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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

Consider a coupled shell-in-tube heat exchange device consisting of two identical heat exchangers \(A\) and \(B\). Air flows on the shell side of heat exchanger A, entering at \(T_{h, i, \mathrm{~A}}=520 \mathrm{~K}\) and \(\dot{m}_{h, \mathrm{~A}}=10 \mathrm{~kg} / \mathrm{s}\). Ammonia flows in the shell of heat exchanger \(B\), entering at \(T_{c, i \mathrm{~B}}=280 \mathrm{~K}, m_{c, \mathrm{~B}}=5 \mathrm{~kg} / \mathrm{s}\). The tube-side flow is common to both heat exchangers and consists of water at a flow rate \(\dot{m}_{c, \mathrm{~A}}=\dot{m}_{\hat{h, B}}\) with two tube passes. The UA product increases with water flow rate for heat exchanger A as expressed by the relation \(U A_{\mathrm{A}}=a+\) \(b \dot{m}_{c, \mathrm{~A}}\) where \(a=6000 \mathrm{~W} / \mathrm{K}\) and \(b=100 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). For heat exchanger \(\mathrm{B}, U A_{\mathrm{B}}=1.2 U A_{\mathrm{A}}\). (a) For \(\dot{m}_{c, \mathrm{~A}}=\dot{m}_{k, \mathrm{~B}}=1 \mathrm{~kg} / \mathrm{s}\), determine the outlet air and ammonia temperatures, as well as the heat transfer rate. (b) The plant engineer wishes to fine-tune the heat exchanger performance by installing a variablespeed pump to allow adjustment of the water flow rate. Plot the outlet air and outlet ammonia temperatures versus the water flow rate over the range \(0 \mathrm{~kg} / \mathrm{s} \leq \dot{m}_{c \mathrm{~A}}=m_{\mathrm{h}, \mathrm{B}} \leq 2 \mathrm{~kg} / \mathrm{s}\).

A two-fluid heat exchanger has inlet and outlet temperatures of 65 and \(40^{\circ} \mathrm{C}\) for the hot fluid and 15 and \(30^{\circ} \mathrm{C}\) for the cold fluid. Can you tell whether this exchanger is operating under counterflow or parallelflow conditions? Determine the effectiveness of the heat exchanger.

A plate-fin heat exchanger is used to condense a saturated refrigerant vapor in an air-conditioning system. The vapor has a saturation temperature of \(45^{\circ} \mathrm{C}\), and a condensation rate of \(0.015 \mathrm{~kg} / \mathrm{s}\) is dictated by system performance requirements. The frontal area of the condenser is fixed at \(A_{\mathrm{fr}}=0.25 \mathrm{~m}^{2}\) by installation requirements, and a value of \(h_{f g}=135 \mathrm{~kJ} / \mathrm{kg}\) may be assumed for the refrigerant. (a) The condenser design is to be based on a nominal air inlet temperature of \(T_{c, i}=30^{\circ} \mathrm{C}\) and nominal air inlet velocity of \(V=2 \mathrm{~m} / \mathrm{s}\) for which the manufacturer of the heat exchanger core indicates an overall coefficient of \(U=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the corresponding value of the heat transfer surface area required to achieve the prescribed condensation rate? What is the air outlet temperature? (b) From the manufacturer of the heat exchanger core, it is also known that \(U \propto V^{0 . t}\). During daily operation the air inlet temperature is not controllable and may vary from 27 to \(38^{\circ} \mathrm{C}\). If the heat exchanger area is fixed by the result of part (a), what is the range of air velocities needed to maintain the prescribed condensation rate? Plot the velocity as a function of the air inlet temperature.

A recuperator is a heat exchanger that heats the air used in a combustion process by extracting energy from the products of combustion (the flue gas). Consider using a single-pass, cross-flow heat exchanger as a recuperator. Eighty \((80)\) silicon carbide ceramic tubes \((k=20\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\) of inner and outer diameters equal to 55 and \(80 \mathrm{~mm}\), respectively, and of length \(L=1.4 \mathrm{~m}\) are arranged as an aligned tube bank of longitudinal and transverse pitches \(S_{L}=100 \mathrm{~mm}\) and \(S_{T}=120 \mathrm{~mm}\), respectively. Cold air is in cross flow over the tube bank with upstream conditions of \(V=1 \mathrm{~m} / \mathrm{s}\) and \(T_{c i}=300 \mathrm{~K}\), while hot flue gases of inlet temperature \(T_{\mathrm{h}, \mathrm{I}}=1400 \mathrm{~K}\) pass through the tubes. The tube outer surface is clean, while the inner surface is characterized by a fouling factor of \(R_{f}^{N \prime}=2 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The air and flue gas flow rates are \(\dot{m}_{c}=1.0 \mathrm{~kg} / \mathrm{s}\) and \(m_{\mathrm{h}}=1.05 \mathrm{~kg} / \mathrm{s}\), respectively. As first approximations, (1) evaluate all required air properties at \(1 \mathrm{~atm}\) and \(300 \mathrm{~K},(2)\) assume the flue gas to have the properties of air at \(1 \mathrm{~atm}\) and \(1400 \mathrm{~K}\), and (3) assume the tube wall temperature to be at \(800 \mathrm{~K}\) for the purpose of treating the effect of variable properties on convection heat transfer. (a) If there is a \(1 \%\) fuel savings associated with each \(10^{\circ} \mathrm{C}\) increase in the temperature of the combustion air \(\left(T_{c o}\right)\) above \(300 \mathrm{~K}\), what is the percentage fuel savings for the prescribed conditions? (b) The performance of the recuperator is strongly influenced by the product of the overall heat transfer coefficient and the total surface area, UA. Compute and plot \(T_{c, \infty}\) and the percentage fuel savings as a function of UA for \(300 \leq U A \leq 600 \mathrm{~W} / \mathrm{K}\). Without changing the flow rates, what measures may be taken to increase \(U A^{*}\) ?

In analyzing thermodynamic cycles involving heat exchangers, it is useful to express the heat rate in terms of an overall thermal resistance \(R_{r}\) and the inlet temperatures of the hot and cold fluids, $$ q=\frac{\left(T_{\mathrm{h}, i}-T_{c, j}\right)}{R_{t}} $$ The heat transfer rate can also be expressed in terms of the rate equations, $$ q=U A \Delta T_{\mathrm{lm}}=\frac{1}{R_{\mathrm{lm}}} \Delta T_{\mathrm{lm}} $$ (a) Derive a relation for \(R_{\mathrm{lm}} / R_{\mathrm{t}}\) for a parallel- \(d w\) heat exchanger in terms of a single dimensionless parameter \(B\), which does not involve any fluid temperatures but only \(U, A, C_{\mathrm{h}}, C_{c}\) (or \(C_{\min }, C_{\max }\) ). (b) Calculate and plot \(R_{\operatorname{lm}} / R_{T}\) for values of \(B=0.1\), \(1.0\), and 5.0. What conclusions can be drawn from the plot?

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