# From fluid energy conservation equation to the continuity equation

• happyparticle
happyparticle
Homework Statement
Derive the continuity equation from the energy conservation equation
Relevant Equations
##\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p ##
##\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0##
##\frac{D}{Dt} \frac{p}{\rho^{\gamma}} = 0##
##e = \frac{1}{\gamma -1} \frac{p}{\rho}##
Hey there,

First of all, all energy conservation equations for a fluid I found on google hadn't the ##\gamma## coefficient. What exactly is the difference?

Secondly, by substituting e by ##e = \frac{1}{\gamma -1} \frac{p}{\rho}## in the following equation ##\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p ## I got ##\frac{D}{Dt}(\frac{1}{\gamma -1} \frac{p}{\rho}) + (\gamma - 1)(\frac{1}{\gamma -1} \frac{p}{\rho}) \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p ##

=> ## \frac{D}{Dt}(\frac{1}{\gamma -1}) p + p \nabla \cdot \vec{u} = - (\vec{u} \cdot \nabla)p ##

Then, I'm wondering if in this case the pressure is constant to have the right hand side equal to 0?
Anyway, I don't get the expecting result.

Last edited:
That's the energy-balance and momentum-balance equation for adiabatic processes. So you deal with ideal hydrodynamics. The task is to derive the continuity equation
$$\partial_t \rho + \vec{\nabla} \cdot (\rho \vec{u})=0.$$

If I understood
$$\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p \hspace{10pt} (1)$$
Is the energy-balance equation

and
$$\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0 \hspace{10pt} (2)$$
is the momentum-balanced equation

I'm should be able to get (2) from (1) by replacing ##e## in (1).

No, why?

vanhees71 said:
No, why?
I have a problem that I have get equation (2) and ##
\frac{D}{Dt} \frac{p}{\rho^{\gamma}} = 0
##

From equation (1) by replacing e with ##
\frac{1}{\gamma -1} \frac{p}{\rho}
##

happyparticle said:
Homework Statement: Derive the continuity equation from the energy conservation equation
Relevant Equations: ##\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p ##
##\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0##
##\frac{D}{Dt} \frac{p}{\rho^{\gamma}} = 0##
##e = \frac{1}{\gamma -1} \frac{p}{\rho}##

Hey there,

Secondly, by substituting e by ##e = \frac{1}{\gamma -1} \frac{p}{\rho}## in the following equation ##\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p ## I got ##\frac{D}{Dt}(\frac{1}{\gamma -1} \frac{p}{\rho}) + (\gamma - 1)(\frac{1}{\gamma -1} \frac{p}{\rho}) \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p ##

=> ## \frac{D}{Dt}(\frac{1}{\gamma -1}) p + p \nabla \cdot \vec{u} = - (\vec{u} \cdot \nabla)p ##

Then, I'm wondering if in this case the pressure is constant to have the right hand side equal to 0?
Anyway, I don't get the expecting result.
For an ideal gas, you just need a constitutive equation, the speed of sound will do. So from:
$$\frac{dp}{d\rho}=a^{2}=\frac{\gamma p}{\rho}$$
You write:
$$dp=\frac{\gamma p}{\rho}d\rho\Rightarrow\frac{Dp}{Dt}=\frac{\gamma p}{\rho}\frac{D\rho}{Dt}$$

From here, you can manipulate it however you want.

## What is the fluid energy conservation equation?

The fluid energy conservation equation, often referred to as the Bernoulli equation in fluid dynamics, states that the total mechanical energy of the fluid remains constant along a streamline. It combines kinetic energy, potential energy, and pressure energy, and is given by: $$P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant}$$, where $$P$$ is the pressure, $$\rho$$ is the fluid density, $$v$$ is the flow velocity, and $$h$$ is the height above a reference level.

## What is the continuity equation in fluid dynamics?

The continuity equation in fluid dynamics is a mathematical statement that expresses the principle of mass conservation. For an incompressible fluid, it states that the mass flow rate must remain constant from one cross-section of a pipe to another. Mathematically, it is expressed as $$A_1 v_1 = A_2 v_2$$, where $$A$$ is the cross-sectional area and $$v$$ is the flow velocity. For a compressible fluid, the equation is $$\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbf{v}) = 0$$, where $$\rho$$ is the fluid density and $$\mathbf{v}$$ is the velocity field.

## How do you derive the continuity equation from the fluid energy conservation equation?

Deriving the continuity equation from the fluid energy conservation equation involves recognizing that while the Bernoulli equation deals with energy conservation along a streamline, the continuity equation deals with mass conservation. To derive the continuity equation, you start from the principle of mass conservation and apply it to a differential volume element in the fluid. By considering the mass flux in and out of this element and setting up a balance, you arrive at the continuity equation. The Bernoulli equation, on the other hand, is derived from the work-energy principle and is not directly used to derive the continuity equation.

## What assumptions are made in fluid energy conservation and continuity equations?

For the fluid energy conservation (Bernoulli) equation, the main assumptions are that the fluid is incompressible, non-viscous, and the flow is steady and along a streamline. For the continuity equation, the primary assumption is the conservation of mass. For incompressible fluids, it assumes that the fluid density remains constant. For compressible fluids, it takes into account changes in fluid density and requires the flow field to be differentiable.

## Can the continuity equation be applied to both compressible and incompressible flows?

Yes, the continuity equation can be applied to both compressible and incompressible flows

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