3: CHAPTER STEADY STATE 1-D CONDUCTION
We already mentioned the parallels between heat flow and electrical current flow and the idea of k and h being measures of the resistance to heat transfer through a medium For electricity, resistance is related to the flow (of electrons) by Ohm's Law:
Flow (of electrons)
V I R
T q R
Voltage Driving Force Electrical Resistance to Flow
In parallel, for heat flow:
Flow (of heat) Temperature Difference Driving Force Thermal Resistance to Flow
We will generate expressions for thermal resistance R in different cases
Conduction Through Plane Walls In general:
T T T T k k k q c p t x x y y z z
For a slab or wall at steady state with no internal heat generation (applying the semi-infinite assumption) T 2T x k x 0 2 0 if k=constant x Integrating this expression: T(x) = C1x + C2 y=mx+b linear T profile inside slabs with no heat generation
T q kA Since x , therefore q kAC1 where C1=slope of temp. profile
T1
If C1 is the slope,
L
T2 T1 T C1 x2 x1 x
T2
qx
x
T T q kA So, x R
Thus, thermal resistance is defined as x R kA (slab, 1-D, steady state) We could also get this result by applying our boundary conditions: In general, T(x) = C1x + C2 B.C. #1 Dirichlet - T=T1 at x=0 T1=C1(0)+C2 T1= C2 B.C. #2 Dirichlet - T=T2 at x=L T2=C1L+C2 = C1L+T1
C1 T2 T1 T x L
Conduction Through Cylinders For a tube or hollow cylinder, it is harder to give an equation since the semi-infinite assumption may not hold i.e. the axis over which the heat flow is occurring may be different Radial flow is most common: 1 T r r kr r 0 (no heat generation) T r r r 0 if k is constant
T r C1 Integrating, r ,
T C1 r r
Therefore, the temperature profile is: T C1 ln r C 2 Thus, in a cylinder, the temperature profile is logarithmic over the radius.
Applying boundary conditions: T = Ti @ r = ri T = To @ r = ro
ri
Ti ro
To Solve for C , C via 1 2 substitution:
r Ti ln i Ti To lnri r o C2 r ln i r o
Ti To C1 r ln i r o
Substituting into T C1 ln r C 2 : ln r r ln r r T To o T T (Ti To ) o o ri or Ti To ln ri ln r ro o
T 1 Ti To and r r ln ri r o
(since d/dx(ln ax)=1/x)
For a cylinder, the rate of heat flow is:
A(r)=radial cross-sectional area= 2rL
T q kA(r ) r
T q k 2rL Therefore, r
Ti To q 2kL r ln o Substituting dT/dr: r i
Again in parallel to Ohm's Law, with q as the flow term and T = Ti To as the driving force, the measure of thermal resistance for this example is:
ro ln ri (cylinder, 1D radial, R steady state) 2kL
Q: What if ri = 0?
Conduction Through Spheres For a hollow sphere, again it is harder to give an equation since the semiinfinite assumption may not hold i.e. the axis over which the heat flow is occurring may be different Again, radial flow is most common:
1 2 T r 2 r kr r 0
(no heat generation)
2 T r r r 0 For constant k: 2 T r C1 T C1r 1 C 2 and Thus, r
B.C. T = Ti @ r = ri Ti C1ri C 2 1 To C1ro C 2 B.C. T = To @ r = ro
1
Combining, Ti C1 ri
1
To C1 ro1
1 1 Rearranging, r r o i Ti To 1 C2 T 1 1 r Substituting, r r o i
Using B.C. T = To @ r = ro again to solve for C2 and substituting, we get,
C1
Ti To
T (Ti To ) 1
ri
1 r
r1o r1o
T To To or Ti To
1 r 1 ri
r1o r1o
T 1 2 1 1 (Ti To ) r (r r ) and r
i o
For a sphere, the heat flow rate is:
A(r) = radial cross-sectional area = 4r2 Therefore,
T q kA(r ) r
T q k 4r r
2
Substituting dT/dr,
Ti To q 4k 1 1 ri ro
Again in parallel to Ohm's Law, with q as the flow term and T = Ti To as the driving force, the measure of thermal resistance for this example is:
R
1 ri
r1o
4k
(sphere, 1D radial, steady state)
Just as with conduction, we can derive resistive terms to express heat transfer for convection and radiation Convection:
q hA(Ts T )
T q R
Ts T 1 R Therefore, q hA
Radiation:
qrad hr ATs T where 2 hr Ts T Ts2 T
Ts T 1 R Therefore, q hr A
T q R
NOTE: hr is not a true heat transfer coefficient but represents a grouped, T-sensitive term which allows us to linearize the radiation equation
Contact Resistance at Joints: In the real world, surfaces are NEVER smooth, so when two objects are pressed together, there will be an irregular gap filled with air, oil, etc. also: adhesives, joints, welds
Temp. Profile
The temperature profile is therefore discontinuous there is a thermal contact resistance (Rtc).
TA TB 1 Rtc q hc A
See Tables 3.1,3.2
NOTE: text values are per unit area (Rtc") and thus represent 1/hc
Summary of Resistance Terms Conduction: 1D, steady (Table 3.3)
x R kA
Wall/slab
ro ln ri R 2kL
Cylinder
R
1 ri
r1o
Sphere
4k
Ts T 1 R Convection: q hA
Ts T 1 R q hr A
" Rtc TA TB 1 Rtc q hc A A
Radiation: Contact:
So... How do these equations help us solve heat transfer problems?
OVERALL HEAT TRANSFER COEFFICIENT
The heat flow of an overall thermal circuit may be defined as:
qx UAT
Thus, in the case of the glass window problem solved (example 1):
1 1 UA 1 1 L Rtotal hi A kA ho A
NOTE: For constant cross-sectional area, As = Ac (U constant at all x) If the cross-sectional area differs in the direction of heat transfer, you must specify where U applies, i.e. Ui (inside wall) or Uo (outside wall) UiAi =UoAo (conservation of energy)
EQUIVALENT THERMAL CIRCUITS
Electric circuits can be analyzed as a series of resistive elements
R3 R1 R2
Resistors in series Total resistance = sum of resistances R1 + R2 + R3 = Rtotal Resistors in parallel Total resistance = reciprocal sum of resistances
R1 R1 R2 R3
1 1 1 1 R1 R2 R3 Rtotal
The same method can be used to solve heat flow problems (but NOT temperature distribution problems)
Consider the analogy with fluid flow: Flow=Pressure Gradient (driving force) Frictional Drag (resistance) Compare the resistance in the following scenarios (hollow tubes): (a)
(b) (c) (d)
(e)
EXAMPLE 1: Glass window at steady state, with convective heat transfer inside and outside and conduction through the glass find the heat flow:
We can solve this problem directly using conservation of energy i.e. q(out) of any layer must equal q(in):
T,i Ts ,i Ts ,i Ts ,o Ts ,o T,o qx 1 1 x hi A ho A k glass A
Need to know at least one surface temperature to solve the problem
Instead, write an equivalent thermal circuit accounting for each of these heat transfer events:
1/hoA
L/kA
1/hiA
q
Thus, we can reduce the problem to a set of resistors in series, such that the total heat transfer resistance is:
Rtotal
L 1 1 hi A kA ho A
qx T,i T,o Rtotal
and the total heat flow is:
Ignore temperatures in the interior of circuit system of equations simple algebra
EXAMPLE 2: Heat flows from left to right in the composite material below
q
Insulation
B C
2/3y 1/3y
A
L
D
LA kA A
k B 2 A 3
LB
LD kD A
q
kC 1 A 3
LC
Q: How will the heat get distributed between materials B and C? In this case, the heat transfer is a combination of units in series (A,B/C, and D) and parallel (B and C)
LA LD 1 1 Rtot Rt LC k A A kD A LB 1 k 2 A k C 3 A B 3
One branch of the circuit One branch of the circuit
1
Q: What would happen in this case?
q A B D L C E
Insulation
2/3y 1/3y
EXAMPLE 3: Critical Insulation A pipe with containing stationary hot water at Tw is insulated using a layer of foam. Find the heat flow through the insulation as a function of the radius and the insulation thickness which gives the maximum heat flow.
ho, T rp Tp ri Ti ro To
Do the heat analysis using both heat balances and equivalent circuits and find U.
ANSWER: Heat Balances Water at rest: Tw = Tp (no convection) Conduction ri 1 ln through pipe: T p Ti q
2kL rp Conduction ro 1 ln through insulation: Ti To q 2kL ri 1 Convection from T T q o outer surface: ho 2ro L
Add all equations inner T cancel:
ro 1 1 ln Tw T q q 2kL ri ho 2ro L
q
Tw T
ri 1 q ln 2kL rp
ri ro ln rp ln ri 1 2kL 2kL ho 2ro L
Equivalent Circuits:
Ti Tw To T
qtotal
r ln i rp 2kL
r ln o r i 2kL
1 hA
surface convection
pipe insulation conduction
Tw T qx Rtotal
Tw T r ln i ln ro r rp 1 i 2kL 2kL ho 2ro L
same as heat balance approach!
UA
1 Rtotal
1 ri ln ln ro r rp 1 i 2kL 2kL ho 2ro L
To maximize q, we must minimize the denominator by changing ro:
dq 0 (condition for maximization) dro 1 q x (Tw T )(2L) r Constant ln i ln ro r rp i 1 k k ho ro
dq d 1 0 (Tw T )(2L) dro dro ri ln ln ro r rp 1 i k k ho ro
Cancelling the constant terms and taking the derivative (quotient rule):
1 1 1 2 0 k ro hro
ro,crit
k h
What does this result mean? Heat flow through an insulating layer has a maximum at a thickness ro
q
ri
ro,
r
Why? This rather unexpected result is due to the fact that the addition of insulation significantly increases the effective (outside) heat transfer area of small cylinders (i.e. wires) Applied to cool high voltage power transmission wires Typical value, ro,crit ~ 0.05/5 = 1 mm
THERMAL CIRCUITS WITH HEAT INPUTS
If a thermal circuit includes a heat generating element, add that element into the circuit and perform an energy balance at the point of generation EXAMPLE: Rear window of an automobile with a thin film heater on the inside and a tinted layer on the outside. Draw the thermal circuit and find the electrical power required to keep the inside surface at 15C
Inside air T,i = 25C hi = 65W/m2K Thin heater qh" Tinted Window Ltw = 0.002m ktw = 0.5W/mK
Ts,i = 15C
Window Lw=0.01m kw = 1.4W/mK
Outside air T = -10C ho = 100W/m2K
DEALING WITH VOLUMETRIC HEAT GENERATION
Resistance approaches are not directly useful in cases in which a volumetric (i.e. not "thin") body in the system is generating heat. Instead: Develop an algebraic expression for T(x) by integrating the heat diffusion equation (as we did in deriving the resistance terms) Find the surface temperatures around the generating body Use these temperatures as the initial or final nodes of your resistance circuit Two typical cases are plane wall (reaction vessel) and cylinder (wire electrical resistance) heat transfer
For Plane Walls:
2T q T x k x q 0 x 2 k 0
q 2 Integrate: T 2k x C1 x C 2
-L +L Ts1 x=0 Ts2
Boundary Conditions: T(-L) = Ts1; T(L) = Ts2
q L2 T ( x) 2k x 2 T s 2 T s1 x T s1 T s 2 1 2 L 2 2 L q L2 T ( x) 2k x2 1 2 T s L
If Ts = Ts1 = Ts2, then
the midplane (x=0) temperature To is
To qL Ts 2k
2
T ( x ) To x T s To L
2
If Ts is unknown, do surface energy balance: Egenerated = Elost by convection
q A( 2 L ) h ( 2 A)(T s T ) q ( L ) h (T s T )
For Cylinders: 1 T 1 T q r r kr r q 0 r r r r k 0 q 2 T r C1 ln r C 2 Integrate: 4k
ro
Boundary Conditions:
L Ts
T(ro) = Ts;
dT dr
0
r 0
(symmetry)
0
2 qro . r 2 1 2 T (r ) 4k ro
Ts
Centreline (ro=0) temperature To is:
r T (r ) Ts 1 r To T s o
2
Again, if Ts is not known, do surface energy balance: Egenerated = Econvection
q (ro2 L ) h ( 2ro L )(Ts T ) q ro Ts T 2h
EXAMPLE: Find the heat generation rate in section B given the surface temperatures Ts1 = 80C and Ts2 = 60C. Assume the system is at steady state. Heat is dissipated via convection only at the edges of objects A and E
Ts1 h, T B q L Ts2 C D E 2/3y 1/3y h, T
A
.
Q: What if you did not know the surface temperatures of B? (a) use a T(x) function you were given to solve for Ts1 = T(-1/2LB) and Ts2 = T(+1/2LB) any function such as T(x) = ax2 + bx + c is essentially the solution to the heat diffusion equation for this situation (T(x,y,z,t)) (b) solve the heat diffusion equation to find the x value within component B where q = 0 (adiabatic point) by locating the adiabatic point, you can estimate what fraction of heat is dissipated out each side of the object.
THERMAL CIRCUITS TIPS: 1) Nodes (locations of known temperatures) assigned to interfaces, resistors assigned to heat transfer pathways 2) Circuits must be drawn so that a single node has the same temperature no matter the path of heat flow in or out of the node 3) You need to know either (a) both temperatures at either end of your circuit OR (b) the heat flow/flux through your circuit and one of the terminal node temperatures to solve a heat transfer problem 4) Point sources of heat ("thin") can be drawn as vectors into a given node (do energy balance at node) 5) Circuits can NOT be drawn through heat-generating volumes start nodal network at interface.
THIN FILM EFFECTS
The use of thin solid or fluid films has significantly increased with the development of new nanotechnology techniques to make small-dimensional functional devices Examples: Microfluidic (lab-on-achip) reactors for chemistry, detection, separations, etc. Flexible, thin-film solar panels for energy collection/conversion Thin-film insulators or electrodes for compact transistors/chips
When the thickness of a solid film or the gap between two solids (air, fluidfilled) is extremely thin (m nm scale), molecular-scale effects must also be considered in conduction Thick gas layer: Gas molecules collide with each other much more than with either solid surface thermal energy based on bulk gas Thin gas layer: Probability of gas molecule collisions with wall becomes large wall changes thermal energy The impact of the surface on the kinetic energy (temperature) of the thin film gas is described by thermal accommodation coefficient (t): Ti Tsc Ti = T just before collision with surface t Ti Ts Tsc = T just after collision with surface
Ts = T of solid surface
High t: solid surface significantly changes gas T (e.g. air-steel ~ 0.97) Low t: surface has minimal effect on gas T (e.g. helium-metal ~0.02) The resistance to heat transfer across the thin gas film is a combination of the resistances associated with gasgas (m-m) collisions and gas-solid surface (m-s) collisions:
q
R
Ts ,1 Ts , 2
t, m m
Rt , m s
Gas-gas collisions: conventional resistance across a conducting slab: L L = distance between two surfaces Rt , m m kf = thermal conductivity of gas kf A A = cross-sectional area of contact Gas-surface collisions: must consider molecular-scale effects collision frequency of gas molecules
For ideal gas:
Rt , m s
mfp 2 t 9 5 kA t 1
mfp = mean free path distance average distance travelled by an energy carrier (electron or phonons vibrations in lattice) before a collision For an ideal gas:
mfp
k BT 2d 2 p
kB = Boltzmann's constant (1.381 x 10-23 J/K) d = diameter of gas molecule (see Fig. 2.8) p = pressure (assume 1 atm if no info given)
= of ratio cp/cv (specific heat capacities at constant pressure and constant volume respectively)
Monoatomic gases (Ar, He): ~1.6 Diatomic gases (H2, O2, N2, CO): ~1.4 Triatomic gases (SO2, CO2): ~1.3
Thus, for thin gas films:
Rtotal Rt , m m Rt , m s
mfp 2 t 9 5 L k f A k f A t 1
Note: If L/mfp is large and t 0, P(m-s collisions) << P(m-m collisions) Rt,m-s << Rt,m-m and Rtotal = L/kA Similarly, for solid films:
k 1 mfp A R 3L Ts,1 Ts ,2 q L
L k 1 mfp A 3L
L = thickness of thin film mfp = mean free path of solid film material see Table 2.1 for sample materials (nm) A = cross-sectional area of contact between thin film and surrounding material Ts,1 and Ts,2 = temperatures at thin film surface interfaces
As with a thin gas layer, if L/mfp is kA large, q L Ts ,1 Ts , 2 (like any slab) Use thin film resistances if L/mfp is <10 as a rule of thumb
FINS
Fins are extended surfaces attached to heat transfer equipment for increasing the rate of heat exchange. e.g. heat sink mounted on a CPU microchip
Cold Fluid
Why? qx = UAT U,T can be changed, but within limits More area = faster heat transfer
HOT
Many designs are possible:
Spines
Baffles Platters
Straight, Uniform A
Straight, Non-Uniform A
Annular
Pin
In general, heat transfer from fins is perpendicular to the principal direction of heat transfer within the solid.
q conv
X=0 X=L
q
qcond
FIN EFFECTIVENESS
The fin effectiveness f is the ratio of the fin heat transfer rate to the heat transfer rate that would exist if the fin was not present:
qf f hAc ,b (Tb T )
qf = actual heat from fin Ac,b = fin cross-sectional area at the base Tb = T at base of fin T = convective fluid T
Usually, installing fins is not worthwhile unless f > 2 (usually >50) For an infinite fin (Case I): kP Q: What does this mean for f hA c designing effective fins? To evaluate f, we need to be able to calculate qf, the fin heat dissipation
1 2
FIN ENERGY BALANCES
For any extended surface at steady state (assume constant k, h, no heat generation, negligible radiation):
dAs Ac(x) dx
qx
qx+dx
dqconv
Energy balance: q x q x dx dq conv dT q x kAc ( x) qx conduction: dx Ac = cross-sectional area of conductive heat flow up fin (may or may not be a function of x)
qx+dx conduction at x - heat losses dq x q x dx q x dq x q x dx (Taylor exp.) dx Substituting and differentiating: d dT q x dx q x k Ac dx dx dx dqconv convective heat loss over dx dqconv hdAs (T T ) Substitute into energy balance:
q x q x dx dq conv
dq x qx dx dqconv dx
dq x dx dqconv 0 Therefore, dx
Substituting and dividing by k dx: d dT h dAs (T T ) 0 Ac dx dx k dx Performing the d/dx differentiation: d 2T 1 dAc dT 1 h dAs 2 A dx dx A k dx (T T ) 0 dx c c
FINS WITH UNIFORM CROSSSECTIONAL AREA
The previous equation is the general solution to a generic fin problem. In cases in which the cross-sectional area is constant as a function of x, the problem can be greatly simplified: For a rectangular fin,
dx T,f
t
Tb L x
w
Ac = wt constant with respect to x No change in Ac over x dAc/dx = 0 As = Px, where P= perimeter (2w+2t) Therefore dAs/dx = P
Substitute these simplifications into the general fin equation d 2T 1 dAc dT 1 h dAs 2 A dx dx A k dx (T T ) 0 dx c c
P h d 2T 2 A k (T T ) 0 dx c
The same equation applies to uniform diameter pin fins, in which case: Ac = r2 and As = 2rx To solve the differential equation, we introduce the variable T T d 2 hP 0 2 (T constant) Substituting: dx kAc hP m2 Define kAc (all constants)
d 2 m 2 0 Therefore: dx 2
This is a linear, homogeneous secondorder differential equation with constant coefficients, the general solution of which is:
C1e mx C2 e mx
The integration constants C1 and C2 can be determined by substituting the appropriate boundary conditions. At x = 0, T = Tb (base temperature of the fin) ALWAYS applies At x = L, different conditions may apply according to the fin properties CASE 1 Fin has an infinite length CASE 2 Fin tip is adiabatic CASE 3 Convective heat transfer occurs at fin tip CASE 4 Fin tip is at a defined T We will develop solutions for each.
CASE I: Fin is Infinite Length Assume L approaches infinity for a long, thin fin. Thus, the temperature at the end of the fin should be approximately equal to the temperature of the surrounding fluid. The boundary conditions are: Tb T b at x = 0, T = Tb T T 0 at x = L, T = T Determination of C1 and C2 leads to:
b exp mx exp mx b
hP T Tb T exp x T kAc
or
Heat flow (loss) from surface (x=0) is T q kAc x x 0 q hPkAc Tb T
CASE II: Fin Tip is Adiabatic The fin has a finite length, L, and the end is insulated. T q 0 kAc Thus, at x = L, x x L The boundary conditions are: Tb T b at x = 0, T = Tb dT d 0, 0 at x = L, dx (adiabatic) dx The integration constants are:
1 exp2mL and 1 exp 2mL T T coshmL x Therefore, b Tb T coshmL Heat flow (loss) from surface (x=0) is hP L q hPkAc Tb T tanh kA c
C1
b
C2
b
CASE III: Fin Tip Loses Heat Via Convection at Surface The fin has a finite length, L, and losses heat by convection from its end Therefore, at x=L, by energy balance, qcond,L= qconv,L
kAc dT dx hL Ac TL T
xL
dT k or dx
hL TL T
xL
where hL is the convection coefficent at the end of the fin Need to know fluid velocity at tip of fin in order to accurately use The boundary conditions are: Tb T b at x = 0, T = Tb d k hL T T at x = L, L L dx
Substituting to find C1 and C2 and then rearranging: sinhmL x coshmL x hL mk T T Tb T coshmL hL mk sinhmL Heat flow (loss) from surface (x=0) is
sinhmL hL mk coshmL q hPkAc Tb T coshmL hL mk sinhmL
NOTE: To evaluate the hyperbolic functions in these expressions, interpolate solutions based on Appendix B1 or substitute definitions: 1 x 1 x x sinh(x) (e e ) cosh(x) (e e x ) 2 2 sinh(x) e x e x tanh(x) x cosh(x) e e x
CASE IV. Fin Tip at Defined T The fin has a finite length, L, and loses heat from its end T(L) = TL The boundary conditions are: Tb T b at x = 0, T = Tb TL T L at x = L, T = TL Substituting to find C1 and C2 and then rearranging:
T T TL T sinhmx sinhmL x Tb T Tb T coshmL
Heat flow (loss) from surface (x=0) is TL T coshmL Tb T q hPkAc Tb T sinhmL See summary of results in Table 3.4
EXAMPLE: Fin Design Equations How long does a rectangular fin have to be to provide 99% of the heat transfer provided by an infinite fin? Assume an adiabatic tip. What is the temperature distribution in this fin? What is the midpoint temperature?
dx
Tb = T 80C b k=100W/mK T = 20C T,f h = 250W/m2K t=2mm w=20mm
L L x
FINS OF ARBITRARY SHAPE
Annular fins and triangular fins are other popular geometries for extended surfaces. In this case, since one or both of the cross-sectional area or surface area change over x, we must retain terms in the overall fin equation d 2T 1 dAc dT 1 h dAs 2 A dx dx A k dx (T T ) 0 dx c c ...and life becomes more difficult! For example, consider an annular fin both the surface area and crosssectional area vary as a function of r, i.e. As 2 (r 2 r12 ) Ac 2rt where t = thickness of the fin r1 = radius of the inner cylinder
Substituting into the general fin equation (replacing x with r): d 2T 1 dAc dT 1 h dAs 2 A dr dr A k dr (T T ) 0 dr c c
d 2T 1 dT 1 h 4r (T T ) 0 ( 2t ) 2 dr dr 2rt k 2rt
Simplifying, d 2T 1 dT 2h (T T ) 0 2 dr r dr kt d 2 1 d 2h m 2 0 m2 or dr 2 r dr if kt The solution to this problem is given in your textbook (see section 3.6.4). However, even for this relatively simple geometry, these calculations (done algebraically) get extremely complicated! Instead, you can use tabular fin efficiency data.
FIN EFFICIENCY METHOD
The fin efficiency f is the ratio between the actual heat transfer from a fin relative to the heat transfer which would be achieved if the entire fin was at the base temperature the maximum possible convective heat loss occurs if Tfin = Tb (highest overall T differential)
qf f hA f b
qf = actual heat from fin Af = total fin surface area = PL (flat surface fin) = DL (cylindrical fin)
Q: What is the range of possible f? We can use the fin efficiency f to calculate the heat dissipation from fins using either geometry-specific efficiency equations or diagrams.
1) Fin Efficiency Equations To develop efficiency equations for each fin geometry, substitute the derived analytical heat expressions as qf in the efficiency equation. Example: for an adiabatic (Case II) fin hPkAc Tb T tanhmL tanhmL f hPLTb T mL The actual heat flow from the fin can be calculated by substituting the qf resulting efficiency value into f hA f b Expressions for the fin efficiencies for other geometries, derived based on the analytical solutions for those geometries, are given in Table 3.5 for straight, circular, and pin fins (rectangular, triangular, or parabolic)
CAUTION: Equations given in Table 3.5 apply directly to adiabatic fin tips. We can use the same expression for fins with convection at the tip by substituting Lc = L+t/2 for L, equating heat transfer from the actual fin to that of a longer fin with an adiabatic tip (accurate if ht/k 0.0625)
t w Lc= L+t/2
t
=
L
w
Tip with convection
Adiabatic Tip
Similarly, for a cylindrical fin, convection from the end area can be compensated for by setting Lad= Lc = L + r/2 (accurate if hr/2k 0.0625)
2) Fin Efficiency Diagrams For straight fins, if w>>t, P~2w
2h 2 3 hP 2 Lc 2 Lc mLc kA kA c p
Lc = L+t/2 (convective) or Lc = L (adiabatic) Ap = profile area of fin = Lct
1 1
Figure 3.18 (rectangular, triangular, and parabolic fins) and Figure 3.19 (annular fins) show plots of f vs. mLc
qf f hA f b
FIN PROBLEMS METHODS
1) Analytical specific solution to the fin heat diffusion equation for the geometry straight or cylindrical fins 2) Fin Efficiency Equations find the appropriate f expression in Table 3.5, calculate the fin efficiency, and find heat flow using qf = f qmax all geometries except annular fins 3) Fin Efficiency Diagrams
h calculate kA p 2 3 Lc 2 and read the fin
1
efficiency from Figure 3.18 (rectangular, triangular, or parabolic slab fins) or Figure 3.19 (annular fins); find heat flow using qf = f qmax NOTE: In methods 2 and 3, use L if tip is adiabatic, Lc if tip is convective
EXAMPLE: Using Fin Efficiencies For the square fin pictured, find the heat flux from the surface of the fin using (a) the design equations (b) the thermal efficiency approach and (c) the graphical approach. Assume the fin tip undergoes convection.
dx
Tb = T 80C b k=100W/m2K T = 20C T,f h = 250W/m2K t=2mm w=20mm
L L=50mm x
CHOOSING THE RIGHT FIN
In design applications, we try to maximize performance with respect to weight (e.g. minimum weight for maximum heat flow). For minimum weight and maximum heat loss, the optimal fin shape is a parabolic pin:
parabolic
Other considerations include the cost of machining the fins, the cost of the fin material, fouling, the geometry of the device, etc. the fin used is not necessarily always the most efficient one!
FIN THERMAL RESISTANCE
The fin thermal resistance term can be defined as: Tb T 1 qf Rt , f note: f q qf hA f f
max
The fin area Af for several common geometries is given in Table 3.5 This is useful if a contact resistance is present at the fin-base interface, since the total heat transfer resistance offered by the fin assembly may be estimated using a thermal circuit:. 1 1 Rtot Rt ,c Rt , f hc At ,c hA f f Thus, heat flow from the surface is Tb T q f qmax f hA f Rtot
OVERALL HEAT FLOW
The calculations up to this point apply to heat flow over a single fin. However, a heat sink is typically comprised of multiple fins positioned around areas with no fins
Example: Temperature profile of an aluminum rod with thin circular aluminum fins. The outer boundary of the rod is at Tb = 120oC, and the fluid (h = 100 W/m2K) is at T = 20oC
Consider the total surface area of the extended surface, fins and base area.
qf qb qf Tb qb T
At NA f Ab
N = number of fins Ab = base surface area exposed to the convective fluid (Ab = Asurf NWtfin for rectangular or annular fins) The total heat transfer by convection from the exposed surface is: qt N f hA f Tb T hAb Tb T
qt hTb T N f A f Ab
Or, since At = Ab + Af qt hTb T N f A f At NA f
Rearranging:
NA f 1 f qt hAt Tb T 1 At
From this, we can get the overall efficiency of the finned surface in the same manner as the fin efficiency: NA f qt o 1 1 f hAt Tb T At
Thus, to calculate the total heat flow from the entire heat transfer surface,
qt o qmax o hAt (Tb T )
This approach can be used to calculate the total heat dissipation from any finned surface.
To account for contact resistance at the point of fin attachment, use an equivalent thermal circuit approach. In the fin thermal circuit: Rt",c 1 Rt , f Rc, f Rconv, f NAc ,b f hNAf
R"t,c = contact resistance of fins per unit contact A Rc,f = contact resistance of a single fin; Ac,b = total contact area between fins and base Ab = exposed area of base (=Ab,t NAc,b)
In the free surface thermal circuit: 1 1 Rt ,s Rconv,s hAb h( At NA f ) Since these resistances are in parallel: qt o hAt Tb T 1 1 1 Rtotal Rt , f Rt ,s Tb T Tb T Solving:
Rt,c << Rt,f !
o ,c
NA f f 1 1 At Rt",c 1 f hA f Ac ,b
EXAMPLE: Resistance Analysis
A fuel cell of Wc = 50mm and Lc = 50mm is equipped with 10 rectangular aluminum fins, each of length 8mm, thickness 1mm, and width 50mm, which are mounted on a tb = 2mm aluminum coating surrounding the fuel cell (k = 200 W/mK). The entire fuel cell assembly is then encased inside an insulated chamber in contact with the fins. A contact resistance of R"t,c = 10-3 m2K/W exists at the fuel cell-aluminum interface. If the fuel cell is maintained at 70C, what is the heat flow out of the fuel cell? Air is used as the convective coolant (h = 50 W/m2K, T = 20C).
HEURISTIC FOR SOLVING HEAT FLOW PROBLEMS FOR FINS
Rectangular or Rod Fin (uniform Ac)? YES Single Fin? NO Use Fin Efficiency Expression @ L=Lc (Table 3.5) YES Use analytical solutions @ L for appropriate tip condition (Table 3.4) NO Annular? YES Use Efficiency Graph @ L=Lc (Figure 3.19) YES NO Triangular? NO Parabolic?
Single Fin? NO
o 1
NA f At
1
f
YES
Use Figure 3.18 or YES Table 3.5 @ L=Lc NO Use Figure 3.18 or Table 3.5 @ L=Lc
Total heat flow (fins+bare surface): qt o hAt (Tb T )
Total heat from single fin q f f hA f (Tb T )
Use Table 3.5 @ L=Lc for appropriate geometry