Question

combustion gases passing through a 5 -cm-internaldiameter circular tube are used to vaporize wastewater at atmospheric pressure. Hot gases enter the tube at \(115 \mathrm{kPa}\) and \(250^{\circ} \mathrm{C}\) at a mean velocity of $5 \mathrm{~m} / \mathrm{s}\( and leave at \)150^{\circ} \mathrm{C}$. If the average heat transfer coefficient is $120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\( and the inner surface temperature of the tube is \)110^{\circ} \mathrm{C}\(, determine \)(a)\( the tube length and \)(b)$ the rate of evaporation of water. Use air properties for the combustion gases.

Short answer

Answer: The tube length is 0.528 m, and the rate of evaporation of water is 3.95 x 10^-4 kg/s.

Step-by-step solution

Calculate the heat transfer rate between the gases and the tube wall

We are given the average heat transfer coefficient as \(h = 120 \frac{W}{m^2\cdot K}\). The temperature difference between the inner surface of the tube \((110^{\circ}C)\) and the average temperature of the gases can be calculated as \(\Delta T_m = (\frac{T_{i}+T_{o}}{2}) - T_{wall} = (\frac{250+150}{2}) - 110 = 90^{\circ}C\). We can now calculate the heat transfer rate per unit area using the formula: \(q = h \Delta T_m\). Substituting the values, we get: \(q = 120 \cdot 90 = 10800 \frac{W}{m^2}\).

Calculate the mass flow rate of the combustion gases

Given the mean velocity of hot gases as \(v = 5 \frac{m}{s}\). Using the ideal gas law, we can find the gas density at the given pressure \((P_{i} = 115\,\mathrm{kPa})\) and temperature \((T_{i} = 250^{\circ}C = 523\,K)\), assuming air to be an ideal gas with the gas constant \(R = 287\,\frac{J}{kg\cdot K}\). We have: \(\rho = \frac{P_{i}}{RT_{i}} = \frac{115000}{287 \cdot 523} = 0.906\,\frac{kg}{m^3}\). Now, we can find the mass flow rate of the combustion gases using the formula: \(\dot{m} = \rho \cdot A_i \cdot v\), where \(A_i\) is the inner area of the tube, and can be found as \(A_i = \pi \cdot (\frac{d}{2})^2 = \pi \cdot (\frac{0.05}{2})^2 = 0.0019635\, m^2\). Substituting the values, we get: \(\dot{m} = 0.906 \cdot 0.0019635 \cdot 5 = 0.00892\,\frac{kg}{s}\).

Calculate the tube length

The heat transfer rate per unit length can be calculated as: \(\dot{Q} = \dot{m} \cdot c_p \cdot (T_{i}-T_{o})\), where \(T_{o} = 150^{\circ}C\) is the exit gas temperature. Substituting the values, we get: \(\dot{Q} = 0.00892 \cdot 1000 \cdot (250-150) = 893.6\,W\). Now, we can find the tube length using the formula: \(L = \frac{\dot{Q}}{q \cdot P_i}\), where \(P_i\) is the inner perimeter of the tube, and can be found as \(P_i = \pi \cdot d = \pi \cdot 0.05 = 0.15708\,m\). Substituting the values, we get: \(L = \frac{893.6}{10800 \cdot 0.15708} = 0.528\,m\). Thus, the tube length is \(0.528\,m\).

Calculate the rate of evaporation of water

To calculate the rate of evaporation of water, we can use the energy balance equation: \(\dot{Q} = \dot{m}_w \cdot h_{fg}\), where \(\dot{m}_w\) is the mass flow rate of evaporated water, and \(h_{fg}\) is the heat of vaporization of water at atmospheric pressure, which we can assume to be \(h_{fg} = 2.26 \times 10^6\,\frac{J}{kg}\). Substituting the values, we get: \(\dot{m}_w = \frac{\dot{Q}}{h_{fg}} = \frac{893.6}{2.26 \times 10^6} = 3.95 \times 10^{-4}\,\frac{kg}{s}\). Thus, the rate of evaporation of water is \(3.95 \times 10^{-4}\,\frac{kg}{s}\). In conclusion: (a) The tube length is \(0.528\,m\). (b) The rate of evaporation of water is \(3.95 \times 10^{-4}\,\frac{kg}{s}\).