Maths problem when deriving conservation of energy equations.

Demonwolf · June 15, 2009, 12:08

Hi,
Ive been reading John Andersons book 'Computational Fluid Dynamics the basics with applications'. Im upto the part where the energy equation is being derived.
The energy equation is derived in nonconservation form and is in terms of the total energy e+ $V^{2}$ /2. This is on page 70.
It then goes on to describe other ways that the equation can be written in different energy terms and begins to do some manipulations to rewrite the energy equations in terms of internal energy only.

It starts by multiplying the momentum equations in nonconversation form by u, v, and w respectively. The original momentum equations are given below:

$\rho$ $\frac{Du}{Dt}$ = - $\frac{\partial p}{\partial x}$ + $\frac{\partial \tau_{xx}}{\partial x}$ + $\frac{\partial \tau_{yx}}{\partial y}$ + $\frac{\partial \tau_{zx}}{\partial z}$ + $\rho$

-- (2.50a)

$\rho$ $\frac{Dv}{Dt}$ = - $\frac{\partial p}{\partial y}$ + $\frac{\partial \tau_{xy}}{\partial x}$ + $\frac{\partial \tau_{yy}}{\partial y}$ + $\frac{\partial \tau_{zy}}{\partial z}$ + $\rho$

-- (2.50b)

$\rho$ $\frac{Dw}{Dt}$ = - $\frac{\partial p}{\partial z}$ + $\frac{\partial \tau_{xz}}{\partial x}$ + $\frac{\partial \tau_{yz}}{\partial y}$ + $\frac{\partial \tau_{zz}}{\partial z}$ + $\rho$

-- (2.50c)

The first part of the manipulation involves multiplying the equations 2.50a, 2.50b, and 2.50c by u, v and w respectively. The book gives the results as shown below:

$\rho$ $\frac{D(u^{2}/2)}{Dt}$ = - $u\frac{\partial p}{\partial x}$ + $u\frac{\partial \tau_{xx}}{\partial x}$ + $u\frac{\partial \tau_{yx}}{\partial y}$ + $u\frac{\partial \tau_{zx}}{\partial z}$ + $\rho$

-- (2.67)

$\rho$ $\frac{D(v^{2}/2)}{Dt}$ = - $v\frac{\partial p}{\partial y}$ + $v\frac{\partial \tau_{xy}}{\partial x}$ + $v\frac{\partial \tau_{yy}}{\partial y}$ + $v\frac{\partial \tau_{zy}}{\partial z}$ + $\rho$

-- (2.68)

$\rho$ $\frac{D(w^{2}/2)}{Dt}$ = - $w\frac{\partial p}{\partial z}$ + $w\frac{\partial \tau_{xz}}{\partial x}$ + $w\frac{\partial \tau_{yz}}{\partial y}$ + $w\frac{\partial \tau_{zz}}{\partial z}$ + $\rho$

-- (2.69)

I can understand the right hand sides of the above three equations. But I dont understand how to get the left hand sides. I am confused how the following is equal:

$\rho$ $\frac{Du}{Dt}\times u$ = $\rho$ $\frac{D(u^{2}/2)}{Dt}$

Can someone show me how this is done, and also tell me what area of differential mathematics I need to read to plug the gap in my knowledge. Thank you.

Demonwolf · June 15, 2009, 15:21

Ive just worked out the solution to this myself, which Ill give below.. so no need to post. It was pretty simple in the end and Im surprised I didnt get it sooner.

Use substitution and then the chain rule to show how $\frac{Du}{Dt}\times u$ = $\frac{D(u^2/2)}{Dt}$

let

Therefore $\frac{Dy}{Du}$ =

Then the chain rule states :

$\frac{Dy}{Dt}$ = $\frac{Dy}{Du}$ $\times$ $\frac{Du}{Dt}$

$\Rightarrow$ $\frac{Dy}{Dt}$ = $u\times\frac{Du}{Dt}$

June 15, 2009, 12:08	Maths problem when deriving conservation of energy equations.	#1
Demonwolf New Member Abid Join Date: Jun 2009 Posts: 4 Rep Power: 17	Hi, Ive been reading John Andersons book 'Computational Fluid Dynamics the basics with applications'. Im upto the part where the energy equation is being derived. The energy equation is derived in nonconservation form and is in terms of the total energy e+ $V^{2}$ /2. This is on page 70. It then goes on to describe other ways that the equation can be written in different energy terms and begins to do some manipulations to rewrite the energy equations in terms of internal energy only. It starts by multiplying the momentum equations in nonconversation form by u, v, and w respectively. The original momentum equations are given below: $\rho$ $\frac{Du}{Dt}$ = - $\frac{\partial p}{\partial x}$ + $\frac{\partial \tau_{xx}}{\partial x}$ + $\frac{\partial \tau_{yx}}{\partial y}$ + $\frac{\partial \tau_{zx}}{\partial z}$ + $\rho$ -- (2.50a) $\rho$ $\frac{Dv}{Dt}$ = - $\frac{\partial p}{\partial y}$ + $\frac{\partial \tau_{xy}}{\partial x}$ + $\frac{\partial \tau_{yy}}{\partial y}$ + $\frac{\partial \tau_{zy}}{\partial z}$ + $\rho$ -- (2.50b) $\rho$ $\frac{Dw}{Dt}$ = - $\frac{\partial p}{\partial z}$ + $\frac{\partial \tau_{xz}}{\partial x}$ + $\frac{\partial \tau_{yz}}{\partial y}$ + $\frac{\partial \tau_{zz}}{\partial z}$ + $\rho$ -- (2.50c) The first part of the manipulation involves multiplying the equations 2.50a, 2.50b, and 2.50c by u, v and w respectively. The book gives the results as shown below: $\rho$ $\frac{D(u^{2}/2)}{Dt}$ = - $u\frac{\partial p}{\partial x}$ + $u\frac{\partial \tau_{xx}}{\partial x}$ + $u\frac{\partial \tau_{yx}}{\partial y}$ + $u\frac{\partial \tau_{zx}}{\partial z}$ + $\rho$ -- (2.67) $\rho$ $\frac{D(v^{2}/2)}{Dt}$ = - $v\frac{\partial p}{\partial y}$ + $v\frac{\partial \tau_{xy}}{\partial x}$ + $v\frac{\partial \tau_{yy}}{\partial y}$ + $v\frac{\partial \tau_{zy}}{\partial z}$ + $\rho$ -- (2.68) $\rho$ $\frac{D(w^{2}/2)}{Dt}$ = - $w\frac{\partial p}{\partial z}$ + $w\frac{\partial \tau_{xz}}{\partial x}$ + $w\frac{\partial \tau_{yz}}{\partial y}$ + $w\frac{\partial \tau_{zz}}{\partial z}$ + $\rho$ -- (2.69) I can understand the right hand sides of the above three equations. But I dont understand how to get the left hand sides. I am confused how the following is equal: $\rho$ $\frac{Du}{Dt}\times u$ = $\rho$ $\frac{D(u^{2}/2)}{Dt}$ Can someone show me how this is done, and also tell me what area of differential mathematics I need to read to plug the gap in my knowledge. Thank you.

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June 15, 2009, 15:21		#2
Demonwolf New Member Abid Join Date: Jun 2009 Posts: 4 Rep Power: 17	Ive just worked out the solution to this myself, which Ill give below.. so no need to post. It was pretty simple in the end and Im surprised I didnt get it sooner. Use substitution and then the chain rule to show how $\frac{Du}{Dt}\times u$ = $\frac{D(u^2/2)}{Dt}$ let Therefore $\frac{Dy}{Du}$ = Then the chain rule states : $\frac{Dy}{Dt}$ = $\frac{Dy}{Du}$ $\times$ $\frac{Du}{Dt}$ $\Rightarrow$ $\frac{Dy}{Dt}$ = $u\times\frac{Du}{Dt}$