Calculate the convective heat transfer
We can use the following formula to calculate the convective heat transfer:
$$
q_{conv} = hA\left(T_{1} - T_{2}\right)
$$
where \(h\) is the convective heat transfer coefficient. Since the pipe is horizontal, we can use Churchill and Chu's correlation for natural convection from a horizontal cylinder. The Nusselt number is given by:
$$
\mathrm{Nu} = 0.60 + 0.387\mathrm{Ra}^{1/6}\left[1+\left(\frac{0.559}{\mathrm{Pr}}\right)^{9/16}\right]^{-8/27}
$$
where \(\mathrm{Ra}\) is the Rayleigh number and \(\mathrm{Pr}\) is the Prandtl number. The Rayleigh number is given by:
$$
\mathrm{Ra} = \frac{g\beta(T_{1} - T_{2})D^3}{\nu\alpha}
$$
where \(g\) is the gravitational acceleration, \(\beta\) is the coefficient of thermal expansion (\(\beta \approx 1/T_{2}\) for ideal gases), \(D\) is the pipe diameter, and \(\nu\) and \(\alpha\) are the kinematic viscosity and thermal diffusivity, respectively.
First, we calculate the kinematic viscosity of air:
$$
\nu = 1.382 \times 10^{-5}\,\mathrm{m^2/s}
$$
From the given information, we can calculate the thermal conductivity of air:
$$
k = 0.02401\,\mathrm{W/m\cdot K}
$$
Now, we calculate the thermal diffusivity \(\alpha\):
$$
\alpha = \frac{k}{\rho c_{p}}
$$
Using the ideal gas law and specific heat capacity of air at constant pressure, \(c_{p} = 1005\,\mathrm{J/kg\cdot K}\), we can find \(\rho = 1.20\,\mathrm{kg/m^3}\). Hence, the thermal diffusivity is calculated as:
$$
\alpha = \frac{0.02401\,\mathrm{W/m\cdot K}}{1.20\,\mathrm{kg/m^3}(1005\,\mathrm{J/kg\cdot K})} = 1.996\times10^{-5}\,\mathrm{m^2/s}
$$
Next, calculate the Rayleigh number:
$$
\mathrm{Ra} = \frac{9.81\,\mathrm{m/s^2}\left(\frac{1}{288.15\,\mathrm{K}}\right)(263.15\,\mathrm{K} - 288.15\,\mathrm{K})(0.05\,\mathrm{m})^3}{1.382\times10^{-5}\,\mathrm{m^2/s}(1.996\times10^{-5}\,\mathrm{m^2/s})} = 2.543\times10^5
$$
Now, compute the Nusselt number:
$$
\mathrm{Nu} = 0.60 + 0.387(2.543\times10^5 )^{1/6}\left[1+\left(\frac{0.559}{0.735}\right)^{9/16}\right]^{-8/27} = 3.61
$$
Finally, we can find the convective heat transfer coefficient and calculate the convective heat transfer:
$$
h = \frac{\mathrm{Nu}\cdot k}{D} = \frac{3.61(0.02401\,\mathrm{W/m\cdot K})}{0.05\,\mathrm{m}} = 1.73\,\mathrm{W/m^2\cdot K}
$$
$$
q_{conv} = (1.73\,\mathrm{W/m^2\cdot K})(0.628\,\mathrm{m^2})(263.15\,\mathrm{K} - 288.15\,\mathrm{K}) = -27.32\,\mathrm{W}
$$
