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Heat Transfer Problem :
Radial temperature distribution in annular chemical reactor

Problem.

An annular chemical reactor consists of a packed bed of catalyst between two coaxial cylinders. The inner and outer cylinders have radii of r0 and r1, respectively. It is reasonable to assume that there is no heat transfer through the surface of the inner cylinder, which is at a constant temperature T0. The catalytic reaction releases heat at a uniform volumetric rate S throughout the reactor, whose effective thermal conductivity k may be considered constant. Neglect the temperature gradients in the axial direction.

a) Derive a second-order differential equation to describe the radial temperature distribution in the annular reactor starting with a shell thermal energy balance.

b) Establish the radial temperature distribution by solving the differential equation.

c) What viscous flow problem is analogous to this heat conduction problem?

d) Derive an expression for the volumetric average temperature in the reactor.

e) Develop an expression for the temperature at the outer cylindrical wall of the reactor. What will be the outer wall temperature if both the inner and outer radii are tripled?

Solution.

a)

From a thermal energy balance over a thin cylindrical shell of thickness symbol : Deltar in the annular reactor, we get

Rate of Heat           In - Out + Generation = Accumulation

At steady-state, the accumulation term will be zero. So,

equation : (2 pi r L q_r)|_r - (2 pi r L q_r)|_r+Delta_r + S (2 pi r Delta_r L) = 0 (1)

where S is the rate of generation of heat by chemical reaction per unit volume and qr is the heat flux in the radial direction.

Dividing by 2symbol : pi symbol : Deltar L 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 = S r (2)

equation : d/dr (r q_r) = S r (3)

Since the effective thermal conductivity k of the reactor bed may be considered constant, on substituting Fourier's law (equation : q_r = -k dT/dr) we get

equation : -k d/dr (r dT/dr) = S r (4)

b)

On integrating,

equation : r dT/dr = -S r^2/(2k) + C_1 or T = - S r^2/(4k) + C_1 ln r + C_2 (5)

The integration constants are determined using the boundary conditions:

equation : BC 1: r = r_0, q_r = dT/dr = 0 or C_1 = S r_0^2/(2k) (6)

equation : BC 2: r = r_0, T = T_0 or C_2 = T_0 + S r_0^2/(4k) - S r_0^2/(2k) ln r_0 (7)

The first boundary condition suggests no heat transfer through the inner cylindrical wall of the annulus.

On substituting the integration constants, the temperature profile is

T - T_0 = S r_0^2/(4k) [1 - (r/r_0)^2 +2 ln (r/r_0)] (8)

c)

figure : velocity profile in falling film on circular tube is analogous to temperature profile in annular chemical reactor
Figure. Velocity profile in falling film on circular tube is analogous to temperature profile in annular chemical reactor.

The velocity profile for the falling film on the outside of a circular tube (see Figure) is given by:

equation : v_z = rho g R^2/(4 mu) [1 - (r/R)^2 + 2a^2 ln(r/R)] (9)

Substituting aR = r0 and R = r1,

equation : v_z = rho g r_1^2/(4 mu) [1 - (r/r_1)^2 + 2 (r_0/r_1)^2 ln(r/r_1)] (10)

The maximum velocity (which occurs at r = r0) is

equation : v_z,max = rho g r_1^2/(4 mu) [1 - (r_0/r_1)^2 + 2 (r_0/r_1)^2 ln(r_0/r_1)] (11)

The difference between the above two equations yields

equation : v_z - v_z,max = ... (12)

Equations (8) and (12) are identical in form. Thus, the analogous viscous flow problem is the laminar flow of a falling film on the inside of a circular tube. The equivalent quantities are

T - T_0 == v_z - v_z,max, S == rho g, and k == mu (13)

d)

The volumetric average temperature in the reactor may be defined as

equation : volumetric average temperature definition (14)

On substituting the temperature profile in the above expression and integrating [using equation : result on integrating by parts], we get

equation : volumetric average temperature expression (15)

e)

The temperature at the outer cylindrical wall (r = r1) of the reactor is given by

equation : outer wall temperature (16)

When both the inner and outer radii are n times their original values, the term in square brackets gets multiplied by n2 and the outer wall temperature is thus given by

equation : outer wall temperature when radii are n times (17)

For the case when both the radii are tripled, n = 3 in the above expression.

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