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Chapter 3: Problem 76

A spherical tank of \(3-\mathrm{m}\) diameter contains a liquifiedpetroleum gas at \(-60^{\circ} \mathrm{C}\). Insulation with a thermal conductivity of \(0.06 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thickness \(250 \mathrm{~mm}\) is applied to the tank to reduce the heat gain. (a) Determine the radial position in the insulation layer at which the temperature is \(0^{\circ} \mathrm{C}\) when the ambient air temperature is \(20^{\circ} \mathrm{C}\) and the convection coefficient on the outer surface is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) If the insulation is pervious to moisture from the atmospheric air, what conclusions can you reach about the formation of ice in the insulation? What effect will ice formation have on heat gain to the LP gas? How could this situation be avoided?

Short Answer

Expert verified
In summary, the radial position within the insulation layer where temperature is \(0^{\circ} C\) is found to be \(r \approx 1.539 \, \mathrm{m}\). At this position, there is potential for ice formation due to moisture penetration, which can lead to increased heat gain for the LP gas. To avoid this, insulation materials that are resistant to moisture penetration, such as closed-cell foam insulation, should be used.

Step by step solution

01

Define Variables and Constants

Let's define the variables and constants in the problem. \(T_i = -60^{\circ} \mathrm{C}\): temperature of the LPG inside the tank \(T_a = 20^{\circ} \mathrm{C}\): Ambient air temperature \(k = 0.06 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\): Thermal conductivity of the insulation \(t = 0.25 \mathrm{~m}\): Thickness of the insulation layer \(h = 6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\): Convection coefficient of the outer surface \(R_1 = 1.5 \mathrm{~m}\): Inner radius of the insulation layer (half of the diameter) \(R_2 = R_1 + t = 1.75 \mathrm{~m}\): Outer radius of the insulation layer
02

Determine the Temperature Distribution of the Insulation Layer

To determine the temperature distribution in the insulation layer, we use the heat transfer equation for a spherical geometry: \(q = \frac{4 \pi R_1 R_2}{R_2 - R_1}\cdot \frac{1}{1 + \frac{R_1(R_2 - R_1)}{R_2k} + \frac{R_1}{h(R_2 - R_1)}}\cdot (T_i - T_a)\)
03

Calculate the Heat Gain Through the Insulation

Plug in the given values and calculate the heat gain through the insulation. \(q = \frac{4 \pi (1.5)(1.75)}{1.75 - 1.5}\cdot \frac{1}{1 + \frac{(1.5)(1.75 - 1.5)}{(1.75)(0.06)} + \frac{1.5}{(6)(1.75 - 1.5)}}\cdot [(-60) - (20)]\) \(q \approx -864.97 \, \mathrm{W}\)
04

Determine the Radial Position Where the Temperature is \(0^{\circ} C\)

By knowing the heat gain 'q', we can derive the temperature distribution in the insulation as follows: \(T(r) = T_i - \frac{q(R_2 - R_1)}{4 \pi R_1 R_2}\cdot (1 - \frac{R_1}{r})\) Now, we want to know when \(T(r) = 0^{\circ} C\). So, solving for 'r': \(0 = -60 - \frac{(-864.97)(1.75 - 1.5)}{4 \pi (1.5) (1.75)}\cdot (1 - \frac{1.5}{r})\) Once we solve for 'r', we can find the radial position of the insulation layer where the temperature is \(0^{\circ} C\): \(r \approx 1.539 \, \mathrm{m}\)
05

Evaluate the Formation of Ice and Potential Solutions

The radial position where the temperature is \(0^{\circ} C\) indicates that there is a possibility of ice formation in the insulation layer, which could lead to increased heat gain to the LP gas due to ice's higher thermal conductivity compared to the insulation material. Such a problem could be avoided by using insulation materials that do not allow moisture penetration, such as closed-cell foam insulation. In conclusion, we calculated the radial position where the temperature is \(0^{\circ} C\) within the insulation layer, and we considered the possible formation of ice and its effect on heat gain to the LP gas. A potential solution for avoiding ice formation can be using insulation materials that are resistant to moisture penetration.

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Key Concepts

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

Thermal Insulation
Thermal insulation plays a crucial role in controlling the heat transfer between different areas with varying temperatures. It is especially important in industrial contexts such as maintaining the low temperature of liquified petroleum gas (LPG) in tanks. In our exercise example, a spherical tank containing LPG at -60掳C is wrapped with insulation material of specific thickness and thermal conductivity to minimize heat gain.

The effectiveness of this insulation is determined by its thermal resistance, which in turn depends on the material's thickness and thermal conductivity. A higher thermal conductivity means heat can travel through the material more rapidly, reducing its insulating capability. Therefore, selecting appropriate insulation with low thermal conductivity (0.06 W/m路K in the exercise) is vital for maintaining the LPG's low temperature.

Additionally, it's important to consider the geometry of the system. Spherical tanks, like the one in our example, have a unique temperature distribution due to their shape. The curvature affects how heat diffuses through the insulation. For instance, the area of the spherical layers increases with radius, which influences the thermal resistance as you move outward from the center of the sphere.
Temperature Distribution
Understanding temperature distribution within a given system, such as our spherical geometry case, allows us to predict how heat will spread over time. In spherical coordinates, heat transfer isn't uniform across the insulation; it varies based on the radial distance from the center. To determine temperature changes at different points, the heat transfer throughout the spherical layers must be calculated using specific equations relevant to the geometry.

In the exercise solution, we used the heat transfer equation for the spherical system to find that at a certain radial position within the insulation layer, the temperature dropped to 0掳C. This information is critical as it indicates potential areas where condensation and subsequent ice formation could occur within the insulation, changing the insulative properties and potentially threatening the integrity of the LPG storage.
Ice Formation Prevention
Preventing ice formation within insulation is pivotal in maintaining the efficiency and effectiveness of the insulation material. If the temperature within the insulation layer reaches 0掳C, as calculated for a certain radial position in our exercise, moisture present in the insulation could freeze. This frozen moisture, or ice, has a higher thermal conductivity than the original insulation material, which would lead to an increase in heat gain and could compromise the stored LPG's temperature.

To prevent this from happening, it's advisable to use insulation materials that are resistant to moisture penetration, such as closed-cell foam or those with vapor barriers. Another approach is to ensure the insulation's outer surface temperature never reaches the dew point by adjusting the insulation thickness or improving the ambient conditions. These measures can negate the condensation process that leads to ice formation and maintain the insulation's integrity.

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Most popular questions from this chapter

Radioactive wastes \(\left(k_{\mathrm{rw}}=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) are stored in a spherical, stainless steel \(\left(k_{\mathrm{ss}}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) container of inner and outer radii equal to \(r_{i}=0.5 \mathrm{~m}\) and \(r_{o}=0.6 \mathrm{~m}\). Heat is generated volumetrically within the wastes at a uniform rate of \(\dot{q}=10^{5} \mathrm{~W} / \mathrm{m}^{3}\), and the outer surface of the container is exposed to a water flow for which \(h=\) \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=25^{\circ} \mathrm{C}\). (a) Evaluate the steady-state outer surface temperature, \(T_{s, o}\) (b) Evaluate the steady-state inner surface temperature, \(T_{s, i^{*}}\) (c) Obtain an expression for the temperature distribution, \(T(r)\), in the radioactive wastes. Express your result in terms of \(r_{i}, T_{s, i}, k_{\mathrm{rw}}\), and \(\dot{q}\). Evaluate the temperature at \(r=0\). (d) A proposed extension of the foregoing design involves storing waste materials having the same thermal conductivity but twice the heat generation \(\left(\dot{q}=2 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\right)\) in a stainless steel container of equivalent inner radius \(\left(r_{i}=0.5 \mathrm{~m}\right)\). Safety considerations dictate that the maximum system temperature not exceed \(475^{\circ} \mathrm{C}\) and that the container wall thickness be no less than \(t=0.04 \mathrm{~m}\) and preferably at or close to the original design \((t=0.1 \mathrm{~m})\). Assess the effect of varying the outside convection coefficient to a maximum achievable value of \(h=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (by increasing the water velocity) and the container wall thickness. Is the proposed extension feasible? If so, recommend suitable operating and design conditions for \(h\) and \(t\), respectively.

An experimental arrangement for measuring the thermal conductivity of solid materials involves the use of two long rods that are equivalent in every respect, except that one is fabricated from a standard material of known thermal conductivity \(k_{\mathrm{A}}\) while the other is fabricated from the material whose thermal conductivity \(k_{\mathrm{B}}\) is desired. Both rods are attached at one end to a heat source of fixed temperature \(T_{b}\), are exposed to a fluid of temperature \(T_{\infty}\), and are instrumented with thermocouples to measure the temperature at a fixed distance \(x_{1}\) from the heat source. If the standard material is aluminum, with \(k_{\mathrm{A}}=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and measurements reveal values of \(T_{\mathrm{A}}=75^{\circ} \mathrm{C}\) and \(T_{\mathrm{B}}=60^{\circ} \mathrm{C}\) at \(x_{1}\) for \(T_{b}=100^{\circ} \mathrm{C}\) and \(T_{\infty}=25^{\circ} \mathrm{C}\), what is the thermal conductivity \(k_{\mathrm{B}}\) of the test material?

Consider cylindrical and spherical shells with inner and outer surfaces at \(r_{1}\) and \(r_{2}\) maintained at uniform temperatures \(T_{s, 1}\) and \(T_{s, 2}\), respectively. If there is uniform heat generation within the shells, obtain expressions for the steady-state, one-dimensional radial distributions of the temperature, heat flux, and heat rate. Contrast your results with those summarized in Appendix C.

A commercial grade cubical freezer, \(3 \mathrm{~m}\) on a side, has a composite wall consisting of an exterior sheet of \(6.35-\mathrm{mm}\)-thick plain carbon steel, an intermediate layer of \(100-\mathrm{mm}\)-thick cork insulation, and an inner sheet of \(6.35\)-mm-thick aluminum alloy (2024). Adhesive interfaces between the insulation and the metallic strips are each characterized by a thermal contact resistance of \(R_{t, c}^{\prime \prime}=2.5 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). What is the steady-state cooling load that must be maintained by the refrigerator under conditions for which the outer and inner surface temperatures are \(22^{\circ} \mathrm{C}\) and \(-6^{\circ} \mathrm{C}\), respectively?

A 40-mm-long, 2-mm-diameter pin fin is fabricated of an aluminum alloy \((k=140 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). (a) Determine the fin heat transfer rate for \(T_{b}=50^{\circ} \mathrm{C}\), \(T_{\infty}=25^{\circ} \mathrm{C}, h=1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and an adiabatic tip condition. (b) An engineer suggests that by holding the fin tip at a low temperature, the fin heat transfer rate can be increased. For \(T(x=L)=0^{\circ} \mathrm{C}\), determine the new fin heat transfer rate. Other conditions are as in part (a). (c) Plot the temperature distribution, \(T(x)\), over the range \(0 \leq x \leq L\) for the adiabatic tip case and the prescribed tip temperature case. Also show the ambient temperature in your graph. Discuss relevant features of the temperature distribution. (d) Plot the fin heat transfer rate over the range \(0 \leq h \leq 1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for the adiabatic tip case and the prescribed tip temperature case. For the prescribed tip temperature case, what would the
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