Question

The heat that is conducted through a body must frequently be removed by other heat transfer processes. For example, the heat generated in an electronic device must be dissipated to the surroundings through convection by means of fins. Consider the one-dimensional aluminum fin (thickness t 3.0 mm, width 20 cm, length L) shown in Figure 1, that is exposed to a surrounding fluid at a temperature T. The conductivity of the aluminum fin (k) and coefficient of heat convection of surrounding fluid (h) are given in Figure 1. The defining equation for the conduction and convection heat-transfer coefficients are recalled as gá = -k A dT/dx and gook = hA' (T-T) where the area in these equations (A and A') are the surface area for conduction and convection, respectively. Z a) The temperature of the base of the fin is To. Make an energy balance on the element of the fin of thickness dx as shown in the figure 1 and develop a linear ODE as the mathematical model for this heat transfer problem. h 10 W/m2 C Temperature T Convection Insulated surface A Conduction Temperature To k 200 W/mC +| dx|^- Temperature Te Temperature T Insulated surface Figure 1: Sketch illustrating one-dimensional conduction and convection through a rectangular fin. b) Consider three following cases and determine the boundary conditions for each case. Find the solution of the linear ODE for each case in the form of temperature distribution along the fin if To 300 C and T 50 °C. CASE 1: The fin is very long, and the temperature at the end of the fin is essentially that of the surrounding fluid. What is the temperature at x L/2? CASE2: The fin is of finite length (7.5 cm long) and loses heat by convection from its end. What are the temperatures at the tip of the fin (x L) and at x = L/2? CASE 3: The end of the fin is insulated so that dT/dx at x L/2? Compare the results and make your conclusion. 0 at x= L. What is the temperature at this point and c) What will be the boundary conditions for the same mathematical model when the base plate contains a uniform heat source rather than just being at a constant temperature T? Repeat the urce calculations of the above three cases if the heat source is an electrical element of 100 W and the only possible heat transfer mechanism is through the fin.

Answer 1

a) To develop a governing differential equation of heat flow within the fun take element of length dx at x from the base of the fin.

Fig 1 shows the element and how heat is flowing through the element.

(ain) Too =Convection heat tbrem fin to ain x,Qx+dz=Conduchion heat Sionv 1 eng tn SLorEnergy balane + Eume dr ax-4atdr eonv Spom Qx id Pconv

Now, Finding the value of Qx that is the rate of conduction along the fun at x using conduction formula we have,

T -KACS -kAcs dr

Where Acs is the cross sectional area of the fin.

Now,for Qx+dx we can write as below,

TAT -k Acs JL+dx 자주자 자

For temperature gradient

dT dx x+dax

We can expand it using the Taylor's expansion for its simplest form,

We can write as,

2 T T + xp xp dxdx2) dx x4dx d Jx So1 dT dT T d Jx tdx dx

Now, modified form of Qx+dx become,

Using squD Qx+dxkAcs dx -K Aes + dT Qx+dx - KAcs KACS dx

For convective heat transfer Qconv from the fin surface to the surrounding air,

QenyhAe T-T P.de CT-Too Conv Cony 9anyhPTTodr whene PE Peni meter ef the fin Cony

Writing the energy balance equation for the element, we have

Conv CarKAcs dx +te(T- T)d -kAsdT -kAs dT - kArs +hP(T-T

Now, further solving we will get,

d2T hP (T-To)-0 KAes 2

This is governing ODE for the fun.

b.) Solution of the above differential equation can be easily derived by substituting,

T-T, = N

We will get,

d2N kAs d2

Solution for N is done by substituting hP/kAcs=m and N=exp(kx) ,We will get

d2N m2N= O M Pat N=e 2 = 0 -m tm, Golution will fe mx N-GeCe

Case1. When fin is very long.

Our boundary conditions will be,

(1りズニ0, N= 300-50-250 2x L Ne 50-50-0

Finding the solution we will have,

m x use d 250- G+C2 use i 0-G() +0 from (i C mast tbe 0 N=GE C2=250 fom T-Too=250e T-50 -250 A mt (ru

For temperature at x=L/2,

Value of m is,

m hp. lov2 (0.26з) kAes 20cx0.003Y0-20 m 33.83 0.38