Bernoulli's equation, Venturi effect, hydraulic head, nozzle

[Mario Rossi]

Hi, I have this problem:

I have a vertical tube 1 meter D, in the bottom end there is a nozzle with 0,5 m D. The tube is full of water. the tube length is 10 meters and the nozzle length is 2 meters. I need to calculate the power of this by this equation:

W = Q * g * h * p

where W is watt, Q is the volumetric flow rate, g is 9.81,h is the head and p is the density (1000 kg/m3 for the water.

The issue is the velocity in the nozzle: for the Venturi effect, in the nozzle the velocity increases. So in the velocity Torricelli's equation: sqrt(2 * g * h) = v the h value changes in h = (v^2) / 2 * g that is greater then the real h value (12 meters). So in the power equations I use this second h value. Is it right?

There is an explanation: https://en.wikipedia.org/wiki/Pelton_wheel#Power but I'm not understanding it.

 

[jack action]

I'm guessing the velocity in the tube $v_t$ is not zero. This is why the velocity in the nozzle $v_n$ gives a head $h_n$ greater than the physical height of the tube $h_t$. The relation between the two should be:
$$\rho g h_t + \frac{\rho v_t^2}{2} = \frac{\rho v_n^2}{2} \equiv \rho g h_n$$
Of course, the velocity of the fluid in the tube does translate into energy in the nozzle, so $h_n$ should be considered in such a case.

 

[Mario Rossi]

I'm guessing the velocity in the tube $v_t$ is not zero. This is why the velocity in the nozzle $v_n$ gives a head $h_n$ greater than the physical height of the tube $h_t$. The relation between the two should be:
$$\rho g h_t + \frac{\rho v_t^2}{2} = \frac{\rho v_n^2}{2} \equiv \rho g h_n$$
Of course, the velocity of the fluid in the tube does translate into energy in the nozzle, so $h_n$ should be considered in such a case.

Thank you!