91Ó°ÊÓ

Waste heat from the exhaust gas of an industrial furnace is recovered by mounting a bank of unfinned tubes in the furnace stack. Pressurized water at a flow rate of \(0.025 \mathrm{~kg} / \mathrm{s}\) makes a single pass through each of the tubes, while the exhaust gas, which has an upstream velocity of \(5.0 \mathrm{~m} / \mathrm{s}\), moves in cross flow over the tubes at \(2.25 \mathrm{~kg} / \mathrm{s}\). The tube bank consists of a square array of 100 thin-walled tubes \((10 \times 10)\), each \(25 \mathrm{~mm}\) in diameter and \(4 \mathrm{~m}\) long. The tubes are aligned with a transverse pitch of \(50 \mathrm{~mm}\). The inlet temperatures of the water and the exhaust gas are 300 and \(800 \mathrm{~K}\), respectively. The water flow is fully developed, and the gas properties may be assumed to be those of atmospheric air. (a) What is the overall heat transfer coefficient? (b) What are the fluid outlet temperatures? (c) Operation of the heat exchanger may vary according to the demand for hot water. For the prescribed heat exchanger design and inlet conditions, compute and plot the rate of heat recovery and the fluid outlet temperatures as a function of water flow rate per tube for \(0.02 \leq \dot{m}_{c, 1} \leq 0.20 \mathrm{~kg} / \mathrm{s}\).

Step by step solution

01

Properties of the fluids

To solve the problem, we first need to know the properties of the fluids. In this case, we can assume that the exhaust gas has similar properties as atmospheric air, and we can use a standard engineering reference for these properties. For this solution, we will use the following values: For air (exhaust gas): density \(\rho_{g} = 1.1614 \, \mathrm{kg/m^3}\), specific heat \(c_{pg} = 1005 \, \mathrm{J/kg \cdot K}\), and thermal conductivity \(k_{g} = 0.0274 \, \mathrm{W/m \cdot K}\). For water: density \(\rho_{w} = 995.2 \, \mathrm{kg/m^3}\), specific heat \(c_{pw} = 4186 \, \mathrm{J/kg \cdot K}\), and thermal conductivity \(k_{w} = 0.6033 \, \mathrm{W/m \cdot K}\). These properties will be employed throughout the following steps. Step 2: Determine the logarithmic temperature difference

02

Logarithmic temperature difference

To determine the overall heat transfer coefficient, we first need to find the logarithmic temperature difference (\(\Delta T_{lm}\)) for the heat exchanger. This can be found using the following equation: \(\Delta T_{lm} = \frac{\Delta T_{1} - \Delta T_{2}}{ln(\frac{\Delta T_{1}}{\Delta T_{2}})}\) where \(\Delta T_{1}\) is the difference between the inlet temperatures of the two fluids (\(800 - 300\)), and \(\Delta T_{2}\) is the difference between the outlet temperatures of the two fluids, to be found in the next step. Step 3: Calculate Q and the fluid outlet temperatures

03

Heat transfer rate and outlet temperatures

Using the mass flow rates of water and gas, and the specific heat values, we can find the overall heat transfer rate \(Q\). \(Q = \dot{m}_{w}c_{pw}(T_{w2} - T_{w1}) = \dot{m}_{g}c_{pg}(T_{g1} - T_{g2})\) Solve for the outlet temperatures \(T_{w2}\) and \(T_{g2}\) using the given flow rates and inlet temperatures for both water and gas. Step 4: Determine the overall heat transfer coefficient

04

Overall heat transfer coefficient

With the logarithmic temperature difference and heat transfer rate, we can now calculate the overall heat transfer coefficient. \(Q = U \cdot A \cdot \Delta T_{lm}\) Here, \(A\) represents the total heat exchange area. In this case, \(A = 4 \pi D L = 4 \pi (0.025)(4) = 1.2566 \, \mathrm{m^2}\) Now, we can solve for \(U\): \(U = \frac{Q}{A \Delta T_{lm}}\) Step 5: Calculate the rate of heat recovery with respect to the water flow rate

05

Rate of heat recovery with respect to the water flow rate

For the prescribed heat exchanger design, we will compute and plot the rate of heat recovery with respect to the water flow rate per tube for \(0.02 \leq \dot{m}_{c, 1} \leq 0.20 \, \mathrm{kg/s}\). To do this, repeat the process of finding the logarithmic temperature difference and the overall heat transfer rate for each value of the water flow rate. Additionally, calculate the outlet temperatures of each fluid to plot against the water flow rate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithmic Temperature Difference

The logarithmic temperature difference, also known as the Logarithmic Mean Temperature Difference (LMTD), is crucial when analyzing heat exchangers. It represents the potential for heat transfer between two fluids. Let's delve into this concept through an example.
In our heat exchanger scenario, exhaust gas and water are involved. Each has a different inlet temperature: 800 K for the exhaust gas and 300 K for the water. The LMTD helps us account for these temperature differences across the length of the heat exchanger.
The formula for LMTD is:
\[ \Delta T_{lm} = \frac{\Delta T_{1} - \Delta T_{2}}{\ln(\frac{\Delta T_{1}}{\Delta T_{2}})}\]

• \(\Delta T_{1}\) is the temperature difference between the fluids' inlets.
• \(\Delta T_{2}\) is the temperature difference between the fluids' outlets.

Calculating the LMTD helps determine other significant values, like the overall heat transfer coefficient. Understanding LMTD is key to mastering heat transfer in engineering contexts.

Overall Heat Transfer Coefficient

The overall heat transfer coefficient (\(U\)) is pivotal in evaluating a heat exchanger's effectiveness. It correlates the heat transfer rate with the temperature difference across the exchanger.
In simple terms, \(U\) reflects how efficiently heat is moving between fluids in the exchanger. The ultimate goal is to optimize this efficiency as much as possible.
To calculate this coefficient, the formula used is:
\[Q = U \cdot A \cdot \Delta T_{lm}\]
Here:

• \(Q\) is the heat transfer rate, which is the energy crossing the system per unit time.
• \(A\) refers to the heat exchange area. The larger the area, the more opportunity exists for heat transfer.
• \(\Delta T_{lm}\) is the logarithmic mean temperature difference that gives the average driving force for heat transfer.

Discovering \(U\) involves understanding all these components and how they work together. Mastery over these includes being able to model complex systems in engineering applications effectively.

Heat Exchanger Design

Heat exchanger design is a nuanced field that combines mathematics, physics, and engineering to optimize the transfer of heat between two fluids. In our example, the setup includes a bank of tubes transferring heat from exhaust gases to water. An ideal design ensures maximal heat transfer efficiency with minimal energy loss.
To achieve this, consider:

• Fluid flow arrangement: The sequence or path the fluids take through an exchanger (counter-flow, co-current, or cross-flow) influences efficiency.
• Material selection: Materials must handle both thermal and stress loads while preventing corrosion and erosion.
• Configuration: The tube layout and tube count, in this case, the square array, significantly impact the heat exchange dynamics.
• Thermal properties: Knowing the thermal conductivity of fluids and the exchanger material allows precise heat transfer predictions.

Creating a successful heat exchanger design means carefully considering these components. By doing so, you can adapt the system to meet varying demand conditions, like changing flow rates in our exercise. Through such strategic designs, heat exchangers become valuable assets in energy recovery and efficiency improvements.