Ó 2001, W. E. Haisler Chapter 7: Heat Transfer Applications in a Solid 30
1-D Heat Conduction through a Single Flat Wall with Convection Boundary Conditions
From previous work for a single flat wall with no heat source, we know that the temperature is a linear function of x:
1-D Heat Conduction through a 2-layer Composite Flat Wall with Convection Boundary Conditions
Again, for no heat sources, temperature field in each layer is a linear function of x. At point A, x=0; at B, x=x1; at C, x=x2.
The heat flux through the boundary between layers 1 and 2 must be continuous. This provides two B.C. at the interface between two layers that must be satisfied.
Consider two bodies that are touching so that they have an interface between them as shown below. Assume that the temperature distribution in bodies 1 and 2 are and , respectively. Thermal conductivity for the two bodies is k1 and k2.
At the interface of two solids, there are two “boundary” conditions that must be met:
1. The temperature of each body at the interface must be equal: .
2. The heat flux leaving body 1 at the interface must equal the heat flux entering body 2, i.e., the heat flux is constant across the interface: .
For the planar wall considered here, the normal .
Temperatures at the interface located at must be equal:
Conservation of heat flux requires that:
Þ Þ Þ
For each layer, steady-state heat transfer (without any heat sources) requires . For layers 1 and 2, this gives
and .
Apply boundary conditions at surfaces A, B, and C.
A: Convection B.C. at x=0 (where )
or
or (1)
B. Interface B.C. at x=x1:
T1(x1)=T2(x1) or (2)
and
or (3)
C: Convection B.C. at x=x2 (where )
or
or (4)
We have the 4 equations above to solve for a1, b1, a2, b2 . Writing these in matrix notation gives:
You can substitute numerical values for all given values like k1, h1, x1, T¥ , etc. Solve for a1, b1, a2, b2, and substitute back into and . You can now calculate temperature, T, and heat flux, , at any point x.
Example: Consider a two-layer composite with 1-D heat transfer through the layers and free convection of air on either side. Assume two layers made of copper [k1=398 J/(sec m oC)], and teflon [k2=0.25 J/(sec m oC)], and thickness of L1=L2=10 cm. Assume the convection coefficient is approximately h1=h2=5 BTU/(hr ft2 oF) and , .
Convert h to SI units:
Thus h1=h2=28.39 J/(sec m2 oC).
Assuming solutions for layers 1 and 2 as and , integrating the ODE and applying boundary conditions, results in the system of equations:
Substituting values of h, k, x, etc. gives the following set of equations:
Solving for the constants gives:
, , and .
Hence, the temperature equations for each layer are given by:
and
(x in meters)
The temperatures at the boundary and interface are given by:
(left boundary)
(interface)
(right boundary)
The heat flux through each layer is given by
Note that the heat flux is positive (indicating heat flow to the right)--as one would expect since higher temperature on left.
To simplify the solution for a composite wall, we seek to develop a simplified relation between the overall heat flux through the composite wall and the given temperature gradient from one side of the composite wall to the other:
or
where U º effective heat transfer coefficient of composite the wall, R=1/U º effective thermal resistance of the composite wall and for this case the known temperature gradient form left to right is given by .
We know from the solution of the ODE for heat transfer that the temperature variation in any layer with no heat source is a linear function of x: T(x) = a + bx where a and b are constants. If the temperature on either side of a wall of thickness L is TA and TB, then T(x)=TA+[(TB-TA)/L] x . For the composite wall, use the following notation:
Note: T1(L1)=T2(L1)=TB. For a convection boundary condition on the left boundary, is usually known (specified) but TA on boundary in unknown. If TA is taken as a known, then must be taken as an unknown. Similarly, on the right boundary, is usually known but TB is unknown.
From conservation of energy, the heat flow rate Qx through a given area A (Qx = qx A) must be constant as it enters on the left and leaves on the right boundary (since we assumed there is no internal heat generation, F). Since heat flow is normal to wall, each layer has same normal area (so area cancels out). Thus,
(q for air on left) = (q for layer 1) = (q for layer 2) = (q for air on right)
Separate into 4 equations:
(1); (2)
(3); (4)
Add the four equations, (1) through (4), to obtain
or
where =
R equivalent thermal resistance of the composite wall. Note that thermal resistance terms (like L/k or 1/h) are additive similar to resistors in electrical theory.
Ó 2001, W. E. Haisler Chapter 7: Heat Transfer Applications in a Solid 30
Evaluate R and solve for qx. Then evaluate temperatures TA, TB, TC, etc. from equations at right.
Two other cases:
TA, T¥,2 specified
TA, TD specified
Ó 2001, W. E. Haisler Chapter 7: Heat Transfer Applications in a Solid 30
Example 1. Consider a two-layer composite with 1-D heat transfer through the layers and free convection of air on either side. Assume two layers made of glass with k1= k2=1.7 J/(sec m oC) and thickness of L1=L2=10 cm. Assume the convection coefficient is approximately h1=h2=5 BTU/(hr ft2 oF) and ,
Convert h to SI units:
Thus h1=h2=28.39 J/(sec m2 oC). The effective thermal resistance is given by
and the heat flux is
2. Same as Case 1 above except copper and teflon layers: (m2 °C)/W
3. Same as Case 2 except BC are TA=200 °C and TC=25 °C:
(m2 °C)/W
Example 4. Same as Case 3 except determine L2 such that
qx=200 W/m2.
m
Refrigerator.
a) Inside air temperature must be maintained at 0 °F, outside air temperature is 75 °F. Three layers: 1/16" aluminum, 2" polyurethane foam insulation (see www.gwsford.com/~fjlawson/matlprops.html#thermal_cond) k=0.026 W/mC, 1/16" aluminum. If the refrigerator is a rectangular box with outside dimensions of 6'x3'x2.5', how much energy in Watts must the cooling device provide to the interior of the refrigerator?
b) For the refrigerator above, if you want to reduce the energy requirement to 1/2 of that required above, what is the required insulation thickness?
b) Required R to reduce heat flux by half:
Determine thickness of insulation:
Heat Conduction through a Cylindrical Wall
For steady-state, Ñ2T=0. Radial flow only, T=T(r). Heat conduction equation in cylindrical coordinates reduces to:
Þ
Assume convection B.C. at interior and exterior surfaces:
Convection B.C. at r=rA (where ):
Convection B.C. at r=rB (where ):
Substitute above B.C. into solution for T(r):
At r=rA:
At r=rB:
We can now solve the following two equations for C1 and C2:
Take the case of air at 50 oF (10 oC) flowing through a cylindrical duct, with an internal radius of 5 inches and a wall thickness of 1 inch. The duct is placed in an environment where the temperature is 140 oF (60 oC). Assume that the duct is made of a dense foam material with k=3 BTU/(hr ft oF). Assume that h1=10 BTU/(Hr ft2 oF) [value depends on flow velocity of air, interior surface, etc] and h2= 5 BTU/(hr ft2 oF). Solve for T(r) and the heat flow through the exterior boundary per unit length of duct.
Substituting these values into the two equations above and solving for C1 and C2 gives C1= 42.84 and C2=118.29 so that
(°F)
Thus, the temperature at the interior and exterior surfaces is T(rA)=80.8 oF and T(rB)=88.6 oF.
The heat flux per unit length of duct on interior surface is
= 308.19 BTU/hr/ft2
and on the exterior surface
= -257.03 BTU/hr/ft2
Multiplying each heat flux term above by circumference gives the heat flux per unit length of pipe (Note: energy flow rate at A = energy flow rate at B):
Qr = radial energy flow rate
= 308.19 BTU/hr/ft2 x (2p x 0.417 ft) (inside boundary)
= -257.03 BTU/hr/ft2 x (2p x 0.5 ft) (outside boundary)
= 807.5 BTU/hr/ft (same on both boundaries)
Solution Summary
for Steady-State 1-D Heat Transfer in a Solid
What equations apply where for 1-D steady-state heat transfer in a plane wall?
There are two basic approaches that have been discussed in this class for the solving 1-D steady-state heat transfer in a solid:
I. Solution of the governing differential equations.
1. ODE (for constant k) is given by
uniaxial (x) heat flow:
radial heat flow:
2. Integrate ODE and obtain two constants of integration, C1 and C2 for each layer: (plane) or (cylindrical).
3. Evaluate constants by applying specified boundary conditions. For a planar problem:
exterior surface temperature, T=given
exterior surface heat flux, or
exterior surface convection,
If there is more than one layer, apply interface boundary conditions by equating temperatures (T) and heat flux at each interface between two layers.
4. From solution for T(x) or T(r): evaluate T at any points desired, evaluate heat flux q, evaluate total energy flow from Q=q (area) (time).
II. Application of Effective Resistance Method
(only for planar wall or uniaxial flow with no heat source)
1. Find effective resistance for the n-layered composite wall from
where < > means to include the h term only if there is convection on left (h1) or right (h2).
2. Determine heat flux from where DT=temperature difference between known right and left boundary conditions.
3. Knowing heat flux from step 2, calculate temperature at boundary of each wall starting at left (or right) where a boundary temperature (or heat flux condition) is given, and continuing for each layer in a composite wall. For example, if the first layer has convection on left with fluid/air temperature T¥,1 and the unknown wall temperature on left is TA (see composite wall derivation) then solve for TA from . Once you know TA, then solve for temperature TB on right side of the first layer from . Continue in same manner until you have temperatures at boundary of each layer.
Summary of boundary conditions for heat transfer. Use the following figure showing a plane wall with two layers.
For no heat sources, the temperature in each layer is given by linear functions of x as shown in the figure: T1(x)=a1+a2x and T2(x)=b1+b2x.
At left boundary, point A (x=0, outward normal points left, nx=-1), choice of 3 B.C.
1. Temperature B.C., T(x=0) = TA(given) = T1(x=0) = a1
2. Heat flux B.C., qx(x=0)=qA(given) = -k1(dT1/dx) = -k1b1
3. Convection B.C.,
qx(x=0) in the fluid = qx(x=0) in the solid
(nx) [-k1 (dT1/dx)] = h1 [T1(x=0)-T¥,1]
(-1) [-k1 b1] = h1 (a1-T¥,1)
At interface between layers (x=L1), temperature and heat flux must be continuous (must apply 2 B.C.)
1. Temperature B.C., T1(x=L1) = T2(x=L1)
a1+b1L1 = a2+b2L1
2. Flux B.C., qx(x=L1) in layer 1= qx(x=L1) in layer 2
-k1 b1 = -k2 b2
At right boundary, point C (x=L1+L2, nx=+1), choice of 3 B.C.
1. Temperature B.C., T(x=0) = TC(given) = T2(x=L1+L2)
= a2 + b2 (L1+L2)
2. Heat flux B.C., qx(x= L1+L2) = qC(given) = -k2(dT2/dx)
= -k2b2
3. Convection B.C.,
qx(x= L1+L2) in the solid = qx(x= L1+L2) in the fluid
(nx) [-k2 (dT2/dx)] = h2 [T2(x=L1+L2)-T¥,2]
(+1) [-k2 b2] = h2 [a2+b2(L1+L2)-T¥,2]
All of the above assumes you have defined the temperature in layer 1 and 2 to be T1(x)=a1+a2x and T2(x)=b1+b2x. If you have a heat source, then you have to solve the following
differential equation:
k (d2T/dx2) + rf = 0
which will give you
For example, if the heat source f is a constant in the layer so that (rf/k)=C, then the integral above is Cx2/2 and the solution is given by