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Chapter 3: Problem 54

A storage tank consists of a cylindrical secticn that has a length and inner diameter of \(L=2 \mathrm{~m}\) and \(D_{2}=1 \mathrm{~m}\), respectively, and two hemispherical end sections. The tank is constructed from 20-mm-thick glass (Pyrex) and is exposed to ambient air for which the temperature is \(300 \mathrm{~K}\) and the convection coefficient is \(10 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\). The tank is used to store heated oil, which maintains the inser surface at a temperature of \(400 \mathrm{~K}\). Determine the electrical power that must be supplied to a heater submerged in the oil if the prescribed conditions are to be maintained. Radiation effects may be neglected, and the Pyrex may be assumed to have a thermal conductivity of \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Short Answer

Expert verified
The heater must supply approximately 8264 W.

Step by step solution

01

Determine the surface area of the tank

The tank consists of a cylindrical section and two hemispherical ends. The total surface area \( A \) of the cylindrical section is \( A_{cylinder} = \pi D_2 L \), and for the hemispherical section, it is \( A_{hemisphere} = 2\pi \left( \frac{D_2}{2} \right)^2 \) (for the entire sphere, but since there are two hemispheres, it becomes \( 2 \times 2\pi \left( \frac{D_2}{2} \right)^2 \)). Calculate these: \( A_{cylinder} = \pi \times 1 \times 2 = 2\pi \) m虏, \( A_{hemisphere} = 2 \times 2\pi \times \left( \frac{1}{2} \right)^2 = \pi \) m虏. Total area \( A = A_{cylinder} + A_{hemisphere} = 2\pi + \pi = 3\pi \) m虏.
02

Calculate the thermal resistance of the Pyrex shell

Use the formula for thermal resistance \( R_{Pyrex} = \frac{thickness}{k \times A} \), where \( thickness = 0.02 \) m, \( k = 1.4 \) \( \mathrm{W/m \cdot K} \) and \( A = 3\pi \). Thus, \( R_{Pyrex} = \frac{0.02}{1.4 \times 3\pi} \approx 0.0015 \) \( \mathrm{K/W} \).
03

Determine the convection resistance

The convection resistance \( R_{convection} \) is calculated using \( R_{convection} = \frac{1}{h \cdot A} \), where \( h = 10 \) \( \mathrm{W/m}^2 \cdot \mathrm{K} \) and \( A = 3\pi \) \( m^2 \). Therefore, \( R_{convection} = \frac{1}{10 \times 3\pi} \approx 0.0106 \) \( \mathrm{K/W} \).
04

Calculate total thermal resistance

The total thermal resistance \( R_{total} \) is the sum of the individual resistances: \( R_{total} = R_{Pyrex} + R_{convection} \approx 0.0015 + 0.0106 = 0.0121 \) \( \mathrm{K/W} \).
05

Compute the heat transfer rate

Determine the heat transfer rate \( Q \) using the formula \( Q = \frac{T_{inner} - T_{ambient}}{R_{total}} \), where \( T_{inner} = 400 \) \( K \) and \( T_{ambient} = 300 \) \( K \). Thus, \( Q = \frac{400 - 300}{0.0121} \approx 8264 \) \( \mathrm{W} \).

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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 key concept in heat transfer that helps us understand how materials impede the flow of heat. Think of it like how electrical resistance affects the flow of electricity. In heat transfer, thermal resistance quantifies how well a specific material can resist heat flow, keeping the heat contained within or blocking it from entering. This is crucial in engineering applications, such as designing insulation for buildings or tanks, as in our example where Pyrex is used.

When dealing with thermal resistance, we commonly use the formula:
  • \( R = \frac{L}{k \cdot A} \)
Where:
  • \( R \) is the thermal resistance in Kelvin per Watt (K/W),
  • \( L \) is the thickness of the material in meters (m),
  • \( k \) is the thermal conductivity in Watts per meter-Kelvin (W/m路K),
  • \( A \) is the surface area across which heat is being transferred in square meters (m虏).
In the exercise, we calculated the thermal resistance of the Pyrex shell using this formula by plugging in the Pyrex thickness, its thermal conductivity, and the total area of the tank.
Convection
Convection is the process through which heat is transferred by the movement of fluids, which can be liquids or gases. It is why, for example, warm air rises and cool air descends, creating a circulating flow pattern.

In the context of our exercise, convection plays a critical role when we have air (a fluid) interacting with the surface of the tank. The ambient air, with a given temperature, is continuously moving around the tank and affecting its heat retention capabilities.

To calculate the effect of convection on thermal resistance, we use:
  • \( R_{convection} = \frac{1}{h \cdot A} \)
Where:
  • \( R_{convection} \) is the convection resistance in K/W,
  • \( h \) is the convection heat transfer coefficient in W/m虏路K,
  • \( A \) is the surface area in m虏.
In our task, the convection coefficient provided, along with the total area, helps us understand how much the ambient air resists or facilitates the heat transfer from the tank.
Thermal Conductivity
Thermal conductivity is a property of a material that describes its ability to conduct heat. Different materials can either efficiently transfer heat or resist this transfer. It is an inherent material property where higher values indicate better heat conduction capabilities.

For example, metals typically have high thermal conductivity and efficiently conduct heat, while materials like wood or foam have low thermal conductivity and are better at insulating. In our example, Pyrex has a thermal conductivity of \(1.4 \ \mathrm{W/m \cdot K} \), which tells us about its efficiency in conducting heat.

The ability of Pyrex to conduct heat is integral when calculating the thermal resistance of the tank. It forms part of the ratio in the thermal resistance formula. By understanding the thermal conductivity, we can tailor materials to specific applications, ensuring that the correct material choice helps us achieve desired thermal outcomes, whether we want to contain heat like in the storage tank or prevent it from escaping.

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Most popular questions from this chapter

A plane wall of thickness \(0.1 \mathrm{~m}\) and thermal conductiv ity \(25 \mathrm{~W} / \mathrm{m}\), K having uniform volumetric heat generation of \(0.3 \mathrm{MW} / \mathrm{m}^{3}\) is insulated on one side, while the other side is exposed to a fluid at \(92^{\circ} \mathrm{C}\). The convection heat transfer coefficient between the wall and the fluid is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the maximum temperature in the wall.

Radinactive wastes \(\left(k_{1 w}=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) are stored in a spherical, stainless steel \(\left(k_{\mathrm{m}}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) container of inner and outer radii equal to \(r_{i}=0.5 \mathrm{~m}\) and \(r_{e}=0.6 \mathrm{~m}\). Heat is genersed volumetrically within the wastes at a unifom rate of \(\hat{q}=10^{8} \mathrm{~W} / \mathrm{m}^{3}\), and the outer surface of the container is exposed to a water flow for which \(h=\) \(1000 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\) and \(T_{\mathrm{s}}=25^{\circ} \mathrm{C}\).

Consider a plane composite wall that is composed of two naterials of thermal conductivities \(k_{A}=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{B}=0.04 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thicknesses \(L_{A}=10 \mathrm{~mm}\) and \(L_{\mathrm{B}}=20 \mathrm{~mm}\). The contact resistance at the interface between the two materials is known to be \(0.30 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). Material A adjoins a fluid at \(200^{\circ} \mathrm{C}\) for which \(h=10\) W/m虏 \(\cdot \mathrm{K}\), and material B adjoins a fluid at \(40^{\circ} \mathrm{C}\) for which \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What is the rate of heat transfer through a wall that is \(2 \mathrm{~m}\) high by \(2.5 \mathrm{~m}\) wide? (b) Sketch the temperature distribution.

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.

The walls of a refrigerator are typically constructed by sandwiching a layer of insulation between shect metal pancls. Consider a wall made from fiberglass insulation of thermal conductivity \(k_{1}=0.046 \mathrm{~W} / \mathrm{m}-\mathrm{K}\) and thickness \(I_{-}=50 \mathrm{~mm}\) and steel parels, each of thermal conductivity \(k_{p}=60\) W/m \(: \mathrm{K}\) and thickness \(L_{7}=3 \mathrm{~mm}\). If the wall separates refrigerated air at \(T_{x, j}=4^{\circ} \mathrm{C}\) from ambient air at \(T_{2,0}=25^{\circ} \mathrm{C}\), what is the heat gain per unit surface area? Cocfficicnts associated with natural convection at the inner and outer surfaces may be approximated as \(h_{y}=h_{g}=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).
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