Transcript Lecture 17 on Energy Balance

Energy Balance
Steven A. Jones
BIEN 501
Wednesday, April 18, 2008
Louisiana Tech University
Ruston, LA 71272
Slide 1
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 University
Ruston, LA 71272
Slide 2
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 University
Ruston, LA 71272
Slide 3
Heat Conduction Equations
The general equation for heat conduction is:
T x, y, z, t 
C p
   kT  x, y , z, t    Q  x, y , z, t 
t
Increase with time. Difference
between flux in
and flux out.
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Ruston, LA 71272
Source of
heat
Slide 4
Heat Conduction Equations
If thermal conductivity is constant:
T x, y , z, t 
Cp
 k 2 T  x , y , z , t    Q  x , y , z , t 
t
Increase with time. Difference
between flux in
and flux out.
Louisiana Tech University
Ruston, LA 71272
Source of
heat
Slide 5
Differential Form
C p  heat capacity
Describes how much a
volume of material will
increase in temperature
with a given amount of
heat input.
Joules
in
T
vz
x  x
x
x
z
y
I.e., if I add x number of Joules to a volume 1 cm3, it will
increase in temperature by T degrees.
Louisiana Tech University
Ruston, LA 71272
Slide 6
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.
q  kT
q is flux, i.e. the amount of heat passing through a
surface per unit area.
Gradient drives the flux.
Louisiana Tech University
Ruston, LA 71272
Slide 7
Heat Production, Example
Consider the case of a resistor:
V1
V2
The resistor dissipates power according to P  IV,
2

V
(where I is the current) or, equivalently P 
R
The volume of the resistor is  a 2 L . Therefore, at any
spatial location within the resistor, it is generating:
V 2
 a 2 LR
Louisiana Tech University
Ruston, LA 71272
Joules/s/cm3
Slide 8
Boundary Conditions
• Constant temperature (T=T0)
• Constant flux
• Heat transfer:
T
k
 J0
n
q  hTw  T 
• More general:
– Surface Condition T  T x, y, z, t  or J  J x, y, z, t 
on the closed surface.
– Initial condition T x, y, z,0  T0 ( x, y, z ) within the
volume.
Louisiana Tech University
Ruston, LA 71272
Slide 9
Semi-Infinite Slab (of marble)
T=T0 uniform
z2
Flow of
Heat
Boundary Conditions:
T  T0 as z2  
T  T1 at z2  0  t
Initial
Temperature
Profile T  T0 at t  0  z2  0
z2  0
T=T1
How does temperature change with time?
Louisiana Tech University
Ruston, LA 71272
Slide 10
Semi-Infinite Slab (of marble)
Differential Equation (no source term):
T
Cp
   kT
t
Boundary
Conditions:
T  T0 at t  0  z2  0
T  T1 at z2  0  t
Louisiana Tech University
Ruston, LA 71272
Slide 11
Semi-Infinite Slab (of marble)
For 1-dimensional geometry and constant k:
T
 2T
Cp
k 2
t
z 2
T  T0 at t  0  z2  0
T  T1 at z2  0  t
This problem is mathematically identical to the fluid flow
near an infinitely long plate that is suddenly set in motion.
Louisiana Tech University
Ruston, LA 71272
Slide 12
Semi-Infinite Slab (of marble)
Dimensionless Temperature:
T  T0
T 
T1  T0
*
T  T0
Dimensionless temperature describes the
difference between temperature and a reference
temperature with respect to some fixed
temperature difference, in this case T1 – T0.
Louisiana Tech University
Ruston, LA 71272
T1  T0
Slide 13
Semi-Infinite Slab (of marble)
Equations in Terms of Dimensionless
Temperature:
T *
 2T *
k

,
where


t
z 22
Cp
T  T0
Boundary Conditions:
T *  0 for t  0
T *  1 for z2  0, t  0
T1  T0
T* is valuable because it nondimensionalizes the equations and
simplifies the boundary conditions.
Louisiana Tech University
Ruston, LA 71272
Slide 14
Semi-Infinite Slab (of marble)
Assume a similarity solution, and define:

z2
t
. We make use of the following relationships:
z2

 
   z2 







t  t   2 1 2 t 3 2 
2 1 2 t 3 2 

 
  1 
1 




z2  z2   t 
t 
2


z22 z2
  
  1  
1   1   1  2 

 

 

   2 

z

z






t

 t
 t   
2  t
 2
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Ruston, LA 71272
Slide 15
Semi-Infinite Slab (of marble)
With these substitutions, the differential equation becomes:

z2
2 1 2 t 3 2
 1  2T * 
T *
0
  
2 

 t  
12 32
This can be divided by  z2 2 t
T * 2 t  2T *

0
2

z 2 
to yield:
z2
The combination
to give:
.
Louisiana Tech University
Ruston, LA 71272
t
can now be replaced with

T * 2  2T *

0
2
  
Slide 16
Semi-Infinite Slab (of marble)
Instead of trying to solve directly for T*, try to solve for the first derivative:
d 
  0
d 2
The equation is rewritten as:
which is separated as:
.
So:
d



2
d
.
Thus: ln
T *


   14  2  C1
 2
dT *
2

 exp  14   C1   Ce 4 ,
d
1
.
Louisiana Tech University
Ruston, LA 71272
Slide 17
Semi-Infinite Slab (of marble)
Integrate:
 2
dT *
2
 exp  14   C1   Ce 4 ,
d
1
.
To obtain:

c    Ce
0
1
 2
4
d  C0
.
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 University
Ruston, LA 71272
Slide 18
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:
q  n  hTs  T 
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Ruston, LA 71272
Slide 19
Newton’s Law of Cooling
q  n  hTs  T 
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.
Louisiana Tech University
Ruston, LA 71272
Slide 20
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 University
Ruston, LA 71272
Slide 21
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 University
Ruston, LA 71272
Slide 22
Convection vs. Conduction
• Convection enhances flow of heat by
increasing the temperature difference
across the boundary.
Small gradient, small heat transfer.
Large gradient, large heat transfer.
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.
Louisiana Tech University
Ruston, LA 71272
Slide 23