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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

Consider a composite wall that includes an 8-mm-thick hardwood siding, 40 -mm by 130 -mm hardwood studs on \(0.65-\mathrm{m}\) centers with glass fiber insulation (paper faced, \(28 \mathrm{~kg} / \mathrm{m}^{3}\) ), and a 12 -mm layer of gypsum (vermiculite) wall board. What is the thermal resistance associated with a wall that is \(2.5 \mathrm{~m}\) high by \(6.5 \mathrm{~m}\) wide (having 10 studs, each \(2.5 \mathrm{~m}\) high)? Assume surfaces normal to the \(x\)-direction are isothermal.

A truncated solid cone is of circular cross section, and its diameter is related to the axial coordinate by an expression of the form \(D=a x^{3 / 2}\), where \(a=1.0 \mathrm{~m}^{-1 / 2}\). The sides are well insulated, while the top surface of the cone at \(x_{1}\) is maintained at \(T_{1}\) and the bottom surface at \(x_{2}\) is maintained at \(T_{2}\). (a) Obtain an expression for the temperature distribution \(T(x)\). (b) What is the rate of heat transfer across the cone if it is constructed of pure aluminum with \(x_{1}=0.075 \mathrm{~m}\), \(T_{1}=100^{\circ} \mathrm{C}, x_{2}=0.225 \mathrm{~m}\), and \(T_{2}=20^{\circ} \mathrm{C}\) ?

The air inside a chamber at \(T_{\infty, i}=50^{\circ} \mathrm{C}\) is heated convectively with \(h_{i}=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) by a 200 -mm-thick wall having a thermal conductivity of \(4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and a uniform heat generation of \(1000 \mathrm{~W} / \mathrm{m}^{3}\). To prevent any heat generated within the wall from being lost to the outside of the chamber at \(T_{\infty, o}=25^{\circ} \mathrm{C}\) with \(h_{o}=5\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), a very thin electrical strip heater is placed on the outer wall to provide a uniform heat flux, \(q_{\sigma^{\prime}}\) (a) Sketch the temperature distribution in the wall on \(T-x\) coordinates for the condition where no heat generated within the wall is lost to the outside of the chamber. (b) What are the temperatures at the wall boundaries, \(T(0)\) and \(T(L)\), for the conditions of part (a)? (c) Determine the value of \(q_{o}^{\prime \prime}\) that must be supplied by the strip heater so that all heat generated within the wall is transferred to the inside of the chamber. (d) If the heat generation in the wall were switched off while the heat flux to the strip heater remained constant, what would be the steady-state temperature, \(T(0)\), of the outer wall surface?

A particular thermal system involves three objects of fixed shape with conduction resistances of \(R_{1}=1 \mathrm{~K} / \mathrm{W}\), \(R_{2}=2 \mathrm{~K} / \mathrm{W}\) and \(R_{3}=4 \mathrm{~K} / \mathrm{W}\), respectively. An objective is to minimize the total thermal resistance \(R_{\text {tot }}\) associated with a combination of \(R_{1}, R_{2}\), and \(R_{3}\). The chief engineer is willing to invest limited funds to specify an alternative material for just one of the three objects; the alternative material will have a thermal conductivity that is twice its nominal value. Which object (1, 2, or 3 ) should be fabricated of the higher thermal conductivity material to most significantly decrease \(R_{\text {tot }}\) ? Hint: Consider two cases, one for which the three thermal resistances are arranged in series, and the second for which the three resistances are arranged in parallel.

A \(0.20\)-m-diameter, thin-walled steel pipe is used to transport saturated steam at a pressure of 20 bars in a room for which the air temperature is \(25^{\circ} \mathrm{C}\) and the convection heat transfer coefficient at the outer surface of the pipe is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What is the heat loss per unit length from the bare pipe (no insulation)? Estimate the heat loss per unit length if a 50 -mm-thick layer of insulation (magnesia, \(85 \%\) is added. The steel and magnesia may each be assumed to have an emissivity of \(0.8\), and the steam-side convection resistance may be neglected. (b) The costs associated with generating the steam and installing the insulation are known to be \(\$ 4 / 10^{9} \mathrm{~J}\) and \(\$ 100 / \mathrm{m}\) of pipe length, respectively. If the steam line is to operate \(7500 \mathrm{~h} / \mathrm{yr}\), how many years are needed to pay back the initial investment in insulation?
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