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Since heat transfer across the steam line occurs by both radiation and natural convection on a calm day, we should first determine the heat loss due to both radiation and convection heat transfer mechanisms. For radiation, we can use the Stefan-Boltzmann law, which states that the heat transfer rate by radiation is: \(q_{rad} = \sigma \cdot \epsilon \cdot A(T_{s}^{4} - T_{\infty}^{4})\), where \(q_{rad}\) = heat transfer rate by radiation (W), \(\sigma\) = Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W/m^2K^4}\)), \(\epsilon\) = surface emissivity, \(A\) = surface area (m²), \(T_{s}\) = surface temperature (K), and \(T_{\infty}\) = surroundings temperature (K). For natural convection, we use the equation: \(q_{conv} = h_{c} \cdot A \cdot (T_{s} - T_{\infty})\), where \(q_{conv}\) = heat transfer rate by convection (W), \(h_{c}\) = convection heat transfer coefficient (W/m²K), \(A\) = surface area (m²), \(T_{s}\) = surface temperature (K), and \(T_{\infty}\) = surroundings temperature (K).

First, we need to convert the given temperatures from Celsius to Kelvin: \(T_{s} = 200 + 273.15 = 473.15 \mathrm{K}\), \(T_{\infty} = 20 + 273.15 = 293.15 \mathrm{K}\). We know the pipe diameter (\(D = 89 \mathrm{mm} = 0.089 \mathrm{m}\)) and the surface emissivity (\(\epsilon = 0.8\)). To calculate the heat transfer rate per unit length on a calm day, we need the surface area per unit length: \(A = \pi \cdot D = \pi \cdot 0.089 \mathrm{m}\). Now, we can determine the heat loss due to radiation: \(q_{rad} = \sigma \cdot \epsilon \cdot A(T_{s}^{4} - T_{\infty}^{4}) = 5.67 \times 10^{-8} \cdot 0.8 \cdot \pi \cdot 0.089 (473.15^{4} - 293.15^{4}) = 1099.94 \mathrm{W/m}\). Assuming that the natural convection heat transfer coefficient (\(h_{c}\)) is roughly 10 W/m²K, as a general value for natural convection around a cylinder, on average : \(q_{conv} = h_c \cdot A \cdot (T_{s} - T_{\infty}) = 10 \cdot \pi \cdot 0.089 (473.15 - 293.15) = 501.94 \mathrm{W/m}\). Adding both components, the total heat loss per unit length on a calm day is: \(q_{total} = q_{rad} + q_{conv} = 1099.94 + 501.94 = 1601.88 \mathrm{W/m}\). Thus, the heat loss per unit length for a calm day is approximately 1601.88 W/m.

On a windy day, forced convection becomes predominant. The forced convection heat transfer coefficient (\(h_{f}\)) can be estimated using empirical correlations as a function of the wind speed. Assuming that \(h_{f}\) for a wind speed of 8 m/s is approximately 60 W/m²K, we can calculate the heat loss due to forced convection: \(q_{wind} = h_{f} \cdot A \cdot (T_{s} - T_{\infty}) = 60 \cdot \pi \cdot 0.089 (473.15 - 293.15) = 3011.63 \mathrm{W/m}\). Therefore, the total heat loss per unit length on a windy day is: \(q_{total(windy)} = q_{rad} + q_{wind} = 1099.94 + 3011.63 = 4111.57 \mathrm{W/m}\). Thus, the heat loss per unit length for a windy day is approximately 4111.57 W/m.

For the conditions given in part (a), we can calculate the heat loss with a 20 mm-thick layer of insulation. Given the thermal conductivity of the insulation material, \(k = 0.08 \mathrm{W/mK}\), the heat transfer resistance for the insulation is given by: \(R_{ins} = \frac{r_{2} - r_{1}}{2 \pi k (r_{1} + r_{2})}\), where \(r_{1} = 44.5 \mathrm{mm}\) (inner pipe radius) and \(r_{2} = 64.5 \mathrm{mm}\) (outer radius with insulation). \(R_{ins} = \frac{0.02}{2 \pi \cdot 0.08 \cdot (0.0445 + 0.0645)} = 25.24 \mathrm{m^2K/W}\). Now, the total heat transfer coefficient with insulation (\(U_{ins}\)) becomes: \(U_{ins} = \frac{1}{R_{ins}} = \frac{1}{25.24} = 0.0396 \mathrm{W/m^2K}\). So, the heat loss per unit length with insulation under calm conditions is: \(q_{ins} = U_{ins} \cdot A \cdot (T_{s} - T_{\infty}) = 0.0396 \cdot \pi \cdot 0.089 \cdot (473.15 - 293.15) = 22.29 \mathrm{W/m}\). The heat loss with insulation is significantly lower than without insulation, as expected. In terms of the effect of wind on heat loss through the insulated steam line, the insulated layer would be relatively more resistant to forced convection, so the change due to the wind would be smaller compared to the uninsulated case. In conclusion: a) The heat loss per unit length on a calm day is approximately 1601.88 W/m. b) The heat loss per unit length on a windy day is approximately 4111.57 W/m. c) The heat loss per unit length with insulation is approximately 22.29 W/m, and it is less affected by wind speed compared to the uninsulated case.