Question

A 30-cm-diameter, 4.8-m-high cylindrical column of a house made of concrete (k = 0.79 W/m·K, α = 5.94 × 10−7 m2/s, rho = 1600 kg/m3, and cp = 0.84 kJ/kg·K) cooled to 14°C during a cold night is heated again during the day by being exposed to ambient air at an average temperature of 28°C with an average heat transfer coefficient of 14 W/m2·K. Use analytical one-term approximation method (not the Heisler charts). Determine how long it will take for the column surface temperature to rise to 25°C. The time taken for the column surface temperature to rise to 25°C is

Answer 1

The time for the surface temperature to reach 25 ◦C is 3.24 hr

Step-by-step explanation:

Assumptions

1. 1-D heat conduction

2. constant thermal properties

3. the heat transfer coefficient is constant & uniform over the surface.

Properties

k = 0.79 W/m · ◦C

α = 5.94 × 10−7 m^2/s

ρ = 1600 kg/m^3

Cp = 0.84 kJ/kg · ◦C

Step 1: Check the Biot number

`Bi = (hr_0)/(k)`

`= ((14)(0.15 m))/(0.79) = 2.658`

As Bi > 0.1 hence lumped system analysis will not considered.

Check the Fourier number.

From Table 18-1

`\lambda_1 = 1.7240,  A_1 = 1.3915`

From Table 18-2, we can find

`J_0 = 0.3841`

The Fourier number is determined as

`(T (r_0, t) - T_(\infty))/((Ti -T\infty)) = A_1 e^(−\lambda^2 \tau) J_0(\lambda_1 r/r_0)`

`(25 - 28)/(14 - 28) = (1.3915)e^(−(1.7240)^2 \tau) (0.3841 ·1)`

solving τ

`\tau = 0.3082`

Since τ > 0.2, the one term approximation or the Heisler charts can be used.

The time for the surface temperature to reach 25 ◦C is

`t = (\tau r^2)/(\alpha)`

`= ((0.3082)(0.15 m)^2)/((5.94 * 10^(−7)))`

= 11674.24 s = 3.24 hours