Question

Two 3-m-long and \(0.4-\mathrm{cm}\)-thick cast iron \((k=\) \(52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) steam pipes of outer diameter \(10 \mathrm{~cm}\) are connected to each other through two 1 -cm-thick flanges of outer diameter \(20 \mathrm{~cm}\). The steam flows inside the pipe at an average temperature of \(200^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(180 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The outer surface of the pipe is exposed to an ambient at \(12^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Disregarding the flanges, determine the average outer surface temperature of the pipe. (b) Using this temperature for the base of the flange and treating the flanges as the fins, determine the fin efficiency and the rate of heat transfer from the flanges. (c) What length of pipe is the flange section equivalent to for heat transfer purposes?

Short answer

The following are short answers based on the step-by-step solution: (a) The average outer surface temperature of the pipe is 25.73°C. (b) The fin efficiency of the flange is 0.961, and the rate of heat transfer from the flanges is 6.386 W. (c) The equivalent pipe length for heat transfer purposes when considering the flange section is 0.6706 m.

Step-by-step solution

Answer (a) Calculate average outer surface temperature of the pipe

First, we need to calculate the conductive thermal resistance of the pipe, \(R_c\). We can do this using the following formula: \(R_c = \frac{k}{r_o}(r_o - r_i)\), where \(k\) is the thermal conductivity of the pipe material, \(r_o\) is the outer radius of the pipe and \(r_i\) is the inner radius. By substituting given values: \(r_i = r_o - 0.004 = 0.05 - 0.004 = 0.046 m\) \(R_c = \frac{52}{0.05} (0.05 - 0.046) = 0.052 \mathrm{~m} \cdot \mathrm{K} / \mathrm{W}\) Now we can calculate the heat transfer rate. Using Newton's Law of Cooling, we have: \(Q = h A (T_i - T_o)\), where \(h\) is the convective heat transfer coefficient, \(A\) is the surface area, \(T_i\) is the average temperature inside the pipe, and \(T_o\) is the average temperature outside the pipe. Considering the dimensions of the pipe, we can compute the surface area \(A\) as follows: \(A = 2\pi r_o L = 2\pi(0.05)(6) = 1.884 \mathrm{m}^2\) Therefore, the heat transfer rate is: \(Q = 180 (1.884)(200 - T_o)\) Now, let's apply Fourier's law of thermal conduction for the heat transfer through the pipe wall: \(Q = k A \frac{T_i - T_o}{R_c}\) And we equate the two equations above to find the unknown value \(T_o\): \(180 (1.884)(200 - T_o) = 52 (1.884) \frac{200 - T_o}{0.052}\) Solving for \(T_o\), we get: \(T_o = 25.73^{\circ} \mathrm{C}\)

Answer (b) Find flange fin efficiency and rate of heat transfer

Now that we have the average outer surface temperature, let's find the efficiency and rate of heat transfer from the flanges. First, we need to find the fin parameter, \(m\), for the flanges. The formula for \(m\) is as follows: \(m = \sqrt{\frac{2h}{k t_f}}\), where \(t_f\) is the thickness of the flange and \(h\) is the convective heat transfer coefficient. Applying the given values, we get: \(m = \sqrt{\frac{2(25)}{52(0.01)}} = 1.44 \mathrm{m}^{-1}\) Now we can compute the fin efficiency, \(\eta\), using the formula for a radial fin: \(\eta = \coth(ml_f) - \frac{1}{ml_f}\), where \(l_f\) is the length of the fin, calculated as half of the flange's outer radius minus half of the pipe's outer diameter: \(l_f = [(0.20/2) - (0.10/2)] m = 0.05 m\) Running the calculation using the value of \(m\) found previously: \(\eta = \coth(1.44 \cdot 0.05) - \frac{1}{1.44 \cdot 0.05} = 0.961\) The rate of heat transfer from the flanges can now be calculated as follows: \(Q_f = \eta \cdot h \cdot A_f (T_b - T_{\infty})\), where \(A_f\) is the flange's total surface area and \(T_b\) is the base temperature of the flanges (which is the same as the average outer surface temperature of the pipe). We can compute \(A_f\) as \(2 \pi r_o t_f\) for each flange, summing their areas, and then multiply it by 2 because there are 2 flanges connecting the pipes: \(A_f = 2 [2 \pi (0.1)(0.01) + 2 \pi (0.2)(0.01)] = 0.0188 \mathrm{m}^2\) Finally, the rate of heat transfer from the flanges is: \(Q_f = 0.961 \cdot 25 \cdot 0.0188 \cdot (25.73 - 12) = 6.386 \mathrm{W}\)

Answer (c) Determine equivalent pipe length for heat transfer purposes

To determine the equivalent length of pipe for heat transfer purposes, we can use the ratio of heat transfer rate between the flange section and the pipe section. Let's denote the equivalent length by \(L_{eq}\). From parts (a) and (b), we found that the rate of heat transfer from the flanges is \(Q_f = 6.386 \mathrm{W}\), and the heat transfer rate through the pipe section is \(Q = 180 (1.884)(200 - 25.73) = 5908.662 \mathrm{W}\). To find the equivalent length, we can apply the following formula: \(Q_f = \frac{A_f}{A} \cdot Q \cdot L_{eq}\) Then we isolate the variable we want to find: \(L_{eq} = \frac{Q_f \cdot A}{A_f \cdot Q} = \frac{6.386 \cdot 1.884}{0.0188 \cdot 5908.662} = 0.6706 \mathrm{~m}\) So, the flange section is equivalent to a 0.6706 m long pipe section for heat transfer purposes.

Key concepts

Thermal Resistance

Understanding thermal resistance is crucial when analyzing heat transfer problems, as it helps us to evaluate the opposition to heat flow within a material. In essence, thermal resistance is analogous to electrical resistance in that it defines how difficult it is for heat to conduct through an object. The higher the thermal resistance, the better the material is at insulating against heat transfer.

The formula for calculating the thermal resistance caused by conduction, coping with cylindrical objects such as pipes, is represented as: \[ R_c = \frac{{\ln(r_o / r_i)}}{{2 \pi L k}} \], where \( r_i \) and \( r_o \) are the inner and outer radii of the cylinder respectively, \( L \) is the length of the cylinder, and \( k \) is the thermal conductivity of the material.When evaluating heat transfer in pipes, like the steam pipes mentioned in the exercise, the calculation of thermal resistance is fundamental for determining how efficiently heat is being transferred and what the temperature distribution within the system will be. It’s critical to choose materials with suitable thermal resistance properties according to the application demands, in order to ensure optimal thermal management and energy efficiency. Correctly choosing materials and calculating thermal resistance is a step that could not be more emphasized in heat transfer exercises and real-world applications alike.

Conductive Heat Transfer

Conductive heat transfer is the process by which heat energy is transported from the hotter part to a cooler part of a material, without the movement of the material itself. This type of heat transfer is governed by Fourier’s Law, which states that the time rate of heat transfer through a material is proportional to the negative gradient of the temperature and the area through which the heat transfers. The equation for Fourier’s law is: \[ Q = -kA(\frac{{dT}}{{dx}}) \], where \( Q \) is the heat transfer rate, \( k \) is the thermal conductivity of the material, \( A \) is the area perpendicular to the direction of heat transfer, and \( dT/dx \) is the temperature gradient.In the context of the exercise provided, conductive heat transfer is critically analyzed when determining the average outer surface temperature of a steam pipe. Since the pipe is made of cast iron, a material with relatively high thermal conductivity, understanding the principles of conduction is paramount to predicting how effectively the steam’s heat will be conducted to the pipe’s exterior. The solution involves applying Fourier's Law in a manner that considers the geometry and thermal properties of the pipe system, thereby illustrating the direct connection between theory and practical application in thermal system designs. When determining the efficiency of heat transfer in real-world scenarios, such as heating systems or manufacturing processes, conductive heat transfer takes center stage, as it influences the design, material selection, and safety protocols.

Fin Efficiency

Fin efficiency is an evaluation metric that compares the actual heat transfer rate from a fin to the heat transfer rate if the entire fin were at the base temperature. A fin is an extended surface used to increase the heat transfer area and to facilitate a higher rate of heat exchange between an object and its environment. The efficiency can be expressed as: \[ \eta = \frac{{Actual Heat Transfer}}{{Heat Transfer if the entire fin were at base temperature}} \], which frequently assumes values between 0 and 1, with 1 meaning perfectly efficient.To calculate the fin efficiency accurately, it's necessary to take into account the fin's geometry, thermal conductivity of the material, and the convective heat transfer coefficient of the surrounding fluid. For the flanges in the given exercise, they are treated as fins, and their efficiency is influenced by their size, shape, and material properties. In practice, highly efficient fins are desirable as they contribute to the overall effectiveness of the heat transfer process, making systems like radiators, heat exchangers, and electronic cooling systems perform better. Fin efficiency is a concept that students must grasp to optimize and design thermal systems properly, ensuring that they achieve the intended performance without unnecessary costs or material usage. It’s a concept that shows the impact of a seemingly small design choice—the addition of fins—can have on the overall thermal management system.