Combined Convection and Radiation Condition A spherical metal ball of radius r0 is heated in an oven to a temperature of 600 F throughout and is then taken out of the oven and allowed to cool in ambient air at T∞ = 78 F, as shown in Figure. The thermal conductivity of the ball material is k=8.3 Btu/h

Question

Combined Convection and Radiation Condition

A spherical metal ball of radius [latex] r_{0} [/latex] is heated in an oven to a temperature of [latex] 600^{\circ} F [/latex] throughout and is then taken out of the oven and allowed to cool in ambient air at [latex] T_{\infty} = 78^{\circ} F [/latex], as shown in Figure. The thermal conductivity of the ball material is [latex] k = 8.3 Btu / h \cdot ft \cdot{ }^{\circ} F [/latex], and the average convection heat transfer coefficient on the outer surface of the ball is evaluated to be [latex] h = 4.5 Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F [/latex] The emissivity of the outer surface of the ball is [latex] \varepsilon = 0.6 [/latex], and the average temperature of the surrounding surfaces is [latex] T_{ ext {surr }} = 525 R [/latex]. Assuming the ball is cooled uniformly from the entire outer surface, express the initial and boundary conditions for the cooling process of the ball.

Accepted Answer

SOLUTION The ball is initially at a uniform temperature and is cooled uniformly from the entire outer surface. Therefore, this is a one-dimensional transient heat transfer problem since the temperature within the ball will change with the radial distance r and the time t. That is, [latex] T = T(r, t) [/latex], Taking the moment the ball is removed from the oven to be t = 0, the initial condition can be expressed as

[latex] T(r, 0) = T_{i} = 600^{\circ} F [/latex]

The problem possesses symmetry about the midpoint (r = 0) since the isotherms in this case will be concentric spheres, and thus no heat will be crossing the midpoint of the ball. Then the boundary condition at the midpoint can be expressed as

[latex] \frac{\partial T(0, t)}{\partial r} = 0 [/latex]

The heat conducted to the outer surface of the ball is lost to the environment by convection and radiation. Then taking the direction of heat transfer to be the positive r direction, the boundary condition on the outer surface can be expressed as

[latex] - k \frac{\partial T\left(r_{0}, t\right)}{\partial r}=h\left[T\left(r_{0}\right)-T_{\infty}\right] + \varepsilon \sigma\left[T\left(r_{0}\right)^{4} - T_{ surr }^{4}\right] [/latex]

All the quantities in the above relations are known except the temperatures and their derivatives at r = 0 and [latex] r_{0} [/latex]. Also, the radiation part of the boundary condition is often ignored for simplicity by modifying the convection heat transfer coefficient to account for the contribution of radiation. The convection coefficient h in that case becomes the combined heat transfer coefficient.

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