Different forms of fluid energy conservation equations

Click For Summary
SUMMARY

The forum discussion centers on deriving fluid energy conservation equations, specifically the expressions ##\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0## and ##\frac{D}{Dt} \frac{p}{\rho^{\gamma}} = 0## from the initial equation ##\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p##. Participants emphasize the importance of the material derivative and the continuity equation in these derivations. The discussion also highlights the need to incorporate kinetic energy into the energy balance for accurate results, referencing the work of Bird et al. in "Transport Phenomena".

PREREQUISITES
  • Understanding of fluid dynamics and thermodynamics principles
  • Familiarity with material derivatives and continuity equations
  • Knowledge of energy conservation equations in fluid mechanics
  • Proficiency in calculus, particularly in applying the product and quotient rules
NEXT STEPS
  • Study the derivation of the material derivative in fluid dynamics
  • Explore the continuity equation and its applications in fluid mechanics
  • Learn about the role of kinetic energy in energy conservation equations
  • Read "Transport Phenomena" by Bird et al., focusing on Chapter 11 for relevant equations
USEFUL FOR

Students and professionals in fluid dynamics, mechanical engineers, and researchers focused on thermodynamic processes and energy conservation in fluids.

  • #31
Chestermiller said:
This equation is the expression for adiabatic reversible expansion or compression of an ideal gas.

This can be derived from the previous equation using the continuity equation.
I though I had to start with##
\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0

##
The starting point has to be the energy equation and the mass balance equation, also known as continuity equation.
Basically, I have to do what @pasmith says in post #2?
 
Physics news on Phys.org
  • #32
happyparticle said:
I though I had to start with##
\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0

##
I don’t think so. This equation can be derived.
 
  • #33
Chestermiller said:
I don’t think so. This equation can be derived.
I guess I misread post #3

I guess I finally found both expression. I think I was completely wrong since the beginning. More than I thought.
 
Last edited:

Similar threads

Replies
5
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 26 ·
Replies
26
Views
3K
Replies
3
Views
2K
Replies
8
Views
2K