Post on 14-Dec-2015
Louisiana Tech UniversityRuston, LA 71272
Slide 1
Energy Balance
Steven A. Jones
BIEN 501
Wednesday, April 18, 2008
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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 ,,,
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
Slide 11
Semi-Infinite Slab (of marble)
tzatTT
ztatTT
0
00
21
20
Tkt
TC p
Differential Equation (no source term):
Boundary Conditions:
Louisiana Tech UniversityRuston, LA 71272
Slide 12
Semi-Infinite Slab (of marble)
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.
Louisiana Tech UniversityRuston, LA 71272
Slide 13
Semi-Infinite Slab (of marble)
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
Louisiana Tech UniversityRuston, LA 71272
Slide 14
Semi-Infinite Slab (of marble)
pC
k
z
T
t
T
where,
22
*2*
0TT
0,01
00
2*
*
tzT
tT
for
for
Equations in Terms of Dimensionless Temperature:
Boundary Conditions:
01 TT
T* is valuable because it nondimensionalizes the equations and simplifies the boundary conditions.
Louisiana Tech UniversityRuston, LA 71272
Slide 15
Semi-Infinite Slab (of marble)
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
Louisiana Tech UniversityRuston, LA 71272
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
.
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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).
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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.