8 m/s²

Water density 1000 kg/m³

Max power before losses 7 w.

So sticking to orders of magnitude, you need 1 order, 10x times more flow or more head.

Most difficult is to estimate the efficiency of a pelton wheel in that size range. In hydro power, pelton wheels are used for high-head low-flow applications, kaplan turbines for high-flow low-head, and propeller turbines in the intermediate ranges.

In your case, an old fashioned water wheel sounds most promising to me, but 12 RPM would be fast for a water wheel, not 120 RPM. The hand crank thing already has a 2000/120 gearing. Increasing that to 2000/12 or 166:1 gearing would be so extreme that friction losses would eat up all the power. Wind power turbines work with very low blade RPM, but I never heard of a practical one at the 20 w scale.

By the way, I found one site that labels anything under 5kW as pico hydro.

Do you have any chance of adding 10x times more solar panels to use in the low-light season of the year?

 

[gmax137]

@anorlunda did you miss a factor of ten? Or am I?
sanity check:
$Power= \dot m g h = 1 \frac {kg} {sec} 10 \frac {m} {{sec}^2} 7 m = 70 \frac {kg~ m^2}{sec^3} = 70~ watt$

 

[anorlunda]

@anorlunda did you miss a factor of ten? Or am I?
sanity check:
$Power= \dot m g h = 1 \frac {kg} {sec} 10 \frac {m} {{sec}^2} 7 m = 70 \frac {kg~ m^2}{sec^3} = 70~ watt$

My bad, you're correct.

So @GreenPelton , I was wrong. You're in the right zone.

 

[GreenPelton]

I read that's the formula to calculate hydropower:

P = m x g x Hnet x η

1 x 9.81 x 4 x 0.751 = 29.46W That would be enough already! My motor is only 20W max anyway.
4m head should be easily doable, not too big and expensive!
I would be actually already fine with just 10W 24/7! That's already more juice than my 20w solar panel (5h max a day) can produce.
So 2m head is 14.7W. That's plenty! I just need enough charge current for a 12v motorcycle battery. 1.2a is enough!

Thanks a lot to you all for these information, now I know it is feasible!

Nevertheless I still like to know how to calculate the kinetic energy or is it called torque ? of 500g on a 17cm long crank?

 

[Baluncore]

Nevertheless I still like to know how to calculate the kinetic energy or is it called torque ? of 500g on a 17cm long crank?

A 500g force, on a 17cm crank, at 120 RPM, g = 9.8 m/s²
Force; 9.8 * 0.5 kg = 4.9 Newton;
Torque; 4.9 * 0.17 = 0.833 Newton·metre;
Angular velocity; 120 RPM * 2π / 60 sec = 12.566 radian/second;
Power; 0.833 * 12.566 = 10.467 watt;
Rate of energy conversion = 10.467 joule/second.

 

[GreenPelton]

500g force on a 17cm crank, at 120 RPM.
Torque = 9.8 * 0.5 * 0.17 = 0.833 Newton·metre.
Angular velocity = 120 RPM * 2π / 60 sec = 12.566 radians/second.
Power = 0.833 * 12.566 = 10.467 watts.
Energy = 10.467 joules/sec.

Thanks a lot, that's exactly what i needed to know!

 

[willem2]

Thanks a lot, that's exactly what i needed to know!

You should also be able to compute what torque the pelton wheel can deliver.
A pelton wheel works by first converting using the potential energy of the water into kinetic energy, and then lettting the resulting water jet collide with the buckets. In the ideal case the bucket speed is half the speed of the water jet, and the water will have no kinetic energy left after colliding with a bucket.

If you have a head of 7m the speed of the water will be $\sqrt{2gh}$ = 12m/s. You'll have to calculate the diameter of the end of the tube from the speed and the water flow. Losses might be rather high here for a thin tube.

The force on the turbine will be equal to the momentum change of the water per second, so $m(V_i - V_f)/t$, wich is $\rho Q(V_i - V_f)$
where ρ is the density (kg/liter) and Q the water flow (liter/sec). V_i is the speed of the incoming water and V_f the speed of the outgoing water. In the ideal case V_f is 0, and if the wheel stands still V_f = - V_i
If the wheel stands still the force will be $2V_i \rho Q$ which is 24N.
The torque will be 24 * radius = 24* 0.175 = 4.2 Nm.

Maximum power is delivered when the speed of the buckets is half the speed of the water. Torque will be 2.1 Nm, and the RPM will be 60 seconds * 6 m/s / (2π radius) = 330 RPM.
With no load maximum RPM would be 660 RPM.
Of course there will be a lot of losses, but I think you should be able to drive the crank with 7m head and 1 liter/sec waterflow.

Using tap water with a 4m head and 0.3 liter/sec, I get only about 1.6Nm of torque if the wheel is standing still so that may not be enough to drive the crank if you include losses.

 

[Dullard]

Worth mentioning?
This is all 'full load' analysis. You can (if necessary) reduce the electrical load to reduce the mechanical input requirements. It doesn't sound as if you'll need to, but it's important to understand that you don't have to operate a 20W generator at 20W (actual).

 

[GreenPelton]

[QU

You should also be able to compute what torque the pelton wheel can deliver.
A pelton wheel works by first converting using the potential energy of the water into kinetic energy, and then lettting the resulting water jet collide with the buckets. In the ideal case the bucket speed is half the speed of the water jet, and the water will have no kinetic energy left after colliding with a bucket.

If you have a head of 7m the speed of the water will be $\sqrt{2gh}$ = 12m/s. You'll have to calculate the diameter of the end of the tube from the speed and the water flow. Losses might be rather high here for a thin tube.

The force on the turbine will be equal to the momentum change of the water per second, so $m(V_i - V_f)/t$, wich is $\rho Q(V_i - V_f)$
where ρ is the density (kg/liter) and Q the water flow (liter/sec). V_i is the speed of the incoming water and V_f the speed of the outgoing water. In the ideal case V_f is 0, and if the wheel stands still V_f = - V_i
If the wheel stands still the force will be $2V_i \rho Q$ which is 24N.
The torque will be 24 * radius = 24* 0.175 = 4.2 Nm.

Maximum power is delivered when the speed of the buckets is half the speed of the water. Torque will be 2.1 Nm, and the RPM will be 60 seconds * 6 m/s / (2π radius) = 330 RPM.
With no load maximum RPM would be 660 RPM.
Of course there will be a lot of losses, but I think you should be able to drive the crank with 7m head and 1 liter/sec waterflow.

Using tap water with a 4m head and 0.3 liter/sec, I get only about 1.6Nm of torque if the wheel is standing still so that may not be enough to drive the crank if you include losses.

Many thanks for these in depth information! It's amazing, the possibilities one has if one has the know how!

Unfortunately I don't know yet the diameter of the nozzle I will eventually use (haven't got one yet... any recommendations, anything to concern while choosing a nozzle? ) so far we used for testing a simple garden hose. I know a nozzle increases water speed but reduces the kinetic energy (loss in friction? etc).

So if I got that right the 330rpm means I get an output of 20w at 330rpm under load?

Thanks a lot Dullard, yes I was hoping not to have to run it at full capacity but just achieve an output of 10 to 12w. I don't want to unnecessarily overuse that motor (reduce its lifespan !?) as well I simply don't need that much electricity, it would go to waste be unused (sure I could also stop the generator and let it rest once the battery is full) .

I went with a 20w motor (even I only need 10w) knowing to achieve the full output would be rather difficult as well knowing/believing that if a motor only has to work 50 to 70% of its max abilities I can use it over a longer period.

So would be 165rpm generate 10w?

My buckets hold about 50ml. The diameter of the wheel is 35cm.
This template was used (blown up to 35cm diameter)
https://www.thingiverse.com/thing:33720

The idea for the simple nano turbine I got from here



Guess I should get such a fire nozzle like in the video!? He got good results!

 

[GreenPelton]

In the video he had only 0.37l/s of water flow! His stepper motor seems to produce more max watts (bigger) than mine, but produces 13W at 200rpm as he said. So I guess around that RPM I should get my needed 10W with mine hopefully!?