Question

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.

Short answer

Answer: No, the air cannot be heated to the desired temperature of 150°C with the given conditions.

Step-by-step solution

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: \(C_{r} = \frac{\dot{C}_{h}}{\dot{C}_{c}} =\frac{8023.5}{16047}=0.5\)

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: \(U = \frac{1}{\frac{1}{h_{1}} + R_{f1} + R_{f2} + \frac{1}{h_{2}}} = \frac{1}{\frac{1}{750} + 0.0004 + 0.0004 + \frac{1}{300}} = 215.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

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: \(Q=\epsilon*\dot{C}_{c}*(T_{h_{inlet}}-T_{c_{inlet}})=(1.8)*8023.5=14442.3\mathrm{~W}\)

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 \(150^{\circ} \mathrm{C}\), \(\dot{m}_{c} = \frac{Q}{c_{p} * (T_{c_{exit}}-T_{c_{inlet}})} = \frac{14442.3}{1069 * (150 - 30)} = 10.57 \mathrm{~kg} / \mathrm{s}\)

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.

Key concepts

Cross Flow Heat Exchanger

A cross flow heat exchanger is a common configuration in heat exchange systems where two fluids flow perpendicularly across each other. This design is particularly useful for applications where maximizing heat transfer efficiency in compact spaces is crucial.

In a cross flow heat exchanger, one fluid helps to transfer heat to or from another fluid by moving across it at a 90º angle. Unlike parallel or counterflow heat exchangers where the fluids move either with or against each other, cross flow arrangements allow for high heat transfer due to continuous temperature differences across the surfaces.

The efficiency of a cross flow heat exchanger depends on:

• The temperature gradients between the two fluids.
• The heat transfer surfaces available for interaction.
• The flow rates of both fluids involved.

These characteristics make cross flow heat exchangers versatile in different industrial applications, such as in cooling systems of gas turbines, radiators in cars, and various HVAC systems. However, as in the gas turbine system described, fouling can reduce efficiency over time, necessitating maintenance and cleaning.

Recuperative Heat Exchanger

A recuperative heat exchanger, like the one in this exercise, is designed to recover heat from exhaust gases and transfer it to incoming air or another fluid. This type of heat exchanger improves overall system efficiency since it utilizes waste heat that would otherwise be lost.

The main components in a recuperative heat exchanger include channels or tubes arranged to allow maximum surface area contact between the hot and cold streams. Heat moves from the hotter fluid to the colder one, an especially beneficial process in industries like power generation, where maximizing energy use is paramount.

• Energy efficiency: Recuperative exchangers save energy by transferring heat between exhaust gases and incoming air.
• System longevity: By reducing waste heat, they help minimize wear on mechanical parts caused by thermal stress.
• Cost-effective operations: With lowered energy demands, these systems can significantly cut down operational costs.

Thus, recuperative heat exchangers are vital in conserving energy and protecting the environment by reducing the system's overall carbon footprint.

Convective Heat Transfer

Convective heat transfer plays a pivotal role in heat exchangers as it governs how efficiently heat moves between the surfaces and fluids involved. This type of heat transfer relies on fluid motion, where heat is carried by the fluid as it moves into or out of the heat exchange surfaces.

Two key components of convective heat transfer in heat exchangers are:

• Convective heat transfer coefficients (\(h\)): These coefficients quantify the transfer ability of a fluid, with higher values usually indicating better heat movement.
• Surface area: More surface area allows for more contact between the fluid and surface, thus increasing the overall heat transfer rate.

In the gas turbine system example, the convective heat transfer coefficients are given for both the exhaust gas and air side as \(750\) and \(300\) W/m²·K, respectively. These coefficients reflect the efficiency with which each side can conduct heat away or towards the fluid streams.

High-performance heat exchangers will optimize these convective interactions to ensure maximum heat transfer efficiency. This importance extends to designing the size and shape of the exchange surfaces, influencing how fast and efficiently the heat exchange occurs.

Temperature Ratio

In heat exchanger analysis, the temperature ratio is a metric that compares the temperature differences across the hot and cold streams. This ratio is crucial for determining the heat exchanger's effectiveness, particularly in systems like the recuperative cross-flow heat exchanger.

The temperature ratio (\(R\)) is defined as:\[R = \frac{T_{h_{inlet}}-T_{h_{exit}}}{T_{c_{exit}}-T_{c_{inlet}}}\]Where:

• \(T_{h_{inlet}}\) = Inlet temperature of hot fluid
• \(T_{h_{exit}}\) = Exit temperature of hot fluid
• \(T_{c_{exit}}\) = Exit temperature of cold fluid
• \(T_{c_{inlet}}\) = Inlet temperature of cold fluid

In our example, with the values given, the ratio is \(1.8\), indicating that the hot fluid loses more temperature as it transfers heat to the cold fluid, which is desirable for effective heat exchange.

Understanding and calculating the temperature ratio is fundamental. It helps to determine if the heat exchanger can achieve the desired exit temperatures for both fluid streams, highlighting the need for precise control in systems aiming for optimal energy efficiency.