Question

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 \(\mathrm{kg} / \mathrm{h}\).

Short answer

Question: Calculate the average convection heat transfer coefficient of the air and the rate of water evaporation in a tube with given conditions. Answer: The average convection heat transfer coefficient of the air is approximately $18.99 \mathrm{W/m^2K}$, and the rate of water evaporation is $3.956 \mathrm{kg/h}$.

Step-by-step solution

Find the heat transfer rate from the air to the water through the tube

To find the heat transfer rate from the hot air to the water, we will use the mass flow rate of the air, the specific heat capacity of air, and the temperature difference between the inlet and outlet of the air: $$Q = m_{air} \cdot c_p \cdot \Delta T$$ First, let's find the mass flow rate. We will use the equation: $$m_{air} = \rho_{air} \cdot A \cdot V$$ where \(\rho_{air}\) is the density of air (assume it's constant at \(1.2 \mathrm{kg/m^3}\)), \(A\) is the cross-sectional area of the tube, and \(V\) is the mean velocity of air. $$A = \pi \left(\frac{D}{2}\right)^2 = \pi \left(\frac{0.05 \mathrm{m}}{2}\right)^2 = 1.963 \times 10^{-3} \mathrm{m^2}$$ Now, let's find the mass flow rate of the air: $$m_{air} = (1.2 \mathrm{kg/m^3}) \cdot (1.963 \times 10^{-3} \mathrm{m^2}) \cdot (7 \mathrm{m/s}) = 0.01649 \mathrm{kg/s}$$ Now we can find the heat transfer rate using the specific heat capacity of air (\(c_p = 1005 \mathrm{J/kgK}\)) and the temperature difference: $$Q = (0.01649 \mathrm{kg/s}) \cdot (1005 \mathrm{J/kgK}) \cdot (300^{\circ} \mathrm{C} - 150^{\circ} \mathrm{C}) = 2.482 \times 10^3 \mathrm{W} \approx 2.48 \mathrm{kW}$$

Calculate the rate of water evaporation

To calculate the rate of water evaporation, we will relate the heat transfer rate to the latent heat of evaporation (\(L = 2257 \mathrm{kJ/kg}\)) and mass flow rate of water: $$Q = m_{water} \cdot L$$ Now let's find the mass flow rate of water: $$m_{water} = \frac{Q}{L} = \frac{2.482 \times 10^3 \mathrm{W}}{2257 \times 10^3 \mathrm{J/kg}} = 1.0989 \times 10^{-3} \mathrm{kg/s}$$ To find the rate of water evaporation in \(\mathrm{kg/h}\), we need to convert the mass flow rate from \(\mathrm{kg/s}\) to \(\mathrm{kg/h}\): $$m_{water} = 1.0989 \times 10^{-3} \mathrm{kg/s} \cdot \frac{3600 \mathrm{s}}{\mathrm{h}} = 3.956 \mathrm{kg/h}$$

Calculate the average convection heat transfer coefficient

With the heat transfer rate determined, we can now use the following equation to calculate the average convection heat transfer coefficient (\(h\)) for the air with the given information: $$Q = h \cdot A_{surface} \cdot \Delta T_s$$ where \(A_{surface}\) is the surface area of the tube, and \(\Delta T_s\) is the temperature difference between the air inlet and the surface temperature of the tube. First, let's find the surface area of the tube: $$A_{surface} = \pi D L = \pi (0.05 \mathrm{m})(5 \mathrm{m}) = 0.785 \mathrm{m^2}$$ Now, let's find the temperature difference between the air inlet and the surface temperature: $$\Delta T_s = 300^{\circ} \mathrm{C} - 120^{\circ} \mathrm{C} = 180^{\circ} \mathrm{C}$$ Finally, we can find the average convection heat transfer coefficient: $$h = \frac{Q}{A_{surface} \cdot \Delta T_s} = \frac{2.482 \times 10^3 \mathrm{W}}{(0.785 \mathrm{m^2})(180^{\circ} \mathrm{C})} = 18.99 \mathrm{W/m^2K}$$ So, the average convection heat transfer coefficient of the air is approximately \(18.99 \mathrm{W/m^2K}\), and the rate of water evaporation is \(3.956 \mathrm{kg/h}\).