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MEC203 > Chapter 2 Heat Conduction Equation > Flashcards

Chapter 2 Heat Conduction Equation Flashcards

1
Q

Draw a diagram to explain why there is a negative sign for the heat transfer equation

A

Gradient is negative, heat transfer in opposite direction see tablet

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2
Q

Difference between heat transfer and temperature

A

temp has only magnitude, it is a scalar, heat transfer has direction and magnitude it is a vector quantity

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3
Q

What must you always do when describing heat transfer

A

provide a direction

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4
Q

What is the driving force for any heat transfer and what is the relationship

A

temperature difference, larger the temp difference, larger the heat transfer

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5
Q

What are the three prime coordinate systems. draw

A 
rectangular T(x y z t)
cylindrical T(r, phi, z, t)
spherical T(r, phi, theta, t)
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6
Q

what does steady imply

A

no change with time at any point within the medium

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7
Q

what does transient imply

A

variation with time or time dependence

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8
Q

what does the special case of lumped systems refer to

A

when variation with time but not with position, uniform variation with time

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9
Q

why can we assume 1D heat transfer

A

if the temperature in the medium vaires in one direction only and thus heat is transferred in one direction and the variation of temperature and thus heat transfer in other directions are negligible or zero

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10
Q

what dimension of heat transfer do we usually deal with

A

1D

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11
Q

if heat transfer is positive in the x direction what is the temperature gradient

A

negative in the x direction

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12
Q

symbol and units of heat generation

A

e. W/m^3

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13
Q

assumption for 1d heat transfer

A

since heat conduction through these geomtries is dominant in one direction and negligible in other directions

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14
Q

1D heat conduction equation (words)

A

Rate of heat conduction at x - rate of heat conduction at x+deltax +rate of heat generation inside the element = rate of change of the enegy content of the element

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15
Q

Proof of heat conduction equation

A

start from heat conduction equation

see exam prep doc

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16
Q

heat conduction equation with cartesian 1D coords

A

A is constant therefore can be removed from the derivative

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17
Q

For the heat conduction equation what can we assume about k

A

k is constant and though it usually depends on x we are able to remove it from the partial

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18
Q

If temperature is not dependent on time, what does the heat conduction equation show about the relationship between temperature profile and heat generation
(cartesian coords)

A

if heat generation is constant then the equation shows
d^2T/dx^2 = -e.gen
therefore curvature is constant making it a parabola where the temperature profile T(x) = ax^2 + bx + c

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19
Q

How does the heat conduction equation simplify if it is steady state
(cartesian coords)

A

d/dt = 0 therefore d^2T/dx^2 + e.gen/k = 0

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20
Q

How does the heat conduction equation simplify if we can consider conductivity constant
(cartesian coords)

A

d^2T/dx^2 + e.gen/k = 1/alpha dT/dt where alpha is heat diffustivity k/rhoc

21
Q

What happens to area in cartesian coords for heat conduction equation
(cartesian coords)

A

it cancels, area does not change with x therefore can be removed from the partial

22
Q

How does the heat conduction equation simplify if there is no heat generation
(cartesian coords)

A

d^2T/dx^2 = 1/alpha *dT/dt as e.gen = 0

23
Q

How does the heat conduction equation simplify if there is no heat generation and it is steady state and how many BCs are needed
(cartesian coords)

A

d^2T/dx^2 = 0

two BCs are needed

24
Q

Proof Heat conduction equation in cylindrical coord system where only temp gradient is along r

A

almost exact same as in cartesian coord system but replace dx with dr

25
Q

How does the heat conduction equation simplify if it is steady state
(cylindrical coords)

A

d/dt = 0 therefore 1/rd/dr (rdT/dr)+ e.gen/k = 0

26
Q

How does the heat conduction equation simplify if there is no heat generation and it is transient
(cylindrical coords)

A

1 /rd/dr (rdT/dr) = 1/alpha *dT/dt

27
Q

How does the heat conduction equation simplify if there is no heat generation and it is steady state and how many BCs are needed
(cylindrical coords)

A

d/dr (r*dT/dr) = 0

28
Q

How do you integrate d/dr (r*dT/dr) = 0

A

r*d^2T/dr^2 + dT/dr = 0

29
Q

Whats been wrong with all these questions

A

where there is two types of differential dr and dt then then it should be partial derivatives not

30
Q

What the is the heat conduction equation for spherical coords with variable conductivity

A

1/r^2 * d/dr (r^2 *k *dT/dr) + e.gen = rho * c * dT/dt

31
Q

What the is the heat conduction equation for spherical coords with constant conductivity

A

1/r^2 * d/dr (r^2 *dT/dr) + e.gen/l=k = 1/alpha * dT/dt

32
Q

How does the heat conduction equation simplify if it is steady state
(spherical coords)

A

1/r^2 * d/dr (r^2 *dT/dr) + e.gen/k = 0

33
Q

How does the heat conduction equation simplify if there is no heat generation and it is transient
(spherical coords)

A

1/r^2 * d/dr (r^2 *dT/dr) = 1/alpha *dT/dt

34
Q

How does the heat conduction equation simplify if there is no heat generation and it is steady state and how many BCs are needed
(spherical coords)

A

d/dr (r^2 *dT/dr) = 0

35
Q

How do you convert d/dr (r^2 *dT/dr) = 0

A

r * d^2T/dr^2 + 2*dT/dr = 0

36
Q

what is the combined one dimension heat conduction equation and what do the different ns represent

A

1/r^n d/dr (r^n k *dT/dr) + e.gen = rhoc dT/dt
n = 0 for a plane wall (replace r with x)
n = 1 for a cylinder
n = 2 for a sphere

37
Q

Spherical coords what is the function A(r) =

A

4 * pi() * r^2

38
Q

what are boundary conditions

A

the mathematical expression of the thermal conditions at the boundaries

39
Q

what are the different types of boundary conditions (6)

A 
Specified temperature boundary condition
specified heat flux boundary condition
convection boundary condition
radiation boundary condition
interface boundary condition
generalized boundary condition
40
Q

What is a specified temperature boundary condition

A 
will give a temperature at a specific place, can be for a certain time or just any time
ie T(x,t)  T(0,t) = 150, can sub into an equation for T to resolve constants
41
Q

How is a Specified heat flux boundary conditions

A

q. = -k*dT/dx, can be used to find out what dT/dx is

42
Q

What is an insulated insulated boundary (special case)

A

where the heat flux is zero therefore k*dT/dx = 0, thus first derivative with respect to x is zero
it does not mean temperature is zero, just initial gradient is zero

43
Q

How is thermal symmetry used

A

two side of a plate subject to identical thermal conditions, thermal symmetry about centre of plate temeprature gradient is zero => dT(L/2)/dx =0

44
Q

What is thermal symmetry used a lot

A

in spherical and clyindrical coord systems as around the centre line have symmetry giving dT(0)/dx = 0

45
Q

What is a convection boundary condition

A

heat conduction at the surface in a selected direction = heat convection at the surface in the same direction
-kdT(0,t)/dx = h(T.inf - T(0,t)

46
Q

What is a radiation boundary condition

A

heat conduction at the surface in selected direction = radiation exchange at the surface in the same direction
-kdT(0,t)/dx = effsigma(T.surr^4 - T(0,t)^4)

47
Q

What is an interface boundary condition

A

two bodies in contact must have the same temp at the contact and heat flux on the two dies must be the same
Ta(x0,t)=Tb(x0,t)
-kadTa(x0,t)/dx = -kbdTb(x0,t)/dx
this does not mean that dT/dx is the same for both as have ratio of thermal conductivity

48
Q

What is the generalized boundary condition

A

Heat transfer to the surface in all modes = heat transfer from the surface in all modes, as a surface may involve conduction convection and radiation

49
Q

Typical method when dealing with head conduction problems

A

heat transfer problem -> mathematical formulation (differentail equation and BCs) -> general solution of differential equation -> application of boundary condtions -> solution of problem

MEC203 (12 decks)

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