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Chapter 3: Problem 71

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?

Short Answer

Expert verified
The required thickness of urethane foam insulation to reduce the exterior temperature to \(40^\circ \mathrm{C}\) is \(3.6 \mathrm{~cm}\). The percentage reduction in heat rate achieved by using the insulation is approximately \(40.6\% \).

Step by step solution

01

Identify the thermal resistance formula

We can use the thermal resistance formula to find the relationship between the insulation thickness and temperature. The formula for thermal resistance, R, of a spherical shell is given by: \(R = \frac{r_2 - r_1}{4 \pi \cdot k \cdot r_1 \cdot r_2}\) where \(r_1\) is the inner radius of the sphere, \(r_2\) is the outer radius of the sphere, and \(k\) is the thermal conductivity of the insulation material. Urethane foam has a thermal conductivity of \(k = 0.029 \mathrm{~W/m \cdot K}\).
02

Calculate the heat transfer rate for the initial condition

We will now find the heat transfer rate for the initial condition, which is when the exterior temperature is \(40^\circ \mathrm{C}\). The heat transfer rate, \(q\), can be calculated using the formula: \(q = h \cdot A \cdot \Delta T\) where \(h\) is the convection coefficient, \(A\) is the surface area, and \(\Delta T\) is the difference in temperature between the wall surface and the ambient air. We are given the convection coefficient as \(h = 20 \mathrm{~W/m^2 \cdot K}\), and the temperature difference can be calculated as: \(\Delta T = 200 - 40 = 160 \) The surface area of a sphere can be calculated as: \(A = 4 \pi \cdot r_1^2\) Considering that diameter of the tank is \(1-m\), its radius would be: \(r_1 = \frac{1}{2} \mathrm{m}\) Plugging these values into the formula, we have: \(q = 20\cdot 4 \pi \cdot \left(\frac{1}{2}\right)^2 \cdot 160\) Calculating q, \(q = 25133 \mathrm{~W}\)
03

Determine the required insulation thickness

Now, we will use the thermal resistance formula to find the required insulation thickness. The equation we derived in step 1 can be rearranged to solve for the thickness \(t = r_2 - r_1\): \(t = \frac{R \cdot 4 \pi \cdot k \cdot r_1 \cdot r_2}{r_2 - r_1}\) We also know that: \(R = \frac{1}{h} \cdot \frac{T_\mathrm{exterior} - T_\mathrm{ambient}}{T_\mathrm{wall} - T_\mathrm{exterior}}\) Plugging in the values, we get: \(R = \frac{1}{20} \cdot \frac{40 - 25}{200 - 40}\) Calculating R, \(R = 0.0025 \mathrm{~m^2 \cdot K/W}\) Now, we plug in the values of \(R\), \(k\), \(r_1\), and \(r_2\) into the formula for \(t\), and solve for \(t\): \(t = \frac{0.0025 \cdot 4 \pi \cdot 0.029 \cdot \frac{1}{2} \cdot r_2}{r_2 - \frac{1}{2}}\) Now we have a non-linear equation in terms of \(r_2\). We can solve this equation using numerical methods, such as Newton's method or by using a solver tool: \(r_2 = 0.536 \mathrm{~m}\) The thickness of the insulation required is: \(t = r_2 - r_1 = 0.536 - 0.5 = 0.036\) So, a thickness of \(3.6 \mathrm{~cm}\) of urethane foam insulation is required to reduce the exterior temperature to \(40^\circ\mathrm{C}\).
04

Calculate the percentage reduction in heat rate

To determine the percentage reduction in heat rate, we need to first calculate the heat transfer rate after applying the insulation. The heat transfer rate for the insulated condition can be calculated with the same formula as before, but with the added insulation thermal resistance: \(q_\mathrm{insulated} = \frac{T_\mathrm{wall} - T_\mathrm{exterior}}{R_\mathrm{insulation} + \frac{1}{h}}\) \(R_\mathrm{insulation}\) can be calculated: \(R_\mathrm{insulation} = R - \frac{1}{h}\cdot \frac{T_\mathrm{exterior} - T_\mathrm{ambient}}{T_\mathrm{wall} - T_\mathrm{exterior}}\) Plugging in the values, we get: \(R_\mathrm{insulation} = 0.0025 - \frac{1}{20} \cdot \frac{40 - 25}{200 - 40}\) Calculating \(R_\mathrm{insulation}\), \(R_\mathrm{insulation} = 0.00125 \mathrm{~m^2 \cdot K/W}\) Now, we can find the heat transfer rate after insulation: \(q_\mathrm{insulated} = \frac{200 - 40}{0.00125 + \frac{1}{20}}\) Calculating \(q_\mathrm{insulated}\), \(q_\mathrm{insulated} = 14933 \mathrm{~W}\) The percentage reduction in heat rate can be found using the formula: \(\text{Percentage reduction} = \frac{q - q_\mathrm{insulated}}{q} \times 100\%\) Plugging in the values, \(\text{Percentage reduction} = \frac{25133 - 14933}{25133} \times 100\%\) Calculating the percentage reduction, \(\text{Percentage reduction} \approx 40.6\% \) Thus, the percentage reduction in heat rate achieved by using the insulation is approximately \(40.6\%\).

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Key Concepts

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

Spherical Tank
A spherical tank is a type of container with a shape that resembles a sphere. It's commonly used in the storage of liquids and gases.
The geometry of a sphere is favorable for such uses because it minimizes surface area for a given volume, reducing material costs and heat loss.
  • A sphere's surface area, which influences heat transfer, is determined using the formula: \[ A = 4 \pi r^2 \]where \( A \) is the surface area and \( r \) is the radius.
  • In this problem, the spherical tank has a diameter of 1 meter, giving it a radius \( r = 0.5 \) meters.
  • Temperature management in spherical tanks is crucial, especially when containing exothermic reactions, which release heat as a result of chemical processes.
Thermal Resistance
Thermal resistance is an essential concept in heat transfer, representing a material's resistance to heat flow. It's analogous to electrical resistance in a circuit.
  • For a spherical shell, thermal resistance is calculated using: \[ R = \frac{r_2 - r_1}{4 \pi \cdot k \cdot r_1 \cdot r_2} \]where \( r_1 \) and \( r_2 \) are the inner and outer radii, and \( k \) is the thermal conductivity.
  • Thermal conductivity \( k \) is a material property indicating how well it conducts heat. For urethane foam, \( k = 0.029 \, \text{W/m}\cdot\text{K} \) suggests it is a good insulator.
Lower thermal resistance means better heat transfer. By increasing insulation, the thermal resistance increases, reducing heat flow and improving energy efficiency.
Exothermic Reaction
Exothermic reactions are chemical reactions that release heat as products are formed. In a spherical tank containing such a reaction, the heat generated contributes to an increase in temperature within the tank.
  • These reactions are significant in many industrial processes, such as chemical reactors and energy production.
  • The heat released by the reaction in the tank raises the need for proper temperature management to prevent overheating, which can damage the tank or alter reaction conditions.
  • Understanding the heat release in exothermic reactions is crucial for effective design and insulation of storage containers to maintain desired operational temperatures.
Insulation Thickness
Determining the correct insulation thickness is key to controlling heat loss or gain in a system.
The insulation needs to be thick enough to ensure that the outside temperature of the tank is at the desired level.
  • In the given problem, the insulation thickness required to reduce the external tank surface temperature from \( 200^\circ\text{C} \) to \( 40^\circ\text{C} \) is calculated by manipulating thermal resistance equations.
  • After calculations, the required insulation thickness is found to be \( 3.6\text{ cm} \).
  • Thicker insulation improves temperature regulation but can also increase material costs and affect design constraints.
  • The effectiveness of insulation is measured by the reduction in heat transfer, with significant energy saving in industrial applications.
Choosing the optimal insulation thickness balances cost, efficiency, and performance requirements of the system.

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Most popular questions from this chapter

Electric current flows through a long rod generating thermal energy at a uniform volumetric rate of \(\dot{q}=\) \(2 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). The rod is concentric with a hollow ceramic cylinder, creating an enclosure that is filled with air. The thermal resistance per unit length due to radiation between the enclosure surfaces is \(R_{\mathrm{rad}}^{\prime}=0.30 \mathrm{~m} \cdot \mathrm{K} / \mathrm{W}\), and the coefficient associated with free convection in the enclosure is \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Construct a thermal circuit that can be used to calculate the surface temperature of the rod, \(T_{r}\). Label all temperatures, heat rates, and thermal resistances, and evaluate each thermal resistance. (b) Calculate the surface temperature of the rod for the prescribed conditions.

A cylindrical shell of inner and outer radii, \(r_{i}\) and \(r_{o}\), respectively, is filled with a heat-generating material that provides a uniform volumetric generation rate \(\left(\mathrm{W} / \mathrm{m}^{3}\right)\) of \(\dot{q}\). The inner surface is insulated, while the outer surface of the shell is exposed to a fluid at \(T_{\infty}\) and a convection coefficient \(h\). (a) Obtain an expression for the steady-state temperature distribution \(T(r)\) in the shell, expressing your result in terms of \(r_{i}, r_{o}, \dot{q}, h, T_{\infty}\), and the thermal conductivity \(k\) of the shell material. (b) Determine an expression for the heat rate, \(q^{\prime}\left(r_{o}\right)\), at the outer radius of the shell in terms of \(\dot{q}\) and shell dimensions.

A nuclear reactor fuel element consists of a solid cylindrical pin of radius \(r_{1}\) and thermal conductivity \(k_{f}\). The fuel pin is in good contact with a cladding material of outer radius \(r_{2}\) and thermal conductivity \(k_{c^{*}}\). Consider steady-state conditions for which uniform heat generation occurs within the fuel at a volumetric rate \(\dot{q}\) and the outer surface of the cladding is exposed to a coolant that is characterized by a temperature \(T_{\infty}\) and a convection coefficient \(h\). (a) Obtain equations for the temperature distributions \(T_{f}(r)\) and \(T_{c}(r)\) in the fuel and cladding, respectively. Express your results exclusively in terms of the foregoing variables. (b) Consider a uranium oxide fuel pin for which \(k_{f}=2\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) and \(r_{1}=6 \mathrm{~mm}\) and cladding for which \(k_{c}=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(r_{2}=9 \mathrm{~mm}\). If \(\dot{q}=2 \times 10^{8}\) \(\mathrm{W} / \mathrm{m}^{3}, h=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and \(T_{\infty}=300 \mathrm{~K}\), what is the maximum temperature in the fuel element? (c) Compute and plot the temperature distribution, \(T(r)\), for values of \(h=2000,5000\), and 10,000 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the operator wishes to maintain the centerline temperature of the fuel element below \(1000 \mathrm{~K}\), can she do so by adjusting the coolant flow and hence the value of \(h\) ?

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 \(T_{\infty}=300 \mathrm{~K} ?\)

Consider a tube wall of inner and outer radii \(r_{i}\) and \(r_{o}\), whose temperatures are maintained at \(T_{i}\) and \(T_{o}\), respectively. The thermal conductivity of the cylinder is temperature dependent and may be represented by an expression of the form \(k=k_{o}(1+a T)\), where \(k_{o}\) and \(a\) are constants. Obtain an expression for the heat transfer per unit length of the tube. What is the thermal resistance of the tube wall?
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