ME 353 Heat Transfer 1

M.M. Yovanovich

P3MT96.MS

Problem 3 of the Midterm Exam, October, 1996.

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The system resistance consists of several component

resistances:

1) contact resistance: Rcont = 1/(hc Abase);

2) base material resistance: Rbase = L/(k Abase);

3) fin array resistance: Rfins = Rfin/Nf;

4) bare unfinned surface resistance: Rbare = 1/(h Abare).

The input parameters are:

(a) fin diameter: Dp = 1.5 mm; fin length: Lp = 50 mm;

(b) base dimensions: 40 mm by 40 mm;

(c) thermal conductivity of base and fins: k = 100 W/(m K);

(d) contact conductance: hc = 500 W/(m^2 K);

(e) film coefficient: h = 120 W/(m^2 K);

(f) case temperature and fluid temperature: Tcase = 95 C,

Tf = 35 C.

Compute the system heat transfer rate.

Compute all component and system resistances.

Steady-state, constant properties, radiation is negligible.

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> restart:

a) Biot number for the pin fins.

> Bi:= 'Bi':

> Bi:= h*Dp/(2*k);

The system resistance.

> Rsystem:=
Rcont + Rbase + 1/(1/Rfins + 1/Rbare);

The heat tranfer rate through the system.

> Qsystem:= (Tcase - Tf)/Rsystem;

> Rcont:= 1/(hc*Abase);

> Rbase:= L/(k*Abase);

> Rfins:= Rfin/Nf;

> Rbare:= 1/(h*Abare);

> Abare:=
Abase - Nf*(Pi/4)*Dp^2; Abase:= w^2;

> m:= sqrt((h*P)/(k*Af));
mLp:= m*Lp;
P:= Pi*Dp;
Af:= (Pi/4)*Dp^2;

> sys:=
(w = 40/1000, Lp = 3/1000, Dp = 1.5/1000,
L = 50/1000, k = 100, Nf = 50, Tf = 35,
Tcase = 95, h = 120, hc = 500);

a) Calculate the fin Biot number and the parameter m*L

to determine whether the fin is very long.

> Bi:= evalf(subs(sys, Bi), 5);
#The temperature along fins is 1D.

> mL:= evalf(subs(sys, m*L), 4);

Since the parameter mL is greater than 2.65, the fin can

be modeled as though it were "infinitely" long.

Therefore the fin resistance is Rfin = 1/(h*P*k*Af)^(1/2).

> `Rfin`:= 1/sqrt(h*P*k*Af);

c) Calculate the component and system resistances,

and the heat transfer rate through the system.

> R[cont]:= evalf(subs(sys, Rcont), 5)*K/W;

> R[base]:= evalf(subs(sys, Rbase), 5)*K/W;

> R[fin]:= evalf(subs(sys, Rfin), 5)*K/W;

> R[bare]:= evalf(subs(sys, Rbare), 5)*K/W;

> R[system]:= evalf(subs(sys, Rsystem), 4)*K/W;

> Q[system]:= evalf(subs(sys, Qsystem), 4)*W;

>

The heat sink is capable of dissipating the heat from the Pentium.