Question

In a natural gas-fired boiler, combustion gases pass through 6-m-long,15-cm- diameter tubes immersed in water at \(1 \mathrm{~atm}\) pressure. The tube temperature is measured to be \(105^{\circ} \mathrm{C}\), and the emissivity of the inner surfaces of the tubes is estimated to be \(0.9\). Combustion gases enter the tube at \(1 \mathrm{~atm}\) and \(1000 \mathrm{~K}\) at a mean velocity of \(3 \mathrm{~m} / \mathrm{s}\). The mole fractions of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) in combustion gases are 8 percent and 16 percent, respectively. Assuming fully developed flow and using properties of air for combustion gases, determine \((a)\) the rates of heat transfer by convection and by radiation from the combustion gases to the tube wall and \((b)\) the rate of evaporation of water.

Short answer

A: The rates of heat transfer by convection and radiation from the combustion gases to the tube wall are 193,803 W and 24,682 W, respectively. The rate of evaporation of water is 0.0967 kg/s.

Step-by-step solution

Calculate the Reynolds number of the combustion gases

The first step to determine the rate of heat transfer by convection is to calculate the Reynolds number of the combustion gases. The Reynolds number is given by: \(Re = \frac{\rho VD_{i}}{\mu}\) where: \(Re\): Reynolds number \(\rho\): density of the combustion gases \(V\): mean velocity (\(3~m/s\)) \(D_{i}\): inner tube diameter (\(0.15~m\)) \(\mu\): dynamic viscosity of the combustion gases Assuming the properties of air for the combustion gases, at \(1000K\), we have: \(\rho = 0.412~kg / m^{3}\) \(\mu = 4.82 \times 10^{-5}~ kg / (m \cdot s)\) Thus, the Reynolds number is: \(Re = \frac{(0.412~kg/m^{3})(3~m/s)(0.15~m)}{4.82 \times 10^{-5}~kg/(m \cdot s)} = 38528\)

Determine the Nusselt number

For fully developed flow in a tube with constant wall temperature, we can use the Sieder-Tate correlation to determine the Nusselt number, which is given by: \(Nu = 0.023 Re^{0.8} Pr^{0.4}\) where: \(Nu\): Nusselt number \(Re\): Reynolds number (calculated in step 1) \(Pr\): Prandtl number Assuming the properties of air for the combustion gases, at \(1000K\), we have: \(Pr = 0.678\) Thus, the Nusselt number is: \(Nu = 0.023 (38528)^{0.8} (0.678)^{0.4} = 268.1\)

Calculate the convection heat transfer coefficient

We can now calculate the convection heat transfer coefficient, using the equation: \(h = \frac{Nu \cdot k}{D_{i}}\) where: \(h\): convection heat transfer coefficient \(Nu\): Nusselt number (calculated in step 2) \(k\): thermal conductivity of the combustion gases \(D_{i}\): inner tube diameter (\(0.15~m\)) Assuming the properties of air for the combustion gases, at \(1000K\), we have: \(k = 0.062~W/(m \cdot K)\) Thus, the convection heat transfer coefficient is: \(h = \frac{(268.1)(0.062~W/(m \cdot K))}{0.15~m} = 110.3~W/(m^{2} \cdot K)\)

Calculate the rate of heat transfer by convection

Now we can calculate the rate of heat transfer by convection, using the equation: \(q_{conv} = h \cdot A_{s} \cdot \Delta T\) where: \(q_{conv}\): rate of heat transfer by convection \(h\): convection heat transfer coefficient (calculated in step 3) \(A_{s}\): surface area of the tubes \(\Delta T\): temperature difference between the combustion gases and the tube wall We have: \(A_{s} = \pi D_{i} L = \pi (0.15~m)(6~m) = 2.83~m^{2}\) and \(\Delta T = T_{g} - T_{w} = 1000~K - (105 + 273.15)~K = 621.85~K\) Therefore, the rate of heat transfer by convection is: \(q_{conv} = (110.3~W/(m^{2} \cdot K))(2.83~m^{2})(621.85~K) = 193803~W\)

Calculate the rate of heat transfer by radiation

We can now calculate the rate of heat transfer by radiation, using the equation: \(q_{rad} = \epsilon \sigma A_{s}(T_{g}^{4} - T_{w}^{4})\) where: \(q_{rad}\): rate of heat transfer by radiation \(\epsilon\): emissivity of the inner surface of the tubes (\(0.9\)) \(\sigma\): Stefan-Boltzmann constant, \(5.67 \times 10^{-8}~W/(m^{2} \cdot K^{4})\) \(A_{s}\): surface area of the tubes (calculated in step 4) \(T_{g}\): temperature of the combustion gases (\(1000~K\)) \(T_{w}\): temperature of the tube wall (\(105^{\circ}C = 378.15~K\)) Thus, the rate of heat transfer by radiation is: \(q_{rad} = (0.9)(5.67 \times 10^{-8}~W/(m^{2} \cdot K^{4}))(2.83~m^{2})[(1000~K)^{4}-(378.15~K)^{4}] = 24682~W\)

Calculate the total rate of heat transfer to the water

The total rate of heat transfer to the water is the sum of the rates of heat transfer by convection and radiation: \(q_{total} = q_{conv} + q_{rad} = 193803~W + 24682~W = 218485~W\)

Calculate the rate of evaporation of water

Finally, we will calculate the rate of evaporation of water by dividing the total rate of heat transfer by the latent heat of vaporization of water: \(\dot{m} = \frac{q_{total}}{L_{v}}\) where: \(\dot{m}\): rate of evaporation of water \(q_{total}\): total rate of heat transfer to the water (calculated in step 6) \(L_{v}\): latent heat of vaporization of water at \(1~atm\), \(2.26 \times 10^{6}~J/kg\) The rate of evaporation of water is: \(\dot{m} = \frac{218485~W}{2.26 \times 10^{6}~J/kg} = 0.0967~ kg/s\) Therefore, the rates of heat transfer by convection and radiation from the combustion gases to the tube wall are \(193803~W\) and \(24682~W\), respectively, and the rate of evaporation of water is \(0.0967~kg/s\).