Question

A spherical container of inner radius \(r_{1}=2 \mathrm{~m}\), outer radius \(r_{2}=2.1 \mathrm{~m}\), and thermal conductivity $k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\( is filled with iced water at \)0^{\circ} \mathrm{C}$. The container is gaining heat by convection from the surrounding air at \(T_{\infty}=25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=18 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\). Assuming the inner surface temperature of the container to be \(0^{\circ} \mathrm{C},(a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the container, (b) obtain a relation for the variation of temperature in the container by solving the differential equation, and \((c)\) evaluate the rate of heat gain to the iced water.

Short answer

Answer: The rate of heat gain to the iced water inside the spherical container is 1335 W, and the temperature variation relation is T(r) = -66.75(r^3)/3 + 267.

Step-by-step solution

(a) Differential equation and boundary conditions

The heat conduction equation for a spherical container in steady-state and one-dimensional condition is given by: \( \frac{1}{r^2} \frac{d}{dr}\left( r^2 \frac{dT}{dr} \right) = 0 \) To determine the boundary conditions, we have the following: 1. At the inner surface (r = r1), the temperature is 0°C (T = 0). 2. At the outer surface (r = r2), the rate of heat transfer by convection is equal to the rate of heat transfer by conduction. So, we can write: \( -k \frac{dT}{dr}\Big|_{r=r2} = h(T_{\infty} - T)\Big|_{r=r2} \)

(b) Temperature variation relation

To obtain the relation for the variation of temperature, we first solve the differential equation: \( \frac{1}{r^2} \frac{d}{dr}\left( r^2 \frac{dT}{dr} \right) = 0 \) Integrating both sides, we get: \( r^2 \frac{dT}{dr} = C_{1} \) Integrating again, we get: \( T(r) = \frac{C_1}{3}r^3 + C_2 \) Now, we can use the boundary conditions to determine the constants: 1. At r = r1, T = 0: \( 0 = \frac{C_1}{3}r_{1}^{3} + C_2 \) From which, we get: \( C_2 = -\frac{C_1}{3}r_{1}^{3} \) 2. At r = r2: \( -kr_{2}^2 \frac{dT}{dr}\Big|_{r=r2} = h (T_{\infty} - T)\Big|_{r=r2} \) \( -30 (\frac{2.1^3}{3}C_1 - \frac{2^3}{3}C_1) = 18 (25 - (\frac{2.1^3}{3}C_1 - \frac{2^3}{3}C_1)) \) Solving for \(C_1\), we get \(C_1 = -66.75\) and \(C_2 = 267\) Now, we have the temperature variation relation: \( T(r) = -66.75 \frac{r^3}{3} + 267 \)

(c) Heat gain rate

To evaluate the rate of heat gain to the iced water, we need to consider the rate of heat transfer through the container at the inner surface (r = r1). We can use the following formula for heat transfer: \( q = -k\frac{dT}{dr}\Big|_{r=r1} \) Now, calculating the heat transfer rate at r1: \( q = -30 \times -66.75 \times \frac{2^2}{3}\, \frac{1}{2} \) \( q = 1335 \,\mathrm{~W} \) Therefore, the rate of heat gain to the iced water is 1335 W.