I Gravity machine force calculations assistance needed

[NeedFreedom]

TL;DR Summary

Machine prototype needs some physics and math input from those in the know

Im prototyping a machine to lower liquid from a height using a chains and sprockets. Essentially liquid will fill at top, be lowered over a distance of 4.8 metres then emptied at bottom. The liquid needs to be lowered at a slow pace due to resistance on the output shaft, I am considering utlising an alternator to provide resistance and convert rotational torque into electricty. I need to work out the forces involved and need some help with the math!

I have put together a basic model with the variables/inputs and would like some help figuring out the following:

Whats the relationship between weight, gravity, speed, losses and output shaft torque?

To achieve 70NM torque on output shaft, how much weight is needed?

What are the calculations for this problem?

Let me know if questions and more info needed, and thanks for any help in advance!

 

[russ_watters]

This sounds a like a hydroelectric dam and hopefully not a perpetual motion machine. So Googling physics of hydroelectric dams should get you what you need. However the basic relations are fairly easy: energy or work equals force times distance. Power is energy divided by time. Rotational power is torque times rotation rate. And of course power in equals power out times efficiency. Conservation of energy always applies!

 

[anorlunda]

TL;DR Summary: Machine prototype needs some physics and math input from those in the know

Im prototyping a machine to lower liquid from a height using a chains and sprockets. Essentially liquid will fill at top, be lowered over a distance of 4.8 metres then emptied at bottom. The liquid needs to be lowered at a slow pace due to resistance on the output shaft, I am considering utlising an alternator to provide resistance and convert rotational torque into electricty. I need to work out the forces involved and need some help with the math!

What happens after the water is emptied at the bottom? Do you raise the empty bucket to the top and fill it with water again?

Is the purpose of this machine to generate electric power from water going downhill?

 

[Baluncore]

To achieve 70NM torque on output shaft, how much weight is needed?

Assuming a 1 metre radius.
Force required = 70 N;
Gravitational acceleration, g = 9.8 m/s²
Force = mass * acceleration; F = m * g;
∴ m = F / g = 70 / 9.8 = 7.14 kg.

 

[Drakkith]

Whats the relationship between weight, gravity, speed, losses and output shaft torque?

The first two are easy.
Weight (kg force): gravity x mass.
Gravity: 9.81 m/s^2

The last two are not so easy. Well, the last one is easy if you're not doing any kind of gearing. The torque on the sprockets in your setup is simply force x distance, where distance is the distance from the center of the shaft to where the ropes/belts are attached to your weight. That is, $\tau = rFsin(\theta)$, where $r$ is the distance I just mentioned, $F$ is the force, and $sin(\theta)$ is the angle between the force vector and the lever arm. Note that this changes if you introduce any kind of non 1:1 gearing or levers.

As for losses, that depends mostly upon the details of what you're attaching the output to. You'll lose some amount to friction even with no output load, and anything you attach to the output, such as an alternator, will have some amount of losses. But that's impossible to know ahead of time. You'd have to directly test whatever you're using.

Knowing the speed is much more difficult, as you have to consider the rotational inertia (moment of inertia) of every rotating object in the system, the inertia of every linearly moving object in the system, and how your output load behaves.

What are the calculations for this problem?

Use the two equations I provided. It's simple algebra to solve for any of the variables.

 

[NeedFreedom]

What happens after the water is emptied at the bottom? Do you raise the empty bucket to the top and fill it with water again?

Is the purpose of this machine to generate electric power from water going downhill?

I have running spring water on my property and have been looking at methods for converting its energy potential into electricity.
I am assessing a custom built apparatus to lower water from height to convert gravitational mass into rotational force. The idea is the water will fill into containers at altitude then be lowered and emptied at bottom.

Once I am clear on the maths and physics, ill be able to compare to water wheels, turbines and other methods and make a decision on how to proceed.

 

[NeedFreedom]

Assuming a 1 metre radius.
Force required = 70 N;
Gravitational acceleration, g = 9.8 m/s²
Force = mass * acceleration; F = m * g;
∴ m = F / g = 70 / 9.8 = 7.14 kg.

Thanks for this.

I actually provided the wrong details of the output force required which should be 5100NM (not 70!)

So using your calculation I get the following:

m = 5100/9.8 = 520kgSo the 1 metre radius would be my sprocket? How is the sprocket radius accommodated in the calculation?

Also the fact that I want the lowing of the mass to be slowed down (IE not freefall) how would this effect the math?

 

[Drakkith]

Before doing absolutely anything, you need to figure out how much power you can even get out of your spring. What is the flow speed and volume? How far is the possible drop? 1,000 liters of water per hour dropping 4.8 meters in height can provide, at best, about 13 watts of power. After losses you might get around 10 watts. That's barely enough to run a lightbulb.

 

[NeedFreedom]

TL;DR Summary: Machine prototype needs some physics and math input from those in the know

Im prototyping a machine to lower liquid from a height using a chains and sprockets. Essentially liquid will fill at top, be lowered over a distance of 4.8 metres then emptied at bottom. The liquid needs to be lowered at a slow pace due to resistance on the output shaft, I am considering utlising an alternator to provide resistance and convert rotational torque into electricty. I need to work out the forces involved and need some help with the math!

I have put together a basic model with the variables/inputs and would like some help figuring out the following:

Whats the relationship between weight, gravity, speed, losses and output shaft torque?

To achieve 70NM torque on output shaft, how much weight is needed?

What are the calculations for this problem?

Let me know if questions and more info needed, and thanks for any help in advance!

The first two are easy.
Weight (kg force): gravity x mass.
Gravity: 9.81 m/s^2

The last two are not so easy. Well, the last one is easy if you're not doing any kind of gearing. The torque on the sprockets in your setup is simply force x distance, where distance is the distance from the center of the shaft to where the ropes/belts are attached to your weight. That is, $\tau = rFsin(\theta)$, where $r$ is the distance I just mentioned, $F$ is the force, and $sin(\theta)$ is the angle between the force vector and the lever arm. Note that this changes if you introduce any kind of non 1:1 gearing or levers.

As for losses, that depends mostly upon the details of what you're attaching the output to. You'll lose some amount to friction even with no output load, and anything you attach to the output, such as an alternator, will have some amount of losses. But that's impossible to know ahead of time. You'd have to directly test whatever you're using.

Knowing the speed is much more difficult, as you have to consider the rotational inertia (moment of inertia) of every rotating object in the system, the inertia of every linearly moving object in the system, and how your output load behaves.Use the two equations I provided. It's simple algebra to solve for any of the variables.

Thanks for the advice and math for calculating the torque on sprocket. Im a little rusty on trig and not sure how to calculate theta, do you mind showing me the calculation based on my varriables?

We could assume 500kg weight
Distance from centre 225mm

Speed sounds complicated. Is there a general rule which would help estimate the force implications if the failing speed was lowered? IE ignoring subtle variables and resistance, would slowing the rate of lowering from gravitational acceleration (such as 50%) be equivalent to reducing gravity from 9.81m/s to 4.4m/s? Ideally Id like to lower the weight at around 0.1m/s!

 

[Drakkith]

. Im a little rusty on trig and not sure how to calculate theta, do you mind showing me the calculation based on my varriables?

You don't calculate theta, it's built into whatever your design is. In this case, with a chain on a sprocket, theta is always 90 degrees (or pi/2 if you're working in radians). Since sin(90)=1, the torque equation simplifies to: $\tau = rF$

Is there a general rule which would help estimate the force implications if the failing speed was lowered?

In a steady fall, the force will be whatever the weight of the water + container is. This will not change no matter what the speed is. This is because force resists acceleration, not speed. A skydiver falling at terminal velocity is stilling being pulled down with a force equal to their weight, but they are no longer accelerating because the air is pushing back up on them with the same force, so they stop accelerating but continue to fall.

What you can change is the rate of fall, which affects how much power you can generate from the water. The faster the rate of fall (for a given volume of water), the higher the power output.

IE ignoring subtle variables and resistance, would slowing the rate of lowering from gravitational acceleration (such as 50%) be equivalent to reducing gravity from 9.81m/s to 4.4m/s? Ideally Id like to lower the weight at around 0.1m/s!

You can do this, but I question whether it's actually beneficial. The potential energy of a falling mass of water is: $u=mgh$, where u is energy, m is mass, g is gravitational acceleration, and h is the height. So for a 100 kg mass of water falling 4.8 meters we get $u=100*9.81*4.8 = 4708.8 J$.

Assuming this mass of water is being used as a power source, the power output is $p=\frac{E}{t}$, where E is energy and t is time. At 0.1 m/s, it will take 48 seconds to lower our mass of water. The power output is then $p=\frac{4708.8}{48} = 98.1 Watts$. At 1 m/s our power output is instead 981 Watts, and at 5 m/s it is 4905 W.

The problem is that this depends on actually being able to continuously move 100 kg of water every second through your device. Your spring may or may not be able to supply that much water, which is why I suggested you find out how much water is actually flowing before you do anything else. Everything else is mostly pointless until you find out how much water is moving and how much you can feasibly capture and use.

 

[NeedFreedom]

You don't calculate theta, it's built into whatever your design is. In this case, with a chain on a sprocket, theta is always 90 degrees (or pi/2 if you're working in radians). Since sin(90)=1, the torque equation simplifies to: $\tau = rF$In a steady fall, the force will be whatever the weight of the water + container is. This will not change no matter what the speed is. This is because force resists acceleration, not speed. A skydiver falling at terminal velocity is stilling being pulled down with a force equal to their weight, but they are no longer accelerating because the air is pushing back up on them with the same force, so they stop accelerating but continue to fall.

What you can change is the rate of fall, which affects how much power you can generate from the water. The faster the rate of fall (for a given volume of water), the higher the power output.You can do this, but I question whether it's actually beneficial. The potential energy of a falling mass of water is: $u=mgh$, where u is energy, m is mass, g is gravitational acceleration, and h is the height. So for a 100 kg mass of water falling 4.8 meters we get $u=100*9.81*4.8 = 4708.8 J$.

Assuming this mass of water is being used as a power source, the power output is $p=\frac{E}{t}$, where E is energy and t is time. At 0.1 m/s, it will take 48 seconds to lower our mass of water. The power output is then $p=\frac{4708.8}{48} = 98.1 Watts$. At 1 m/s our power output is instead 981 Watts, and at 5 m/s it is 4905 W.

The problem is that this depends on actually being able to continuously move 100 kg of water every second through your device. Your spring may or may not be able to supply that much water, which is why I suggested you find out how much water is actually flowing before you do anything else. Everything else is mostly pointless until you find out how much water is moving and how much you can feasibly capture and use.

Thanks @Drakkith, I appreciate your time and input. That all makes a lot of sense.

An important part of this exercise is me learning the math involved so it has not been pointless or a waste of time! Thanks to you I have a much clearer understanding of the forces involved for several of my ideas :)

My water source flows at 30 litres per second, so not huge. I figure 100 litre containers would be too big as would take over 3 seconds to fill. So perhaps 30L containers are better suited for my water source, taking 1 sec to fill. In 4.8m spacing would allow approx 20 containers of 30L decending so total volume = containers x volume, v=20 x 30 = 600 litres/kg.

600kg energy potential over 4.8m we get u=600x9.81*4.8 = 28,252.8 J
600kg lowing at 0.3m/s over 4.8m we get p={28252.8}{16} = 1,765.75 Watts

Im not exactly sure how now to convert watts into rotational torque on output shaft in Nm, but I can already deduce that if I wanted to generate electricity based on this concept, it will be 1765.75 Watts less losses. There is obviously a lot of engineering and challenges involved such as getting the water into the containers effectively while moving and fabricating such a large machine. Ill think about it some more ;)

 

[Baluncore]

My water source flows at 30 litres per second, so not huge.

Water flow rate = 30 litres per second (=30 kg); 4.8 metre drop; g = 9.8 m/s².
Potential energy, E = m·g·h = 30 * 9.8 * 4.8 = 1411.2 joule/sec = 1.4 kW.

Power (watts) = Torque (Nm) * angular velocity (rad/s),
Angular velocity (rad/s) = RPM * 60 * 2π .

 

[Drakkith]

For those interested in the topic or who may have small water sources on their property, the U.S. Department of Energy has a webpage with some additional info: https://www.energy.gov/energysaver/planning-microhydropower-system

@NeedFreedom be sure to check out your local regulations regarding water rights if you haven't already.

 

[hutchphd]

The classic solution to this mechanism is an overslung water wheel. Why do you want chains and buckets? There is a reason for the usual solution, why do a more complicated mechanism??