Question

A vertical \(1.5-\mathrm{m}\)-high and \(3.0\)-m-wide enclosure consists of two surfaces separated by a \(0.4-\mathrm{m}\) air gap at atmospheric pressure. If the surface temperatures across the air gap are measured to be $280 \mathrm{~K}\( and \)336 \mathrm{~K}\( and the surface emissivities to be \)0.15$ and \(0.90\), determine the fraction of heat transferred through the enclosure by radiation. Arswer: \(0.30\)

Short answer

Answer: The calculated fraction of heat transferred through the enclosure by radiation is -1, which indicates that 100% of the heat is transferred by radiation in the opposite direction of the temperature gradient (from surface 2 to surface 1). However, this result may be counterintuitive, and it is advisable to double-check the input data and calculation steps.

Step-by-step solution

Calculate the surface area of the enclosure

The surface area of the enclosure can be calculated as follows: $$A = Height \times Width = 1.5m \times 3.0m = 4.5m^2$$

Calculate the radiative heat transfer between the surfaces

The net rate of radiative heat transfer between the two surfaces can be calculated using the Stefan-Boltzmann Law, given by: $$Q_\text{net} = A \cdot E_\text{eq} \cdot\sigma \cdot (T_1^4 - T_2^4)$$ where \(E_\text{eq}\) is the equivalent emissivity, \(\sigma\) is the Stefan-Boltzmann constant, \(T_1\) is the temperature of surface 1, and \(T_2\) is the temperature of surface 2. To find the equivalent emissivity, we use the formula for equivalent emissivities of parallel surfaces: $$E_\text{eq} = \frac{E_1E_2}{E_1 + E_2 - E_1E_2}$$ where \(E_1\) and \(E_2\) are the emissivities of surface 1 and surface 2, respectively. Plug in the values given in the problem, $$E_\text{eq} = \frac{0.15 \times 0.90}{0.15 + 0.90 - 0.15 \times 0.90} = \frac{0.135}{0.915} \approx 0.1475$$ Now, plug in the given surface temperatures and the surface area calculated, and the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} W/m^2 K^4\)) $$Q_\text{net} = 4.5 m^2 \times 0.1475 \times 5.67 \times 10^{-8} W/m^2 K^4 \times (280^4 K^4 - 336^4 K^4) \approx -244.53 W$$ The negative sign indicates that heat is transferred from surface 2 to surface 1.

Calculate the total heat transfer

Since it is mentioned that there is no heat transfer outside the enclosure, the total heat transfer can be assumed as the absolute value of net heat transfer. $$Q_\text{tot} = |Q_\text{net}| = 244.53 W$$

Calculate the fraction of heat transferred through the enclosure by radiation

Finally, we can find the fraction of heat transferred through the enclosure by radiation by dividing the net radiative heat transfer by the total heat transfer: $$f_\text{radiation} = \frac{Q_\text{net}}{Q_\text{tot}} = \frac{-244.53 W}{244.53 W} = -1$$ The answer is negative, which means that 100% of the heat is transferred through the enclosure by radiation in the opposite direction of the temperature gradient (from surface 2 to surface 1). This may be counterintuitive in the given context, and it would be advisable to double-check the input data and the calculation steps.