Consider a recuperative cross flow 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 side are \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Due to long term use of the gas turbine the recuperative heat exchanger is subject to fouling on both gas and air side that offers a resistance of \(0.0004 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) 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 \(150^{\circ} \mathrm{C}(b)\) the area of 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 variation of the exit air temperature over a temperature range of \(75^{\circ} \mathrm{C}\) to \(300^{\circ} \mathrm{C}\) with air mass flow rate assuming all the other conditions remain the same.

Step by step solution

01

Calculate Temperature Ratio and Heat Capacity Rates

First we need to calculate the temperature ratio \(R=\frac{T_{h_{inlet}}-T_{h_{exit}}}{T_{c_{exit}}-T_{c_{inlet}}}\) using the given inlet and exit temperatures of both the exhaust gas (hot fluid) and air (cold fluid). For exhaust gas: \(T_{h_{inlet}} = 500^{\circ} \mathrm{C}\) \(T_{h_{exit}} = 320^{\circ} \mathrm{C}\) For air: \(T_{c_{inlet}} = 30^{\circ} \mathrm{C}\) \(T_{c_{exit}} = 150^{\circ} \mathrm{C}\) (desired) Using these temperatures, we can find the temperature ratio R: \(R = \frac{500 - 320}{150 - 30} = 1.8\) Next, we need to find the heat capacity rates for both fluids using their mass flow rates and specific heat capacities: For exhaust gas: \(\dot{m}_{h} = 7.5 \mathrm{~kg} / \mathrm{s}\) \(c_{p_{h}} = 1069 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) For air: \(\dot{m}_{c} = 15 \mathrm{~kg} / \mathrm{s}\) \(c_{p_{c}} = 1069 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) Heat capacity rates are calculated as follows: \(\dot{C}_{h} = \dot{m}_{h} \times c_{p_{h}}=7.5 * 1069= 8023.5 \mathrm{~W} / \mathrm{K}\) \(\dot{C}_{c} = \dot{m}_{c} \times c_{p_{c}}=15 * 1069= 16047 \mathrm{~W} / \mathrm{K}\) And the heat capacity rate ratio Cr: \(C_{r} = \frac{\dot{C}_{h}}{\dot{C}_{c}} =\frac{8023.5}{16047}=0.5\)

02

Calculate Overall Heat Transfer Coefficient (U)

Since we're given the convective heat transfer coefficients on both the exhaust gas and air sides, as well as the fouling resistance for both sides, we can find the overall heat transfer coefficient. Let \(h_{1}\) and \(h_{2}\) be the heat transfer coefficients on the exhaust gas and air sides, respectively, and \(R_{f1}\) and \(R_{f2}\) be the fouling resistances: \(h_{1} = 750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) \(h_{2} = 300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) \(R_{f1} = R_{f2} = 0.0004 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) The overall heat transfer coefficient can be found using the following formula: \(U = \frac{1}{\frac{1}{h_{1}} + R_{f1} + R_{f2} + \frac{1}{h_{2}}} = \frac{1}{\frac{1}{750} + 0.0004 + 0.0004 + \frac{1}{300}} = 215.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

03

Determine the Maximum Effectiveness of Heat Exchanger

With the temperature ratio (R), heat capacity rate ratio (Cr), and overall heat transfer coefficient (U) found, we can determine the maximum effectiveness of the heat exchanger using the NTU-ε method and evaluate whether or not the air can be heated to the desired temperature. From the following equation for effectiveness, we can find the maximum effectiveness ε: \(\epsilon = \frac{1-\exp \left(-NTU\left(1+C_{r}\right)\right)}{1+C_{r}}\) However, we don't know NTU (number of transfer units) yet. So we need to find the relation of NTU with U and A first: \(NTU =\frac{UA}{\dot{C}_c}\) Note that A is the area of the heat exchanger. We need to find the area of the heat exchanger anyway, so we use this equation to find the value of NTU: \(\frac{NTU*U*\dot{C}_c}{U} = \frac{Q_{max}}{\dot{C}_c}= \frac{(1-\exp \left(-NTU\left(1+C_{r}\right)\right)}{1+C_{r}}\) We have all the values except for the area (A) in this equation. So, we can solve for A: \(A = NTU*\frac{\dot{C}_c}{U} = (1.8)\frac{16047}{215.9}=133.5 \mathrm{~m}^{2}\) Now we can find the heat transfer (Q) and check whether the air can be heated to the desired temperature: \(Q=\epsilon*\dot{C}_{c}*(T_{h_{inlet}}-T_{c_{inlet}})=(1.8)*8023.5=14442.3\mathrm{~W}\)

04

Adjust Air Mass Flow Rate for Desired Exit Temperature

Since the exit air temperature is not equal to the desired temperature of \(150^{\circ} \mathrm{C}\), we need to find the required air mass flow rate to attain the desired exit temperature. We know that: \(Q=\dot{C}_{c} * (T_{c_{exit}}-T_{c_{inlet}})\) So in order to reach \(150^{\circ} \mathrm{C}\), \(\dot{m}_{c} = \frac{Q}{c_{p} * (T_{c_{exit}}-T_{c_{inlet}})} = \frac{14442.3}{1069 * (150 - 30)} = 10.57 \mathrm{~kg} / \mathrm{s}\)

05

Plot Exit Air Temperature vs. Air Mass Flow Rate

Finally, we need to make a plot of exit air temperature vs. air mass flow rate over a temperature range of \(75^{\circ} \mathrm{C}\) to \(300^{\circ} \mathrm{C}\), assuming other conditions remain the same. This part of the exercise will require specialized plotting software or a spreadsheet program. To create the plot, you will need to: 1. Generate a range of air exit temperatures between \(75^{\circ} \mathrm{C}\) and \(300^{\circ} \mathrm{C}\). 2. Calculate the corresponding air mass flow rates for each temperature by rearranging the effectiveness equation. 3. Plot the air mass flow rates on the x-axis and their corresponding air exit temperatures on the y-axis. 4. Analyze and compare the trends and differences in mass flow rates needed for various air exit temperatures. In conclusion, to answer the exercise's questions: a) No, the air cannot be heated to the desired temperature of \(150^{\circ} \mathrm{C}\) with the given conditions. b) The area of the heat exchanger is \(133.5 \mathrm{~m}^{2}\). c) The air mass flow rate required to achieve the desired exit temperature is \(10.57 \mathrm{~kg} / \mathrm{s}\). d) A plot of exit air temperature vs. air mass flow rate must be created using specialized software or a spreadsheet program, taking into account the range of exit air temperatures stated in the exercise.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!