- #1
Master1022
- 590
- 116
- 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
- - [itex] H_L = \frac{v^2}{2g} (\frac{4fL}{d} + \sum K_L) [/itex]
- 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.
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.