91Ó°ÊÓ

A scheme for concurrently heating separate water and air streams involves passing them through and over an array of tubes, respectively, while the tube wall is heated electrically. To enhance gas-side heat transfer, annular fins of rectangular profile are attached to the outer tube surface. Attachment is facilitated with a dielectric adhesive that electrically isolates the fins from the current-carrying tube wall. (a) Assuming uniform volumetric heat generation within the tube wall, obtain expressions for the heat rate per unit tube length \((\mathrm{W} / \mathrm{m})\) at the inner \(\left(r_{i}\right)\) and outer \(\left(r_{o}\right)\) surfaces of the wall. Express your results in terms of the tube inner and outer surface temperatures, \(T_{s, i}\) and \(T_{s, e}\), and other pertinent parameters. (b) Obtain expressions that could be used to determine \(T_{s, i}\) and \(T_{s, o}\) in terms of parameters associated with the water- and air-side conditions. (c) Consider conditions for which the water and air are at \(T_{\infty, i}=T_{\infty, o}=300 \mathrm{~K}\), with corresponding convection coefficients of \(h_{i}=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{o}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Heat is uniformly dissipated in a stainless steel tube \(\left(k_{w}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\), having inner and outer radii of \(r_{i}=25 \mathrm{~mm}\) and \(r_{o}=30\) \(\mathrm{mm}\), and aluminum fins \(\left(t=\delta=2 \mathrm{~mm}, r_{t}=55\right.\) \(\mathrm{mm}\) ) are attached to the outer surface, with \(R_{t, c}^{\prime \prime}=\) \(10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). Determine the heat rates and temperatures at the inner and outer surfaces as a function of the rate of volumetric heating \(\dot{q}\). The upper limit to \(\dot{q}\) will be determined by the constraints that \(T_{s, i}\) not exceed the boiling point of water \(\left(100^{\circ} \mathrm{C}\right)\) and \(T_{s, o}\) not exceed the decomposition temperature of the adhesive \(\left(250^{\circ} \mathrm{C}\right)\).

Step by step solution

01

Heat Rate Per Unit Tube Length Expressions

Given that the heat is uniformly generated within the tube wall, we can derive the expressions for the heat rate per unit tube length at the inner and outer surfaces. Using conduction equation: \[q_{cond} = -k_w\frac{dT}{dr}\] Integrating the equation between inner and outer surfaces: \[\int_{q_{i}}^{q_{o}} dq = -k_w \int_{r_{i}}^{r_{o}} \frac{dT}{dr} dr\] where \(q_i\) represents the heat rate at the inner surface, \(q_o\) represents the heat rate at the outer surface, and \(k_w\) represents the conductivity of the wall material. The boundary conditions are as follows: 1. \(T(r_i) = T_{s,i}\) 2. \(T(r_o) = T_{s,o}\) Notice that \(q_i\) is the heat rate at the inner surface, where \[q_i = h_i (T_{s,i} - T_{\infty,i})\] Similarly, \(q_o\) is the heat rate at the outer surface. Hence, \[q_o = h_o (T_{s,o} - T_{\infty,o})\]

02

Temperature Expressions

Now we need to find expressions to determine the inner and outer surface temperatures, \(T_{s, i}\) and \(T_{s, o}\), in terms of the given parameters. Using the heat rate expressions from Step 1, we get: \[T_{s,i} = T_{\infty,i} + \frac{q_i}{h_i}\] \[T_{s,o} = T_{\infty,o} + \frac{q_o}{h_o}\]

03

Calculating Heat Rates and Temperatures

We are given conditions for which the water and air are at \(T_{\infty,i}=T_{\infty,o}=300 \,K\), with their respective convection coefficients. Heat is uniformly dissipated, and the tube and fins' dimensions and materials are given. From the given information in the exercise, the parameters are as follows: 1. \(T_{\infty,i}=T_{\infty,o}=300 \,K\) 2. \(h_i=2000 \,W/m^2\cdot K\) 3. \(h_o=100 \,W/m^2\cdot K\) 4. \(k_w=15 \,W/m\cdot K\) 5. \(r_i=25 \,mm\) 6. \(r_o=30\,\mathrm{mm}\) 7. Aluminum fins: \(t=\delta=2 \,mm\), \(r_t=55\,mm\) 8. \(R_{t,c}^{\prime \prime}=10^{-4} m^2\cdot K/W\) It is also given that the constraints are \(T_{s, i}\) not exceeding the boiling point of water (100°C) and \(T_{s, o}\) not exceeding the decomposition temperature of the adhesive (250°C). We calculate the heat rates and temperatures at the inner and outer surfaces using the expressions obtained in Steps 1 and 2: \[q_i = h_i \left(T_{s,i} - T_{\infty,i}\right) \leq h_i(100°C - 300K)\] \[q_o = h_o \left(T_{s,o} - T_{\infty,o}\right) \leq h_o(250°C - 300K)\] Now, using the conduction equation and the given radii, substitute these values into the expression from Step 1: \[\int_{q_i}^{q_o} dq = -k_w \int_{r_i}^{r_o} \frac{dT}{dr} dr\] Solving for \(\dot{q}\) will give us the rate of volumetric heating as a function of the heat rates at the inner and outer surfaces. Note that the results should respect the constraints imposed by the boiling point of water and the decomposition temperature of the adhesive.

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Key Concepts

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

Conduction Equation

Underlying many heat transfer problems, such as the flow of heat through finned tubes, is the basic conduction equation. This equation is derived from Fourier’s law of heat conduction and expresses how thermal energy moves within solids.

In its simplest one-dimensional form, the conduction equation is represented as:
\[q_{cond} = -k \frac{dT}{dr}\]
Here, \(q_{cond}\) is the heat flux (the rate of heat transfer per unit area), \(k\) is the thermal conductivity of the material, and \(\frac{dT}{dr}\) is the temperature gradient across the material in the radial direction.

The negative sign indicates heat flows from higher to lower temperatures. In cylindrical coordinates like those for a tube, solving this equation involves integration across the radius from the inner surface to the outer surface, which leads to expressions for heat rate per unit length at these surfaces.

In the exercise, uniform volumetric heat generation within the tube wall simplified the equation as it allowed for disregarding any axial heat transfer, thus focusing on radial conduction.

Heat Rate per Unit Length

Calculating the heat rate per unit length is crucial for understanding how efficiently a finned tube can transfer thermal energy. The heat rate per unit length, often denoted by \(q\), measures the amount of heat transferred by a section of the tube per unit length along its axis.

Using the conduction equation provides the mathematical foundation required to derive the heat rate expressions at the inner (\(r_i\)) and outer (\(r_o\)) tube surfaces:
\[q_i = h_i \left(T_{s,i} - T_{\infty,i}\right)\]
\[q_o = h_o \left(T_{s,o} - T_{\infty,o}\right)\]
Here, \(h_i\) and \(h_o\) are the convective heat transfer coefficients for the inner and outer surfaces, respectively, \(T_{s,i}\) and \(T_{s,o}\) are the temperatures at the inner and outer surfaces, and \(T_{\infty,i}\) and \(T_{\infty,o}\) are the temperatures of the fluids (water and air) in close contact with those surfaces.

It's essential to accurately calculate the heat rate per unit length as it impacts the design and operation parameters of heat exchangers and affects the efficiency of the heating or cooling process.

Thermal Conductivity

Thermal conductivity, denoted by \(k\), is a fundamental material property that quantifies a material's ability to conduct heat. It plays a pivotal role in the conduction equation, as it directly affects the rate at which heat is transferred.

In the context of the exercise:
\[k_w = 15\; \text{W/m}\cdot\text{K}\]
This value represents the thermal conductivity of stainless steel, the material of the tube through which heat is being transferred from the electrically heated tube wall to the water and air streams.

Materials with high thermal conductivity, like metals, are efficient at conducting heat, while those with low thermal conductivity, such as plastics or rubbers, are considered insulators. The design of finned tubes and the choice of materials like stainless steel with a known thermal conductivity ensure that sufficient heat is transferred to the fluids while maintaining structural integrity and respecting temperature constraints.

Therefore, understanding and applying the thermal conductivity concept is essential in solving heat transfer problems and designing equipment for thermal applications.