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?
