91Ó°ÊÓ

First, let's calculate the average heat transfer coefficient for the exhaust gas inside the tube. We will use Dittus-Boelter equation for turbulent flow inside the tube: \(h_{i} = 0.023 * Re_{D}^{0.8} * Pr_{m}^{n}\) where \(h_{i}\) is the average heat transfer coefficient, \(Re_{D}\) is the Reynolds number, \(Pr_{m}\) is the Prandtl number, and \(n\) is equal to 0.4 for heating and 0.3 for cooling. We are given that the diameter of the tube is 6 mm and the mass flow rate is \(0.003 \mathrm{ \frac{kg}{s}}\). We need to calculate the Reynolds number using: \(Re_{D} = \frac{4 * m_{g}}{\pi * D * \nu}\) To calculate the Reynolds number, we need to find the kinematic viscosity (\(\nu\)) of the exhaust gas. Assuming the properties of air, we can check the air property table at a temperature of \(200^{\circ}\mathrm{C}\). Since the temperature is very high, we can approximate the value of \(\nu\) as \(3.6 *10^{-5}\) \(\mathrm{m^2/s}\).

Plugging the given values into the above equation: \(Re_{D} = \frac{4 * 0.003}{\pi * 0.006 * 3.6 * 10^{-5}} \approx 20944\) Thus, the flow inside the tube is turbulent. Now we need to find the Prandtl number for air at \(200^{\circ}\mathrm{C}\), which we can approximate as \(Pr_{m} \approx 0.7\).

Now let's calculate the average heat transfer coefficient for the exhaust gas inside the tube using Dittus-Boelter equation: \(h_{i} = 0.023 * 20944^{0.8} * 0.7^{0.4} \approx 791.45 \mathrm{\frac{W}{m^2K}}\) Hence, the average heat transfer coefficient for the exhaust gas inside the tube is approximately \(791.45 \mathrm{\frac{W}{m^2K}}\).

Now we need to calculate the heat transfer coefficient for the air outside the tube. Since we don't have enough information, we will directly use the provided experimental value for air flowing over a cylinder at a velocity of \(5 \mathrm{\frac{m}{s}}\): \(h_{o} \approx 100 \mathrm{\frac{W}{m^2K}}\) Hence, the heat transfer coefficient for the air outside the tube is approximately \(100 \mathrm{\frac{W}{m^2K}}\).

Now let's calculate the overall heat transfer coefficient: \(\frac{1}{U} = \frac{1}{h_{i}} + \frac{1}{h_{o}}\) \(\frac{1}{U} = \frac{1}{791.45} + \frac{1}{100}\) \(U \approx 89.58 \mathrm{\frac{W}{m^2K}}\) Hence, the overall heat transfer coefficient is approximately \(89.58 \mathrm{\frac{W}{m^2K}}\).

Now let's calculate the temperature of the exhaust gas when it reaches the laboratory. We will use the logarithmic mean temperature difference (LMTD) method to calculate the heat transfer: \(Q = U * A * LMTD\) where \(Q\) is the heat transfer, \(A\) is the area of the tube, and \(LMTD\) is the logarithmic mean temperature difference. First, let's calculate \(A\): \(A = \pi * D * L = \pi * 0.006 * 20 \approx 0.37699 \mathrm{m^2}\) To calculate the LMTD, we need to find the inlet and outlet temperatures of the gas. The inlet temperature is given as \(T_{1} = 200^\circ\mathrm{C}\). We can calculate the outlet temperature (\(T_{2}\)) using the energy balance equation: \(m_{g} * c_{p} * (T_{1} - T_{2}) = U * A * LMTD\) Assuming the specific heat of the exhaust gas is close to that of air, we can approximate \(c_{p} \approx 1000 \mathrm{\frac{J}{kgK}}\). Replacing the known values in the above equation: \(0.003 * 1000 * (200 - T_{2}) = 89.58 * 0.37699 * (15 - T_{2})\) Solving this equation, we get: \(T_{2} \approx 188.2^\circ\mathrm{C}\) Hence, the temperature of the exhaust gas when it reaches the laboratory is approximately \(188.2^\circ\mathrm{C}\).