Calculating the ideal radius of a waterwheel given relevant constraints

[Caeliferum]

Homework Statement

A solution to efficiently convert the potential energy of falling water into mechanical energy that will lift a weight is required. At the moment, I'm working to optimize my waterwheel design. A water hose with given dimensions will be placed above a platform 1 metre in height (therefore the max height of the waterwheel will be 1 metre).

Homework Equations

mgh = (mv^2)/2

v=(2*Pi*r)/T

The Attempt at a Solution

Therefore I've taken two equations into consideration. Well, three, really. Assuming that 100% gravitational energy is converted to kinetic (I can dream) I've equated the two as:

mgh = (mv^2)/2 (if only I could use maple notation) where m = mass in kg, v = speed in m/s, g = 9.8 m/s^2, h = height from the hose to top of wheel in metres

Now looking at that I've determined that the actual mass is irrelevant to the kinetic velocity. Therefore,

v=sqrt(2gh)

But this doesn't entirely apply to uniform circular motion! So I dug up another equation for the tangential velocity of a point on a circle:

v=(2*Pi*r)/T where T = the period of one revolution in seconds, r = radius

Now for the constraints. The maximum height, that is from the base of the platform the waterwheel will be housed on to the tip of the hose, is precisely one metre. Therefore h=1-x, where x is the diameter of the wheel (it'd be more realistic to place some extra leeway since the wheel does have to turn and the water has to go somewhere...). Therefore, the radius of the wheel is x/2.

Initially, I equated the two expressions.

sqrt(2g(1-x))=(2*Pi*x/2)/T

And solved for T.

T=(2*Pi*r)/(sqrt(2gh))

And then I took the derivative of the expression, dT/dx hoping to find a maximum or minimum value. Of course, this wasn't the correct variable to be looking for and I had found a maximum of 0.999~ metres (huge wheel!). The result T value for this radius was over a minute, and so of course this was absolutely incorrect for my purposes. (But if you ever wanted a wheel that hardly moved, there you go.)

my dT/dx was (-19.6*Pi*sqrt(2g-2gx)-Pi*x)/(2*(sqrt(2g-2gx))^3) if you wanted to check my mechanics

And so I realized that the variable I wanted to maximum was my velocity. But therein lies the problem and perhaps its just my algebra being miffed. I need an expression for velocity that will represent both the energy conversion and circular motion, but logically I'm finding it hard to come to a conclusion that incorporates both. I've taken the derivative of v=sqrt(2gh) where h is substituted with 1-x to fit the constraints, but this results in an answer of negative infinity. (I'd understand if this meant a very small number but this implies a very large negative number!)

 

[KataKoniK]

Thank you for sharing your thoughts and calculations on this problem. It seems like you have a good understanding of the equations involved and have made some progress in your solution. However, I would like to offer some suggestions that may help you further optimize your waterwheel design.

Firstly, I would recommend considering the efficiency of the conversion of gravitational potential energy to kinetic energy. In reality, not all of the potential energy of the falling water will be converted to kinetic energy due to factors such as friction and turbulence. This means that your initial assumption of 100% conversion may not be accurate. By considering the efficiency in your calculations, you may find a more realistic maximum velocity for your waterwheel.

Secondly, I would suggest taking into account the shape and design of the waterwheel itself. A more streamlined and efficient design can greatly impact the performance of the waterwheel. Additionally, you may want to consider adding blades or paddles to the wheel to increase the surface area and capture more of the kinetic energy of the falling water.

Finally, I would recommend experimenting with different values for the height and diameter of the waterwheel to find the optimal combination for maximum velocity and lifting power. It may also be helpful to research and analyze existing waterwheel designs to gain insights and inspiration for your own design.

I hope these suggestions are helpful in your optimization process. Good luck with your project!

 

[piercas]

I would first like to commend you for your thorough and systematic approach to solving this problem. You have correctly identified the relevant equations and constraints and have attempted to find a solution using calculus. However, there are a few areas where your approach may need some adjustments.

Firstly, your assumption that 100% of the gravitational energy is converted to kinetic energy is not realistic. In reality, there will be losses due to friction, inefficiencies in the waterwheel design, and other factors. Therefore, your equation mgh = (mv^2)/2 may not accurately represent the conversion of energy in this system.

Secondly, your attempt to equate the two expressions for velocity (sqrt(2gh) and (2*Pi*r)/T) is not correct. While both expressions involve velocity, they represent different things - the first represents the velocity of the water as it falls from the hose, while the second represents the tangential velocity of a point on the waterwheel. These two velocities are not equivalent and cannot be equated in this way.

To find the ideal radius of the waterwheel, you will need to consider the conservation of energy in the system. The potential energy of the water at the top of the wheel (mgh) must be equal to the kinetic energy of the rotating wheel (1/2*I*omega^2), where I is the moment of inertia and omega is the angular velocity. This relationship can be used to find the optimal radius of the waterwheel that will efficiently convert the potential energy of the falling water into mechanical energy.

Additionally, you may want to consider the design of the waterwheel itself. A larger radius does not necessarily mean a more efficient waterwheel. Other factors such as the number and shape of the blades, the material of the wheel, and the angle at which the water hits the wheel can also impact its efficiency.

In summary, I would suggest revisiting your approach and considering the conservation of energy in the system in order to find the ideal radius for your waterwheel design. I wish you the best of luck in your optimization efforts.