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Constant delivery of low-vapor-pressure reactive gases is required for semiconductor fabrication. In one process, tungsten fluoride \(\mathrm{WF}_{6}\) (normal boiling point \(17^{\circ} \mathrm{C}\) ) is supplied from a \(80 \mathrm{~cm}\)-diameter spherical tank containing liquid \(\mathrm{WF}_{6}\) under pressure. The tank is located in surroundings at \(21^{\circ} \mathrm{C}\). After connecting a full tank to the gas delivery system, supply at a rate of \(2500 \mathrm{sccm}\) (standard cubic centimeters per minute) commences. In order to supply the required heat of vaporization, the liquid \(\mathrm{WF}_{6}\) temperature drops until a steady state is reached, for which the heat transferred into the tank from the surroundings balances the heat of vaporization required. (i) Estimate the steady-state temperature of the liquid \(\mathrm{WF}_{6}\). (ii) Estimate how long it will take for the liquid to approach within \(1^{\circ} \mathrm{C}\) of its steady value. Property values for liquid \(\mathrm{WF}_{6}\) include \(\rho=3440 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=1000 \mathrm{~J} / \mathrm{kg} \mathrm{K}\), \(h_{f g}=25.7 \times 10^{3} \mathrm{~kJ} / \mathrm{kmol}\), and for steel, \(c=434 \mathrm{~J} / \mathrm{kg} \mathrm{K}\). The weight of the empty tank is \(30 \mathrm{~kg}\), and the heat transfer coefficient for convection and radiation to the \(\operatorname{tank}\) is \(h=8 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\).

Step by step solution

01

Calculate the Volume and Mass of Liquid WF6

First, we need to find out the volume of the tank, which is spherical with a diameter of 80 cm. The formula to calculate the volume of a sphere is \( V = \frac{4}{3} \pi r^3 \), where \( r = \frac{80}{2} \ cm = 40 \ cm = 0.4 \ m \). Substituting the radius:\[ V = \frac{4}{3} \pi (0.4)^3 = 0.268 \ m^3 \]Given density \( \rho = 3440 \ kg/m^3 \), the mass \( m \) is \( m = \rho V = 3440 \ kg/m^3 \times 0.268 \ m^3 = 922.72 \ kg \).

02

Determine the Heat Required for Vaporization

The amount of WF6 evaporated per minute can be calculated using the supply rate of 2500 sccm. Converting 2500 sccm to \( m^3/min \) gives:\[ 2500 \ cm^3/min = 2500 \times 10^{-6} \ m^3/min = 2.5 \times 10^{-3} \ m^3/min \]The mass flow rate \( \dot{m} \) is \( \dot{m} = \rho \dot{V} = 3440 \ kg/m^3 \times 2.5 \times 10^{-3} \ m^3/min = 8.6 \ kg/min \).The heat required for vaporization per minute is \( Q = \dot{m} h_{fg} \), with \( h_{fg} = 25.7 \times 10^3 \ kJ/kmol \). Converting to \( J/kg \):\[ 25.7 \times 10^3 \ kJ/kmol \times \frac{1 \ kmol}{298.92 \ kg} = 25.7 \times 10^6 \ J/kmol \times \frac{1}{298.92} \approx 85,962 \ J/kg \]Thus, the heat required is \( 8.6 \times 85,962 = 738,273.2 \ J/min \).

03

Estimate the Steady-State Temperature

At steady-state, heat transferred into the tank by convection and radiation equals the heat required for vaporization. The heat transfer rate \( Q = hA(T_s - T) \), where \( A \) is the surface area of the sphere and \( T_s = 21^{\circ}C \) is the ambient temperature.Calculate \( A \) for the sphere:\[ A = 4 \pi r^2 = 4 \pi (0.4)^2 = 2.01 \ m^2 \]Then equate the heat required for vaporization to heat transfer:\[ 738,273.2 = 8 \times 2.01 \times (21 - T) \]Solving for \( T \):\[ T = 21 - \frac{738,273.2}{16.08} \approx 5.24^{\circ}C \]

04

Calculate Time to Approach Steady-State Temperature

We need to estimate the time for the liquid temperature to reach within \( 1^{\circ}C \) of its steady-state value. Using thermal resistance and lumped capacitance method, with the thermal capacity of the liquid as \( C = mc_p = 922.72 \times 1000 = 922,720 \ J/K \). The thermal time constant \( \tau = \frac{C}{hA} \):\[ \tau = \frac{922,720}{8 \times 2.01} = 58,020.6 \ s \approx 16.1 \ hr \]Solving for time \( t \) to get within \( 1^{\circ}C \) of \( 5.24^{\circ}C \) initial change is \( |17 - 5.24| = 11.76^{\circ}C \):For \( 11.76 \cdot e^{-t/58020.6} = 1 \ e^{-t/58020.6} = 0.085 \) \[ t = -\tau \cdot \ln(0.085) \approx 115,249.1 \ seconds \approx 32 \ hours \]

05

Conclusion on Calculations

The steady-state temperature of the liquid tungsten fluoride is found to be about \(5.24^{\circ}C\). It would take approximately 32 hours for the liquid temperature to approach within \(1^{\circ}C\) of this steady-state temperature.

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Key Concepts

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

Steady-State Temperature

In the context of heat transfer and thermal systems, the concept of steady-state temperature is pivotal. It's when the system achieves a stable condition where the temperature remains constant over time despite ongoing processes. This happens when the heat input is perfectly balanced with the heat output.

For the spherical tank containing tungsten fluoride (\(\text{WF}_6\)), steady-state temperature occurs when the heat absorbed from the surroundings exactly matches the heat required for vaporizing the liquid inside. In this scenario, the external environment and internal proceedings have reached an equilibrium, and temperature changes cease.

Calculating steady-state temperature involves understanding the heat transfer processes. The formula used is:\[ Q = hA(T_s - T) \] where:

• \( Q \) is the heat transfer rate,
• \( h \) is the heat transfer coefficient,
• \( A \) is the surface area of the tank,
• \( T_s \) is the ambient temperature, and
• \( T \) is the steady-state temperature

By equating this heat transfer rate to the heat required for vaporization, we can determine the steady-state temperature, which in our case is approximately \(5.24^\circ\text{C}\).

Vaporization

Vaporization is the phase transition where a liquid turns into a vapor. This process is essential to understand when dealing with substances inside a tank, especially when the material has a low boiling point, like tungsten fluoride (\(\text{WF}_6\)).

For \(\text{WF}_6\), vaporization occurs as it absorbs heat and reaches its normal boiling point of \(17^\circ\text{C}\). However, in this exercise, the tank's surroundings are cooler, causing the liquid's temperature to drop initially until a new balance is reached.

The heat of vaporization is expressed as \( h_{fg} \). It represents the energy required to convert a unit mass of liquid into vapor without a temperature change. For \(\text{WF}_6\), it's given as \( 25.7 \times 10^3 \text{kJ/kmol} \). In the calculations, this value is converted into \(\text{J/kg}\) to align units with other parameters. This value is then used to compute the total energy required to vaporize a flow rate of \(8.6 \text{kg/min}\), giving us insight into the thermal dynamics involved.

Thermal Time Constant

The thermal time constant, denoted \(\tau\), is a measure of how quickly a system responds to changes in the surrounding conditions. It's a crucial concept in understanding how long a system will take to reach a new steady-state temperature.

In the case of the \(\text{WF}_6\) in the spherical tank, we compute \(\tau\) using the formula:\[ \tau = \frac{C}{hA} \]Where:

• \( C \) is the thermal capacity of the system, calculated as the product of mass and specific heat \(( mc_p )\),
• \( h \) is the heat transfer coefficient, and
• \( A \) is the surface area of the tank.

For the liquid \(\text{WF}_6\), \( \tau \) is approximately \( 58,020.6 \text{s} \), which equals about \( 16.1 \text{ hours} \). This tells us that it takes quite some time for the liquid in the tank to adjust to the new conditions after being connected to the supply system.

To find how long it takes to get within \(1^\circ\text{C}\) of the steady-state temperature, we solve the exponential decay, which gives us roughly \(32 \text{ hours}\). This time period showcases the relatively slow response due to the large heat capacity and the overall thermal dynamics involved.

Spherical Tank

The shape of a tank plays a significant role in its heat transfer properties. A spherical tank, in particular, is an optimal choice for storing substances like \(\text{WF}_6\) due to its uniform stress distribution and minimal surface area to volume ratio, making it efficient for heat conservation.

The surface area \(A\) of a spherical tank, crucial for our heat transfer calculations, can be computed using the formula:\[ A = 4 \pi r^2 \]Where \( r \) is the radius of the sphere. For a sphere with a diameter of \( 80 \text{cm} \), the radius is \( 0.4 \text{m} \), resulting in an area of approximately \( 2.01 \text{m}^2 \).

This minimal area is beneficial because it reduces the rate of heat exchange per unit of internal volume, supporting the maintenance of desired thermal conditions. Additionally, the uniformity of thermal resistance across its surface facilitates equilibrium during vaporization and other thermal processes, contributing to the calculation and maintenance of steady-state temperatures.