In a gas-fired boiler, water is being boiled at \(120^{\circ} \mathrm{C}\) by hot air flowing through a 5 -m-long, 5 -cm-diameter tube submerged in water. Hot air enters the tube at 1 atm and \(300^{\circ} \mathrm{C}\) at a mean velocity of \(7 \mathrm{~m} / \mathrm{s}\), and leaves at \(150^{\circ} \mathrm{C}\). If the surface temperature of the tube is \(120^{\circ} \mathrm{C}\), determine the average convection heat transfer coefficient of the air and the rate of water evaporation, in \(\mathrm{kg} / \mathrm{h}\).

Find the heat capacity rate of the air, given by \(C_{p,a} = \dot{m}_a \, c_{p,a}\), where \(\dot{m}_a\) is the mass flow rate of air and \(c_{p,a}\) is the specific heat of air at constant pressure (can be taken as 1005 J/kg.K). Firstly, we need to assume air behaves as an ideal gas (which is reasonable at these conditions). Using the ideal gas law at the inlet and outlet, and the mass flow rate continuity equation, we can relate the mass flow rate to the given mean velocity: $$\dot{m}_a = \rho_a A v = \frac{P}{RT_a}A v$$ where \(\rho_a\) is the air density, \(A\) is the tube cross-sectional area, \(v\) is the air velocity, \(P\) is the air pressure, \(R\) is the ideal gas constant for air (287 J/kg.K), and \(T_a\) is the air temperature (300°C = 573.15 K). Now plug in the values: $$\dot{m}_a = \frac{101325}{286.9 \times 573.15} \left(\pi \frac{(0.05)^{2}}{4}\right) 7 = 10.14 \,\text{kg/s}$$ Now, we can find the heat capacity rate of air: $$ C_{p,a} = 10.14 * 1005 = 10204.1 \,\text{W/K} $$

Using the given inlet and outlet temperature of the air, we can find the overall heat transfer \(Q\) with the following formula: $$ Q = C_{p,a}(T_{inlet} - T_{outlet}) = 10204.1(573.15 - 423.15) = 1530480 \,\text{W} $$

Now the heat transfer rate can be related to the temperature difference between the tube's surface temperature and the bulk fluid along with the heat transfer area as follows: $$ Q = h A \Delta T $$ where \(h\) is the average convection heat transfer coefficient, \(A\) is the heat transfer area, and \(\Delta T\) is the temperature difference. $$ h = \frac{Q}{A \Delta T} $$ Here, \(A = \pi d L\) where \(d\) is the diameter, and \(L\) is the length of the tube. Also, \(\Delta T = T_{surface} - T_{outlet} = 120 - 150 = -30^\circ \mathrm{C}\) (note that a negative sign indicates heat transfer from the air to the tube, which is expected). Now plug in the values: $$ h = \frac{1530480}{\pi (0.05) (5) \times -30 } = 6465.8 \,\text{W/m^2.K} $$

Given the heat transfer rate and the latent heat of vaporization for water \(L_v = 2257\,\text{kJ/kg}\), we can calculate the evaporation rate as: $$ \dot{m}_{water}=\frac{Q}{L_v} $$ $$ \dot{m}_{water}=\frac{1530480}{2257 \times 1000} = 0.6779 \,\text{kg/s} $$ To get the evaporation rate in \(\text{kg/h}\), convert from seconds to hours: $$ \dot{m}_{water} = 0.6779 \times 3600 = 2440.4 \,\text{kg/h} $$ The average convection heat transfer coefficient of the air is \(h = 6465.8 \,\text{W/m^2.K}\) and the rate of water evaporation is 2440.4 kg/h.