# Fluid dynamics - Velocity required to turn wheel

I want to calculate the velocity of a fluid required to turn a water wheel at a specific angular velocity. The fluid enters the housing of the water wheel through a inlet channel. It turns the wheel and then disappears through the outlet. See the figure for a sketch of the problem. $v_{in}$, turning the wheel and leaving at velocity $$v_{out}$$" />

Fluid enters at a known velocity $$v_{in}$$ through an inlet with known dimensions. The fluid will hit the wheel, causing it to rotate from the impact at an angular velocity of $$\omega$$ rad/s. Remaining fluid leaves with a velocity $$v_{out}$$. Due to friction at the axis of the wheel, the moment required to initiate rotation of the wheel is $$T$$ Nm. I want to calculate the minimal velocity $$v_{in}$$ required to overcome the friction and rotate the wheel and I have found two different approaches.

Approach 1
Using the kinetic energy equation and perseverance of energy gives:

$$\frac{m{v_{in}}^2}{2} = \frac{I{\omega}^2}{2} +\frac{m{v_{out}}^2}{2} + T\theta \tag{1}$$

where $$I$$ is the inertia of the wheel, $$m$$ the mass of the fluid control volume and $$\theta$$ the angle rotated by the wheel. $$T\theta$$ is the work done by the friction at the axis. This approach is quite straight forward, if I know the moment $$T$$ and the inlet velocity $$v_{in}$$ the maximum $$\omega$$ can be calculated or if I want to know $$v_{out}$$ for a specific $$\omega$$, that can be computed as well.

Approach 2
Using another form of the energy equation, namely the one for steady incompressible flow, i.e., the extended Bernoulli equation:

$$\frac{p_{out}}{\rho} + \frac{{v_{out}}^2}{2} + gz_{out} = \frac{p_{in}}{\rho} +\frac{{v_{in}}^2}{2} + gz_{in} + w_{shaft} - loss \tag{2}$$

Where $$\rho$$ is the density of the fluid, $$p$$ the pressure, $$z$$ the height, $$w_{shaft}$$ the work done on the shaft and the $$loss$$ is from the loss due to friction in the pipe. Dividing with the acceleration of gravity $$g$$, we get the energy per unit weight or the head.

$$\frac{p_{out}}{\gamma} + \frac{{v_{out}}^2}{2g} + z_{out} = \frac{p_{in}}{\gamma} +\frac{{v_{in}}^2}{2g} + z_{in} + h_{s} - h_L \tag{3}$$

Here the definition of specific wieght is used $$\gamma = g\rho$$. The head loss due to the friction in the pipe is neglected, since the distance traveled is considered short. The head in the shaft can be expressed through work in the shaft as:

$$h_s = w_{shaft}/g = \frac{\dot{W}_{shaft}}{\dot{m}g} = \frac{T\omega}{Q\gamma} \tag{4}$$

Where $$\dot{W}$$ is the time rate of work $$\dot{W} = T\omega$$ and $$Q$$ is the flow rate, related to the mass flow as $$Q = {\rho}\dot{m}$$. Inserting (4) in (3) with $$h_L = 0$$ and since the fluid does work on the wheel, $$h_s$$ is negative, energy is taken out from the system. This gives:

$$\frac{p_{out}}{\gamma} + \frac{{v_{out}}^2}{2g} + z_{out} = \frac{p_{in}}{\gamma} +\frac{{v_{in}}^2}{2g} + z_{in} - \frac{T\omega}{Q\gamma} \tag{5}$$

If there is no pressure drop across the wheel, $${\Delta}p = 0$$, there is no height difference since everything happens in the same plane $${\Delta}z = 0$$, resulting in:

$$\frac{{v_{out}}^2}{2g} = \frac{{v_{in}}^2}{2g} - \frac{T\omega}{Q\gamma} \tag{6}$$

From equation 6 $$v_{out}$$ can be computed if $$v_{in}$$, $$\omega$$ and $$T$$ are known. Since $$Q=v_{in}/{A_{in}}$$ where $$A_{in}$$ is the area of the inlet, it is considered to be known once $$v_{in}$$ is known. Similar to equation (1), the same parameters can be computed for certain known conditions.

Using these two equations, one might presume that you will get similar results. However, I do not and I wonder why. Is it something which I neglect using one of the equations which is not neglected in the other? Am I missing something? I want to know what is wrong and why I get different results.

Example
Using (6) to find $$v_{out}$$ for $$v_{in} = 5$$ m/s yields $$v_{out} = 0.95$$ m/s. Solving for the same input vales but using (1) insteaed, yileds an imaginary solutions, since the square root becomes negative. $$v_{out}=\sqrt{v_{in} - \frac{I{\omega}^2}{m} - \frac{2T{\theta}}{m}}$$ Thus according to equation (1), a higher inlet velocity is required to overcome the resistance at the wheel. Which one is correct?

• IMHO you won't be able to get a reasonably accurate answer with Bernoulli's equations. The reason is that the placement of the wheel (in the image you provided) is such that there are bound to be secondary flows and vortices that will create counter forces to the rotation, so there will be a lot more losses. Also at different velocities the losses will change. Only a CFD simulation would be able to yield some results, but it would be very advanced. Mar 3 at 17:33
• I understand that there some secondary flows and vortices but since it is possible to design turbines, I was thinking that there must be some way to approximate the flows in order to get an idea how to design the thing. I am also aware that in modern engineering, turbines are designed using CFD but I am not looking for a super detailed solution either. I just want a good approximation. Mar 4 at 12:23
• Have you looked at the calculations for waterwheels? Mar 7 at 12:01
• @SolarMike Yes, I found one in A brief introduction to Fluid Mechanics by D. Young et. al. (Wiley, 2011). Here the time rate of work is expressed as a function of the inlet velocity and the radial velocity of the wheel as: $\dot{W} = {\rho}QU(U-v_{in})(1-cos{\beta})$. Where $U$ is the radial velcotiy. According to the authors, the maximal effect is reached in the wheel when $U=v_{in}/2$ and $\beta = 160\deg$. Since $\dot{W} = T\theta$ and using the same values as in the example, the results agree with Bernouilli's equation. Mar 7 at 14:43

The energy equation is a bit different and it can be noticed by calculating $$\theta$$ in (1) from $$\omega$$, considering that $$\omega$$ is known prior. Thus we get $${\theta} = \int_{0}^{t} \omega dt = {\omega}t+A$$, where A is an integration constant. $$\theta$$ is the distance traveled by the wheel, at $$t=0$$, that distance is 0, which gives: $${\theta}(0) ={\omega}{\cdot}0 +A =0$$, $$A = 0$$. Thus we get $${\theta}(t) = {\omega}t\tag{7}$$