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Chapter 3: Problem 67

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 thickncss \(250 \mathrm{~mm}\) is applied to the tank to roduce the heat gain. (a) Determine the radial position in the insulation layer at which the temperature is \(0^{2} \mathrm{C}\) 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
Radial position: intermediate step needed, effects include increased heat gain. Prevent ice by using vapor barriers.

Step by step solution

01

Understand Heat Conduction Through Insulation

We need to calculate the heat transfer through the insulation layer. Using the formula for heat conduction, we have the conduction equation as \[ q = \frac{dT}{dr} \times (k \cdot A) \] where, - \( k \) is the thermal conductivity,- \( A \) is the surface area of the sphere,- and \( \frac{dT}{dr} \) is the temperature gradient.
02

Set Up Heat Transfer in Steady State

Using Fourier鈥檚 law of heat conduction in spherical coordinates, the equation is simplified for the insulation layer:\[ q = \frac{k}{r^2} \times \frac{dT}{dr} \] \Integrate this equation from the inner radius \( r_1 \) (tank surface) to the desired radius \( r \).
03

Apply Boundary Conditions

Given:- Inner surface \( r_1 = 1.5 \text{ m} \) (radius),- Outer surface \( r_2 = 1.75 \text{ m} \) (radius since thickness = 0.25m),- Temperature at \( r_1 = -60^\circ \text{C} \) and \( r_2 = 20^\circ \text{C} \).Use these conditions to calculate the heat transfer \( q \) through the spherical layer which is constant.
04

Find Radial Position for 0掳C

For a temperature of \( 0^\circ \text{C} \), set up:\[ q = -k \frac{A}{r} \left( \frac{0 + 60}{r - r_1} \right) \]Rearrange and solve for \( r \) when \( T = 0^\circ \text{C} \).
05

Effects of Ice Formation

The formation of ice can significantly affect the insulation鈥檚 thermal conductivity by reducing it, which increases heat transfer. This causes more heat to reach the liquified gas, causing efficiency losses. Keeping insulation dry by using vapor barriers can prevent ice formation.

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Key Concepts

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

Spherical Tank Heat Transfer
When dealing with spherical tank heat transfer, it's crucial to understand how heat moves through the tank and its insulation. This is different from heat transfer in flat surfaces because the surface area of the sphere changes as you move outward. This variation in area means that mathematical treatment must include spherical geometry.To determine heat transfer in a spherical tank, we rely on Fourier's law for thermal conduction, expressed in spherical coordinates. The formula:\[ q = rac{k}{r^2} imes rac{dT}{dr} \]offers insights. Here, the heat transfer rate (q) is linked to the thermal conductivity (k), the radius (r), and the temperature gradient ( \frac{dT}{dr} ). The integration of this equation over the tank's geometry helps us find how heat moves from the ambient exterior into the colder interior.Understanding this process helps us to predict points where specific temperatures will occur鈥攍ike the radial point at 0掳C鈥攁nd manage heat gains within the tank.
Thermal Conductivity
Thermal conductivity is an essential factor in determining how well a material can insulate. It is denoted by (k) and gives a measure of a material's ability to conduct heat. In the case of the tank in question, the insulation has a thermal conductivity of 0.06 \text{W/m}a measure of its efficiency in slowing thermal transfer.Materials with lower thermal conductivities are better insulators. Given the high-temperature difference inside and outside the spherical tank, selecting the correct material is critical. In practice, this means assessing different materials not just by (k), but also other factors like cost and durability.When using insulation in designs, engineers often calculate the heat transfer limit (q) by applying: \[ q = rac{dT}{dr} imes (k \鈭栤垨\ a) \]This helps estimate whether chosen insulations will keep the internal temperatures stable, which is especially crucial in industrial and storage applications.
Insulation Moisture Effects
Moisture's impact on insulation is a critical consideration. If the insulation around your spherical tank allows moisture penetration, several problems may arise. When water vapor enters insulation material and condenses, it can form ice, which negates the insulation's effectiveness. Ice formation can reduce the thermal conductivity. As ice has a very different thermal behavior than dry insulation, it increases heat transfer and compromises the insulation layer's performance. This leads to higher heat gain to the liquefied petroleum gas in the tank, causing inefficiencies and potentially dangerous operating conditions. Preventing moisture ingress is a key strategy. Engineers often utilize vapor barriers or sealants around insulation to keep it dry. These measures help maintain the intended thermal properties and energy efficiency, ensuring safety and cost efficiency in the long term.

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Most popular questions from this chapter

A composite wall separates combustion gases at \(2600^{\circ} \mathrm{C}\) from a liquid coolant at \(100^{\circ} \mathrm{C}\), with gas- and liquid-side convection coefficients of 50 and 1000 \(W / \mathrm{m}^{2} \cdot \mathbf{K}\). The wall is composed of a 10 -mm-thick layer of beryllium oxide on the gas side and a 20 -mm-thick slab of stainless steel (AISI 304) on the liquid side. The contact resistance between the oxide and the steel is \(0.05 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). What is the heat loss per unit surface area of the composite? Sketch the temperature distribution from the gas to the liquid.

A thin flat plate of length \(L\) thickness \(t\), and width \(W=L\) is thermally joined to two large heat sinks that are maintained at a temperature \(T_{e}\). The bottom of the plate is well insulated, while the net heat flux to the top surface of the plate is known to have a uniform value of \(q^{*}\) - (a) Derive the differential equation that determines the steady-state temperature distribution \(T(x)\) in the plate. (b) Solve the foregoing equation for the temperature distribution, and obtain an expression for the rate of heat transfer from the plate to the heat sinks.

An air heater may be fabricated by coiling Nichfome wire and passing air in cross flow over the wire. Consider a heater fabricated from wire of diameter \(D=\) \(1 \mathrm{~mm}\), electrical resistivity \(p_{e}=10^{-6} \mathrm{n} \cdot \mathrm{m}\), thernal conductivity \(k=25 \mathrm{~W} / \mathrm{m}=\mathrm{K}\), and emissivity \(\varepsilon=0.2\). The heater is designed to deliver air at a temperatiue of \(T_{m}=50^{\circ} \mathrm{C}\) under flow conditions that provide a convection coefficient of \(h=250 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\) for the wire. The temperature of the housing that encloses the wire and through which the air flows is \(T_{\text {er }}=50^{\circ} \mathrm{C}\). If the maximum allowable temperature of the wire is \(T_{\max }=1200^{\circ} \mathrm{C}\), what is the maximum allowable clextric current \(n\) ? If the maximum available voltage is \(\Delta E=110 \mathrm{~V}\), what is the corresponding length \(L\) of wire that may be used in the heater and the power raing of the heater? Himt: In your solution, assume negligible temperature variations within the wire, but after obtaining the desired results, assess the validity of this arsumption.

A firefighter's protective clothing, referred to as a humout cout, is typically constructed as an ensemble of three layexs separated by air gaps, as shown schematically. Representative dimensions and thermal conductivities for the layers are as follows. \begin{tabular}{lcc} \hline Layer & Thickness (mu) & \(k(\mathrm{~W} / \mathrm{m}-\mathrm{K})\) \\ \hline Shell (s) & \(0.8\) & \(0.047\) \\ Moisture barricr \((\mathrm{mb})\) & \(0.55\) & \(0.012\) \\ Thermal liner (t) & \(3.5\) & \(0.038\) \\ \hline \end{tabular} The air gaps between the layers are \(1 \mathrm{~mm}\) thick, and heat is transferred by conduction and radiation ex. change through the stagnant air. The linearized radiation coefficient for a gap may be approximated as, \(h_{\text {rat }}=\sigma\left(T_{1}+T_{2}\right)\left(T_{1}^{2}+T_{2}^{2}\right)=4 o T_{m z}^{3}\), where \(T_{\text {ax }}\) represents the average temperature of the surfaces comprising the gap, and the radiation flux across the gap may be expressed as \(q_{n d}^{*}=h_{\text {ad }}\left(T_{1}-T_{2}\right)\). (a) Represent the turnout coat by a thermal circuit, labeling all the thermal resistances. Calculate and tabulate the thermal resistances per unit area \(\left(\mathrm{m}^{2}\right.\). \(\mathrm{K} / \mathrm{W})\) for each of the layers, as well as for the conduction and radiation processes in the gaps. Assume that a value of \(T_{\text {es }}=470 \mathrm{~K}\) may be used to approximate the radiation resistance of both gaps. Comment on the relative magnitudes of the resistances. (b) For a pre-flash-over fire environment in which firefighters often work, the typical radiant heat flux on the fire-side of the tumout coat is \(0.25 \mathrm{~W} / \mathrm{cm}^{2}\). What is the outer surface temperature of the tumout coat if the inner surface temperature is \(66^{\circ} \mathrm{C}\), a condition that would result in bum injury?

In a manufacturing process, as transparent film is being bonded to a substrate as shown in the sketch. To cure the bond at a temperature \(T_{\text {ty }}\) a radiant source is used to provide a heat flux \(q_{i}^{\pi}\left(\mathrm{W} / \mathrm{m}^{2}\right)\), all of which is absorbed at the bonded surface. The back of the substrate is maintained at \(T_{1}\) while the free surface of the film is exposed to air at \(T_{m}\) and a convection heat transfer coefficient \(h\). (a) Show the thermal circuit representing the steady-state heat transfer situation, Be sure to label all elements, nodes, and heat rates. Leave in symbolic form. (b) Assume the following conditions: \(T_{w}=20^{\circ} \mathrm{C}, h=\) \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}_{\text {, }}\) and \(T_{1}=30^{\circ} \mathrm{C}\). Calculate the heat flux \(q_{0}^{\prime}\) that is required to maintain the bonded surface at \(T_{0}=60^{\circ} \mathrm{C}\). (c) Compute and plot the required heat flux as a function of the film thickness for \(0 \leq L_{y} \leq 1 \mathrm{~mm}\). (d) If the film is not transparent and all of the radiant heat flux is absorbed at its upper surface, determine the heat flux required to achieve bonding. Plot your results as a function of \(L_{f}\) for \(0 \leq L_{f} \leq 1 \mathrm{~mm}\).
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