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A steel tube \((k=50 \mathrm{~W} / \mathrm{m}\) - \(\mathrm{K})\) of inner and outer diameters \(D_{i}=20 \mathrm{~mm}\) and \(D_{a}=26 \mathrm{~mm}\), respectively, is used to transfer heat from hot gases flowing over the tube \(\left(h_{\mathrm{h}}=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\) to cold water flowing through the tube \(\left(h_{c}=8000 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\right)\). What is the cold-side overall heat transfer coefficient \(U_{c}\) ? To enhance heat transfer, 16 straight fins of rectangular profile are installed longitudinally along the outer surface of the tube. The fins are equally spaced around the circumference of the tube, each having a thickness of \(2 \mathrm{~mm}\) and a length of \(15 \mathrm{~mm}\). What is the corresponding overall heat transfer coefficient \(U_{c}\) ?

Short Answer

Expert verified
To find the cold-side overall heat transfer coefficient without fins, \(U_c\), first calculate the thermal resistances \(R_h\), \(R_{cond}\), and \(R_c\). Then, find the total thermal resistance, \(R_{tot}\), and use the equation \( U_c = \frac{1}{A_c R_{tot}}\) to obtain \(U_c\). For the case with fins, calculate fin efficiency, \( \eta \), and overall surface efficiency, \( \eta_{o} \). Then, calculate the new thermal resistance, \( R'_c = \frac{R_c}{\eta_{o}} \), and the new total thermal resistance, \(R'_\mathrm{tot}\). Finally, calculate the overall heat transfer coefficient with fins, \( U'_c = \frac{1}{A_c R'_\mathrm{tot}} \).

Step by step solution

01

Calculate thermal resistances

First, we need to calculate the thermal resistances for convection on the hot side, convection on the cold side, and conduction through the tube wall. The equations for these resistances are: 1. \( R_h = \frac{1}{h_h A_h} \) 2. \( R_{cond} = \frac{\ln \frac{D_a}{D_i}}{2 \pi k L} \) 3. \( R_c = \frac{1}{h_c A_c} \)
02

Calculate the overall heat transfer coefficient without fins

Now that we have the thermal resistances, we can calculate the overall heat transfer coefficient without fins using the equations from the analysis: 1. Calculate total thermal resistance, \(R_{tot} = R_h + R_{cond} + R_c\) 2. Calculate overall heat transfer coefficient, \( U_c = \frac{1}{A_c R_{tot}}\)
03

Determine the fin efficiency and overall surface efficiency

For the fins, we need to first find the fin efficiency (\( \eta \)) and overall surface efficiency (\( \eta_{o} \)). 1. Calculate \( m = \sqrt{\frac{2 h_{f}}{k_f t_f}} \) 2. Calculate fin efficiency (\( \eta \)) using the equation \( \eta = \frac{\tanh ( mL ) }{ mL }\) 3. Calculate the overall surface efficiency (\( \eta_{o} \)) For overall surface efficiency, we have to consider that 16 fins are installed, and the finned surface area and unfinned surface area are important parameters. Therefore, the equation for the overall surface efficiency is: \( \eta_{o} = \frac{\text{finned surface area} * \eta + \text{unfinned surface area}} {\text{total outer surface area}} \)
04

Calculate the new thermal resistance and overall heat transfer coefficient with fins

Now that we have the overall surface efficiency, the new thermal resistance and overall heat transfer coefficient can be calculated as follows: 1. Calculate new thermal resistance, \( R'_c = \frac{R_c}{\eta_{o}} \) 2. Calculate new \( R'_\mathrm{tot} = R_h + R_{cond} + R'_c \) 3. Calculate overall heat transfer coefficient with fins, \( U'_c = \frac{1}{A_c R'_\mathrm{tot}} \) The obtained value of \( U'_c \) is the overall heat transfer coefficient with fins installed along the outer surface of the tube.

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Key Concepts

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

Thermal Resistances
To understand how heat flows through a material, it is essential to grasp the concept of thermal resistance. This is similar to electrical resistance, but instead of opposing electrical flow, it opposes heat flow.

Thermal resistance can be broken into three parts when dealing with a pipe transferring heat:
  • Convection thermal resistance on the hot side, often denoted as \(R_h\), which is the resistance to heat flow from hot gases to the pipe. This is calculated as \(R_h = \frac{1}{h_h A_h}\).
  • Conduction thermal resistance through the tube wall, \(R_{cond}\), representing resistance through the pipe's material. Calculated using \(R_{cond} = \frac{\ln \frac{D_a}{D_i}}{2 \pi k L}\), it shows how the material itself may slow down heat transfer.
  • Convection thermal resistance on the cold side, \(R_c\), referring to heat flow from the pipe to the cold water inside. It is expressed as \(R_c = \frac{1}{h_c A_c}\).
Together, these resistances provide the total thermal resistance \(R_{tot}\), which can be summed up as \(R_{tot} = R_h + R_{cond} + R_c\). Calculating these correctly helps in determining the overall heat transfer capability of the system.
Heat Transfer Enhancement
In many engineering systems, merely calculating the natural rate of heat transfer might not suffice, especially under industrial demands for efficiency. Here, enhancing the heat transfer process becomes vital.

Heat transfer enhancement aims to increase the rate of heat transfer without altering the overall flow pathway. One common method is through using fins. By adding fins to a system, especially a steel tube as described in the original exercise, more area becomes available for heat exchange.
  • The fins are designed to extend the surface area, allowing more heat from the hot gases to be transferred to the cold fluid through the tube.
  • When installed, fins align longitudinally along the outer surface, effectively increasing the surface area available for convection.
Therefore, the addition of fins significantly reduces thermal resistance, improving the overall efficiency of the heat transfer process. This enhancement ensures that energy is more effectively utilized within the system, important for both process consistency and energy savings.
Fin Efficiency
Fin efficiency is a measure of how effectively a fin transfers heat relative to a hypothetical perfect fin, which loses its entire surface area potential to the heat transfer process.

When a fin is added to increase surface area, not all parts of the fin contribute equally due to various factors, such as temperature gradients along its length. This is where calculating the fin efficiency becomes crucial. It is defined mathematically as\[ \eta = \frac{\tanh(mL)}{mL}\]Here, \(m\) is derived from the expression \(m = \sqrt{\frac{2 h_f}{k_f t_f}}\), where \(h_f\) is the heat transfer coefficient for the fin, \(k_f\) is the thermal conductivity of the fin material, and \(t_f\) is the thickness of each fin.

  • A perfect fin (\(\eta = 1\)) would use its entire surface area without any temperature gradient. In practice, no fin achieves this.
  • The efficiency factor helps engineers determine just how much of the extra surface area actually contributes to increased heat transfer.
By accurately calculating fin efficiency, systems can be better optimized for thermal performance, ensuring designs make the best use of materials and comply with energy standards.
Surface Efficiency
Surface efficiency differs slightly from fin efficiency as it takes into account the combined efficiency of both finned and unfinned surfaces. The overall surface efficiency is crucial for comparing real-world heat transfer performance to theoretical expectations. This metric demonstrates how effective the entire surface area is in facilitating heat transfer.

To calculate overall surface efficiency \(\eta_o\), it's important to understand the relationship between different areas involved.
  • Consider the finned surface area, which is enhanced by fin efficiency \(\eta\).
  • Include the unfinned surface area, which does not benefit from fins.
  • Combine these to reflect the actual performance over the total outer surface area of the tube.
The formula is given by:\[\eta_o = \frac{\text{finned surface area} \times \eta + \text{unfinned surface area}}{\text{total outer surface area}}\]Calculating surface efficiency allows for a comprehensive understanding of the system’s effectiveness, guiding engineers in making informed decisions on configurations or materials to further enhance heat transfer for particular applications.

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Most popular questions from this chapter

A process fluid having a specific heat of \(3500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and flowing at \(2 \mathrm{~kg} / \mathrm{s}\) is to be cooled from \(80^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) with chilled water, which is supplied at a temperature of \(15^{\circ} \mathrm{C}\) and a flow rate of \(2.5 \mathrm{~kg} / \mathrm{s}\). Assuming an overall heat transfer coefficient of \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), calculate the required heat transfer areas for the following exchanger configurations: (a) parallel flow, (b) counterflow, (c) shell-and-tube, one shell pass and two tube passes, and (d) cross-flow, single pass, both fluids unmixed. Compare the results of your analysis. Your work can be reduced by using IHT.

A shell-and-tube heat exchanger is to heat \(10,000 \mathrm{~kg} / \mathrm{h}\) of water from 16 to \(84^{\circ} \mathrm{C}\) by hot engine oil flowing through the shell. The oil makes a single shell pass, entering at \(160^{\circ} \mathrm{C}\) and leaving at \(94^{\circ} \mathrm{C}\), with an average heat transfer coefficient of \(400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The water flows through 11 brass tubes of \(22.9-\mathrm{mm}\) inside diameter and 25.4-mm outside diameter, with each tube making four passes through the shell. (a) Assuming fully developed flow for the water, determine the required tube length per pass. (b) For the tube length found in part (a), plot the effectiveness, fluid outlet temperatures, and water-side convection coefficient as a function of the water flow rate for \(5000 \leq m_{c} \leq 15,000 \mathrm{~kg} / \mathrm{h}\), with all other conditions remaining the same.

A single-pass, cross-flow heat exchanger uses hot exhaust gases (mixed) to heat water (unmixed) from 30 to \(80^{\circ} \mathrm{C}\) at a rate of \(3 \mathrm{~kg} / \mathrm{s}\). The exhaust gases, having thermophysical properties similar to air, enter and exit the exchanger at 225 and \(100^{\circ} \mathrm{C}\), respectively. If the overall heat transfer coefficient is \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), estimate the required surface area.

Thin-walled aluminum tubes of diameter \(D=10 \mathrm{~mm}\) are used in the condenser of an air conditioner. Under normal operating conditions, a convection coefficient of \(h_{i}=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is associated with condensation on the inner surface of the tubes, while a coefficient of \(h_{o}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained by airflow over the tubes. (a) What is the overall heat transfer coefficient if the tubes are unfinned? (b) What is the overall heat transfer coefficient based on the inner surface, \(U_{i}\), if aluminum annular fins of thickness \(t=1.5 \mathrm{~mm}\), outer diameter \(D_{o}=20 \mathrm{~mm}\), and pitch \(S=3.5 \mathrm{~mm}\) are added to the outer surface? Base your calculations on a 1-m-long section of tube. Subject to the requirements that \(t \geq 1 \mathrm{~mm}\) and \((S-t) \geq 1.5 \mathrm{~mm}\), explore the effect of variations in \(t\) and \(S\) on \(U_{i}\). What combination of \(t\) and \(S\) would yield the best heat transfer performance?

In a dairy operation, milk at a flow rate of \(250 \mathrm{~L} / \mathrm{h}\) and a cow-body temperature of \(38.6^{\circ} \mathrm{C}\) must be chilled to a safe-to-store temperature of \(13^{\circ} \mathrm{C}\) or less. Ground water at \(10^{\circ} \mathrm{C}\) is available at a flow rate of \(0.72 \mathrm{~m}^{3} / \mathrm{h}\). The density and specific heat of milk are \(1030 \mathrm{~kg} / \mathrm{m}^{3}\) and \(3860 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. (a) Determine the UA product of a counterflow heat exchanger required for the chilling process. Determine the length of the exchanger if the inner pipe has a 50 -mm diameter and the overall heat transfer coefficient is \(U=1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) Determine the outlet temperature of the water. (c) Using the value of \(U A\) found in part (a), determine the milk outlet temperature if the water flow rate is doubled. What is the outlet temperature if the flow rate is halved?

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