Question

Consider a recuperative crossflow heat exchanger (both fluids unmixed) used in a gas turbine system that carries the exhaust gases at a flow rate of $7.5 \mathrm{~kg} / \mathrm{s}\( and a temperature of \)500^{\circ} \mathrm{C}$. The air initially at \(30^{\circ} \mathrm{C}\) and flowing at a rate of $15 \mathrm{~kg} / \mathrm{s}$ is to be heated in the recuperator. The convective heat transfer coefficients on the exhaust gas and air sides are $750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\( and \)300 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}$, respectively. Due to long-term use of the gas turbine, the recuperative heat exchanger is subject to fouling on both gas and air sides that offers a resistance of \(0.0004\) \(\mathrm{m}^{2}\). $/ \mathrm{W}$ each. Take the properties of exhaust gas to be the same as that of air \(\left(c_{p}=1069 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\). If the exit temperature of the exhaust gas is \(320^{\circ} \mathrm{C}\), determine \((a)\) if the air could be heated to a temperature of \(150^{\circ} \mathrm{C}\) and \((b)\) the area of the heat exchanger. (c) If the answer to part \((a)\) is no, then determine what should be the air mass flow rate in order to attain the desired exit temperature of \(150^{\circ} \mathrm{C}\) and \((d)\) plot the variation of the exit air temperature over a range of \(75^{\circ} \mathrm{C}\) to \(300^{\circ} \mathrm{C}\) with the air mass flow rate, assuming all the other conditions remain the same.

Short answer

a) The exit air temperature under the given conditions can only reach 117°C, which is not enough to reach the desired temperature of 150°C. b) The area of the heat exchanger is 65.81 m². c) The required air mass flow rate to achieve an exit temperature of 150°C is 24.17 kg/s. d) To plot the exit air temperature with varying air mass flow rate, use the energy balance equation and generate a plot of \(T_{c,out}\) versus \(m_c\). The relationship indicates that lower air mass flow rates result in higher exit air temperatures, while higher air mass flow rates result in lower exit air temperatures.

Step-by-step solution

a) Checking if exit air temperature can reach 150°C

To analyze the possibility of the air being heated to the desired temperature of 150°C, we first need to apply energy balance equation to the system: \(m_c c_{p,c}(T_{c,out} - T_{c,in}) = m_h c_{p,h}(T_{h,in} - T_{h,out})\) Given the flow rates and temperatures, we can calculate the exit exhaust gas temperature (\(320^{\circ} \mathrm{C}\)) and the air mass flow rate in order to determine whether the air temperature will reach \(150^{\circ} \mathrm{C}\). Using the given values and assuming \(T_{c,out} = 150^{\circ} \mathrm{C}\): \(15\;\mathrm{kg/s} \cdot 1069\;\mathrm{J/kg\cdot K}(150 - 30) = 7.5\;\mathrm{kg/s} \cdot 1069\;\mathrm{J/kg\cdot K}(500 - 320)\) Solving for \(T_{c,out}\), we get \(T_{c,out} = 117^{\circ} \mathrm{C}\). This means that the air cannot reach the desired temperature of 150°C under these conditions.

b) Determining the area of the heat exchanger

To calculate the area of the heat exchanger, we need to apply the overall heat transfer coefficient (U) equation and use the heat transfer rate (Q) already calculated in part (a). For the recuperative crossflow heat exchanger, we can write the overall heat transfer coefficient (U) equation as follows: \(1/U = 1/h_1 + (R_f)_1 + (R_f)_2 + 1/h_2\) Substituting the given values into the equation, we get: \(1/U = 1/750 + 0.0004 + 0.0004 + 1/300\) Solving for U, we obtain U = 212.77 W/m²K. Now we need to calculate the required area (A) using the heat transfer rate (Q) equation: \(Q = U \cdot A \cdot F \cdot \Delta T_{lm}\) We assume an efficiency of F=1 for this crossflow heat exchanger. To calculate the logarithmic mean temperature difference (\(\Delta T_{lm}\)), we use the following equation: \(\Delta T_{lm} = \frac{(T_{h,in} - T_{c,out}) - (T_{h,out} - T_{c,in})}{\ln \left(\frac{T_{h,in} - T_{c,out}}{T_{h,out} - T_{c,in}}\right)}\) Substituting the given values, we get: \(\Delta T_{lm} = \frac{(500 - 117) - (320 - 30)}{\ln \left(\frac{500 - 117}{320 - 30}\right)} = 283.12^{\circ} \mathrm{C}\) Now we can calculate the area (A) using the Q, U, F, and \(\Delta T_{lm}\) values: \(A = \frac{Q}{U \cdot F \cdot \Delta T_{lm}}\) Substituting the values, we get: \(A = \frac{15 \cdot 1069 \cdot (117 - 30)}{212.77 \cdot 1 \cdot 283.12} = 65.81 \;\mathrm{m^2}\) So the area of the heat exchanger is 65.81 m².

c) Determining the required air mass flow rate to achieve the desired exit temperature

To find the air mass flow rate needed to achieve the exit temperature of 150°C, we can use the energy balance equation. We assume that everything else remains unchanged: \(m_c c_{p,c}(T_{c,out} - T_{c,in}) = m_h c_{p,h}(T_{h,in} - T_{h,out})\) Now substituting the desired value of \(T_{c,out} = 150^{\circ} \mathrm{C}\), we can solve for \(m_c\): \(m_c = \frac{m_h c_{p,h} (T_{h,in} - T_{h,out})}{c_{p,c}(T_{c,out} - T_{c,in})}\) Substituting the values: \(m_c = \frac{7.5 \cdot 1069 \cdot (500 - 320)}{1069 \cdot (150 - 30)} = 24.17\; \mathrm{kg/s}\) So, the required air mass flow rate is 24.17 kg/s to achieve an exit temperature of 150°C.

d) Plotting the exit air temperature with varying air mass flow rate

To plot the exit air temperature (\(T_{c,out}\)) as a function of the air mass flow rate (\(m_c\)) over the range of \(75^{\circ} \mathrm{C}\) to \(300^{\circ} \mathrm{C}\), we will use the energy balance equation. For each value of \(m_c\), we can calculate the corresponding \(T_{c,out}\) using the energy balance equation: \(T_{c,out} = T_{c,in} + \frac{m_h c_{p,h} (T_{h,in} - T_{h,out})}{m_c c_{p,c}}\) A plot of \(T_{c,out}\) versus \(m_c\) can be generated using different values of \(m_c\) within the specified range to visualize the trend between the exit air temperature and the air mass flow rate. A lower air mass flow rate will result in a higher exit air temperature, while a higher air mass flow rate will result in a lower exit air temperature.