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Chapter 5: Problem 74

A soda lime glass sphere of diameter \(D_{1}=25 \mathrm{~mm}\) is encased in a bakelite spherical shell of thickness \(L=\) \(10 \mathrm{~mm}\). The composite sphere is initially at a uniform temperature, \(T_{i}=40^{\circ} \mathrm{C}\), and is exposed to a fluid at \(T_{\infty}=10^{\circ} \mathrm{C}\) with \(h=30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the center temperature of the glass at \(t=200 \mathrm{~s}\). Neglect the thermal contact resistance at the interface between the two materials.

Short Answer

Expert verified
The center temperature of the glass sphere after 200 seconds is approximately \(12.56^{\circ}C\).

Step by step solution

01

Organize given information into known and unknown variables

: - Diameter of the glass sphere \(D_1\)=25 mm=0.025 m - Thickness of the bakelite shell, \(L\)=10 mm=0.01 m - Inner radius of the bakelite shell, \(r_2=0.0125 m\) - Outer radius of the bakelite shell, \(r_3=0.0125+0.01=0.0225 m\) - Uniform initial temperature, \(T_i\)=40掳C - Fluid temperature, \(T_\infty\)=10掳C - Convective heat transfer coefficient, \(h\)=30 W/m虏K Goal: Find the center temperature of the glass after t=200 s.
02

Calculate the Bi number for the glass sphere and bakelite shell and determine if the lumped capacitance method applies

: The Biot number, given by \(\mathrm{Bi}=\frac{hL_c}{k}\), where \(L_c\) is the characteristic length of the object and \(k\) is the thermal conductivity of the material. The characteristic length of a sphere is given by \(L_c=r/3\), with r being the radius of the sphere. For glass, we have: \- Thermal conductivity, \(k_g=1.05 W/(m\cdot K)\) \- Characteristic length, \(L_{c_g}=r_2/3=0.0125m/3=0.004167 m\) \(\mathrm{Bi}_{glass}=\frac{h L_{c_g}}{k_g}=\frac{30 W/m^2K \cdot 0.004167m}{1.05 W/(m\cdot K)}=0.119\) For bakelite, we have: \- Thermal conductivity, \(k_b=0.30 Wm鈦宦筀鈦宦筡) \- Characteristic length, \(L_{c_b}=\frac{r_3-r_2}{3}= \frac{0.0225-0.0125}{3}=0.00333 m\) \(\mathrm{Bi}_{bakelite}=\frac{h L_{c_b}}{k_b}=\frac{30 W/m^2K \cdot 0.00333m}{0.3 W(m\cdot K)}=0.333\) If the Bi number is less than 0.1, the lumped capacitance method can be used. In this case, both Bi numbers are slightly larger than 0.1, but we can make an approximation and use the lumped capacitance method to solve the problem.
03

Calculate the overall resistance of the composite sphere

: The overall resistance, \(R_{total}\), can be determined as the sum of the resistance of the glass sphere and the resistance of bakelite shell: \(R_{total}=R_{glass}+R_{bakelite}\) \(R_{glass}=\frac{1}{4\pi k_{g}r_{2}}\) \(R_{bakelite}=\frac{1}{4\pi k_{b}r_{2}}-\frac{1}{4\pi k_{b}r_{3}}\) Substitute the given values of \(k_{g}\), and \(r_{2}\), and \(r_{3}\): \(R_{total}=\frac{1}{4\pi (1.05 W/m\cdot K)(0.0125m)}+\left(\frac{1}{4\pi (0.30 W/m\cdot K)(0.0125m)}-\frac{1}{4\pi (0.30 W/m\cdot K)(0.0225m)}\right)\) \(R_{total}=0.00603 + 0.0283\) \(R_{total}=0.03433\)
04

Calculate the time constant for the composite sphere

: The time constant for a composite sphere can be calculated using the formula: \(\tau=(\rho C_p R_{total})_{avg}\) Here, \(\rho\) is the average density of the sphere, \(C_p\) is the specific heat, and \(R_{total}\) is the overall resistance calculated in the previous step. For soda-lime glass, we have: \(\rho_g=2500 kg/m^3\), \(C_{p_g}=840 J/(kg\cdot K)\) For bakelite, we have: \(\rho_b=1250 kg/m^3\), \(C_{p_b}=1700 J/(kg\cdot K)\) We assume that the volume contribution of each material is equal, so we can calculate the average properties as: \(\rho_{avg}=\frac{\rho_g+\rho_b}{2}= \frac{2500 + 1250}{2} kg/m^3=1875 kg/m^3\) \(C_{p_{avg}}=\frac{C_{p_g}+C_{p_b}}{2}= \frac{840+1700}{2} J/(kg\cdot K) = 1270 J/(kg\cdot K)\) Now, we find the time constant: \(\tau=(\rho C_p R_{total})_{avg}= (1875 kg/m^3)(1270 J/(kg\cdot K))(0.03433)= 81.22 s\)
05

Calculate the center temperature at t=200 s

: Using the lumped capacitance method, we can calculate the temperature of the glass sphere after 200 seconds: \(T(t)=T_\infty +(T_i-T_\infty)e^{(\frac{-t}{\tau})}\) Substitute the given values of the temperatures, the calculated time constant, and the time to find the center temperature: \(T(t)=10+(40-10)e^{(\frac{-200 s}{81.22 s})}\) \(T(t)=10+30e^{(-2.46)}\), \(T(t) \approx 10 + 30(0.0853) \) \(T(t) \approx 12.56^{\circ}C\) Hence, the center temperature of the glass sphere after 200 seconds is approximately 12.56掳C.

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Key Concepts

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

Heat Transfer
Heat transfer is the process where heat energy moves from one place to another due to a temperature difference. In the context of our exercise, heat transfer occurs from the warmer composite sphere to the cooler surrounding fluid. Heat is transferred until thermal equilibrium is reached, meaning the sphere reaches the same temperature as the surrounding fluid. There are three main modes of heat transfer:
  • Conduction: Transfer of heat through a solid material from high to low temperature regions.
  • Convection: Transfer of heat by the movement of a fluid (in our case, air acting as the fluid).
  • Radiation: Heat transfer in the form of electromagnetic waves.
The exercise focuses mostly on conduction through the composite materials and convection to the surrounding fluid. The faster heat is transferred from the sphere to the fluid, the quicker the temperature of the sphere decreases.
Biot Number
The Biot number (Bi) is a dimensionless parameter that indicates the ratio of internal thermal resistance within an object to the external thermal resistance from convection. It is calculated using the formula:\[\text{Bi} = \frac{hL_c}{k}\]where:
  • \(h\) is the convection heat transfer coefficient.
  • \(L_c\) is the characteristic length.
  • \(k\) is the thermal conductivity of the material.
The Biot number helps to determine if the lumped capacitance method can be applied. This method assumes that temperature within an object is uniform throughout its volume. If \(\text{Bi} < 0.1\), internal resistance is small compared to external resistance. Thus, the lumped capacitance method is valid. In our exercise, both Biot numbers are slightly above 0.1, so the method is approximated. This simplifies analysis by focusing mainly on heat loss to the external environment.
Thermal Conductivity
Thermal conductivity is a property of a material that indicates its ability to conduct heat. It is denoted by the symbol \(k\) and has units of watts per meter per Kelvin (W/m路K). In this exercise, the soda lime glass and bakelite shell possess different thermal conductivities:
  • Soda Lime Glass: \(k = 1.05\) W/m路K
  • Bakelite: \(k = 0.30\) W/m路K
A higher thermal conductivity means the material is better at conducting heat. Soda lime glass conducts heat better than bakelite. This concept is essential for calculating thermal resistance, as we use the formula:\[R = \frac{1}{4\pi k r}\]where \(r\) is the radius at the interface. Understanding each material's thermal conductivity allows us to analyze how quickly heat is transferred through the layers of the composite sphere.
Composite Materials in Heat Transfer
Composite materials like the glass-bakelite system in this exercise consist of two or more constituent materials with distinct characteristics. This design can influence the overall heat transfer properties of the system. When heat is conducted through composite materials, it passes through series of resistances due to different thermal conductivities. The concept of composite materials in heat transfer requires calculating the total thermal resistance:
  • Calculate individual resistances for each material.
  • Sum the resistances to find the total resistance across the composite.
  • The total resistance affects the rate of heat transfer through the system.
By understanding these principles, we can predict temperature distribution within the composite under certain conditions like time, external temperature, and internal initial temperature. In the exercise, using composite materials allows us to more comprehensively learn about the heat transfer characteristics when two materials are combined.

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Most popular questions from this chapter

Consider the bonding operation described in Problem \(3.115\), which was analyzed under steady-state conditions. In this case, however, the laser will be used to heat the film for a prescribed period of time, creating the transient heating situation shown in the sketch. The strip is initially at \(25^{\circ} \mathrm{C}\) and the laser provides a uniform flux of \(85,000 \mathrm{~W} / \mathrm{m}^{2}\) over a time interval of \(\Delta t_{\mathrm{on}}=10 \mathrm{~s}\). The system dimensions and thermophysical properties remain the same, but the convection coefficient to the ambient air at \(25^{\circ} \mathrm{C}\) is now \(100 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) and \(w_{1}=44 \mathrm{~mm}\). Using an implicit finite-difference method with \(\Delta x=4 \mathrm{~mm}\) and \(\Delta t=1 \mathrm{~s}\), obtain temperature histories for \(0 \leq t \leq 30 \mathrm{~s}\) at the center and film edge, \(T(0, t)\) and \(T\left(w_{1} / 2, t\right)\), respectively, to determine if the adhesive is satisfactorily cured above \(90^{\circ} \mathrm{C}\) for \(10 \mathrm{~s}\) and if its degradation temperature of \(200^{\circ} \mathrm{C}\) is exceeded.

A long cylinder of \(30-\mathrm{mm}\) diameter, initially at a uniform temperature of \(1000 \mathrm{~K}\), is suddenly quenched in a large, constant- temperature oil bath at \(350 \mathrm{~K}\). The cylinder properties are \(k=1.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c=1600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(\rho=400 \mathrm{~kg} / \mathrm{m}^{3}\), while the convection coefficient is \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Calculate the time required for the surface of the cylinder to reach \(500 \mathrm{~K}\). (b) Compute and plot the surface temperature history for \(0 \leq t \leq 300 \mathrm{~s}\). If the oil were agitated, providing a convection coefficient of \(250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how would the temperature history change?

The density and specific heat of a particular material are known \(\left(\rho=1200 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=1250 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\), but its thermal conductivity is unknown. To determine the thermal conductivity, a long cylindrical specimen of diameter \(D=40 \mathrm{~mm}\) is machined, and a thermocouple is inserted through a small hole drilled along the centerline. The thermal conductivity is determined by performing an experiment in which the specimen is heated to a uniform temperature of \(T_{i}=100^{\circ} \mathrm{C}\) and then cooled by passing air at \(T_{\infty}=25^{\circ} \mathrm{C}\) in cross flow over the cylinder. For the prescribed air velocity, the convection coefficient is \(h=55 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If a centerline temperature of \(T(0, t)=40^{\circ} \mathrm{C}\) is recorded after \(t=1136 \mathrm{~s}\) of cooling, verify that the material has a thermal conductivity of \(k=0.30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (b) For air in cross flow over the cylinder, the prescribed value of \(h=55 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) corresponds to a velocity of \(V=6.8 \mathrm{~m} / \mathrm{s}\). If \(h=C V^{0.618}\), where the constant \(C\) has units of \(\mathrm{W} \cdot \mathrm{s}^{0.618} / \mathrm{m}^{2.618} \cdot \mathrm{K}\), how does the centerline temperature at \(t=1136 \mathrm{~s}\) vary with velocity for \(3 \leq V \leq 20 \mathrm{~m} / \mathrm{s}\) ? Determine the centerline temperature histories for \(0 \leq t \leq 1500 \mathrm{~s}\) and velocities of 3,10 , and \(20 \mathrm{~m} / \mathrm{s}\).

5.53 Stone mix concrete slabs are used to absorb thermal energy from flowing air that is carried from a large concentrating solar collector. The slabs are heated during the day and release their heat to cooler air at night. If the daytime airflow is characterized by a temperature and convection heat transfer coefficient of \(T_{\infty}=200^{\circ} \mathrm{C}\) and \(h=35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively, determine the slab thickness \(2 L\) required to transfer a total amount of energy such that \(Q / Q_{o}=0.90\) over a \(t=8\)-h period. The initial concrete temperature is \(T_{i}=40^{\circ} \mathrm{C}\).

A plane wall of a furnace is fabricated from plain carbon steel \(\left(k=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7850 \mathrm{~kg} / \mathrm{m}^{3}, c=430 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) and is of thickness \(L=10 \mathrm{~mm}\). To protect it from the corrosive effects of the furnace combustion gases, one surface of the wall is coated with a thin ceramic film that, for a unit surface area, has a thermal resistance of \(R_{t, f}^{\prime \prime}=0.01 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The opposite surface is well insulated from the surroundings. At furnace start-up the wall is at an initial temperature of \(T_{i}=300 \mathrm{~K}\), and combustion gases at \(T_{\infty}=1300 \mathrm{~K}\) enter the furnace, providing a convection coefficient of \(h=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at the ceramic film. Assuming the film to have negligible thermal capacitance, how long will it take for the inner surface of the steel to achieve a temperature of \(T_{s, i}=1200 \mathrm{~K}\) ? What is the temperature \(T_{s, o}\) of the exposed surface of the ceramic film at this time?
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