Question

A thick casting with a thermal diffusivity of 5 x 10-6 m2/s is initially at a uniform temperature of 150oC. One surface of the casting is suddenly exposed to a high-speed water jet at 20oC, resulting in a very large convective heat transfer coefficient (Hint: assume imposed surface temperature). The thermal conductivity of the casting is 20 W/m-K. Determine the temperature 20 mm in from the surface after 40 seconds. Check your answer using an alternative technique. Your final answer should be about 101.3 oC.

Answer 1

`T_o = 141.81 ^0C`

Step-by-step explanation:

Given that;

Thermal diffusivity
`\alpha = 5 * 10 ^(-6) m^2/s`

Thermal conductivity
`k = 20 \ W/m.K`

Heat transfer coefficient h = ( we are to assume the imposed surface temperature ) = 20 W/m².K

Initial temperature = 150 ° C = (150+273) K = 423 K

Then coolant temperature with which the casting is exposed to = 20° C = (20+273)K = 293 K

Time = 40 seconds

Length = 20mm = 0.02 m

The objective is to determine the temperature at the surface at a depth of 20 mm after 40 seconds.

`Bi = (hL)/(k)`

`Bi = (20*0.02)/(20)`

Bi == 0.02

`\tau = (\alpha t)/(L^2)`

`\tau= (5*10^(-6 )* 40)/(0.020^2)`

`\tau = 0.5`

For a wall at 0.2 Bi

`A_1 = 1.0311`

`\lambda _1 = 0.4328`

Therefore;

`(T_o - T_(\infty))/(T_i - T_(\infty))= A_1 e ^{-( \lambda_1^2 \ \tau)`

`(T_o - 293 )/(423 - 293)= 1.0311 * e ^{-( 0.438^2 * 0.5 )`

`(T_o - 293 )/(423 - 293)= 1.0311 * e ^{-( 0.0959 )`

`(T_o - 293 )/(130)= 1.0311 * 0.9085`

`(T_o - 293 )/(130)= 0.937`

`T_o - 293= 0.937 * 130`

`T_o - 293= 121.81`

`T_o = 121.81+ 293`

`T_o = 414.81 \ K`

`T_o = (414.81 - 273)^0C`

`T_o = 141.81 ^0C`