Question

A long cylindrical rod of diameter 240 mm with thermal conductivity of 0.6 W/m ⋅ K experiences uniform volumetric heat generation of 24,000 W/m3 . The rod is encapsulated by a circular sleeve having an outer diameter of 440 mm and a thermal conductivity of 6 W/m ⋅ K. The outer surface of the sleeve is exposed to cross flow of air at 27°C with a convection coefficient of 25 W/m2 ⋅ K. (a) Find the temperature at the interface between the rod and sleeve and on the outer surface. (b) What is the temperature at the center of the rod?

Answer 1

Answer:

temperature at the interface between the rod and sleeve: 75.87 °C

temperature on sleeve's  the outer surface: 58.42 °C

temperature at the center of the rod: 219.87 °C

Explanation:

Data

rod radius, $r_R = 0.12 m$

rod thermal conductivity, $k_R = 0.6 W/(m \, K)$

heat generation, $q_{gen} = 24,000 W/m^3$

sleeve outer radius, $r_S = 0.22 m$

sleeve thermal conductivity, $k_S = 6 W/(m \, K)$

air temperature, $T_a = 27 \, C$

convection coefficient, $h = 25 W/(m^2 \, K)$

From figure attached, the unknowns are

temperature at the interface between the rod and sleeve, $T_{o,R}$

temperature at the outer surface, $T_{o,S}$

temperature at the center of the rod, $T_C$

Assumptions: steady state, heat flux in radius direction only

(a)

Heat flux from the inner face of the sleeve to air

Thermal resistance (R_t) combines conduction through sleeve and convection to air (L is the length of the sleeve and the rod)

$R_t = \frac{ln(r_S/r_R)}{2 \pi k_S L} + \frac{1}{h A}$

$R_t = \frac{ln(r_S/r_R)}{2 \pi k_S L} + \frac{1}{h 2 \pi r_S L}$

$R_t \times L = \frac{ln(0.22 m/0.12 m)}{2 \pi 6 W/(m \, K)} + \frac{1}{25 W/(m^2 \, K) 2 \pi 0.220 m}$

$R_t \times L = 0.045 (m \, K)/W$

Rod's Volume is calculated as

$V_r = \pi \times r_R^2 \times L$

$V_r = \pi \times (0.12 m)^2 \times L$

$V_r = 0.045 m^2 \times L$

Energy balance for the rod gives

$q_{in} - q_{out} + q_{gen} = \frac{dq}{dt}$

where $\frac{dq}{dt} = 0$ from steady state assumption, also, $q_{in} = 0$ for this system. So

$q_{out} = q_{gen} \times V_r$

$q_{out} = 24,000 W/m^3 \times 0.045 m^2 \times L$

$q_{out} = 1085.73 W/m^2 \times L$

Energy balance for the sleeve gives

$q_{in} - q_{out} + q_{gen} = \frac{dq}{dt}$

where $\frac{dq}{dt} = 0$ from steady state assumption, also, $q_{gen} = 0$ for this system. So

$q_{in} = q_{out} = 1085.73 W/m^2 \times L$

Heat flux is computed as

$q = \frac{\Delta T}{R}$

For the case inner face of the sleeve to air, the equation is

$q = \frac{T_{o,R} - T_a}{R_t}$

$T_{o,R} = q \times R_t + T_a$  

$T_{o,R} = 1085.73 W/m^2 \times L \times \frac{0.045 (m \, C)/W}{L} + 27 \, C$

$T_{o,R} = 75.87 \, C$  

Heat flux from sleeve's outer face to air has a thermal resistance called R_{conv}

$R_{conv} = \frac{1}{h 2 \pi r_S L}$

$R_{conv} = \frac{1}{25 W/(m^2 \, K) 2 \pi 0.220 m \, L}$

$R_{conv} = 0.029 (m \, K)/W \times 1/L$

For the case outer face of the sleeve to air, heat flux the equation is

$q = \frac{T_{o,S} - T_a}{R_{conv}}$

$T_{o,S} = q \times R_{conv} + T_a$  

$T_{o,S} = 1085.73 W/m^2 \times L \times 0.029 (m \, C)/W \times 1/L + 27 \, C$

$T_{o,S} = 58.42 \, C$

(b)

Rod's temperature as function of radius can be computed as

$T(r) = \frac{q_{gen} \, r_R^2}{4 \, k_R} (1 - \frac{r^2}{r_R^2}) + T_{o,R}$

For r = 0 (the center)

$T(r = 0) = \frac{q_{gen} \, r_R^2}{4 \, k_R} + T_{o,R}$

$T(r = 0) = \frac{24,000 W/m^3 \, (0.12 m)^2}{4 \, 0.6 W/(m \, C)} + 75.87 \, C$

$T(r = 0) = 219.87 C$