/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 122 An air-cooled R-22 condenser has... [FREE SOLUTION] | 91Ó°ÊÓ

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An air-cooled R-22 condenser has 10 mm-O.D., 9 mm-I.D. aluminum tubes with \(50 \mathrm{~mm}-\mathrm{O} . \mathrm{D} .\) annular rectangular aluminum fins of thickness \(0.2 \mathrm{~mm}\) at a pitch of \(2 \mathrm{~mm}\). The heat transfer coefficient inside the tubes is \(800 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), and on the fins it is \(20 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). If the \(\mathrm{R}-22\) is at \(320 \mathrm{~K}\) and the air is at \(300 \mathrm{~K}\), calculate the heat transfer per unit length of tube. Take \(k=180 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the aluminum.

Short Answer

Expert verified
The heat transfer per unit length of the tube is approximately 92.1 W/m.

Step by step solution

01

Calculate the Area for Heat Transfer Inside the Tube

The internal area per unit length for heat transfer inside the tube is calculated using the internal diameter:\[ A_i = \pi \times D_{in} \times L = \pi \times 9 \times 10^{-3} \text{ m} \times 1 \text{ m} = 0.02827 \text{ m}^2 \]
02

Calculate the Internal Thermal Resistance

The heat transfer coefficient inside the tubes is given as \( h_i = 800 \text{ W/m}^2\text{K} \). Thus, the internal thermal resistance per unit length is:\[ R_{i} = \frac{1}{h_i \times A_i} = \frac{1}{800 \times 0.02827} \approx 0.0441 \text{ K/W} \]
03

Calculate the External Area for Heat Transfer Including Fins

The finned external tube diameter with fins is calculated as follows. The finned diameter (including the two thicknesses of fins) comes to:\[D_{finned} = 10 \text{ mm} + 2(50 \text{ mm} - 9 \text{ mm}) = 92 \text{ mm} = 0.092 \text{ m}\] The pitch is 2 mm. The fin area for heat transfer per unit length is then:\[ A_o = \pi \times D_{o} \times 1 \text{ m} = \pi \times 0.092 \approx 0.289 \text{ m}^2 \]
04

Calculate the External Thermal Resistance

The external thermal resistance including the effect on fins \( R_o \) is calculated with:\[ R_o = \frac{1}{h_o \times A_o} = \frac{1}{20 \times 0.289} \approx 0.173 \text{ K/W}\]
05

Calculate the Total Thermal Resistance

The total thermal resistance \( R_{total} \) is the sum of internal and external resistances since it treats them as two resistors in series:\[ R_{total} = R_i + R_o = 0.0441 + 0.173 = 0.2171 \text{ K/W} \]
06

Calculate the Heat Transfer per Unit Length

Using the total thermal resistance, the net heat transfer rate per unit length is:\[ q' = \frac{T_{R-22} - T_{air}}{R_{total}} = \frac{320 - 300}{0.2171} \approx 92.1 \text{ W/m} \]

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Key Concepts

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

Thermal Resistance
Thermal resistance is a crucial concept when analyzing heat transfer processes. It is the measure of a material's ability to resist heat flow. The higher the thermal resistance, the less heat passes through the material. Thermal resistance is formulated in terms of the temperature difference per unit of heat flow. In this exercise, thermal resistance is crucial to determine the efficiency of the heat exchanger. We calculated two kinds of thermal resistance: internal and external. The internal thermal resistance is calculated from the heat transfer coefficient inside the tubes and the area available for heat transfer. This resistance gives us insight into how effectively heat is transferred from the refrigerant inside the tube to the tube wall. The external thermal resistance involves the area provided by the fins as well as the convection heat transfer coefficient. By assessing both internal and external resistances and adding them, we find the total thermal resistance. This cumulative resistance quantifies the entire setup's ability to minimize heat flow from inside the tube through to external air. Lower thermal resistance signifies higher efficiency in heat transfer.
Finned Heat Exchangers
Finned heat exchangers are a common design feature utilized to enhance heat dissipation. Fins increase the surface area available for heat transfer, thus amplifying the amount of heat that can be extracted from the hotter medium inside the tubes. The fins in this exercise are rectangular, with specific dimensions and arrangements. By having aluminum fins with a described thickness and a specific pitch, the heat transfer coefficient on the outside is efficiently utilized. The finned diameter of the tube showcases the physical extent to which heat exchange is heightened. In the calculation, the increased area due to fins is accounted for in computing the external thermal resistance. The more extensive the surface area, the lower the external resistance, and thus, the greater the capacity for heat removal. Optimizing the size and arrangement of the fins is vital for ensuring effective performance of the heat exchanger.
Convection Heat Transfer
Convection heat transfer plays a pivotal role when heat is transferred via fluid movement. There are typically two types: natural and forced convection. In many applications, the mode of convection can significantly impact the rate of heat exchange. Inside the tubes of our exercise, the refrigerant moving through dictates the internal heat transfer coefficient. The rate of heat transfer is heavily influenced by the motion of the R-22 refrigerant. Similarly, on the outside, air moves past the fins to facilitate heat removal. The external convection coefficient tells us about the interaction between the air and the finned surface. Understanding these coefficients is essential as they directly relate to how effectively heat is exchanged through convection. The higher these coefficients, the better the setup is at transferring heat due to the rapid movement of molecules carrying thermal energy away.
R-22 Refrigerant
R-22 refrigerant, also known as chlorodifluoromethane, is commonly utilized in air conditioning and refrigeration applications. It is known for its cooling properties and efficiency in heat absorption and release. In this exercise, understanding the temperature of the R-22 is vital as it impacts the heat transfer process. Operating at 320 K, the refrigerant's high capability for heat absorption plays a role in calculating the internal thermal resistance. R-22 is being phased out and replaced with more environmentally friendly options. However, it still provides valuable insights into the mechanisms of refrigeration systems. Knowing the properties of refrigerants like R-22 helps in devising better heat exchanger systems, even when switching to newer alternatives.

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Most popular questions from this chapter

A \(60 \mathrm{~cm}\)-long, \(3 \mathrm{~cm}\)-diameter AISI 1010 steel rod is welded to a furnace wall and passes through \(20 \mathrm{~cm}\) of insulation before emerging into the surrounding air. The furnace wall is at \(300^{\circ} \mathrm{C}\), and the air temperature is \(20^{\circ} \mathrm{C}\). Estimate the temperature of the bar tip if the heat transfer coefficient between the rod and air is taken to be \(13 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\).

(i) Derive an expression for the relation between heat loss and temperature difference across the inner and outer surfaces of a hollow sphere, the conductivity of which varies with temperature in manner given by \(k=k_{0}\left[1+a\left(T-T_{0}\right)\right]\), where \(T_{0}\) is a reference temperature. (ii) Find the corresponding result for a hollow cylinder. (iii) Compare the expressions derived for parts (i) and (ii) for the special case of the outside radius becoming infinite. Explain the different values obtained.

A spherical metal tank has a \(2.5 \mathrm{~m}\) outside diameter and is insulated with a \(0.5\) \(\mathrm{m}\)-thick cork layer. The tank contains liquefied gas at \(-60^{\circ} \mathrm{C}\) and the ambient air is at \(20^{\circ} \mathrm{C}\). The inside heat transfer coefficient can be assumed to be large, and the combined convection and radiation outside heat transfer coefficient is estimated to be \(8 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). Atmospheric water vapor diffuses into the cork, and a layer of ice forms adjacent to the tank wall. Determine the thickness of the layer. Assume that the cork thermal conductivity of \(0.06 \mathrm{~W} / \mathrm{m} \mathrm{K}\) is unaffected by the ice and water, but comment on the validity of this assumption.

A \(5 \mathrm{~kW}\) electric heater using Nichrome wire is to be designed to heat air to \(400 \mathrm{~K}\). The maximum allowable wire temperature is \(1500 \mathrm{~K}\), and a minimum heat transfer coefficient of \(600 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) is expected. A variable voltage power supply up to \(130 \mathrm{~V}\) is available. Determine the length of \(1.0 \mathrm{~mm}\)-diameter wire required. Also check the current and voltage. Take the electrical resistivity of Nichrome wire as \(100 \mu \Omega \mathrm{cm}\) and its thermal conductivity as \(30 \mathrm{~W} / \mathrm{m} \mathrm{K}\).

Calcium silicate has replaced asbestos as the preferred insulation for steam lines in power plants. Consider a \(40 \mathrm{~cm}-\mathrm{O} . \mathrm{D} ., 4 \mathrm{~cm}\)-wall-thickness steel steam line insulated with a \(12 \mathrm{~cm}\) thickness of calcium silicate. The insulation is protected from damage by an aluminum sheet lagging that is \(2.5 \mathrm{~mm}\) thick. The steam temperature is \(565^{\circ} \mathrm{C}\) and ambient air temperature in the power plant is \(26^{\circ} \mathrm{C}\). The inside convective resistance is negligible, the outside convective heat transfer coefficient can be taken as \(11 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), and the emittance of the aluminum is \(0.1\). Calculate the rate of heat loss per meter. The thermal conductivity of the pipe wall is \(40 \mathrm{~W} / \mathrm{m} \mathrm{K}\), and the table gives values for Calsilite calcium silicate insulation blocks. $$ \begin{array}{l|ccccc} \mathrm{T}, \mathrm{K} & 500 & 600 & 700 & 800 & 900 \\ \hline \mathrm{k}, \mathrm{W} / \mathrm{m} \mathrm{K} & 0.074 & 0.096 & 0.142 & 0.211 & 0.303 \end{array} $$

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