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A stainless steel (AISI 304) twbe used to transport a chilled pharmaceutical has an inner diameter of \(36 \mathrm{~mm}\) and a wall thickness of \(2 \mathrm{~mm}\). The pharmaceutical and ambient air are at temperatures of \(6^{\circ} \mathrm{C}\) and \(23^{\circ} \mathrm{C}\). respectively, while the corresponding inner and outer convection coefficients are \(400 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\) and \(6 \mathrm{~W} / \mathrm{m}^{2}\). \(K_{\text {, respectively. }}\) (a) What is the heat gain per unit tube length? (b) What is the heat gain per unit length if a \(10-m m-\) thick layer of calcium silicate insulation \(\left(k_{\text {mes }}=\right.\) \(0.050 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) is applied to the tube?

Short Answer

Expert verified
(a) Calculate without insulation: \(q' = U \times 17\). (b) With insulation: Add resistance, then find \(q' = U_{new} \times 17\).

Step by step solution

01

Understanding the Problem

We need to determine the heat gain per unit length of a stainless steel tube and then reassess it after adding insulation. The fundamental principle involves calculating the overall heat transfer coefficient and then using it to find the heat gain using the temperature difference between the inner and outer surfaces.
02

Calculate the Inner Heat Transfer Resistance

The heat transfer resistance inside the tube due to convection is given by: \(R_{i} = \frac{1}{h_i A_i}\). The inner surface area per unit length, \(A_i\), is \(\pi D_i\), where \(D_i = 36 \text{ mm} = 0.036 \text{ m}\). Thus, \(R_i = \frac{1}{400 \times \pi \times 0.036}\).
03

Calculate the Heat Transfer Resistance of the Tube Wall

The heat transfer resistance of the tube wall is \(R_w = \frac{t}{kA_m}\), where \(t = 2 \text{ mm} = 0.002 \text{ m}\) is the thickness, \(k\) is the thermal conductivity of stainless steel (roughly \(15 \text{ W/m.K}\)), and \(A_m\) is the logarithmic mean area: \(A_m = \pi \times (D_i + \frac{t}{2})\). Calculating this, \(R_w = \frac{0.002}{15 \times \pi \times 0.037}\).
04

Calculate the Outer Heat Transfer Resistance

The heat transfer resistance outside the tube is \(R_o = \frac{1}{h_o A_o}\), where \(A_o = \pi D_o\), and \(D_o = 40 \text{ mm} = 0.04 \text{ m}\). Thus, \(R_o = \frac{1}{6 \times \pi \times 0.04}\).
05

Determine the Overall Heat Transfer Coefficient

Summing all resistances gives the total resistance: \(R_{total} = R_i + R_w + R_o\). The overall heat transfer coefficient \(U\) is then \(\frac{1}{R_{total}}\).
06

Calculate Heat Gain without Insulation

The heat gain per unit length is given by \(q' = U \times \Delta T\), where \(\Delta T = 23 - 6\). Substitute the values: \(q' = U \times 17\).
07

Calculate Heat Gain with Insulation

Add the insulation layer by including its thermal resistance: \(R_{ins} = \frac{t_{ins}}{k_{ins} A_{ins}}\), where \(t_{ins} = 10 \text{ mm} = 0.01 \text{ m}\), \(k_{ins} = 0.050 \text{ W/m.K}\), and \(A_{ins} = \pi \times (D_o + 2t_{ins})\). Recalculate \(R_{total}\) with insulation and find the new heat gain \(q' = U_{new} \times 17\).

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Key Concepts

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

Convection
Convection is a crucial concept in heat transfer that deals with the movement of heat through fluids, which can be either liquids or gases. This process involves the transfer of heat from a surface to a fluid or vice versa. There are two types of convection:
  • Natural Convection: Occurs due to temperature differences within the fluid, leading to natural circulation and heat transfer. For example, warm air rising and cool air descending in a room.
  • Forced Convection: Involves external forces like fans or pumps to enhance the fluid movement, thereby improving heat transfer efficiency.
In the context of the problem, convection plays a pivotal role in transferring heat between the fluid inside the tube and the ambient air outside. The convection coefficients given ( therelationship between convective heat transfer and surface area. The higher the coefficient, the more efficiently heat is transferred. This concept is fundamental when calculating thermal resistance and the overall heat transfer coefficient.
Thermal Resistance
Thermal resistance is a measure of an object's ability to resist the flow of heat. It is comparable to how electrical resistance works in a circuit but instead of electricity, it's about heat. The more resistant a material or interface is, the less efficient it is at transferring heat.Thermal resistance is essential in calculating the heat transfer rate. It considers three main types of resistances:
  • Inner Convective Resistance: This occurs within the fluid inside the pipe and is determined using the formula \( R_i = \frac{1}{h_i A_i} \), where \( h_i \) is the inner heat transfer coefficient and \( A_i \) is the inner surface area.
  • Wall Conduction Resistance: The resistance in the pipe wall, calculated as \( R_w = \frac{t}{kA_m} \), where \( t \) is the wall thickness, \( k \) is the material's thermal conductivity, and \( A_m \) is the mean surface area.
  • Outer Convective Resistance: This accounts for heat transfer to the surrounding air, given by \( R_o = \frac{1}{h_o A_o} \).
The sum of these resistances provides the total thermal resistance, which helps in determining the overall efficiency of the heat transfer process.
Thermal Insulation
Thermal insulation is a material applied to reduce heat transfer between objects that are not in thermal equilibrium. By adding a layer of insulation, you can effectively decrease the rate of heat gain or loss. This is especially important in applications where energy efficiency is a priority.In the exercise, a layer of calcium silicate insulation was applied to the steel tube. By doing so, an additional thermal resistance was introduced, calculated using the formula \( R_{ins} = \frac{t_{ins}}{k_{ins} A_{ins}} \), where \( t_{ins} \) represents the insulation thickness, \( k_{ins} \) is the insulation's thermal conductivity, and \( A_{ins} \) is the surface area covered by the insulation.Adding insulation is beneficial for:
  • Reducing energy consumption by minimizing heat loss/gain.
  • Maintaining desired operational temperatures within systems.
  • Protecting against thermal wear and extending the life of equipment.
  • Improving process efficiency and cost-effectiveness.
Understanding how to apply thermal insulation effectively can save energy and costs in industrial processes.
Overall Heat Transfer Coefficient
The overall heat transfer coefficient, denoted as \( U \), is a measure of the total heat transfer capability across multiple layers and interfaces within a system, accounting for conduction and convection resistances. It is crucial for calculating how effectively heat can move from one medium to another.To find the overall heat transfer coefficient, you take the reciprocal of the total thermal resistance: \( U = \frac{1}{R_{total}} \). The total resistance \( R_{total} \) includes:
  • Inner convective resistance \( R_i \).
  • Wall conduction resistance \( R_w \).
  • Outer convective resistance \( R_o \).
  • With insulation, the insulation resistance \( R_{ins} \) is also included.
Understanding \( U \) is essential, as it allows engineers to predict how temperature differentials across a system result in heat transfer rates. With higher values of \( U \), heat can transfer more quickly, which might be desirable in applications like heat exchangers but less so in systems where retaining heat or cold is the goal.

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Most popular questions from this chapter

Consider two long. slender rods of the same diameter but different materials. One end of each rod is attached to a base surface maintained at \(100^{\circ} \mathrm{C}\), while the surfaces of the rods are exposed to ambient air at \(20^{\circ} \mathrm{C}\). By traversing the length of each rod with a thermocoeple, it was observed that the temperatures of the rods were equal at the positions \(x_{A}=0.15 \mathrm{~m}\) and \(x_{\mathrm{A}}=\) \(0.075 \mathrm{~m}\), where \(x\) is measured from the base surface. If the thermal conductivity of rod \(A\) is known to be \(k_{A}=\) \(70 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), determine the value of \(k_{\mathrm{m}}\) for rod \(\mathrm{B}\).

Aluminum fins of triangular profile are attached to a plane wall whose surface temperature is \(250^{\circ} \mathrm{C}\). The fin base thickness is \(2 \mathrm{~mm}\), and its length is \(6 \mathrm{~mm}\). The system is in ambient air at a temperature of \(20^{\circ} \mathrm{C}\), and the surface convection coefficient is \(40 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). (a) What are the fin efficiency and effectiveness? (b) What is the heat dissipated per unit width by a single fin?

A commercial grade cubical freezer, \(3 \mathrm{~m}\) on a side, has a compesite wall consisting of an exterior sheet of 6.35-mm-thick plain carbon steel, an intermediate layer of 100 -rnm-thick cork insulation, and an inner sheet of 6.35-mm-thick aluminum alloy (2024). Adhesive interfaces berween the insulation and the metallic strips are each characterized by a thermul contact resistance of \(R_{L A}^{\prime \prime}=2.5 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). What is the steady-state cooling load that must be maintained by the refrigerawor under conditions for which the outer and inner surface temperatures are \(2^{\circ} 2^{\circ} \mathrm{C}\) and \(-6^{\circ} \mathrm{C}\) respectively?

A hollow aluminem sphere, with an clectrical heater in the center, is used in tests to determine the thermal conductivity of insulating materials. The inner and outer radii of the sphere are \(0.15\) and \(0.18 \mathrm{~m}\), respectively. and testing is done under steady-state conditions with the inner surface of the aluminum maintained at \(250^{\circ} \mathrm{C}\). In a particular test, a spherical shell of insulation is cast on the outer surface of the sphere to a thickness of \(0.12 \mathrm{~m}\). The system is in a roorn for which the air temperature is \(20^{\circ} \mathrm{C}\) and the convection cocfficient at the outer surface of the insulation is \(30 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). If \(80 \mathrm{~W}\) are dissipated by the heater under steady-state conditions, what is the thermal conductivity of the insulation?

As a means of enhancing heat transfer from highperformance logic chips, it is common to attach a heot sink to the chip surface in order to increase the surface area available for convection heat transfer. Because of the ease with which it may be manufactured (by taking orthogonal sawcuts in a block of material), an attmctive option is to use a heat sink consisting of an atmy of square fins of width w on s side. The spacing between adjoining fins would be determined by the width of the sawblade, with the sum of this spacing and the fan width designated as the fin pitch \(S\). The method by which the hent sink is joined to the chip would detesmine the interfacial contact resistance, \(R_{\mathcal{U}^{*}}^{*}\) Consider a square chip of width \(W_{c}=16 \mathrm{~mm}\) and conditions for which cooling is provided by a diclectric liquid with \(T_{w}=25^{\circ} \mathrm{C}\) and \(h=1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat sink is fahricated from copper \((k=400\) Whe \(\mathrm{K})\), and its characteristic dimensions are \(w=0.25\) \(\operatorname{mmn}_{,} S=0.50 \mathrm{~mm}, L_{y}=6 \mathrm{~mm}\), and \(L_{1}=3 \mathrm{~mm}\). The preccribed values of \(w\) and \(S\) represent minima imposed by manufacturing constraints and the need to maintain adequate flow in the passages between fins. (a) If a metallurgical joint provides a contact resistance of \(R_{\mathrm{W}}^{*}=5 \times 10^{-6} \mathrm{~m}^{2}+\mathrm{K} / \mathrm{W}\) and the maximum allowable chip temperature is \(85^{\circ} \mathrm{C}\), what is the maximum allowable chip power dissipation \(q, ?\) Assume all of the heat to be transferred through the heat sink. (b) It may be possible to increase the heat dissipation by increasing \(w\), subject to the constraint that \((S-w) \geq 0.25 \mathrm{~mm}\), and/or increasing \(L_{f}\) (suhject to manufacturing constraints that \(L_{y} \leq 10 \mathrm{~mm}\). Assess the effect of such changes.

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