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Chapter 2: Problem 119

A steel heat exchanger tube of \(2 \mathrm{~cm}\) outer diameter is fitted with a steel spiral annular fin of \(4 \mathrm{~cm}\) outer diameter, thickness \(0.4 \mathrm{~mm}\), and wound at a pitch of 3 \(\mathrm{mm}\). If the outside heat transfer coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) on the original bare tube and \(15 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) on the finned tube, determine the reduction in the outside thermal resistance achieved by adding the fins. Take \(k_{\text {steel }}=42 \mathrm{~W} / \mathrm{m} \mathrm{K}\).

Short Answer

Expert verified
The reduction in thermal resistance is the difference between the bare tube resistance and the finned tube resistance, calculated using the specified parameters.

Step by step solution

01

Calculate the Bare Tube Resistance

First, compute the thermal resistance without fins. The formula for thermal resistance of a cylindrical surface with an external heat transfer coefficient \(h\) is given by \( R_{bare} = \frac{1}{hA} \). Here, the area \(A\) of the bare tube with diameter \(d_o\) and length \(L\) is \( A = \pi d_o L \), where \( d_o = 0.02 \text{ m} \). Thus, \( R_{bare} = \frac{1}{20 \cdot \pi \cdot 0.02 \cdot L } \).
02

Calculate the Finned Tube Resistance

Now, determine the resistance of the finned tube. The thermal resistance for a finned surface is \( R_{fin} = \frac{1}{h_{fin}A_{fin} + \eta A_{fin}} \), where \(h_{fin} = 15 \text{ W/m}^2\) K and \(\eta\) is the fin efficiency. First, calculate the total fin area considering both the fins and the bare part: \(A_{fin} = N \cdot A_{per\ fin} + \text{bare area} \). Since \(\text{bare area} = \pi d_o L \) and \(A_{per\ fin} = \pi d_t \cdot \text{thickness} \), determine \(N = \frac{L}{\text{pitch}}\). Finally, substitute all these values back to find \( R_{fin} = \frac{1}{15 \cdot (N \cdot A_{per\ fin} + \pi d_o L )} \).
03

Calculate the Reduction in Thermal Resistance

Subtract the finned resistance from the bare tube resistance to get the reduction in thermal resistance: \(\Delta R = R_{bare} - R_{fin}\). This value will indicate how much more efficient the tube has become with the addition of fins.

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Key Concepts

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

Finned Tube Efficiency
When we talk about finned tube efficiency, we're focusing on how effectively the fins attached to a heat exchanger tube enhance the heat transfer process. Finned tubes are commonly used to increase the surface area available for heat exchange, which can significantly improve the performance of the device.

The efficiency of the fin is denoted as \(\eta\) and refers to the ratio of the actual heat transfer to the maximum possible heat transfer if the entire fin were at the base temperature. High efficiency means the fins are effectively transferring heat, whereas low efficiency indicates otherwise.

There are several factors affecting fin efficiency:
  • Material Thermal Conductivity: Materials with high thermal conductivity, like steel, usually lead to better fin efficiency.
  • Fin Dimensions: The thickness, length, and shape can influence how well the fins conduct and dissipate heat.
  • Fin Arrangement: The spacing and orientation also affect efficiency.
Understanding these factors helps in designing fins that can fulfill the specific requirements of any heat exchange application, leading to better performance and energy savings.
Thermal Resistance Calculation
Thermal resistance is a crucial concept in understanding how effectively heat is transferred through a material or system. It acts like a barrier to heat flow, and in the context of heat exchangers, it typically limits the system's efficiency.

For a bare tube, the thermal resistance can be calculated using the formula:\[R_{bare} = \frac{1}{hA},\]where \(h\) represents the heat transfer coefficient and \(A\) is the surface area available for heat exchange. The lower the thermal resistance, the better the heat transfer.

When fins are added to the tube, the thermal resistance changes since the additional area provided by the fins impacts the heat transfer capabilities. The finned tube resistance \(R_{fin}\) is slightly more complex and incorporates fin efficiency \(\eta\) and the total fin area \(A_{fin}\):\[R_{fin} = \frac{1}{h_{fin}A_{fin} + \eta A_{fin}}.\]This formula shows how both the efficiency and additional area contribute to a reduced total thermal resistance, enhancing the heat exchanger's performance.
Heat Transfer Coefficient
The heat transfer coefficient \(h\) is a vital part of understanding thermal systems. It measures the heat exchanged between a surface and a fluid per unit area due to convection. This coefficient is influences factors like fluid velocity, material properties, and temperature differences.

In the context of a heat exchanger, the heat transfer coefficient determines how effectively heat can be transferred from the surface to the surrounding fluid or vice versa. For instance, a bare tube and a finned tube will have different coefficients due to their differing surface areas and designs.

  • Bare Tube: Before adding fins, the heat transfer coefficient helps determine how well the cylindrical surface alone transfers heat. The initial coefficient, as provided, is \(20 \text{ W/m}^2 \text{K}\).
  • Finned Tube: With fins added, the coefficient changes, often due to the increased surface area and altered fluid dynamics around the fins. The finned tube here has a reduced coefficient of \(15 \text{ W/m}^2 \text{K}\) due to the complexities introduced by the fins.
By understanding the role of the heat transfer coefficient, one can predict and improve the efficiency of the heat exchanger in different operational conditions.

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Most popular questions from this chapter

A \(4 \mathrm{~mm}\)-diameter, \(25 \mathrm{~cm}\)-long aluminum alloy rod has an electric heater wound over the central \(5 \mathrm{~cm}\) length. The outside of the heater is well insulated. The two \(10 \mathrm{~cm}\)-long exposed portions of rod are cooled by an air stream at \(300 \mathrm{~K}\) giving an average convective heat transfer coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). If the power input to the heater is \(10 \mathrm{~W}\), determine the temperature at the ends of the rod. Take \(k=190\) \(\mathrm{W} / \mathrm{m} \mathrm{K}\) for the aluminum alloy.

A \(4 \mathrm{~cm}\)-O.D., \(2 \mathrm{~mm}\)-wall-thickness stainless steel tube is insulated with a 5 \(\mathrm{cm}\)-thick layer of cork. Chilled milk flows through the tube. At a given location the milk temperature is \(5^{\circ} \mathrm{C}\) when the ambient temperature is \(25^{\circ} \mathrm{C}\). If the inside and outside heat transfer coefficients are estimated to be 50 and \(5 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), respectively, calculate the rate of heat gain per meter length of tube.

A \(1 \mathrm{~m}\)-diameter liquid oxygen (LOX) tank is insulated with a \(10 \mathrm{~cm}\)-thick blanket of fiberglass insulation having a thermal conductivity of \(0.022 \mathrm{~W} / \mathrm{m} \mathrm{K}\). The tank is vented to the atmosphere. Determine the boil-off rate if the ambient air is at \(310 \mathrm{~K}\) and the outside convective and radiative heat transfer coefficients are \(3 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) and \(2 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), respectively. The boiling point of oxygen is \(90 \mathrm{~K}\), and its enthalpy of vaporization is \(0.213 \times 10^{6} \mathrm{~J} / \mathrm{kg}\).

Consider steady conduction across a cylindrical or spherical shell. It is sometimes convenient to express the heat flow as $$ \dot{Q}=\frac{k A_{\mathrm{eff}}\left(T_{1}-T_{2}\right)}{r_{2}-r_{1}} $$ that is, in the same form as for a plane slab where \(A_{\text {eff }}\) is an effective crosssectional area. (i) Determine \(A_{\text {eff }}\) for a cylindrical shell. (ii) Repeat for a spherical shell. (iii) If the arithmetic mean area \(A_{m}=\frac{1}{2}\left(A_{1}+A_{2}\right)\) is used instead of \(A_{\text {eff }}\), determine the error in heat flow for \(r_{2} / r_{1}=1.5,3\), and 5 .

A test rig for the measurement of interfacial conductance is used to determine the effect of surface anodization treatment on the interfacial conductance for aluminum-aluminum contact. The specimens themselves form the heat flux meters because they are each fitted with a pair of thermocouples as shown. The heat flux is determined from the measured temperature gradient and a known thermal conductivity of the aluminum of \(185 \mathrm{~W} / \mathrm{m} \mathrm{K}\) as $$ q_{1}=\frac{k\left(T_{1}-T_{2}\right)}{L_{1}} ; \quad q_{2}=\frac{k\left(T_{3}-T_{4}\right)}{L_{2}} ; \quad q=\frac{1}{2}\left(q_{1}+q_{2}\right) $$ and the interfacial temperatures obtained by linear extrapolation. $$ \begin{gathered} T_{5}=\frac{1}{2}\left(T_{1}+T_{2}\right)-\left[\frac{L_{3}+\frac{L_{1}}{2}}{L_{1}}\right]\left(T_{1}-T_{2}\right) ; \\ T_{6}=\frac{1}{2}\left(T_{3}+T_{4}\right)+\left[\frac{L_{4}+\frac{L_{2}}{2}}{L_{2}}\right]\left(T_{3}-T_{4}\right) \end{gathered} $$ The interfacial conductance is then obtained as $$ h_{i}=\frac{q}{\left(T_{5}-T_{6}\right)} $$ The main possible sources of error in the values of \(h_{i}\) so determined are due to uncertainty in temperature measurement and thermocouple locations. Since only temperature differences are involved, the absolute uncertainty in the individual temperature measurements is not of concern; rather it is the relative uncertainties. Previous calibrations of similar type and grade thermocouples indicate that the relative uncertainties are \(\pm 0.2^{\circ} \mathrm{C}\). Also, the techinician who drilled the thermocouple holes and installed the thermocouples estimates an uncertainty of \(\pm 0.5 \mathrm{~mm}\) in the thermocouple junction locations. At a particular pressure, the temperatures recorded are \(T_{1}=338.7 \mathrm{~K}, T_{2}=328.7 \mathrm{~K}, T_{3}=305.3 \mathrm{~K}\), \(T_{4}=295.0 \mathrm{~K}\). (i) Assuming that the uncertainties in temperature measurement and thermocouple location can be treated as random errors, estimate the uncertainty in the interfacial conductance. (ii) In reality, the uncertainties in temperature and location are bias errors (for example, the location of a thermocouple does not vary from test to test). Thus it is more appropriate to determine bounds on the possible error in \(h_{i}\), by considering best and worst cases. Determine these bounds.
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