Question

One surface of a thick aluminum block (*α* = 97.1 ×
10^{−6} m^{2}/s, k= 237 W/m K) initially at 20°C is
subjected to an energy pulse of 5000 kJ/m^{2}. Determine
the temperature (in ^{o}C 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**

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