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

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

Step-by-step 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 = (ln(r_S/r_R))/(2 \pi k_S L) + (1)/(h A)`

`R_t = (ln(r_S/r_R))/(2 \pi k_S L) + (1)/(h 2 \pi r_S L)`

`R_t * L = (ln(0.22 m/0.12 m))/(2 \pi 6 W/(m \, K)) + (1)/(25 W/(m^2 \, K) 2 \pi 0.220 m)`

`R_t * L = 0.045 (m \, K)/W`

Rod's Volume is calculated as

`V_r = \pi * r_R^2 * L`

`V_r = \pi * (0.12 m)^2 * L`

`V_r = 0.045 m^2 * L`

Energy balance for the rod gives

`q_(in) - q_(out) + q_(gen) = (dq)/(dt)`

where
`(dq)/(dt) = 0` from steady state assumption, also,
`q_(in) = 0` for this system. So

`q_(out) = q_(gen) * V_r`

`q_(out) = 24,000 W/m^3 * 0.045 m^2 * L`

`q_(out) = 1085.73 W/m^2 * L`

Energy balance for the sleeve gives

`q_(in) - q_(out) + q_(gen) = (dq)/(dt)`

where
`(dq)/(dt) = 0` from steady state assumption, also,
`q_(gen) = 0` for this system. So

`q_(in) = q_(out) = 1085.73 W/m^2 * L`

Heat flux is computed as

`q = (\Delta T)/(R)`

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

`q = (T_(o,R) - T_a)/(R_t)`

`T_(o,R) = q * R_t + T_a`

`T_(o,R) = 1085.73 W/m^2 * L * (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) = (1)/(h 2 \pi r_S L)`

`R_(conv) = (1)/(25 W/(m^2 \, K) 2 \pi 0.220 m \, L)`

`R_(conv) = 0.029 (m \, K)/W * 1/L`

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

`q = (T_(o,S) - T_a)/(R_(conv))`

`T_(o,S) = q * R_(conv) + T_a`

`T_(o,S) = 1085.73 W/m^2 * L * 0.029 (m \, C)/W * 1/L + 27 \, C`

`T_(o,S) = 58.42 \, C`

(b)

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

`T(r) = (q_(gen) \, r_R^2)/(4 \, k_R) (1 - (r^2)/(r_R^2)) + T_(o,R)`

For r = 0 (the center)

`T(r = 0) = (q_(gen) \, r_R^2)/(4 \, k_R) + T_(o,R)`

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

`T(r = 0) = 219.87 C`