louisiana tech university ruston, la 71272 slide 1 energy balance steven a. jones bien 501...

Louisiana Tech UniversityRuston, LA 71272

Slide 1

Energy Balance

Steven A. Jones

BIEN 501

Wednesday, April 18, 2008


Slide 2

Energy Balance

Major Learning Objectives:1. Provide the equations for the different

modes of energy transfer.2. Describe typical boundary conditions for

heat transfer problems.3. Derive complete solutions for heat tranfer

problems without flow.4. Derive complete solutions for specific

heat transfer problems involving flow.


Slide 3

Energy Balance

Minor Learning Objectives:1. Use Newton’s law of cooling.2. Derive the complete solution for Couette flow of a

compessible Newtonian fluid.3. Describe the dimensional and non-dimensional

parameters involved in heat transfer problems.4. Distinguish between convection and conduction.5. Obtain more facility with the separation of variables

method for solving partial differential equations.6. Examine the coupling between the energy equation and

the momentum equation caused by viscous heating.


Slide 4

Heat Conduction Equations

tzyxQtzyxTkt

tzyxTC p ,,,,,,

,,,

The general equation for heat conduction is:

Increase with time. Source of heat

Difference between flux in and flux out.


Slide 5

Heat Conduction Equations

tzyxQtzyxTkt

tzyxTC p ,,,,,,

,,, 2

If thermal conductivity is constant:

Increase with time. Source of heat

Difference between flux in and flux out.


Slide 6

Differential Form

xx

capacityheatpC

xz

y

x

zvDescribes how much a volume of material will increase in temperature with a given amount of heat input.

I.e., if I add x number of Joules to a volume 1 cm3, it will increase in temperature by T degrees.

TJoules in


Slide 7

Thermal Conductivity

k – defines the rate at which heat “flows” through a material.

Fourier’s law (A hot cup of coffee will become cold). Fourier’s law is strictly analogous to Fick’s law for diffusion.

Tkq

q is flux, i.e. the amount of heat passing through a surface per unit area.

Gradient drives the flux.


Slide 8

Heat Production, Example

VIP

R

VP

2

2a L

Consider the case of a resistor:

V1 V2

The resistor dissipates power according to ,

(where I is the current) or, equivalently

The volume of the resistor is . Therefore, at any spatial location within the resistor, it is generating:

2

2

V

a LR

Joules/s/cm3


Slide 9

Boundary Conditions

• Constant temperature (T=T0)

• Constant flux

• Heat transfer:

• More general:– Surface Condition or

on the closed surface.– Initial condition within the

volume.

0JT

k n

TThq w

),,(0,,, 0 zyxTzyxT

tzyxTT ,,, tzyxJJ ,,,


Slide 10

Semi-Infinite Slab (of marble)

2z

02 z

Flow of Heat

T=T1

T=T0 uniform

How does temperature change with time?

Initial Temperature Profile

tzTT

zTT

021

20

at

as

Boundary Conditions:

00 20 ztTT at


Slide 11


tzatTT

ztatTT

0

00

21

20

Tkt

TC p

Differential Equation (no source term):



Slide 12


tzTT

ztTT

0

00

21

20

at

at

22

2

z

Tk

t

TC p

For 1-dimensional geometry and constant k:

This problem is mathematically identical to the fluid flow near an infinitely long plate that is suddenly set in motion.


Slide 13


01

0*

TT

TTT

0TT

Dimensionless Temperature:

Dimensionless temperature describes the difference between temperature and a reference temperature with respect to some fixed temperature difference, in this case T1 – T0. 01 TT


Slide 14


pC

k

z

T

t

T

where,

22

*2*

0TT

0,01

00

2*

*

tzT

tT

for

for

Equations in Terms of Dimensionless Temperature:


01 TT

T* is valuable because it nondimensionalizes the equations and simplifies the boundary conditions.


Slide 15


t

z

2

Assume a similarity solution, and define:

. We make use of the following relationships:

23212

23212

22 t

z

t

z

tt

ttzz

11

22

2

2

22222

2 1111

ttttzzzz


Slide 16

Semi-Infinite Slab (of marble)With these substitutions, the differential equation becomes:

01

2 2

*2*

23212

T

t

T

t

z

This can be divided by 2321

2 2 tz

to yield:0

22

*2

2

*

T

z

tT

The combination tz 2 can now be replaced with

to give: 02

2

*2*

TT

.


Slide 17

Semi-Infinite Slab (of marble)Instead of trying to solve directly for T*, try to solve for the first derivative:

*T

.

02

d

d

,exp2

4

1

12

41

*

CeC

d

dT

The equation is rewritten as:

which is separated as:

dd

2

.

Thus:1

241ln C

.

So:


Slide 18

Semi-Infinite Slab (of marble)Integrate:

.

,exp2

4

1

12

41

*

CeCd

dT

.

To obtain:

00

4

1 2

CdCec

.

This integral cannot be evaluated in closed form by standard methods. However, it is tabulated in handbooks and it can be evaluated under standard software. The integral is called the Error function (because of it’s origins in probability theory, where Gaussian functions are important).


Slide 19

Newton’s Law of Cooling

Often we must evaluate the heat transfer in a body that is in contact with a fluid (e.g. heat dissipation from a jet engine, cooling of an engine by a radiator system, heat loss from a cannonball that is shot through the air). The boundary between the solid and fluid conforms neither to a constant temperature, nor to a constant flux. We make the assumption that the rate of heat loss per unit area is governed by:

TTh snq


Slide 20

Newton’s Law of Cooling

The heat transfer coefficent, h, is a function of the velocity of the cannonball. I.e. the higher the velocity, the more rapidly heat is extracted from the cannonball.

One generally assumes that the heat transfer coefficient does not depend on temperature. However, in free convection problems, it can be a strong function of temperature because the velocity of the fluid depends on the fluid viscosity, which depends on temperature.

TTh snq


Slide 21

Free Convection

In free convection, the fluid moves as a result of heating of the fluid near the body in question.

Fluid becomes less dense near the body. Bouyency causes it to move up, enhancing transfer of heat.


Slide 22

Forced Convection

• In forced convection (vs. free convection), the velocity is better controlled because it does not depend strongly on the heat flow itself. A fan, for example, controls the velocity of the fluid.


Slide 23

Convection vs. Conduction

• Convection enhances flow of heat by increasing the temperature difference across the boundary.

Because “hot” fluid is removed from near the body, fluid near the body is colder, therefore the temperature gradient is higher and heat transfer is higher, by Fourier’s law. In other words, Fourier’s law still holds at the boundary.

Small gradient, small heat transfer.

Large gradient, large heat transfer.

louisiana tech university ruston, la 71272 slide 1 energy balance steven a. jones bien 501...

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