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Heat Transfer Problem :
Heat transfer from a radial circular fin


Problem.

A pipe of radius R0 has a circular fin of radius R1 and thickness 2B on it (as shown in the figure below). The outside wall temperature of the pipe is Tw and the ambient air temperature is Ta. Neglect the heat loss from the edge of the fin (of thickness 2B). Assume heat is transferred to the ambient air by surface convection with a constant heat transfer coefficient h.

figure : radial circular fin on heated pipe Figure. Radial circular fin on heated pipe.

a) Starting with a shell thermal energy balance, derive the differential equation that describes the radial temperature distribution in the fin.

b) Obtain the radial temperature distribution in the circular fin.

c) Develop an expression for the total heat loss from the fin.


Solution.

A circular fin is also called a radial fin or circumferential fin.

a)

From a thermal energy balance over a thin cylindrical ring of width symbol : Deltar in the circular fin, we get

Rate of Heat           In - Out + Generation = Accumulation

The accumulation term (at steady-state) and the generation term will be zero. So,

equation : (2 pi r 2B q_r)|_r - (2 pi r 2B q_r)|_r+Delta_r - 2 (2 pi r Delta_r) h (T - T_a) = 0 (1)

where h is the (constant) heat transfer coefficient for surface convection to the ambient air and qr is the heat flux for conduction in the radial direction.

Dividing by 4symbol : piB symbol : Deltar and taking the limit as symbol : Deltar tends to zero,

equation : lim Delta-r --> O [(r q_r)|_r+Delta_r - (r q_r)|r]/r = -h/B r (T - T_a) (2)

equation : d/dr (r q_r) = -h/B r (T - T_a) (3)

If the thermal conductivity k of the fin material is considered constant, on substituting Fourier's law (equation : q_r = -k dT/dr) we get

equation : d/dr (r dT/dr) = h/(kB) r (T - T_a) (4)

Let the dimensionless excess temperature be denoted by symbol : theta = (T - Ta)/(Tw - Ta). Then,

equation : d/dr (r d theta/dr) = h/(kB) r theta (5)

Using the chain rule on the left-hand side of the above equation and dividing throughout by r,

equation : d^2 theta/dr^2 + 1/r d theta/dr - h/(kB) theta = 0 (6)

Alternatively, starting with the general expression for the cooling fin, we have

equation : d/dz (kA d theta/dz) = hP theta (7)

For the radial fin, the cross-sectional area (for conduction) is A = 2symbol : pir 2B and the perimeter (for surface convection) is P = 4symbol : pir. Also, r = R0 + z, where z is the coordinate measuring distance from the outside wall of the pipe. Substituting these expressions for A and P with dr = dz, we get

equation : k d/dr (2 pi r 2B d theta/dr) = h 4 pi r theta (8)

The above equation on simplifying is identical to equation (5).

b)

Equation (6) is a modified Bessel equation of order zero. Its solution is

equation : theta = C_1 I_0 (c r) + C_2 K_0 (c r) (9)

where c2 = h/(kB). Note that I0 and K0 are modified Bessel functions (of order zero) of first and second kind, respectively.

The integration constants C1 and C2 are determined using the boundary conditions:

equation : BC 1: r = R_0, theta = 1 implies 1 = C_1 I_0(c R_0) + C_2 K_0(c R_0) (10)

equation : BC 2: r = R_1, d theta/dr = 0 implies 0 = C_1 c I_1(c R_1) - C_2 c K_1(c R_1) (11)

The second boundary condition suggests no heat loss through the edge of the circular fin (of thickness 2B), and requires the evaluation of derivatives of Bessel functions as given below:

equation : derivatives of Bessel functions (12)

Equations (10) and (11) may be solved to yield C1 and C2. Thus,

equation : expressions for C_1 and C_2 (13)

On substituting the integration constants, the dimensionless temperature profile is

equation : final expression for temperature profile (14)

c)

The heat flux at the base of the fin (r = R0) is given by

equation : q_r| r=R_0 = ... (15)

On multiplying the heat flux by the cross-sectional area for heat conduction, the total heat loss from the fin is obtained as

equation : Total heat loss from fin = ... (16)


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