What Equation Is This? (Fluid Mechanics)

[askor]

Can someone please tell me what is the name of below formula?

##H = z + \frac{v^2}{2g} + \frac{p}{γ}##

 

[FEAnalyst]

That's the head form of Bernoulli equation:

https://learn.lboro.ac.uk/pluginfile.php/504743/mod_resource/content/1/Fluid_Mechanics_5.pdf

 

[askor]

That's the head form of Bernoulli equation:

https://learn.lboro.ac.uk/pluginfile.php/504743/mod_resource/content/1/Fluid_Mechanics_5.pdf

What is "head form"?

Isn't Bernoulli's equation should look like below?

##P + \frac{1}{2}ρv^2 + ρgh = \text{constant}##

 

[FEAnalyst]

Usually Bernoulli's equation is given in pressure form (the one that you used). However in some cases it is expressed using hydraulic head. The constant ##H## stands for total head.

Here's another reference: https://en.wikipedia.org/wiki/Bernoulli's_principle

The equation from your first post is given there right after "The constant in the Bernoulli equation can be normalised."

 

[askor]

How do you obtain the head form?
Can the head form derived from pressure form? If yes, can you show me how to do it?

 

[FEAnalyst]

Here's the pressure form that you've given in previous post (I just replaced ##h## with ##z##): $$p + \frac{1}{2} \rho v^{2} + \rho g z = const$$ If we divide both sides by ##\rho g## we will get: $$\frac{p}{\rho g} + \frac{v^{2}}{2g}+z=const$$ We can also replace ##\rho g## with specific weight ##\gamma## so that the equation becomes: $$\frac{p}{\gamma} + \frac{v^{2}}{2g}+z=const$$ Now just name the constant as total head ##H## and here's the equation from your first post.

 

[askor]

Here's the pressure form that you've given in previous post (I just replaced ##h## with ##z##): $$p + \frac{1}{2} \rho v^{2} + \rho g z = const$$ If we divide both sides by ##\rho g## we will get: $$\frac{p}{\rho g} + \frac{v^{2}}{2g}+z=const$$ We can also replace ##\rho g## with specific weight ##\gamma## so that the equation becomes: $$\frac{p}{\gamma} + \frac{v^{2}}{2g}+z=const$$ Now just name the constant as total head ##H## and here's the equation from your first post.

Thank you very much for the explanation, now I understand.