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

Answer:

$T_o = 141.81 ^0C$

Explanation:

Given that;

Thermal diffusivity $\alpha = 5 \times 10 ^{-6} m^2/s$

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.

$Bi = \dfrac{hL}{k}$

$Bi = \dfrac{20*0.02}{20}$

Bi == 0.02

$\tau = \dfrac{\alpha t}{L^2}$

$\tau= \dfrac{5*10^{-6 }* 40}{0.020^2}$

$\tau = 0.5$

For a wall at 0.2 Bi

$A_1 = 1.0311$

$\lambda _1 = 0.4328$

Therefore;

$\dfrac{T_o - T_{\infty}}{T_i - T_{\infty}}= A_1 e ^{-( \lambda_1^2 \ \tau)$

$\dfrac{T_o - 293 }{423 - 293}= 1.0311 \times e ^{-( 0.438^2 \times 0.5 )$

$\dfrac{T_o - 293 }{423 - 293}= 1.0311 \times e ^{-( 0.0959 )$

$\dfrac{T_o - 293 }{130}= 1.0311 \times 0.9085$

$\dfrac{T_o - 293 }{130}= 0.937$

$T_o - 293= 0.937 \times 130$

$T_o - 293= 121.81$

$T_o = 121.81+ 293$

$T_o = 414.81 \ K$

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

$T_o = 141.81 ^0C$