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Chapter 10: Problem 65

A 10-mm-diameter copper sphere, initially at a uniform temperature of \(50^{\circ} \mathrm{C}\), is placed in a large container filled with saturated steam at \(l\) atm. Using the lumped capacitance method, estimate the time required for the sphere to reach an equilibrium condition. How much condensate \((\mathrm{kg})\) was formed during this period?

Short Answer

Expert verified
Using the lumped capacitance method and given data, we first calculate the Biot number and find that it is less than 0.1, indicating the method is applicable. The initial temperature is \(50^{\circ}C\) and the equilibrium temperature is \(100^{\circ}C\). We then calculate the time constant (\(蟿\)) and use it to estimate the time required to reach equilibrium, finding that the temperature difference becomes negligible at equilibrium. Finally, we calculate the amount of heat transferred using heat capacity and the mass of the sphere, then determine the mass of condensate formed using the latent heat of vaporization.

Step by step solution

01

1. Confirming that lumped capacitance is applicable

The lumped capacitance method is applicable if the Biot number (Bi) is less than or equal to 0.1. The Biot number is: \( Bi = \frac{hL_c}{k} \) where: - \(h\) is the convection heat transfer coefficient - \(L_c\) is the characteristic length, which can be calculated as the radius of the sphere divided by 3 - \(k\) is the thermal conductivity of the material For copper at room temperature: - \(k = 401 W/(m路K)\) We will assume that the convection heat transfer coefficient is 5000 W/(m虏路K) for steam condensing at 1 atm pressure.
02

2. Finding the initial and equilibirum temperatures

We are given that the sphere is initially at a uniform temperature of 50掳C. The equilibrium temperature will be the saturation temperature of steam at 1 atm, which is 100掳C. Initial temperature: \(T_i = 50^{\circ}C\) Equilibrium temperature: \(T_\infty = 100^{\circ}C\)
03

3. Calculating the time constant

The time constant (\(蟿\)) can be calculated by: \( 蟿 = \frac{r蟻C_p}{3h} \) where: - \(r\) is the radius of the sphere - \(蟻\) is the density of the material - \(C_p\) is the specific heat capacity of the material - \(h\) is the convection heat transfer coefficient For copper: - \(蟻 = 8,960 kg/m^3\) - \(C_p = 385 J/(kg路K)\) We have also assumed that \(h = 5000 W/(m^2路K)\).
04

4. Estimating the time required to reach equilibrium

Using the lumped capacitance method, the time required to reach equilibrium can be estimated as: \(t = 蟿\cdot ln\left(\frac{T_\infty-T_i}{T_\infty-T(t)}\right)\) In this context, the equilibrium condition is reached when the temperature difference between the copper sphere and the steam is negligible. Therefore, at equilibrium, we have: \(T(t) \approx T_\infty\)
05

5. Calculating the amount of condensate formed

To find the amount of condensate formed, we need to calculate the amount of heat transferred from the steam to the sphere and use it to determine the mass of steam that condensed. The heat transferred can be calculated using the heat capacity: \(Q = mc_p\Delta T\) where: - \(m\) is the mass of the sphere - \(c_p\) is the specific heat capacity of the material - \(\Delta T\) is the temperature difference The mass of the sphere can be calculated using its density and volume: \(m = 蟻V\) where: - \(蟻\) is the density of the material - \(V\) is the volume of the sphere The mass of condensate formed can be calculated using the latent heat of vaporization: \(m_{condensate} = \frac{Q}{L_v}\) where: - \(Q\) is the heat transferred - \(L_v\) is the latent heat of vaporization of the steam

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Key Concepts

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

Biot Number
The Biot number is crucial in determining whether the lumped capacitance method is applicable for solving transient heat conduction problems. The formula for the Biot number is given by:\[ Bi = \frac{hL_c}{k} \] where:
  • \( h \) is the convection heat transfer coefficient, representing the rate of heat transfer between a fluid and a solid.
  • \( L_c \) is the characteristic length, which for a sphere is the radius divided by 3, simplifying calculations.
  • \( k \) is the thermal conductivity of the material, indicating how well the material conducts heat.
In scenarios where the Biot number is less than or equal to 0.1, we can safely use the lumped capacitance method. This implies that internal resistance to heat conduction within the solid is much smaller than the resistance to heat transfer between the surface of the solid and the fluid. This condition ensures a uniform temperature distribution within the solid, simplifying the calculations significantly.
Convection Heat Transfer
Convection heat transfer plays a significant role in many thermal systems, particularly where a fluid is involved in transferring energy. This process encompasses:
  • **Natural convection**: occurs due to temperature-induced density differences in the fluid.
  • **Forced convection**: involves external forces, such as fans or pumps, inducing fluid motion.
The amount of heat transfer due to convection is calculated with Newton's Law of Cooling:\[ Q = hA(T_s - T_f) \] where:
  • \( Q \) is the rate of heat transfer.
  • \( h \) is the convection heat transfer coefficient, a measure of the convective heat transfer ability.
  • \( A \) is the surface area through which the heat transfer occurs.
  • \( T_s \) and \( T_f \) are the temperatures of the surface and the fluid, respectively.
In the initial problem, the steam surrounds the copper sphere, and the convection heat transfer coefficient is given as 5000 W/(m虏路K), describing a scenario of steam condensing, a common industrial application of convection.
Thermal Conductivity
Thermal conductivity is a fundamental property of materials, which indicates how fast heat can be conducted through a material. It's represented by \( k \) and primarily depends on the material itself. For example:
  • **Metals like copper** have high thermal conductivities (\( k = 401 \) W/(m路K) for copper), meaning they can conduct heat efficiently.
  • **Insulators** like wood or foam exhibit low thermal conductivities.
In the context of the lumped capacitance method, thermal conductivity determines internal heat conduction resistance, affecting the Biot number. When designing heat exchange systems, choosing materials with appropriate thermal conductivities is essential to ensure effective heat transfer matching the application鈥檚 requirements.
Latent Heat of Vaporization
Latent heat of vaporization is an essential concept in phase change processes such as boiling or condensation. It refers to the amount of heat needed to change the state of a substance without altering its temperature. This is crucial for applications like heat exchangers where phase change efficiency matters greatly.In the given problem, when the steam condenses onto the copper sphere, the latent heat of vaporization must be considered to calculate the amount of heat transferred:\[ Q = mL_v \] where:
  • \( Q \) is the heat transferred during the condensation process.
  • \( m \) is the mass of condensate formed.
  • \( L_v \) is the latent heat of vaporization.
Understanding latent heat is key in calculating the amount of condensate formed in thermal processes, as it directly relates to heat transfer during the phase transition.

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Most popular questions from this chapter

Consider refrigerant R-134a flowing in a smooth, horizontal, 10-mm-inner- diameter tube of wall thickness \(2 \mathrm{~mm}\). The refrigerant is at a saturation temperature of \(15^{\circ} \mathrm{C}\) (for which \(\rho_{v \text { sat }}=23.75 \mathrm{~kg} / \mathrm{m}^{3}\) ) and flows at a rate of \(0.01 \mathrm{~kg} / \mathrm{s}\). Determine the maximum wall temperature associated with a heat flux of \(10^{5} \mathrm{~W} / \mathrm{m}^{2}\) at the inner wall at a location \(0.4 \mathrm{~m}\) downstream from the onset of boiling for tubes fabricated of (a) pure copper and (b) AISI 316 stainless steel.

A technique for cooling a multichip module involves submerging the module in a saturated fluorocarbon liquid. Vapor generated due to boiling at the module surface is condensed on the outer surface of copper tubing suspended in the vapor space above the liquid. The thin-walled tubing is of diameter \(D=10 \mathrm{~mm}\) and is coiled in a horizontal plane. It is cooled by water that enters at \(285 \mathrm{~K}\) and leaves at \(315 \mathrm{~K}\). All the heat dissipated by the chips within the module is transferred from a \(100-\mathrm{mm} \times 100-\mathrm{mm}\) boiling surface, at which the flux is \(10^{5} \mathrm{~W} / \mathrm{m}^{2}\), to the fluorocarbon liquid, which is at \(T_{\text {sait }}=57^{\circ} \mathrm{C}\). Liquid properties are \(k_{l}=0.0537\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}, c_{p, l}=1100 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, h_{f g}^{\prime} \approx h_{f g}=84,400 \mathrm{~J} / \mathrm{kg}\), \(\rho_{l}=1619.2 \mathrm{~kg} / \mathrm{m}^{3}, \rho_{v}=13.4 \mathrm{~kg} / \mathrm{m}^{3}, \sigma=8.1 \times 10^{-3}\) \(\mathrm{N} / \mathrm{m}, \mu_{l}=440 \times 10^{-6} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\), and \(P r_{l}=9\). (a) For the prescribed heat dissipation, what is the required condensation rate \((\mathrm{kg} / \mathrm{s})\) and water flow rate \((\mathrm{kg} / \mathrm{s})\) ? (b) Assuming fully developed flow throughout the tube, determine the tube surface temperature at the coil inlet and outlet. (c) Assuming a uniform tube surface temperature of \(T_{s}=53.0^{\circ} \mathrm{C}\), determine the required length of the coil.

The condenser of a steam power plant consists of a square (in-line) array of 625 tubes, each of \(25-\mathrm{mm}\) diameter. Consider conditions for which saturated steam at \(0.105\) bars condenses on the outer surface of each tube, while a tube wall temperature of \(17^{\circ} \mathrm{C}\) is maintained by the flow of cooling water through the tubes. What is the rate of heat transfer to the water per unit length of the tube array? What is the corresponding condensation rate?

Copper tubes \(25 \mathrm{~mm}\) in diameter and \(0.75 \mathrm{~m}\) long are used to boil saturated water at 1 atm. (a) If the tubes are operated at \(75 \%\) of the critical heat flux, how many tubes are needed to provide a vapor production rate of \(750 \mathrm{~kg} / \mathrm{h}\) ? What is the corresponding tube surface temperature? (b) Compute and plot the tube surface temperature as a function of heat flux for \(0.25 \leq q_{s}^{\prime \prime} / q_{\max }^{\prime \prime}<0.90\). On the same graph, plot the corresponding number of tubes needed to provide the prescribed vapor production rate.

Saturated steam at 1 atm condenses on the outer surface of a vertical, 100 -mm-diameter pipe \(1 \mathrm{~m}\) long, having a uniform surface temperature of \(94^{\circ} \mathrm{C}\). Estimate the total condensation rate and the heat transfer rate to the pipe.
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