2 CHAPTER CONDUCTION
Consider a wall where the surface temperatures are such that T1>T2 T1 L qx = flow rate of heat in the x direction (vector!) [units of Watts] T2 qx
x
qx is dependent on THREE factors:
1. A - surface area perpendicular to the flow of heat (i.e. y x z), 2. k the capacity of material to transfer heat across its body (thermal conductivity) T 3. L temperature gradient across the thickness of the wall, L
These three factors are related together through FOURIER'S LAW:
Q: why the negative sign? Or, in terms of heat flux q" (q" = q/A)
dT q x kA dx
dT q x " k dx
Although we are only considering 1D heat flow in this case, Fourier's Law may be expressed in 3 dimensions:
dT dT dT q" kT k i dx j dy k dz
where is the del operator NOTE: Fourier's law is typically written individually for each dimension over which heat flow occurs.
The driving force of conduction is the temperature gradient
dT [oC/m or K/m] dx
T1
Average gradient:
dT dx
T2
rise run dT
T2 T1 L
L
here dx
0
, thus qx>0
T1
Instantaneous gradient: may be some function of x
dT dx
T2
If T(x)=Px2+Qx+R qx = -kA(dT/dx) = -kA(2Px+Q)
How can you determine if the temperature gradient is linear? Is there an internal source of heat? Is the material through which heat is flowing homogeneous? o Is k constant? Does the surface area perpendicular to the heat flow change with dx? o Increasing cross-sectional area = lower flux, more uniform temperature as a function of x
T T wall
r x q q
HEAT TRANSFER PROPERTIES
Thermal Conductivity (k): (W/mK) High k values are good conductors Low k values are better insulators
e.g.
Material Temp k Material Temp k (K) (W/mK) (K) (W/mK) Brick 300 0.72 Air 300 0.026 Cork 300 0.039 Copper 300 401 Glass 300 1.4 Aluminum 300 237
More values available in Appendix A
k is often given as a constant, but it does vary with T get a value valid for the T range you are looking at.
A more accurate relationship is:
k ko 1 aT
where ko and a are constants. a can be negative or positive depending on the material.
For blends of materials A and B, use a weighted average to determine k:
k eff ak A 1 a k B
(a = fraction of material A in the blend)
For porous media consisting of a mixture of a stationary solid and a fluid, k depends on the geometry of the porous media (size distribution of pores, pore shape, etc.) Minimum k: heat conducts sequentially through a fluid region of length L and a solid region of length (1- )L: (fluid forms long, random fingers in solid)
q Solid ks L Pore (fluid) kf
L
k eff ,min
1
1 ks kf
Maximum k: heat can conduct through either a continuous fluid region of cross-section w or a continuous solid region of crosssection (1-)w: (interconnected solid)
q Solid - ks Pore (fluid) kf L
L
k eff , min k f 1 k s
Choose best approximation based on the material you are using. For solids with uniformly distributed, non-contacting spherical pores (<0.25), use Maxwell equation:
k eff , min k f 2k s 2 k s k f k s k f 2k s k s k f
Thermal Diffusivity (): (m2/s) k c p
k = thermal conductivity (W/mK) cp = heat capacity of material (J/kgK) = density of material (kg/m3)
Heat Capacity = the ability of a material to store heat (per kg) Density = volume per kg cp is the volumetric heat capacity (per m3 instead of per kg) Q: What would cp and be for a good heat conductor? A bad heat conductor?
HEAT DIFFUSION EQUATION
Consider a control volume (CV) as shown below (no bulk motion):
qz+dz dz qx qy qz dx dy qy+dy qx+dx
Heat balance on the control volume:
rate of heat rate of heat rate of heat flow in flow out generation rate of heat accumulation
Ein Eout E g E st or,
dT q k dy dz where x dx
First, consider transfer (Ein and Eout) To determine qx+dx, we must use the Taylor Expansion:
F 2F 2 F ( x x) F ( x) x 2 x ... x x
Taking only the first term of the expansion (since x = dx is small): q x dx q x q x dx x Thus, in the x direction, the difference
E rate of heat rate of heat Ein out flow in flow out
is given as:
T q x q x dx q x dx k xyz x x x T q y q y dy k y xyz In parallel, y q z q z dz T k xyz z z
Summing up over all dimensions: Ein Eout q x q y q z q x dx q y dy q z dz
T T T k k y z k z dxdydz x x y
The energy source term Eg can be written as:
(heat content ) q dxdydz Eg (time)
where q is the rate of heat generation per unit volume (units W/m3)
Heat generation may occur via several different mechanisms, including: Electrical resistance, P = I2R [W] Nuclear reaction Chemical reaction ( H rxn ) Direct thermal heating (furnaces)
Assuming no phase changes (i.e. only sensible heat effects are present), the energy storage (or accumulation) term is given as (again in parallel to the other derivations):
T (heat content ) c p dxdydz E st t (time)
Substituting each of the derived terms in the energy balance Ein Eout E g E st and dividing out the xyz term:
T T T T k k q c p k x x y y z z t
General Heat Diffusion Equation This equation can be used to solve any conduction problem in Cartesian (x,y,z or rectangular) coordinates.
Solutions for other coordinates are: Cylindrical Coordinates (r, , z)
r
Z
T 1 T 1 T T r r kr r r 2 k z k z q c p t
r
Spherical Coordinates (r, , )
1 2 T T 1 r 2 r kr r r 2 sin2 k
1 T T q c p k sin 2 r sin t
SIMPLIFYING ASSUMPTIONS
1. Semi-Infinite Assumption Heat flow is always a vector quantity it has a magnitude and direction
dT q x kA dx
(x-axis)
dT q y kA dy
(y-axis)
dT q z kA dz
(z-axis)
But, do all the dimensions always matter to a given problem? NO! In cases where heat transfer in one or more dimension is negligible relative to heat transfer in another dimension, the semi-infinite assumption be may used to simplify the problem neglect edges of large slabs neglect small contact areas neglect smaller cross-section areas
Example: Consider our wall thought experiment but add the floor and the ceiling. Find the wall temperature.
Ceiling = 16C air = 20C Floor = 18C wall outside = -10C
Over 98% of the wall's cross-sectional area, the wall is in contact with only the inside and outside air, T = 30C At the ceiling contact (1% of the wall's cross-sectional area), T = 4C At the floor contact (1% of the wall's cross-sectional area), T = 2C
Does heat flow between the ceiling, the floor, the wall, and the air? YES Is it significant in the context of the problem NOT REALLY
2. Steady State Assumption This assumption is useful in situations in which there is a steady heat flow through a plane wall or slab INSIDE
air = 20C wall
OUTSIDE
air = -10C
While the inside wall feels cool to our hand, the temperature is constant. Outside the wall will be >-10C since the wall is heated from the inside. The process is at steady-state since none of the temperatures vary as a function of time. In this case, T/t = 0 such that T T T q 0 y z z c p x x y
There are two conditions under which you can apply this assumption: 1) An external driving force is acting to either add or remove heat from the surfaces of interest. In the case of the wall, the furnace adds heat as it flows out of the room to keep the room temperature constant while the air flow continuously replaces the air adjacent to the wall to keep the outdoor temperature constant TRUE STEADY STATE 2) When T1 and T2 do change over time, if the rate of change is sufficiently slow or the time period we are considering is sufficiently short, we can assume T1 and T2 to be constant over this period "PSEUDO" STEADY STATE
KEY PHRASES TO CHECK FOR
FOR STEADY STATE ASSUMPTION TRUE STEADY STATE: "with a uniform temperature gradient" "with a constant heat flux" "with constant surface temperatures" PSEUDO-STEADY STATE: "find the instantaneous temperature" "find the instantaneous heat flow" "find the temperature after x minutes" FOR SEMI-INFINITE ASSUMPTION "thin" "slab" "wall" "narrow" "plane" "negligible thickness" "neglect edge effects"
PROBLEM ASSESSMENT
T T T T k k k q c p x x y y z z t
the result will be T(x,y,z,t) (T profile) ASK YOURSELF: 1. 2. 3. 4. Steady State?
No Heat Generation? q 0 T 0 2-D Heat Flow? z T T , 0 1-D Heat Flow? y z
T 0 t
T 2T x k x k x 2 5. k constant? Then, solve the simplified equation for
dT q kA T and solve x dx for heat flow.
MUST USE Heat Diffusion Equation To determine the temperature profile of a system (T(x,y,z,t)) integrate over significant variable(s) To determine heat flows if k or A vary along a dimensional axis Use Heat Diffusion Equation OR Energy Balance Ein Eout + Eg = Est To determine heat flows in cases with volumetric heat generation To account for non steady-state heat flow within the object (Est 0) You only need Fourier's Law if: Heat flow is steady state AND Cross-sectional area is constant AND k is constant over full volume AND Temperature profile is known (i.e. T(x)=ax2 + bx + cz) or assumed to be a certain shape (line, parabola...) T(x,y,z,t) solved for you!
BOUNDARY CONDITIONS
From calculus, remember that every time you integrate a function, one integration constant is generated. 2T T 0 C1 If x 2 x T C1 x C 2 To define the temperature profile (i.e. to find the integration constants C1 and C2), we need to know T(x) at two points at the boundary of our object (i.e. at x=0 and x=L in a wall or x=0 and x=ro in a cylinder) 2 equations in 2 unknowns e.g. If T=T1 at x=0 and T=T2 at x=L x=0 T(x=0)=T1= C2 T2 T1 x=L T(x=L)=T2=C1L+C2 C1 L Therefore, the temperature profile is:
T2 T1 T ( x) C1 x C2 x T1 L
The heat equation has three 2nd derivatives as a function of position (d2T/dx2, d2T/dy2, d2T/dz2) and one 1st derivative as a function of time(dT/dt)
2 T 2 T 2 T k 2 2 2 q c p T x y z t
Thus, to solve for T(t), we need 1 initial condition (generally, T or q at t=0, the start of transient heat flow) to find 1 integration constant Initial Conditions time o Defined in problem statement To solve for T(x,y,z), we need 2 boundary conditions per dimension of heat flow (T or q defined at x=0 and x=L) to find 2 integration constants Boundary Conditions position o Defined in problem but typically one of three common conditions
Typical Boundary Conditions 1) Constant surface temperature (Dirichlet condition) T
s
T(0,t) = Ts
x
T(x,t)
2) Constant surface heat flux (Neumann condition)
k dT dx
" qs x0
(a) Finite heat flux at surface - heater
qs" x T(x,t)
(b) Adiabiatic/insulated surface
dT dx 0
x0
x
T(x,t)
3) Convection surface condition
T(0,t)
k
dT dx
h T T ( 0, t )
x 0
T , h
x
T(x,t)
EXAMPLE
A plane wall has constant properties, no internal heat generation, and is initially at a uniform temperature of 20C. The back of the wall is attached to a thin electric heater, backed by insulation, which is initially off. Suddenly, the outer surface is heated by a fluid at 50C with a convection coefficient of 1000W/m2K. 1) Write the simplified heat diffusion equation to find the temperature distribution in the wall a few seconds after the fluid flow begins. 2) Specify the boundary and initial conditions. 3) Sketch the temperature profile throughout the system (a) before heat or fluid flow; (b) with fluid flow but no heater; (c) with heater and fluid flow (assume the q" of heater is higher than q" of fluid). Draw curves just after the heater/flow is started, in the middle of the heating, and at steady state. 4) Sketch the heat flux at x=L as a function of time under the same conditions.