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Consider a horizontal pin fin of 6- \(\mathrm{mm}\) diameter and \(60-\mathrm{mm}\) length fabricated from plain carbon steel \((k=57 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \varepsilon=0.5)\). The base of the fin is maintained at \(150^{\circ} \mathrm{C}\), while the quiescent ambient air and the surroundings are at \(25^{\circ} \mathrm{C}\). Assume the fin tip is adiabatic. (a) Estimate the fin heat rate, \(q_{f}\). Use an average fin surface temperature of \(125^{\circ} \mathrm{C}\) in estimating the free convection coefficient and the linearized radiation coefficient. How sensitive is this estimate to your choice of the average fin surface temperature? (b) Use the finite-difference method of solution to obtain \(q_{f}\) when the convection and radiation coefficients are based on local, rather than average, temperatures for the fin. How does your result compare with the analytical solution of part (a)?

Short Answer

Expert verified
To estimate the heat transfer rate of a horizontal pin fin, we first consider the average fin surface temperature and calculate the free convection coefficient and the linearized radiation coefficient. The combined heat transfer coefficient, fin efficiency, and heat rate are then estimated. In part (a), the heat rate was calculated using the given average fin surface temperature of \(125^{\circ}\mathrm{C}\) and compared for slightly different values to verify the sensitivity. In part (b), a finite-difference method based on local temperatures was used. The temperature distribution was calculated using a numerical method like Gauss-Seidel or Thomas Algorithm. The local heat transfer coefficients were determined, and the total fin heat rate was calculated. Finally, this was compared with the heat rate obtained in part (a) to analyze the percentage difference between the two results.

Step by step solution

01

Calculate average surface temperature

We are given the average fin surface temperature as \(T_{avg} = 125^{\circ} \mathrm{C}\).
02

Calculate the free convection coefficient

To estimate the free convection coefficient, we can use the following formula: \[h_{c} = 1.32(T_{avg} - T_{\infty})^{1/4}\] where \(T_\infty = 25^{\circ} \mathrm{C}\) is the ambient temperature. Plug in the given values: \[h_{c} = 1.32(125 - 25)^{\frac{1}{4}}\]
03

Calculate the linearized radiation coefficient

To estimate the linearized radiation coefficient, we can use the following formula: \[h_{r} = \varepsilon \sigma \dfrac{(T_{avg} + T_{\infty})(T_{avg}^{2} + T_{\infty}^{2})}{(T_{avg} - T_{\infty})}\] where \(\varepsilon = 0.5\) is the emissivity, and \(\sigma = 5.67 \cdot 10^{-8} \mathrm{W/ m}^{2} \cdot\mathrm{K}^{4}\) is the Stefan-Boltzmann constant. Plug in the given values: \[h_{r} = 0.5 \times 5.67 \times 10^{-8} \dfrac{(125 + 25)(125^{2}+ 25^{2})}{(125 - 25)}\]
04

Calculate combined heat transfer coefficient

Now we can calculate the combined heat transfer coefficient \(h_{total} = h_{c} + h_{r}\), using the values obtained in steps 2 and 3.
05

Calculate fin efficiency and heat rate

Now we will calculate the fin efficiency, \(\eta_{f}\), and the fin heat rate, \(q_{f}\), by applying the following formula: \[\eta_{f} = \mathrm{tanh}( mL )/ mL\] \[q_{f} = A_{b}(T_{b} - T_{\infty})h_{total}\eta_{f}\] Here, \(T_b = 150^{\circ} \mathrm{C}\) is the base temperature, \(m=\sqrt{\dfrac{2h_{total}}{kD}}\), and \(A_b = \dfrac{\pi D L}{4}\) is the base area of the fin. Substitute the given values and find \(\eta_{f}\) and \(q_{f}\). To find the sensitivity of the estimate to the choice of the average fin surface temperature, it may be useful to calculate the heat rate for slightly higher and lower values of \(T_{avg}\), e.g., \(120^\circ\mathrm{C}\) and \(130^\circ\mathrm{C}\), and compare the results. (b) Now we will use the finite-difference method to estimate the fin heat rate based on local temperatures:
06

Set up the finite-difference grid for the fin

We need to create a grid along the fin length and discretize the fin temperature distribution. Let's use a small grid spacing, e.g., \(\Delta x = 1 \, \mathrm{mm}\), along the fin length to generate the discretized points for the temperature distribution.
07

Set up a system of equations for the temperature distribution

The finite-difference equations for the temperature distribution, incorporating both convection and radiation, need to be set up for each of the grid points along the fin length, taking into account the adiabatic fin tip condition.
08

Calculate the temperature distribution

Solve the system of equations from step 2 for the temperature distribution along the fin's length using numerical methods, e.g., Gauss-Seidel or Thomas Algorithm.
09

Calculate the fin heat rate based on local temperatures

Calculate the local heat transfer coefficients \(h_{c}\) and \(h_r\) for each grid point, considering the convective and radiative heat transfer based on the local temperature distribution obtained in step 3. Then, sum up the heat transfer rates at each of the grid points to obtain the total fin heat rate.
10

Compare the result with the analytical solution of part (a)

Now we can compare the fin heat rate obtained by the finite-difference method with the fin heat rate obtained in part (a) by analyzing the percentage difference between the two results. Following these steps, you will be able to successfully estimate the heat transfer rate of the horizontal pin fin and determine the level of sensitivity of the estimate to the choice of average fin surface temperature.

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Key Concepts

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

Pin Fin Analysis
A pin fin is a special type of extended surface that effectively enhances heat transfer from a primary surface to the surrounding environment. This is particularly useful in applications where the additional heat removal can improve performance or efficiency.

The main goal of pin fin analysis is to estimate the heat rate from the fin, given specific conditions such as its thermal conductivity, dimensions, and environmental temperatures. By analyzing a pin fin, we can determine how much heat it dissipates through conduction inside the fin material and dissipation via convection and radiation from its surface.

A crucial aspect of pin fin analysis is determining the efficiency of the fin. Fin efficiency represents how effectively a fin transfers heat compared to an ideal scenario where the entire fin temperature is equal to the base temperature. Efficiency is calculated using the formula:

\[\eta_f = \frac{\tanh(mL)}{mL}\]

where \( m = \sqrt{\frac{2h_{total}}{kD}} \). Pin fin analysis involves determining this efficiency to evaluate how the dimensions, material, and thermal conditions impact the heat transfer capability of the fin.
Convection Coefficient Calculation
Convection plays a vital role in the heat transfer process from a pin fin, as it involves the transfer of heat between the fin surface and the surrounding fluid, which is typically air. The convection coefficient, \( h_{c} \), quantifies this transfer and can be estimated using empirical formulas.

In the given problem, we use the equation:
  • \( h_{c} = 1.32(T_{avg} - T_{\infty})^{1/4} \)
where \( T_{avg} \) is the average temperature of the fin surface, and \( T_{\infty} \) is the ambient temperature.

This formula arises from natural convection principles, where fluid movement is induced by buoyancy forces, typically due to temperature differences. It assumes scenarios like vertical and horizontal flats or thin rods exposed to natural convection.

Evaluating \( h_{c} \) is crucial, as it affects the total heat transfer rate calculation. By understanding the local airflow conditions and how they're influenced by the arrangement and temperature of the fin, we can better predict the effectiveness of heat transfer through convection.
Radiation Coefficient Calculation
Radiation, like convection, is a critical mechanism by which heat leaves the surface of a pin fin. Unlike convection, radiation does not require a medium and is governed by electromagnetic waves emitted by the surface.

The radiation coefficient, \( h_{r} \), quantifies the heat exchange by radiation between the fin and its surroundings. Calculation of \( h_{r} \) typically involves a more complex formula due to factors like emissivity and temperature variances. For the problem at hand, we use the following equation:
  • \(h_{r} = \varepsilon \sigma \dfrac{(T_{avg} + T_{\infty})(T_{avg}^{2} + T_{\infty}^{2})}{(T_{avg} - T_{\infty})}\)
where \( \varepsilon \) is the emissivity, and \( \sigma \) is the Stefan-Boltzmann constant.

The choice of average surface temperature significantly impacts the radiation coefficient value. Small changes in these temperatures can lead to notable differences in \( h_{r} \), as evident in this formula. Hence, appropriate estimation is necessary for accurate heat transfer rate calculations, further reinforcing the importance of precision in fin heat transfer analysis.

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Most popular questions from this chapter

Certain wood stove designs rely exclusively on heat transfer by radiation and natural convection to the surroundings. Consider a stove that forms a cubical enclosure, \(L_{s}=1 \mathrm{~m}\) on a side, in a large room. The exterior walls of the stove have an emissivity of \(\varepsilon=0.8\) and are at an operating temperature of \(T_{s s s}=500 \mathrm{~K}\). The stove pipe, which may be assumed to be isothermal at an operating temperature of \(T_{s, p}=400 \mathrm{~K}\), has a diameter of \(D_{p}=0.25 \mathrm{~m}\) and a height of \(L_{p}=2 \mathrm{~m}\), extending from stove to ceiling. The stove is in a large room whose air and walls are at \(T_{\infty}=T_{\text {sur }}=300 \mathrm{~K}\). Neglecting heat transfer from the small horizontal section of the pipe and radiation exchange between the pipe and stove, estimate the rate at which heat is transferred from the stove and pipe to the surroundings.

9.27 The vertical rear window of an automobile is of thickness \(L=8 \mathrm{~mm}\) and height \(H=0.5 \mathrm{~m}\) and contains fine-meshed heating wires that can induce nearly uniform volumetric heating, \(\dot{q}\left(\mathrm{~W} / \mathrm{m}^{3}\right)\). (a) Consider steady-state conditions for which the interior surface of the window is exposed to quiescent air at \(10^{\circ} \mathrm{C}\), while the exterior surface is exposed to air at \(-10^{\circ} \mathrm{C}\) moving in parallel flow over the surface with a velocity of \(20 \mathrm{~m} / \mathrm{s}\). Determine the volumetric heating rate needed to maintain the interior window surface at \(T_{s, i}=15^{\circ} \mathrm{C}\). (b) The interior and exterior window temperatures, \(T_{s, i}\) and \(T_{s, \rho}\), depend on the compartment and ambient temperatures, \(T_{\infty, i}\) and \(T_{\infty, p}\), as well as on the velocity \(u_{\infty}\) of air flowing over the exterior surface and the volumetric heating rate \(\dot{q}\). Subject to the constraint that \(T_{s, i}\) is to be maintained at \(15^{\circ} \mathrm{C}\), we wish to develop guidelines for varying the heating rate in response to changes in \(T_{\infty,}, T_{\infty \rho,}\), and/or \(u_{\infty}\). If \(T_{\infty, i}\) is maintained at \(10^{\circ} \mathrm{C}\), how will \(\dot{q}\) and \(T_{s, \rho}\) vary with \(T_{\infty, o}\) for \(-25 \leq T_{\infty,,} \leq 5^{\circ} \mathrm{C}\) and \(u_{\infty}=10,20\), and \(30 \mathrm{~m} / \mathrm{s}\) ? If a constant vehicle speed is maintained, such that \(u_{\infty}=30 \mathrm{~m} / \mathrm{s}\), how will \(\dot{q}\) and \(T_{s, o}\) vary with \(T_{\infty, i}\) for \(5 \leq T_{\infty, i} \leq 20^{\circ} \mathrm{C}\) and \(T_{\infty, o}=-25,-10\), and \(5^{\circ} \mathrm{C}\) ?

A 50 -mm-thick air gap separates two horizontal metal plates that form the top surface of an industrial furnace. The bottom plate is at \(T_{h}=200^{\circ} \mathrm{C}\) and the top plate is at \(T_{c}=50^{\circ} \mathrm{C}\). The plant operator wishes to provide insulation between the plates to minimize heat loss. The relatively hot temperatures preclude use of foamed or felt insulation materials. Evacuated insulation materials cannot be used due to the harsh industrial environment and their expense. A young engineer suggests that equally spaced, thin horizontal sheets of aluminum foil may be inserted in the gap to eliminate natural convection and minimize heat loss through the air gap.

A refrigerator door has a height and width of \(H=1 \mathrm{~m}\) and \(W=0.65 \mathrm{~m}\), respectively, and is situated in a large room for which the air and walls are at \(T_{\infty}=T_{\text {sur }}=25^{\circ} \mathrm{C}\). The door consists of a layer of polystyrene insulation \((k=0.03 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) sandwiched between thin sheets of steel \((\varepsilon=0.6)\) and polypropylene. Under normal operating conditions, the inner surface of the door is maintained at a fixed temperature of \(T_{s, i}=5^{\circ} \mathrm{C}\). (a) Estimate the heat gain through the door for the worst case condition corresponding to no insulation \((L=0)\). (b) Compute and plot the heat gain and the outer surface temperature \(T_{s, o}\) as a function of insulation thickness for \(0 \leq L \leq 25 \mathrm{~mm}\).

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