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Old May 8th 2017, 01:37 AM   #1
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project research

Hi guys,

I am taking a class about heat transfer this semester. (physics is not my major)

I found this equation in my textbook.
has anyone encountered this equation before?
This is the energy differential equation through a rectangular tube


uρ(∂i/∂x)+vρ(∂i/∂y)+wρ(∂i/∂x)-[∂/∂x(k∂t/∂x)+∂/∂y(k∂t/∂y)+∂/∂z(k∂t/∂x)]=0

I would like to derive this equation in order to make a presentation.
k I believe is the heat transfer coefficient. And i don t really know how to proceed.
any help would be greatly appreciated
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Old May 8th 2017, 05:10 AM   #2
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First and foremost can you check this equation as it does not seem to include time.

Quite scary for a rate of heat flow equation.

It seems to be a version of the steady state Fourier conduction equation with no internal sources, in three dimensions.


$\displaystyle c\rho \frac{{\partial u}}{{\partial t}} - k{\nabla ^2}u = 0$


Where c is the specific heat, rho is the density, u is the temperature, t is time and k is the thermal conductivity.

It would be very helpful if you provided a list of what your symbols stand for.
Perhaps you can also provide a reference to the book and page no.

When we have got this straightened out I can help with derivations.
Are you more interested from a physics or a maths point of view?
The equation I quoted is the vector version, are you happy with this?

Last edited by studiot; May 8th 2017 at 06:46 AM.
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Old May 9th 2017, 02:16 AM   #3
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hi,

I found out, in the equation
t is the time
i is the enthalpy

the flow is laminar so the terms with t can be ignored

k the thermal conductivity

u v et w advection

I would be more interested in a physics point of view. ( but any help is appreciated)
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Last edited by chrismdn; May 9th 2017 at 02:24 AM.
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Old May 9th 2017, 11:55 AM   #4
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In order to make sense of your information I offer the following reasoning.

Rewriting your equation as follows


$\displaystyle u\rho \frac{{\partial i}}{{\partial x}} + v\rho \frac{{\partial i}}{{\partial y}} + w\rho \frac{{\partial i}}{{\partial z}} = \frac{\partial }{{\partial x}}\left( {k\frac{{\partial t}}{{\partial x}}} \right) + \frac{\partial }{{\partial y}}\left( {k\frac{{\partial t}}{{\partial y}}} \right) + \frac{\partial }{{\partial z}}\left( {k\frac{{\partial t}}{{\partial z}}} \right)$

and taking the symbols to mean the following we have

u, v and w are velocities with dimensions $\displaystyle L{T^{ - 1}}$

p (rho) is a density with dimensions $\displaystyle M{L^{ - 3}}$

i cannot be just enthalpy but if we take it as specific enthalpy we have dimensions of energy per kilogramme or $\displaystyle E{M^{ - 1}}$
(I am keeping energy lumped together as a new dimension E which will make things easier)

So the first distance derivative of specific enthalpy has dimensions $\displaystyle E{M^{ - 1}}{L^{ - 1}}$

So the left hand side of the equation now reads

velocity x density x spatial variation of specific enthalpy


$\displaystyle \left( {L{T^{ - 1}}} \right)\left( {M{L^{ - 3}}} \right)\left( {E{M^{ - 1}}{L^{ - 1}}} \right)$

Simplifying


$\displaystyle \left( {E{L^{ - 3}}} \right)\left( {{T^{ - 1}}} \right)$

We have (energy/volume) per second or rate of change of energy density.

So on the right hand side we must end up with this.

So again we have three terms all the same form and if k is the thermal conductivity it has units Energy per metre-second per degree K (or C)

Now I cannot make t in the equation work as time but if it represents temperature (t) then it makes sense because

The second spatial derivative of temperature has dimensions $\displaystyle t{L^{ - 2}}$

and k has dimensions $\displaystyle E{L^{ - 1}}{T^{ - 1}}{t^{ - 1}}$

So each term on the right hand side has dimensions


$\displaystyle \left( {t{L^{ - 2}}} \right)\left( {E{L^{ - 1}}{T^{ - 1}}{t^{ - 1}}} \right)$

or rate of change of energy density as the LHS.

I think that it is reasonable to call this an energy equation.

Last edited by studiot; May 9th 2017 at 01:27 PM.
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Old May 10th 2017, 09:17 AM   #5
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Thanks or your reply,

I got your point but this is the dimentional analysis of the equation.
How does this help in deriving the equation? should I be using this dimentional analysis?

Chris
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Old May 10th 2017, 09:56 AM   #6
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Well I'm glad you understand dimensional analysis.

You did say that you were interested in the Physics of the situation and I would like to ensure that we have the correct equation, that is we understand what all the terms mean.

That is the physics and you did tell me that t is time which is just not possible or meaningful.

I do not know what you are studying or what your project entails, but that equation is an unusual way of going about the subject.

The advantage of using the standard method is that all the standard data is tabulated and available for looking up and plugging in. The authors must, surely, have also presented this, there must be some context with the equation as presented.
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Old May 12th 2017, 08:57 AM   #7
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maybe

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Old May 12th 2017, 12:27 PM   #8
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OK there seems to be a continued interest in this question so here is more.

Mattlock, Welcome but I'm afraid that your MIT lecture doesn't cover this topic.

Let us start with a control volume

$\displaystyle {U_1} + {P_1}{V_1} + K{E_1} + P{E_1} + Q = {U_2} + {P_2}{V_2} + K{E_2} + P{E_2} + W$

What the given equation is telling us is that as a parcel of fluid passes through this control volume in a time $\displaystyle \partial T$ and in a steady state condition, it looses some energy by conduction. This outflow of energy from the control volume is replenished by new fluid entering the control volume.

This situation is covered by the open system version of the First Law of Thermodynamics, otherwise known as the steady flow equation.


$\displaystyle {U_1} + {P_1}{V_1} + K{E_1} + P{E_1} + Q = {U_2} + {P_2}{V_2} + K{E_2} + P{E_2} + W$

Where U is the internal energy of the entering fluid, PE is the (gravitational) potnetial energy and KE is the kinetic enrgy of the fluid parcel.
Q is the external heat supplied and W is the external work performed.

Now we simplify this as follows:

We assume no change to potential energy

$\displaystyle P{E_1} = P{E_2}$

Since the given equation shows u, v and w constant there is not change to the kinetic energy


$\displaystyle K{E_1} = K{E_2}$

We assume there is no external heat supplied at the boundary walls


$\displaystyle Q = 0$

And that no external work is performed.


$\displaystyle W = 0$


this leaves


$\displaystyle {U_1} + {P_1}{V_1} = {U_2} + {P_2}{V_2}$

But the definition of enthalpy is


$\displaystyle H = U + PV$


or


$\displaystyle \partial H = {U_2} + {P_2}{V_2} - \left( {{U_1} + {P_1}V} \right)$


The equation given uses i = specific enthapy and operates in 3 dimensions so we have the first term in the equation


$\displaystyle \left[ {u\rho \frac{{\partial i}}{{\partial x}} + v\rho \frac{{\partial i}}{{\partial y}} + w\rho \frac{{\partial i}}{{\partial z}}} \right]$


This is equated to the Heat diffusion equation with no sources or sinks in a steady state, which forms the second term in the given equation, after the minus sign.

This equation links the heat flow to the extenal temperature distribution by way of the specific conductivity as already noted.

Because there is a steady state the mass inflow of enthalpy exactly balances the conduction outflow of heat so one the first is considered positive and the second negative and their difference is zero.

Last edited by studiot; May 12th 2017 at 12:33 PM.
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