Exhaust gases at \(1 \mathrm{~atm}\) and \(300^{\circ} \mathrm{C}\) are used to preheat water in an industrial facility by passing them over a bank of tubes through which water is flowing at a rate of \(6 \mathrm{~kg} / \mathrm{s}\). The mean tube wall temperature is \(80^{\circ} \mathrm{C}\). Exhaust gases approach the tube bank in normal direction at \(4.5 \mathrm{~m} / \mathrm{s}\). The outer diameter of the tubes is \(2.1 \mathrm{~cm}\), and the tubes are arranged in- line with longitudinal and transverse pitches of \(S_{L}=S_{T}=8 \mathrm{~cm}\). There are 16 rows in the flow direction with eight tubes in each row. Using the properties of air for exhaust gases, determine \((a)\) the rate of heat transfer per unit length of tubes, \((b)\) and pressure drop across the tube bank, and \((c)\) the temperature rise of water flowing through the tubes per unit length of tubes. Evaluate the air properties at an assumed mean temperature of \(250^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\). Is this a good assumption?

To begin, we need to calculate the Reynolds number for the exhaust gasses, which determines whether the flow is laminar or turbulent. We can find the Reynolds number using the formula: \(Re = \frac{VD}{\nu}\) where \(V\) is the gas velocity, \(D\) is the tube outer diameter, and \(\nu\) is the kinematic viscosity of air at the assumed mean temperature. We should first convert the temperature from Celsius to Kelvin. \(T_{mean} = 250^{\circ}C + 273.15 = 523.15\mathrm{K}\) Now, using the properties of air at \(250^{\circ}C\) and \(1 \mathrm{atm}\), we can find the kinematic viscosity, \(\nu = 5.93 \times 10^{-5} \mathrm{m^2/s}\). \(Re = \frac{4.5\mathrm{m/s}\times 0.021\mathrm{m}}{5.93\times10^{-5}\mathrm{m^2/s}} = 1.7\times10^4\) Since the Reynolds number is greater than 2300, the flow is turbulent.

With the Reynolds number determined, we can now calculate the convective heat transfer coefficient using the Dittus-Boelter equation for turbulent flow: \(h = 0.023\cdot Re^\frac{4}{5}\cdot Pr^\frac{1}{3}\cdot C_p\cdot\frac{k}{\mu}\) where \(Re\) is the Reynolds number, \(Pr\) is the Prandtl number, \(C_p\) is the specific heat capacity, \(k\) is the thermal conductivity, and \(\mu\) is the dynamic viscosity. From the properties of air at the given temperature and pressure, we find that \(Pr = 0.7\), \(C_p = 1005\mathrm{J/(kg\cdot K)}\), \(k = 0.0374 \mathrm{W/(m\cdot K)}\), and \(\mu = 3.55\times10^{-5}\mathrm{kg/(m\cdot s)}\). \(h = 0.023\cdot (1.7\times10^4)^\frac{4}{5}\cdot 0.7^\frac{1}{3}\cdot 1005\frac{0.0374}{3.55\times10^{-5}\mathrm{kg/(m\cdot s)}} = 123\mathrm{W/(m^2\cdot K)}\)

Now that we have the convective heat transfer coefficient, we can calculate the heat transfer rate per unit length of tubes, \(q'\), using the following formula: \(q' = h\cdot A_s\cdot\Delta T\) where \(A_s = \pi D L = \pi(0.021\mathrm{m})L\) is the surface area of the tubes per unit length and \(\Delta T\) is the temperature difference between the mean tube wall temperature and the exhaust gas temperature, \(\Delta T = 300^{\circ}C - 80^{\circ}C = 220^{\circ}C = 220\mathrm{K}\). \(q' = 123\mathrm{W/(m^2\cdot K)}\cdot \pi(0.021\mathrm{m})L\cdot 220\mathrm{K} = 1822\mathrm{W/m}\cdot L\) The heat transfer rate per unit length of tubes is \(1822 \mathrm{W/m}\cdot L\).

To calculate the pressure drop across the tube bank, we use the following equation: \(\Delta P = \frac{1}{2}\rho V^2 \xi C\) where \(\rho\) is the density of air, \(V\) is the velocity of the exhaust gases, \(\xi\) is the loss coefficient, and \(C\) is the number of rows in the flow direction. From the properties of air at the given temperature and pressure, we find that \(\rho = 0.841\mathrm{kg/m^3}\). The loss coefficient, \(\xi\), can be calculated using the method proposed by Idelchik: \(\xi = 4\left(\frac{S_T - D}{S_L}\right)^{2/3}（1 - C_{eq, tube}C_{eq, area}）\) For turbulence flow through in-line tube banks, we can find the equivalent drag coefficient \(C_{eq, tube} = 1\) , and the equivalent flow area coefficient \(C_{eq, area} = 68 \cdot (S_T - D)/S_L\). The longitudinal and transverse pitches are given as \(S_L = S_T = 8\mathrm{cm} = 0.08\mathrm{m}\), and we have: \(C_{eq, area} = 68\cdot \frac{0.08 - 0.021}{0.08} = 51\) \(\xi = 4\left(\frac{0.08 - 0.021}{0.08}\right)^{2/3}(1 - 1\times 51) = 592\) Now we can plug in the values and calculate the pressure drop: \(\Delta P = \frac{1}{2}(0.841\mathrm{kg/m^3})(4.5\mathrm{m/s})^2 \times 592 \times 16 = 578\mathrm{Pa}\) The pressure drop across the tube bank is \(578\mathrm{Pa}\).

The temperature rise of water per unit length of tubes, \(dT_{water}\), can be calculated using the following formula: \(dT_{water} = \frac{q'}{m_w\cdot C_{p,w}}\) where \(m_w = 6\mathrm{kg/s}\) is the water flow rate and \(C_{p,w} = 4186\mathrm{J/(kg\cdot K)}\) is the specific heat capacity of water. \(dT_{water} = \frac{1822\mathrm{W/m}\cdot L}{6\mathrm{kg/s}\cdot 4186\mathrm{J/(kg\cdot K)}} = 0.048\frac{\mathrm{K}}{L}\) The temperature rise of water per unit length of tubes is \(0.048\mathrm{K/m}\cdot L\).

We assumed that the air properties are evaluated at a mean temperature of \(250^{\circ}C\) and \(1\mathrm{atm}\). Since the actual exhaust gas temperature is \(300^{\circ}C\) and the water's mean tube wall temperature is \(80^{\circ}C\), the chosen mean temperature of \(250^{\circ}C\) is a reasonable approximation, as it lies between these two temperatures.