Question

One surface of a thick aluminum block (α = 97.1 × 10⁻⁶ m²/s, k= 237 W/m K) initially at 20°C is subjected to an energy pulse of 5000 kJ/m². Determine the temperature (in ^oC to the nearest degree) at a depth of 5 cm from the surface of the block 15 seconds after the pulse

Answer 1

Data given : for thick aluminum block

Thermal diffusivity α = 97.1*10^-6 m2/s

Thermal conductivity k = 237 W/m.k

Constant Energy pulse at the surface, Es = 5000 kJ/m2

Initial temperature of block, Ti = 20 C

Depth at which temperature required, x = 5 cm = 0.05 m from the surface of the block

Time required t = 15 sec

formula for energy pulse at surface ( Es = constant) :

T(x,t) - Ti = Es/k√(πt/α) * exp(-x^2/4αt)

√(πt/α) = √(3.14*15sec/ 97.1*10^-6m2/sec) = 696.46 s/m

x^2/4αt = (0.05m)^2/(4*97.1*10^-6 m2/s * 15s) = 0.429

T(5cm, 15s) - 20 = (5*10^6j/m2)/{(237J/s.m.k)*696.46s/m)} * exp(-0.429)

T(5cm,15s) = 20 + 30.29*exp(-0.429)

T(5cm,15s) = 39.72 C

temperature after 15 sec at the 5 cm depth from the surface of the block, T(5cm,15s) = 40 C.

ans: 40 C