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Chapter 3: Problem 63

A storage tank consists of a cylindrical section that has a length and inner diameter of \(L=2 \mathrm{~m}\) and \(D_{i}=1 \mathrm{~m}\), 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 \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The tank is used to store heated oil, which maintains the inner 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 electrical power required to maintain the prescribed conditions inside the tank can be calculated using the following steps: 1. Calculate the surface areas of the cylinder and the hemispheres: \(A_{cylinder} = 2\pi r h = 2\pi (0.5)(2) = 2\pi \mathrm{m}^2\) \(A_{hemisphere} = 2\pi r^2 = 2\pi (0.5^2) = 0.5\pi \mathrm{m}^2\) \(A_{hemispheres} = 2 * A_{hemisphere} = \pi \mathrm{m}^2\) \(A_{total} = A_{cylinder} + A_{hemispheres} = 2\pi + \pi = 3\pi \mathrm{m}^2\) 2. Calculate the temperature on the outer surface \(T_o\): \(T_o = T_i - \frac{h(T_s - T_\infty)t}{k} = 400 - \frac{10(400 - 300)(0.02)}{1.4} \approx 386.43\mathrm{~K}\) 3. Finally, calculate the electrical power required \(P_{electric}\): \(P_{electric} = hA_{total}(T_o - T_\infty) = 10(3\pi)(386.43 - 300) \approx 8153.35\mathrm{~W}\) Therefore, the electrical power that must be supplied to the heater submerged in the oil to maintain the prescribed conditions is approximately \(8153.35\mathrm{~W}\).

Step by step solution

01

Calculate the surface area of the cylinder and the hemispheres

To determine the heat transfer from the tank, we will first need to calculate the surface area of the longitudinal cylindrical section and the hemispherical end sections of the tank. The surface area of the cylinder can be calculated using the formula for the lateral surface area of the cylinder: \(A_{cylinder} = 2\pi r h\) where 'r' is the radius of the cylinder and 'h' is the height (or length, L) of the cylinder. The surface area of one hemisphere can be calculated using the formula: \(A_{hemisphere} = 2\pi r^2\) Since there are two hemispherical end sections, the total surface area of the hemispheres will be: \(A_{hemispheres} = 2 * A_{hemisphere}\) Finally, the total surface area of the tank will be the sum of the surface areas of the cylinder and the hemispheres: \(A_{total} = A_{cylinder} + A_{hemispheres}\)
02

Calculate the heat transfer through the tank walls

Now that we have the total surface area of the tank, we need to calculate the heat transfer through the tank walls using conduction and convection principles. For the cylindrical portion and the hemispherical sections, we can use the following equation for heat transfer: \(Q = hA(T_s-T_\infty)\) where 'h' is the convection coefficient, 'A' is the surface area, 'T_s' is the surface temperature, and 'T_鈭�' is the air temperature. We also need to account for the heat transfer through the thickness of the Pyrex. Using the thermal conduction equation, we can calculate the heat transfer through the thickness of the Pyrex: \(Q = kA\frac{(T_i - T_o)}{t}\) where 'k' is the thermal conductivity of Pyrex, 'A' is the surface area, 'T_i' is the temperature on the inner surface, 'T_o' is the temperature on the outer surface, and 't' is the thickness of Pyrex. Combining the two equations, we have: \(Q = hA(T_s-T_\infty) = kA\frac{(T_i-T_o)}{t}\) We can solve for 'T_o' (temperature on the outer surface): \(T_o = T_i - \frac{h(T_s - T_\infty)t}{k}\)
03

Calculate the electrical power

Now that we have calculated the heat transfer through the tank walls, we can determine the electrical power required to maintain the prescribed conditions inside the tank. The electrical power will be equal to the total heat transfer through the tank walls: \(P_{electric} = Q_{total}\) where \(Q_{total} = Q_{cylinder} + Q_{hemispheres}\) Plug in the values and solve for the electrical power required. Using the given values in the problem, we have: Length of the cylinder (L) = 2 m, Diameter of the cylinder (D_i) = 1 m, Ambient air temperature (T_鈭�) = 300 K, Convection coefficient (h) = 10 W/m^2鈭橩, Inner surface temperature (T_i) = 400 K, Pyrex thickness (t) = 20 mm = 0.02 m, Pyrex thermal conductivity (k) = 1.4 W/m鈭橩. Perform the calculations in steps 1 to 3 to determine the electrical power required to maintain the prescribed conditions inside the tank.

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Key Concepts

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

Conduction
Conduction is a fundamental concept in heat transfer where thermal energy flows through a material without the movement of the material itself.
In the context of the storage tank described in the exercise, conduction occurs through the thickness of the Pyrex glass that forms the tank walls.
When heat is conducted, it moves from the hotter interior surface, where the heated oil is, to the cooler exterior surface exposed to the ambient air.
To understand conduction in this problem, we must use the equation for conductive heat transfer:
  • The equation is given by \( Q = kA \frac{(T_i - T_o)}{t} \)
  • Where:
    • \(Q\) is the heat transfer rate
    • \(k\) is the thermal conductivity of Pyrex
    • \(A\) is the surface area through which heat is conducted
    • \(T_i\) and \(T_o\) are the temperatures at the interior and exterior surfaces, respectively
    • \(t\) is the thickness of the material
The thermal conductivity \(k\) represents the glass's ability to conduct heat. Higher \(k\) values mean better heat conduction.
Using the formula, you can solve for the outward temperature or determine the quantity of heat transferred through the Pyrex.
Convection
Convection is another essential mechanism of heat transfer, where heat is carried away by a fluid, such as air or liquid, moving over a surface.
In this exercise, convection occurs on the outer surface of the tank, where ambient air moves past and carries heat away from the Pyrex glass.
This is known as convection heating.The rate of heat transfer in convection can be calculated using the equation:
  • \( Q = hA(T_s - T_鈭�) \)
  • Where:
    • \(Q\) is the heat being transferred
    • \(h\) is the convection coefficient
    • \(A\) is the surface area in contact with the fluid (air, in this case)
    • \(T_s\) is the surface temperature
    • \(T_鈭瀄) represents the ambient air temperature
The convection coefficient \(h\) relates to how effectively the air carries heat away. For instance, \(h = 10 \text{ W/m}^2\text{ K}\) indicates how much heat is transferred per unit area per temperature difference.
To complete this part of the exercise, you'd calculate the convective heat loss using this formula once the tank's surface temperature is known.
Thermal Conductivity
Thermal conductivity is a property of materials that indicates how well they conduct heat.
In this exercise, we are working with Pyrex glass, which has a specific thermal conductivity value.
The thermal conductivity, denoted by \(k\), is essential in determining how much heat passes through the tank walls under steady-state conditions.
  • The value given for Pyrex here is \(k = 1.4 \text{ W/m} \cdot \text{K}\).
  • This value tells us that for each meter thickness of Pyrex, a heat of \(1.4\) Watts per square meter per Kelvin temperature difference will be conducted through the glass.
In practical terms, if Pyrex had higher thermal conductivity, it would transfer heat more efficiently, possibly requiring more insulation to maintain the temperature of the hot oil inside.
In contrast, lower thermal conductivity would mean better insulation properties, reducing heat loss.
Understanding thermal conductivity helps in designing tanks and other systems where heat preservation is crucial.

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Most popular questions from this chapter

The wall of a spherical tank of \(1-m\) diameter contains an exothermic chemical reaction and is at \(200^{\circ} \mathrm{C}\) when the ambient air temperature is \(25^{\circ} \mathrm{C}\). What thickness of urethane foam is required to reduce the exterior temperature to \(40^{\circ} \mathrm{C}\), assuming the convection coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for both situations? What is the percentage reduction in heat rate achieved by using the insulation?

Finned passages are frequently formed between parallel plates to enhance convection heat transfer in compact heat exchanger cores. An important application is in electronic equipment cooling, where one or more air-cooled stacks are placed between heat-dissipating electrical components. Consider a single stack of rectangular fins of length \(L\) and thickness \(t\), with convection conditions corresponding to \(h\) and \(T_{\infty}\). (a) Obtain expressions for the fin heat transfer rates, \(q_{f, o}\) and \(q_{f, L}\), in terms of the base temperatures, \(T_{o}\) and \(T_{L}\). (b) In a specific application, a stack that is \(200 \mathrm{~mm}\) wide and \(100 \mathrm{~mm}\) deep contains 50 fins, each of length \(L=12 \mathrm{~mm}\). The entire stack is made from aluminum, which is everywhere \(1.0 \mathrm{~mm}\) thick. If temperature limitations associated with electrical components joined to opposite plates dictate maximum allowable plate temperatures of \(T_{o}=400 \mathrm{~K}\) and \(T_{L}=350 \mathrm{~K}\), what are the corresponding maximum power dissipations if \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=300 \mathrm{~K} ?\)

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 \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What are the fin efficiency and effectiveness? (b) What is the heat dissipated per unit width by a single fin?

When raised to very high temperatures, many conventional liquid fuels dissociate into hydrogen and other components. Thus the advantage of a solid oxide fuel cell is that such a device can internally reform readily available liquid fuels into hydrogen that can then be used to produce electrical power in a manner similar to Example 1.5. Consider a portable solid oxide fuel cell, operating at a temperature of \(T_{\mathrm{fc}}=800^{\circ} \mathrm{C}\). The fuel cell is housed within a cylindrical canister of diameter \(D=\) \(75 \mathrm{~mm}\) and length \(L=120 \mathrm{~mm}\). The outer surface of the canister is insulated with a low-thermal-conductivity material. For a particular application, it is desired that the thermal signature of the canister be small, to avoid its detection by infrared sensors. The degree to which the canister can be detected with an infrared sensor may be estimated by equating the radiation heat flux emitted from the exterior surface of the canister (Equation 1.5; \(E_{s}=\varepsilon_{s} \sigma T_{s}^{4}\) ) to the heat flux emitted from an equivalent black surface, \(\left(E_{b}=\sigma T_{b}^{4}\right)\). If the equivalent black surface temperature \(T_{b}\) is near the surroundings temperature, the thermal signature of the canister is too small to be detected-the canister is indistinguishable from the surroundings. (a) Determine the required thickness of insulation to be applied to the cylindrical wall of the canister to ensure that the canister does not become highly visible to an infrared sensor (i.e., \(T_{b}-T_{\text {sur }}<5 \mathrm{~K}\) ). Consider cases where (i) the outer surface is covered with a very thin layer of \(\operatorname{dirt}\left(\varepsilon_{s}=0.90\right)\) and (ii) the outer surface is comprised of a very thin polished aluminum sheet \(\left(\varepsilon_{s}=0.08\right)\). Calculate the required thicknesses for two types of insulating material, calcium silicate \((k=0.09 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and aerogel \((k=0.006 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The temperatures of the surroundings and the ambient are \(T_{\text {sur }}=300 \mathrm{~K}\) and \(T_{\infty}=298 \mathrm{~K}\), respectively. The outer surface is characterized by a convective heat transfer coefficient of \(h=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) Calculate the outer surface temperature of the canister for the four cases (high and low thermal conductivity; high and low surface emissivity). (c) Calculate the heat loss from the cylindrical walls of the canister for the four cases.

An air heater consists of a steel tube \((k=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), with inner and outer radii of \(r_{1}=13 \mathrm{~mm}\) and \(r_{2}=16\) \(\mathrm{mm}\), respectively, and eight integrally machined longitudinal fins, each of thickness \(t=3 \mathrm{~mm}\). The fins extend to a concentric tube, which is of radius \(r_{3}=\) \(40 \mathrm{~mm}\) and insulated on its outer surface. Water at a temperature \(T_{\infty, i}=90^{\circ} \mathrm{C}\) flows through the inner tube, while air at \(T_{\infty, o}=25^{\circ} \mathrm{C}\) flows through the annular region formed by the larger concentric tube. (a) Sketch the equivalent thermal circuit of the heater and relate each thermal resistance to appropriate system parameters. (b) If \(h_{i}=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{o}=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the heat rate per unit length? (c) Assess the effect of increasing the number of fins \(N\) and/or the fin thickness \(t\) on the heat rate, subject to the constraint that \(N t<50 \mathrm{~mm}\).
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