Fluid Mechanics: Head Loss Question (Conceptual)

[Master1022]

Homework Statement

A pipe (with a known length, diameter, roughness ratio k/d) carries water (with known kinematic viscosity) is 800 metres long and carries water from a reservoir to a point 6 m below the reservoir surface. If the pipe inlet is sharp-edged, includes a standard 90 degree elbow, and the exit is to the atmosphere, show that the head loss in metres due to the layout of the pipework, but excluding friction is given by.... (some expression with Q)

Relevant Equations

- $H_L = \frac{v^2}{2g} (\frac{4fL}{d} + \sum K_L)$
- Conservation of Energy

I am not really worried about the numbers, but more about the simple concepts with head loss in these pipe flow questions. I want to confirm that head loss just means the change in static head, right?

I have been advised that for a problem like this, it is nothing more than the conservation of energy, and the questions are within the working below.

The solution lays out the working as follows: Defining point 1 at the top of the reservoir (not moving and at atmospheric pressure) and point 2 at the exit of the pipe

Conservation of energy yields: $E_1 = E_2 + E_{lost}$
$$p_{1} + \frac{1}{2} \rho u_{1}^2 + \rho g z_{1} = p_{2} + \frac{1}{2} \rho u_{2}^2 + \rho g z_{2} + \frac{4fL}{d}\times \frac{1}{2}\rho u_{2}^2 + \frac{1}{2} \rho u_{2}^2 \sum_{}^{} K_{L}$$

from here, it jumps to end of the working, with no commentary, thus leaving me to fill in the gaps (though I am not sure whether I have done this correctly or not).

Given that we can ignore friction, we can ignore that term, giving:
$$p_{1} + \frac{1}{2} \rho u_{1}^2 + \rho g z_{1} = p_{2} + \frac{1}{2} \rho u_{2}^2 + \rho g z_{2} + \frac{1}{2} \rho u_{2}^2 \sum_{}^{} K_{L}$$
Then, if we set our datum at point 2 and assume velocity = 0 at point 1, we can remove two more terms:
$$p_{1} + \rho g z_{1} = p_{2} + \frac{1}{2} \rho u_{2}^2 + \frac{1}{2} \rho u_{2}^2 \sum_{}^{} K_{L}$$
Here are couple more questions and I apologise if these are really obvious, but I cannot seem to understand this topic properly:
1. Is it correct to say $p_1 = p_2 = p_{atm}$? How do I know whether $p_{atm}$ is just the static pressure or the total pressure? Does it just depend on the point where we are talking about on our streamline? (for this example, I think that it is just the static pressure. For $p_{atm}$ to be the total pressure, we would need to take a point above the water level?) This would make the equation (and dividing by $\rho g$):
$$z_{1} = \frac{1}{2g} u_{2}^2 + \frac{1}{2g} u_{2}^2 \sum_{}^{} K_{L}$$

2. From here, I don't know what happens to the dynamic head term at point 2. Looking back, the only thing I can guess is that it has something to do with this energy loss term? Our professor made a point of telling us not to 'double count' the $\frac{u_{2}^2}{2g}$, but I cannot see how to get rid of it. Would anyone be able to help explain this to me?

3. Looking back at that energy lost (per unit volume?) due to friction term, is it in that form because of $head loss = \frac{4fL}{d} \frac{V^2}{2g}$ and thus $p_L = \rho g h_L = \frac{4fL}{d} \frac{\rho V^2}{2}$?

After this, I can read the $K_L$ values out of a data table and thus get the required expression, but I struggle to get here.

Thanks for any help.

 

[Chestermiller]

I am not really worried about the numbers, but more about the simple concepts with head loss in these pipe flow questions. I want to confirm that head loss just means the change in static head, right?

I have been advised that for a problem like this, it is nothing more than the conservation of energy, and the questions are within the working below.

The solution lays out the working as follows: Defining point 1 at the top of the reservoir (not moving and at atmospheric pressure) and point 2 at the exit of the pipe

Conservation of energy yields: $E_1 = E_2 + E_{lost}$
$$p_{1} + \frac{1}{2} \rho u_{1}^2 + \rho g z_{1} = p_{2} + \frac{1}{2} \rho u_{2}^2 + \rho g z_{2} + \frac{4fL}{d}\times \frac{1}{2}\rho u_{2}^2 + \frac{1}{2} \rho u_{2}^2 \sum_{}^{} K_{L}$$

from here, it jumps to end of the working, with no commentary, thus leaving me to fill in the gaps (though I am not sure whether I have done this correctly or not).

Given that we can ignore friction, we can ignore that term, giving:
$$p_{1} + \frac{1}{2} \rho u_{1}^2 + \rho g z_{1} = p_{2} + \frac{1}{2} \rho u_{2}^2 + \rho g z_{2} + \frac{1}{2} \rho u_{2}^2 \sum_{}^{} K_{L}$$
Then, if we set our datum at point 2 and assume velocity = 0 at point 1, we can remove two more terms:
$$p_{1} + \rho g z_{1} = p_{2} + \frac{1}{2} \rho u_{2}^2 + \frac{1}{2} \rho u_{2}^2 \sum_{}^{} K_{L}$$
Here are couple more questions and I apologise if these are really obvious, but I cannot seem to understand this topic properly:
1. Is it correct to say $p_1 = p_2 = p_{atm}$? How do I know whether $p_{atm}$ is just the static pressure or the total pressure? Does it just depend on the point where we are talking about on our streamline? (for this example, I think that it is just the static pressure. For $p_{atm}$ to be the total pressure, we would need to take a point above the water level?) This would make the equation (and dividing by $\rho g$):
$$z_{1} = \frac{1}{2g} u_{2}^2 + \frac{1}{2g} u_{2}^2 \sum_{}^{} K_{L}$$

These equations are based on the static pressure. As far as I'm concerned, this is the total pressure and also the just-plain pressure. The pressure at point 1 at the surface of the upper reservoir is atmospheric. The exit pressure from the pipe at the lower point is also atmospheric.

2. From here, I don't know what happens to the dynamic head term at point 2. Looking back, the only thing I can guess is that it has something to do with this energy loss term? Our professor made a point of telling us not to 'double count' the $\frac{u_{2}^2}{2g}$, but I cannot see how to get rid of it. Would anyone be able to help explain this to me?

The velocity of the water at point 1 at the surface of the upper reservoir is essentially zero. The velocity of the fluid coming out of the pipe below is certainly not zero.

3. Looking back at that energy lost (per unit volume?) due to friction term, is it in that form because of $head loss = \frac{4fL}{d} \frac{V^2}{2g}$ and thus $p_L = \rho g h_L = \frac{4fL}{d} \frac{\rho V^2}{2}$?

I'm not quite sure I fully understand what you are asking, but I think you are correct.

Chet

 

[Master1022]

Thank you for your response.

The velocity of the water at point 1 at the surface of the upper reservoir is essentially zero. The velocity of the fluid coming out of the pipe below is certainly not zero.

Yes, but there is still an extra term for the dynamic head at point 2 on the RHS which doesn't feature in the answer. In the solution that term is no longer there, but I am unable to find the reason for doing so.

I'm not quite sure I fully understand what you are asking, but I think you are correct.
Chet

I think I understand it now, I was just used to seeing the Darcy equation in the head loss form.

Thank you once again.

 

[Chestermiller]

Yes, but there is still an extra term for the dynamic head at point 2 on the RHS which doesn't feature in the answer. In the solution that term is no longer there, but I am unable to find the reason for doing so.

In my judgment, that term belongs there. If your professor disagrees with that, then, in my judgment, he is incorrect.