Question

A cylindrical concrete (r = 1495 kg/m3; Cp = 880 J/kg*K; k = 1.5 W/m*K) beam is exposed to a hot gas flow at 500 °C. The convection coefficient of the flow is 24 W/m2*K. The beam is 0.5 meters in diameter and its initial temperature is 20 °C. Determine the centerline temperature of the concrete beam after 46 minutes in the hot gas flow.

Answer 1

Answer:

The center line temperature of the beam is $164^{circ}C$

Solution:

As per the question:

Diameter of the cylinder, D = 0.5 m

Radius of the cylinder, r' = $\frac{D}{2} = \frac{0.5}{2} = 0.25\ m$

Temperature, $T_{infty} = 500^{\circ}C$

Initial temperature, $T_{o} = 20^{\circ}C$

Convection coefficient of heat flow, $h = 24 W/m^{2}$

Time, t = 46 min

k = $1.5\ W/mK$

Now,

Biot no. is given by:

$B_{i} = \frac{hr'}{2k}$

$B_{i} = \frac{24\times 0.25}{2\times 1.5} = 2$

Now, Fourier no. is given by:

$\frac{\alpha t}{r^{2}} = \frac{k}{C}\times t$

$\frac{\alpha t}{r^{2}} = \frac{k\times t}{rC_{p}r'^{2}} = \frac{1.5\times 46\times 60}{1495\times 880} = 0.05$

At $B_{i} = 2$, $\frac{\alpha t}{r^{2}} = 0.05$

Now, using Heisler chart, the temperature of the beam is given by:

$\frac{T - T_{infty}}{T_{o} - T_{infty}} = 0.7$

$\frac{T - 500}{20 - 500} = 0.7$

$T = 164^{circ}C$