Question

8. Calculate the heat transfer rate per m2 area by radiation between the surfaces

of two long cylinders having radii 100 mm and 50 mm respectively. The smaller cylinder being in the

larger cylinder. The axes of the cylinders are parallel to each other and separated by a distance of

20 mm. The surfaces of inner and outer cylinders are maintained at 127°C and 27°C respectively. The

emissivity of both the surfaces is 0.5.

Assume the medium between the two cylinders is non-absorbing.

Answer 1

$r_1=50 \;mm = 0.05 \;m \\ r_2= 100 \;mm = 0.1 \;m \\ T_1 = 127+273 = 400 \;K \\ T_2 = 27+273=300 \;K \\ \varepsilon_1 = \varepsilon_1 = 0.5$

The heat transfer between two concentric or eccentric cylinders is given by

$(Q_{12})_{net} = \frac{A_1 \sigma (T^4_1 -T^4_2)}{(\frac{1 - \varepsilon_1}{\varepsilon_1}) + \frac{1}{F_{1-2}} + (\frac{1 -\varepsilon_2}{\varepsilon_2}) \frac{A_1}{A_2}} \\ F_{1-2} = 1 \\ \frac{A_1}{A_2} = \frac{2 \pi r_1L}{2 \pi r_2L} = \frac{r_1}{r_2}$

Substituting the values, we have

$(Q_{12})_{net} = \frac{1 \times 5.67 ((\frac{400}{100})^4 -(\frac{300}{100})^4)}{(\frac{1 – 0.5}{0.5}) + 1 + (\frac{1 -0.5}{0.5}) \frac{0.05}{0.1}} \\ = \frac{992.25}{2.5} \\ = 396.9 \;W/m^2$