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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

One method that is used to grow nanowires (nanotubes with solid cores) is to initially deposit a small droplet of a liquid catalyst onto a flat surface. The surface and catalyst are heated and simultaneously exposed to a higher- temperature, low-pressure gas that contains a mixture of chemical species from which the nanowire is to be formed. The catalytic liquid slowly absorbs the species from the gas through its top surface and converts these to a solid material that is deposited onto the underlying liquid-solid interface, resulting in construction of the nanowire. The liquid catalyst remains suspended at the tip of the nanowire. Consider the growth of a 15 -nm-diameter silicon carbide nanowire onto a silicon carbide surface. The surface is maintained at a temperature of \(T_{s}=2400 \mathrm{~K}\), and the particular liquid catalyst that is used must be maintained in the range \(2400 \mathrm{~K} \leq T_{c} \leq 3000 \mathrm{~K}\) to perform its function. Determine the maximum length of a nanowire that may be grown for conditions characterized by \(h=10^{5} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=8000 \mathrm{~K}\). Assume properties of the nanowire are the same as for bulk silicon carbide.

Measurements show that steady-state conduction through a plane wall without heat generation produced a convex temperature distribution such that the midpoint temperature was \(\Delta T_{o}\) higher than expected for a linear temperature distribution. Assuming that the thermal conductivity has a linear dependence on temperature, \(k=k_{o}(1+\alpha T)\), where \(\alpha\) is a constant, develop a relationship to evaluate \(\alpha\) in terms of \(\Delta T_{o}, T_{1}\), and \(T_{2}\).

The outer surface of a hollow sphere of radius \(r_{2}\) is subjected to a uniform heat flux \(q_{2}^{\prime \prime}\). The inner surface at \(r_{1}\) is held at a constant temperature \(T_{s, 1}\). (a) Develop an expression for the temperature distribution \(T(r)\) in the sphere wall in terms of \(q_{2}^{\prime \prime}, T_{s, 1}, r_{1}, r_{2}\), and the thermal conductivity of the wall material \(k\). (b) If the inner and outer tube radii are \(r_{1}=50 \mathrm{~mm}\) and \(r_{2}=100 \mathrm{~mm}\), what heat flux \(q_{2}^{\prime \prime}\) is required to maintain the outer surface at \(T_{s, 2}=50^{\circ} \mathrm{C}\), while the inner surface is at \(T_{s, 1}=20^{\circ} \mathrm{C}\) ? The thermal conductivity of the wall material is \(k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

The energy transferred from the anterior chamber of the eye through the cornea varies considerably depending on whether a contact lens is worn. Treat the eye as a spherical system and assume the system to be at steady state. The convection coefficient \(h_{o}\) is unchanged with and without the contact lens in place. The cornea and the lens cover one-third of the spherical surface area. Values of the parameters representing this situation are as follows: \(\begin{array}{ll}r_{1}=10.2 \mathrm{~mm} & r_{2}=12.7 \mathrm{~mm} \\\ r_{3}=16.5 \mathrm{~mm} & T_{\infty, o}=21^{\circ} \mathrm{C} \\ T_{\infty \infty, i}=37^{\circ} \mathrm{C} & k_{2}=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ k_{1}=0.35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} & h_{o}=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \\ h_{i}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} & \end{array}\) (a) Construct the thermal circuits, labeling all potentials and flows for the systems excluding the contact lens and including the contact lens. Write resistance elements in terms of appropriate parameters. (b) Determine the heat loss from the anterior chamber with and without the contact lens in place. (c) Discuss the implication of your results.

A very long rod of \(5-\mathrm{mm}\) diameter and uniform thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is subjected to a heat treatment process. The center, 30 -mm-long portion of the rod within the induction heating coil experiences uniform volumetric heat generation of \(7.5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). The unheated portions of the rod, which protrude from the heating coil on either side, experience convection with the ambient air at \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assume that there is no convection from the surface of the rod within the coil. (a) Calculate the steady-state temperature \(T_{o}\) of the rod at the midpoint of the heated portion in the coil. (b) Calculate the temperature of the rod \(T_{b}\) at the edge of the heated portion.
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