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

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' =
`(D)/(2) = (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) = (hr')/(2k)`

`B_(i) = (24* 0.25)/(2* 1.5) = 2`

Now, Fourier no. is given by:

`(\alpha t)/(r^(2)) = (k)/(C)* t`

`(\alpha t)/(r^(2)) = (k* t)/(rC_(p)r'^(2)) = (1.5* 46* 60)/(1495* 880) = 0.05`

At
`B_(i) = 2`,
`(\alpha t)/(r^(2)) = 0.05`

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

`(T - T_(infty))/(T_(o) - T_(infty)) = 0.7`

`(T - 500)/(20 - 500) = 0.7`

`T = 164^(circ)C`