Difference between pipe flow and open channel flow in reference to egl and hgl
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How can I understand hydraulic gradient line and total energy line?
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Roland Yonaba, Engineer in Hydraulics
Answered Feb 20, 2016
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Just assume those lines are simple graphical representation of how the flow energy behaves along a direction. The rough idea that lies behind flow energy is unless a flow is inviscid (zero viscosity), the whole amount of available energy will decrease because of friction. Friction create head losses, thus energy decrease.
For practical purposes, I will assume energy is expressed in terms of head, expressed in meters. The total energy available in a flow, at a specific position, noted [math]H[/math] can be written as follows :
[math]H = \frac{P}{\rho g} + z + \frac{V^2}{2g}[/math]
As you can see, this total energy head is a sum of three energy heads :
A pressure head, [math]\frac{P}{\rho g}[/math]A potential head, [math]z[/math]A kinetic head, [math]\frac{V^2}{2g}[/math]
The Total Energy Line, also commonly referred as the Energy Grade Line (EGL) is a graphical representation of the aforementioned total head H.
Now consider a flow standing still. Which is not moving at all. In that case, the kinetic head becomes zero. The remaining head is somehow referred as "static head". It is just the total energy head minus the velocity head.
[math]H_{static} = \frac{P}{\rho g} + z = H - \frac{V^2}{2g}[/math]
The Hydraulic Grade Line (HGL) is just a graphical representation of this static head. Therefore one can make the following deductions :
The Hydraulic Grade Line (HGL) lies one velocity head below the the Energy Grade Line (EGL)In case the cross section is constant along the streamline and the flow steady, discharge remains constant and also so does velocity. Therefore, velocity head remains the same aswell. In that case, HGL will always remain parallel to EGL.

Of course, in case the fluid goes through a pump, energy increases suddenly...

And it will decrease suddenly if it goes through a turbine.

Hope this helps.
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Ankush Panwar, Mechanical Engineer at Indian Institute of Technology, Kanpur
Answered Feb 7, 2017
If you know what stream lines are, then its very easy to understand. Lets take Bernoulli's equation which is valid along stream lines (unless flow is irrotational, then it is valid every where),
H =[math] P/ρg + v^2/2g + z [/math],
where,
H= total head ,
Pressure head = [math]P/ρg[/math]
Kinetic head = [math]v^2/2g[/math]
Potential head = [math]z [/math]
These stream lines which contains all the mechanical energy (all 3 heads) would be called total energy lines.
Now subtract the kinetic head from the total energy (H) , then the lines you would get are Hydraulic Gradient lines (you can also say that these lines contains piezometric head )
This question previously had details. They are now in a comment.
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Roland Yonaba, Engineer in Hydraulics
Answered Feb 20, 2016
Continue Reading
Just assume those lines are simple graphical representation of how the flow energy behaves along a direction. The rough idea that lies behind flow energy is unless a flow is inviscid (zero viscosity), the whole amount of available energy will decrease because of friction. Friction create head losses, thus energy decrease.
For practical purposes, I will assume energy is expressed in terms of head, expressed in meters. The total energy available in a flow, at a specific position, noted [math]H[/math] can be written as follows :
[math]H = \frac{P}{\rho g} + z + \frac{V^2}{2g}[/math]
As you can see, this total energy head is a sum of three energy heads :
A pressure head, [math]\frac{P}{\rho g}[/math]A potential head, [math]z[/math]A kinetic head, [math]\frac{V^2}{2g}[/math]
The Total Energy Line, also commonly referred as the Energy Grade Line (EGL) is a graphical representation of the aforementioned total head H.
Now consider a flow standing still. Which is not moving at all. In that case, the kinetic head becomes zero. The remaining head is somehow referred as "static head". It is just the total energy head minus the velocity head.
[math]H_{static} = \frac{P}{\rho g} + z = H - \frac{V^2}{2g}[/math]
The Hydraulic Grade Line (HGL) is just a graphical representation of this static head. Therefore one can make the following deductions :
The Hydraulic Grade Line (HGL) lies one velocity head below the the Energy Grade Line (EGL)In case the cross section is constant along the streamline and the flow steady, discharge remains constant and also so does velocity. Therefore, velocity head remains the same aswell. In that case, HGL will always remain parallel to EGL.

Of course, in case the fluid goes through a pump, energy increases suddenly...

And it will decrease suddenly if it goes through a turbine.

Hope this helps.
32.9k Views · View Upvoters
Your response is private.
Is this answer still relevant and up to date?
Upvote · 114
Comment...
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Ankush Panwar, Mechanical Engineer at Indian Institute of Technology, Kanpur
Answered Feb 7, 2017
If you know what stream lines are, then its very easy to understand. Lets take Bernoulli's equation which is valid along stream lines (unless flow is irrotational, then it is valid every where),
H =[math] P/ρg + v^2/2g + z [/math],
where,
H= total head ,
Pressure head = [math]P/ρg[/math]
Kinetic head = [math]v^2/2g[/math]
Potential head = [math]z [/math]
These stream lines which contains all the mechanical energy (all 3 heads) would be called total energy lines.
Now subtract the kinetic head from the total energy (H) , then the lines you would get are Hydraulic Gradient lines (you can also say that these lines contains piezometric head )
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