Consider steady state cooling of a cylindrical fin (shown in Figure 1) by means

Question

Consider steady state cooling of a cylindrical fin (shown in Figure 1) by means ofonly radiative heat transfer along its length. Cross sectional area of the fin is uniform and the base is at a temperature of TB-500 K. The fin is exposed to an ambient temperature of Tambien,-25 K. Assume constant thermophysical and transport properties. The length of the fin is lm, fin emissivity is 1 and PIkA-1W/K. Here, k is the fin thermal conductivity, P is the fin perimeter, A is the fin cross-sectional area. Using source term linearization and finite volume method, compute the steady state temperature distribution in the fin. Show the FVM coefficients and matrix form AT=b for 5 nodes. Use Jacobi and Gauss-Seidel method for numerical computation. Perform grid independence test to test the accuracy of your numerical results. How many iterations does it take for the solution to converge for both methods with an initial guess of T 500 K?

Explanation / Answer

Differential equation for this problem is given by

d/ dx (k dT /dx ) + h P (Tf - T)/A + q = 0

S = q+ hf P Tf/A – hfPT/A

Ae=K(1*1)/Sxe

To derive a physically correct interfacial conductivity,

Ke=KpKE(Sx)e/((Ke(Sx)e+Kp(Sx)e).

Substituting ke , aE becomes,

Ae=KEKp(1*1)/((Sx)eKe+(Sx)eKp)

bi=ae=Ki+18Ki(1*1)/((0.5XiKi-1)+(0.5Xi+1Ki))

         = 2KiKi+1(1*1)/((XiKi+1)+Xi+1Ki))

Ci=aw=2Ki-1Ki(1*1)/(Xi-1Ki+XiKi-1))

I=2,3,4,5.

b2=2k2K3(1*1)/((X2K3)+(X3k2))

    p/(KA)=1,

length =1

Area=4.71 m2, Permiter = 2*(3.14 D +H)= 8.28 mtr (d=1,H=1)

K= 1.75 W/Mk

   B2=2*1.75*1.75(1*1)/((0.5*1.75)+(0.5*1.75))===3.5

   C2=2ki*J2(1*1)/((X1K2)+(X2K1))

X1=0

   C2=4

B3=5.25

C3=8

B4=6.12

C4=5.25

B5=12

C5=6.12

SPxI=– hfPxI(1*1)/A=-1.75

AI=BI+Ci-SpiXi

   A2=b2+c2-Sp2X2=3.5+4-(-1.75)=9.25

A3=11.5

A4=9.3

A5=14.5

Di=SciXi(1*1)=((q+ hf P Tf/A)Xi(1*1)

Energy is being generated uniformly at 100 W/m3 .

S = hfPT/A=1.75*25hf=43.75.W/m3

Hf=1

=(100+43.75)(0.5)=287.5

Thus the simultaneous equations are

9.25T2=8t3+4T1+287.5

11.5T3=5.25T4+8T2+287.5

9.3T4=6.12T5+5.25T3+287.5

14.5T5=12T6+6.12T4+287.5

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