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Chapter 3: Problem 70

A spherical vessel used as a reactor for producing pharmaceuticals has a 10 -mm-thick stainless steel wall \((k=17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and an inner diameter of \(1 \mathrm{~m}\). The exterior surface of the vessel is exposed to ambient air \(\left(T_{\infty}=25^{\circ} \mathrm{C}\right)\) for which a convection coefficient of \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) may be assumed. (a) During steady-state operation, an inner surface temperature of \(50^{\circ} \mathrm{C}\) is maintained by energy generated within the reactor. What is the heat loss from the vessel? (b) If a 20 -mm-thick layer of fiberglass insulation \((k=0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is applied to the exterior of the vessel and the rate of thermal energy generation is unchanged, what is the inner surface temperature of the vessel?

Short Answer

Expert verified
The heat loss from the vessel during steady-state operation is approximately \(2042.49\,\text{W}\). After adding a \(20\,\text{mm}\)-thick layer of fiberglass insulation and keeping the rate of thermal energy generation unchanged, the new inner surface temperature of the vessel is approximately \(123.25^{\circ} \mathrm{C}\).

Step by step solution

01

Identify and write the formula for heat transfer through a sphere

The formula for heat transfer through a sphere is given by: \[q = 4\pi R_{1}R_{2} \frac{ (T_{1} - T_{\infty}) }{ ( ({R_{2}} - {R_{1}}) / k ) } \] Where: - \(q\) is the heat loss, - \(R_{1}\) and \(R_{2}\) are the inner and outer radius of the sphere, - \(T_{1}\) is the inner surface temperature, - \(T_{ \infty}\) is the ambient temperature, - \(k\) is the thermal conductivity of the material.
02

Calculate the inner and outer radius of the sphere

We know that the inner diameter of the sphere is \(1\,\text{m}\), and wall thickness is \(10\,\text{mm}\) or \(0.01\,\text{m}\). Thus, we can calculate the inner and outer radius as: \(R_{1} = \frac{1}{2} = 0.5\,\text{m}\) \(R_{2} = R_{1} + 0.01 = 0.51\,\text{m}\)
03

Substitute the values into the formula and calculate the heat loss

Now we can substitute the given values into the formula: \(q = 4\pi (0.5)(0.51) \frac{( 50 - 25 )}{ ( (0.51 - 0.5) / 17 ) }\) After calculating, we get: \(q \approx 2042.49 \, \text{W}\) Therefore, the heat loss from the vessel is approximately \(2042.49 \,\text{W}\). ## Problem (b) ##
04

Updated heat transfer formula with insulation

When insulation is applied, the total resistance to heat transfer consists of two parts, conduction through the stainless steel and the insulation. We can modify the formula for heat transfer through a sphere to take into account the insulation: \[q = 4\pi R_{1}R_{3} \frac{(T_{1}' - T_{\infty})}{( ( { R_{3} } - { R_{1} } ) / k_1 + ( { R_{2} } - { R_{3} } ) / k_2 )}\] Where: - \(q\) is the heat loss - \(R_{1}\) is the inner radius of the vessel - \(R_{2}\) is the outer radius of the vessel - \(R_{3}\) is the outer radius of the insulation - \(T_{1}'\) is the new inner surface temperature - \(T_{ \infty}\) is the ambient temperature - \(k_1\) is the thermal conductivity of the stainless steel - \(k_2\) is the thermal conductivity of the fiberglass insulation.
05

Calculate the outer radius of the insulation

We know that the insulation thickness is \(20\,\text{mm}\) or \(0.02\,\text{m}\). Thus, we can calculate the outer radius of the insulation as: \(R_{3} = R_{2} + 0.02 = 0.53 \,\text{m}\)
06

Apply steady state condition to derive the formula for the new inner surface temperature

For the steady-state condition, the heat loss before and after adding insulation remains the same. By equating the expressions for heat loss in both cases, we can derive the formula for the new inner surface temperature: \( q = 4\pi R_{1}R_{3} \frac{(T_{1}' - T_{\infty})}{( ( { R_{3} } - { R_{1} } ) / k_1 + ( { R_{2} } - { R_{3} } ) / k_2 )} \) Now we need to solve for \(T_{1}'\): \(T_{1}' = T_{\infty} + \frac{ q ( ( { R_{3} } - { R_{1} } ) / k_1 + ( { R_{2} } - { R_{3} } ) / k_{2} ) }{ 4\pi R_{1}R_{3} }\)
07

Calculate the new inner surface temperature

Substitute the known values and calculate the new inner surface temperature: \(T_{1}' = 25 + \frac{ 2042.49 ( ( 0.53 - 0.5 ) / 17 + ( 0.51 - 0.53 ) / 0.040 ) }{ 4\pi (0.5)(0.53) }\) After calculating, we get: \(T_{1}' \approx 123.25^{\circ}\,\mathrm{C}\) Therefore, the new inner surface temperature of the vessel is approximately \(123.25^{\circ}\,\mathrm{C}\).

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a fundamental property that measures a material's ability to conduct heat. It represents how easily heat passes through a material and is denoted by the symbol 'k'. Its units are Watts per meter Kelvin (\(W/m\text{K}\)). Materials with high thermal conductivity, like metals, are excellent heat conductors, while those with low thermal conductivity, such as fiberglass insulation, are used for thermal insulation because they inhibit heat flow.

When solving problems involving heat transfer, it's crucial to understand that the rate of heat loss or gain through a material is directly proportional to its thermal conductivity. So, higher thermal conductivity means more heat transfer and vice versa. The exercise provided presents us with two situations, one where the spherical vessel is only made of stainless steel (\(k = 17 W/m\text{K}\)) and another where it's layered with fiberglass insulation (\(k = 0.040 W/m\text{K}\)), essentially setting up a comparison of thermal conductivities in real-life applications.
Steady-State Operation
Steady-state operation refers to a condition where the variables (such as temperature, pressure, or flow rate) in a process remain constant over time, despite energy or mass being added or removed from the system. It's an assumption that simplifies calculations in many engineering problems.

In the context of the exercise, when the reactor reaches steady-state operation, the heat loss through its walls matches the energy generated inside, maintaining a consistent inner surface temperature. This assumption allows us to use the formula for heat transfer through a spherical vessel to calculate heat loss without considering the time variable, leading to a simplified and solvable equation for steady-state conditions.
Spherical Vessel Insulation
Spherical vessel insulation is implemented to reduce the rate of heat transfer to or from the contents of the vessel. Insulating materials are chosen based on their low thermal conductivity to minimize heat loss or gain. The thicker the insulation, the greater the resistance to heat flow.

In the exercise, insulation is added to a reactor to reduce heat loss. The addition of a 20-mm-thick layer of fiberglass drastically alters the heat transfer dynamics of the vessel. Mathematically, this is depicted as an additional resistance term in the heat transfer formula. A careful balance must be struck to ensure that without changing the rate of thermal energy generation, the new insulation does not lead to undesirable or unsafe internal temperatures, underscoring the importance of calculating the new inner surface temperature after insulation has been applied.

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Most popular questions from this chapter

A plane wall of thickness \(0.1 \mathrm{~m}\) and thermal conductivity \(25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) 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.

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 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The wall is composed of a \(10-\mathrm{mm}\)-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 nanolaminated material is fabricated with an atomic layer deposition process, resulting in a series of stacked, alternating layers of tungsten and aluminum oxide, each layer being \(\delta=0.5 \mathrm{~nm}\) thick. Each tungsten-aluminum oxide interface is associated with a thermal resistance of \(R_{t, i}^{\prime \prime}=3.85 \times 10^{-9} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The theoretical values of the thermal conductivities of the thin aluminum oxide and tungsten layers are \(k_{\mathrm{A}}=1.65 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{T}}=6.10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively. The properties are evaluated at \(T=300 \mathrm{~K}\). (a) Determine the effective thermal conductivity of the nanolaminated material. Compare the value of the effective thermal conductivity to the bulk thermal conductivities of aluminum oxide and tungsten, given in Tables A.1 and A.2. (b) Determine the effective thermal conductivity of the nanolaminated material assuming that the thermal conductivities of the tungsten and aluminum oxide layers are equal to their bulk values.

A steam pipe of \(0.12-\mathrm{m}\) outside diameter is insulated with a layer of calcium silicate. (a) If the insulation is \(20 \mathrm{~mm}\) thick and its inner and outer surfaces are maintained at \(T_{s, 1}=800 \mathrm{~K}\) and \(T_{s, 2}=490 \mathrm{~K}\), respectively, what is the heat loss per unit length \(\left(q^{\prime}\right)\) of the pipe? (b) We wish to explore the effect of insulation thickness on the heat loss \(q^{\prime}\) and outer surface temperature \(T_{s, 2}\), with the inner surface temperature fixed at \(T_{s, 1}=\) \(800 \mathrm{~K}\). The outer surface is exposed to an airflow \(\left(T_{\infty}=25^{\circ} \mathrm{C}\right)\) that maintains a convection coefficient of \(h=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and to large surroundings for which \(T_{\text {sur }}=T_{\infty}=25^{\circ} \mathrm{C}\). The surface emissivity of calcium silicate is approximately \(0.8\). Compute and plot the temperature distribution in the insulation as a function of the dimensionless radial coordinate, \(\left(r-r_{1}\right) /\left(r_{2}-r_{1}\right)\), where \(r_{1}=0.06 \mathrm{~m}\) and \(r_{2}\) is a variable \(\left(0.06

Consider the flat plate of Problem \(3.112\), but with the heat sinks at different temperatures, \(T(0)=T_{o}\) and \(T(L)=T_{L}\), and with the bottom surface no longer insulated. Convection heat transfer is now allowed to occur between this surface and a fluid at \(T_{\infty}\), with a convection coefficient \(h\). (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. (c) For \(q_{o}^{\prime \prime}=20,000 \mathrm{~W} / \mathrm{m}^{2}, T_{o}=100^{\circ} \mathrm{C}, T_{L}=35^{\circ} \mathrm{C}\), \(T_{\infty}=25^{\circ} \mathrm{C}, k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(L=100 \mathrm{~mm}, t=5 \mathrm{~mm}\), and a plate width of \(W=\) \(30 \mathrm{~mm}\), plot the temperature distribution and determine the sink heat rates, \(q_{x}(0)\) and \(q_{x}(L)\). On the same graph, plot three additional temperature distributions corresponding to changes in the following parameters, with the remaining parameters unchanged: (i) \(q_{o}^{\prime \prime}=30,000 \mathrm{~W} / \mathrm{m}^{2}\), (ii) \(h=200\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and (iii) the value of \(q_{o}^{\prime \prime}\) for which \(q_{x}(0)=0\) when \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).
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