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1. Video Lectures for MBA
By:
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2. Appendices 10.A & 10.B:
An Educational Presentation
Presented By:
Joseph Ash
Jordan Baldwin
Justin Hirt
Andrea Lance
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3. History of Heat Conduction
Jean Baptiste Biot
 (1774-1862)
 French Physicist
 Worked on analysis of
heat conduction
 Unsuccessful at dealing
with the problem of
incorporating external
convection effects in heat
conduction analysis
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4. History of Heat Conduction
Jean Baptiste Joseph Fourier
 (1768 – 1830)
 Read Biot’s work
 1807 determined how to solve the
problem
 Fourier’s Law
 Time rate of heat flow (Q) through a
slab is proportional to the gradient of
temperature difference
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5. History of Heat Conduction
Ernst Schmidt
 German scientist
 Pioneer in Engineering
Thermodynamics
 Published paper “Graphical Difference
Method for Unsteady Heat Conduction”
 First to measure velocity and
temperature field in free convection
boundary layer and large heat transfer
coefficients
 Schmidt Number
 Analogy between heat and mass
transfer that causes a dimensionless
quantity
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6. Derivation of the Heat
Conduction Equation
A first approximation of the equations
that govern the conduction of heat in a
solid rod.
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7. Consider the following:
 A uniform rod is insulated on both lateral
ends.
 Heat can now only flow in the axial direction.
 It is proven that heat per unit time will pass
from the warmer section to the cooler one.
 The amount of heat is proportional to the
area, A, and to the temperature difference
T2-T1, and is inversely proportional to the
separation distance, d.
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8.  The final consideration can be expressed as the
following:
is a proportionality factor called the thermal
conductivity and is determined by material properties
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9. Assumptions
 The bar has a length L so x=0 and x=L
 Perfectly insulated
 Temperature, u, depends only on position, x,
and time, t
 Usually valid when the lateral dimensions are
small compared to the total length.
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10.  The differential equation governing
the temperature of the bar is a
physical balance between two rates:
 Flux/Flow term
 Absorption term
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11. Flux
 The instantaneous rate of heat transfer from left to
right across the cross sections x=x0 where x0 is arbitrary
can be defined as:
 The negative is needed in order to show a positive
rate from left to right (hot to cold)
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12. Flux
 Similarly, the instantaneous rate of heat transfer
from right to left across the cross section x=x0+Δx
where Δx is small can be defined as:
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13. Flux
 The amount of heat entering the bar in a time span
of Δt is found by subtracting the previous two
equations and then multiplying the result by Δt:
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14. Heat Absorption
 The average change in temperature, Δu, can be
written in terms of the heat introduced, Q Δt and
the mass Δm of the element as:
where s = specific heat of the material
ρ = density
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15. Heat Absorption
 The actual temperature change of the bar is simply
the actual change in temperature at some
intermediate point, so the above equation can also
be written as:
This is the heat absorption equation.
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16. Heat Equation
 Equating the QΔt in the flux and absorption
terms, we find the heat absorption equation to
be:
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17.  If we divide the above equation by ΔxΔt and allow
both Δx and Δt to both go to 0, we will obtain the
heat conduction or diffusion equation:
where
and has the dimensions of length^2/time and called
the thermal diffusivity
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18. Boundary Conditions
 Certain boundary conditions may apply to the
specific heat conduction problem, for
example:
 If one end is maintained at some constant
temperature value, then the boundary condition
for that end is u = T.
 If one end is perfectly insulated, then the
boundary condition stipulates ux = 0.
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19. Generalized Boundary Conditions
 Consider the end where x=0 and the rate of flow of
heat is proportional to the temperature at the end of
the bar.
 Recall that the rate of flow will be given, from left to right, as
 With this said, the rate of heat flow out of the bar from right to
left will be
 Therefore, the boundary condition at x=0 is
where h1 is a proportionality constant
if h1=0, then it corresponds to an insulated end
if h1 goes to infinity, then the end is held at 0 temp.
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20. Generalized Boundary Conditions
 Similarly, if heat flow occurs at the end x = L, then the
boundary condition is as follows:
where, again, h2 is a nonzero proportionality
factor
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21. Initial Boundary Condition
 Finally, the temperature distribution at one
fixed instant – usually taken at t = 0, takes the
form:
occurring throughout the bar
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22. Generalizations
 Sometimes, the thermal conductivity, density,
specific heat, or area may change as the axial
position changes. The rate of heat transfer under
such conditions at x=x0 is now:
 The heat equation then becomes a partial
differential equation in the form:
or
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23. Generalizations
 Other ways for heat to enter or leave a bar must
also be taken into consideration.
 Assume G(x,t,u) is a rate per unit per time.
 Source
 G(x,t,u) is added to the bar
 G(x,t,u) is positive, non-zero, linear, and u does not depend on t
 G(x,t,u) must be added to the left side of the heat equation
yielding the following differential equation
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24. Generalizations
 Similarly,
 Sink
 G(x,t,u) is subtracted from the bar
 G(x,t,u) is positive, non-zero, linear, and u does not
depend on t
 G(x,t,u) then under this sink condition takes the
form:
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25. Generalizations
 Putting the source and sink equations together
in the heat equation yields
which is commonly called the generalized
heat conduction equation
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26. Multi-dimensional space
 Now consider a bar in which the temperature is
a function of more than just the axial x-direction.
Then the heat conduction equation
can then be written:
 2-D:
 3-D:
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27. Example 1: Section 10.6, Problem 9
Let an aluminum rod of length 20 cm be initially
at the uniform temperature 25°C. Suppose that
at time t=0, the end x=0 is cooled to 0°C while
the end x=20 is heated to 60°C, and both are
thereafter maintained at those temperatures.
Find the temperature distribution in
the rod at any time t
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28. Example 1: Section 10.6, Problem 9
Find the temperature distribution, u(x,t)
a2uxx=ut, 0<x<20, t<0
u(0,t)=0 u(20,t)=60, t<0
u(x,0)=25, 0<x<20
From the initial equation we find that:
L=20, T1=0, T2=60, f(x)=25
We look up the Thermal Diffusivity of aluminum→a2=0.86
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29. Example 1: Section 10.6, Problem 9
Using Equations 16 and 17 found on page 614, we
find that
where
u x t T T x p p a
T c e n x
( ) ( ) å¥
n t
2 1 1 , sin 2
=
-
ö çè
÷ø
= - + + æ
1
2 2 2
n
L
n L
L
T n x
f x T T x
= é - - - L
n dx
çè
æ
ö ù
( ) ( ) ò ÷ø
úû
êë
L
L
L
c
0 2 1 1 2 sin p
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30. Example 1: Section 10.6, Problem 9
Evaluating cn, we find that
c x n x dx
= ò é - ( - )
-
ö çè
ù
( ( ) ( ) )
2
é - + =
c n n n n
p p p p
10 7 cos 12sin 5
( )
( ( ) )
n
c n
p
p
p
p
n
n
n
L
n
70cos 50
20
0 sin
20
25 60 0
20
2
0
= +
ù
úû
êë
÷ø
æ
úû
êë
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31. Example 1: Section 10.6, Problem 9
Now we can solve for u(x,t)
( ) ( ) ( ( ) )
( ) ( ( ) ) ( )
u x t x n
ö çè= - + + æ +
u x t x n
å
å
¥
=
-
¥
=
-
2 2 2
0.86
ö çè
÷ø
p
ö çè
p
0.86
æ ÷ø
= + æ +
ö çè
÷ø
æ ÷ø
1
400
1
20
20
, 3 70cos 50 sin
20
0 70cos 50 sin
20
, 60 0
2
2
n
n t
n
n t
e n x
n
e n x
n
p
p
p
p
p
p
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32. Example 1: Section 10.6, Problem 9
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33. Derivation of the Wave Equation
Applicable for:
•One space dimension, transverse vibrations on elastic string
•Endpoints at x = 0 and x = L along the x-axis
•Set in motion at t = 0 and then left undisturbed
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34. Schematic of String in Tension
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35. Equation Derivation
Since there is no acceleration in the horizontal direction
T(x + Dx, t) cos(q + Dq ) - T(x, t) cosq = 0
However the vertical components must satisfy
T(x x,t) sin( ) T(x,t)sin xu (x,t) tt +D q + Dq - q = rD
x
where is the coordinate to the center of mass and the
weight is neglected
Replacing T with V the and rearranging the equation becomes
V x + D x t -
V x t
( , ) ( , ) u (x, t)
x
tt = r
D
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36. Derivation continued
Letting , the equation becomes
Dx®0
V (x, t) u (x, t) x tt = r
To express this in terms of only terms of u we note that
V (x,t) H(t) tan H(t)u (x,t) x = q =
The resulting equation in terms of u is
x x tt (Hu ) = ru
and since H(t) is not dependant on x the resulting equation is
xx tt Hu = ru
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37. Derivation Continued
For small motions of the string, it is approximated that
H = T cosq » T
using the substitution that
a2 = T / r
the wave equation takes its customary form of
xx tt a2u = u
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38. Wave Equation Generalizations
The telegraph equation
u cu ku a2u F(x,t) tt t xx + + = +
where c and k are nonnegative constants
cut arises from a viscous damping force
ku arises from an elastic restoring force
F(x,t) arises from an external force
The differences between this telegraph equation and the customary
wave equation are due to the consideration of internal elastic
forces. This equation also governs flow of voltage or current in a
transmission line, where the coefficients are related to the electrical
parameters in the line.
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39. Wave Equations in Additional Dimensions
For a vibrating system with more than on significant space
coordinate it may be necessary to consider the wave equation in
more than one dimension.
For two dimensions the wave equation becomes
xx yy tt a2 (u + u ) = u
For three dimensions the wave equation becomes
xx yy zz tt a2 (u + u + u ) = u
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40. Example 2: Section 10.7, Problem 6
Consider an elastic string of length L whose ends
are held fixed. The string is set in motion from
its equilibrium position with an initial velocity
g(x). Let L=10 and a=1. Find the string
displacement for any time t.
( )
ì
ï ï
x
4 ,
1,
í
( L -
x
ï ) ï
î
=
4 ,
L
L
g x
x L
£ £
L < x <
3
L
L £ x £
L
3
4
4
4
4
0
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41. Example 2: Section 10.7, Problem 6
From equations 35 and 36 on page 631, we find
that
where
u x t k npx p
( ) å¥ =
n at
ö çè
÷ø
ö çè
æ ÷ø
= æ
, sin sin
1
n
n L
L
g x n x
p 2 sin p
= æ L
n dx
ö çè
ò ÷ø
( ) L
L
n a
k
L
0
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42. Example 2: Section 10.7, Problem 6
Solving for kn, we find:
( )
é
3
n x
dx L x
dx n x
n x
x
n a
p p p p
2 4 sin sin 4 sin
= ò æ ò ò
( )
( )
æ + ÷ø
ö çè
ö
æ
ö çè
æ
æ ÷ø
ö çè
n
k L
ö çè= æ
( ) ÷ ÷ø
ç çè
÷ø
æ + ÷ø
ö
÷ ÷ø
ç çè
- ÷ø
æ + ÷ø
= æ
ù
úû
êë
ö çè
÷ø
æ - + ÷ø
ö çè
ö çè
ö çè
4
sin
4
8 sin 3
sin
4
sin
4
2 4 sin 3
3
2
4
0
4
3
4 4
p p
p
p p p
p p
n n
a n
n n n
n
L
n a
k
dx
L
L
L
L
L
L
k
L
n
L L
L
L
n L
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43. Example 2: Section 10.7, Problem 6
Now we can solve for u(x,t)
u x t L
( )
( )
¥
æ
( )
å
( ) å
å
3 3
=
1
n
¥
3 3
=
¥
=
n at
ö çè
n at
ö çè
ö çè
÷ø
n n n x
ö çè
n n n x
ö çè
ö çè
æ ÷ø
æ
ö
ö
ö
÷ ÷ø
æ
æ
ç çè
ö çè
ö
ö
÷ ÷ø
æ
ç çè
ö çè
ö çè
÷ø
ö çè
ö çè
ö çè
æ + ÷ø
æ
= æ
÷ø
æ ÷ø
æ
÷ ÷ø
ç çè
÷ ÷ø
ç çèæ
÷ø
æ + ÷ø
= æ
÷ø
æ ÷ø
æ
ö
÷ ÷ø
ç çè
÷ ÷ø
ç çè
÷ø
æ + ÷ø
= æ
1
1
3
10
sin
10
sin
4
sin
4
, 80 1 sin 3
sin sin
4
sin
4
, 8 1 sin 3
sin sin
4
sin
4
, 8 sin 3
n
n
n n n x n t
n
u x t
L
L
n
u x t L
L
L
a n
p p p p
p
p p p p
p
p p p p
p
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44. THE END
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