Another alternative refrigerant (see preceding problem) is \(1,1,1,2\) tetrafluoroethane (refrigerant HFC-134a). Tillner-Roth and Baehr published a compendium of thermophysical properties of this substance (R. Tillner-Roth and H.D. Baehr, \(J\). Phys. Chem. Ref. Data 23, 657 (1994) from which properties such as the Joule-Thomson coefficient \(\mu\) can be computed. (a) Compute \(\mu\) at \(0.100 \mathrm{MPa}\) and \(300 \mathrm{~K}\) from the following data (all referring to \(300 \mathrm{~K}\) ): \(\begin{array}{llll} \text { P/MPa } & 0.080 & 0.100 & 0.12 \\ \text { Specific enthalpy/(k] } \mathrm{kg}^{-1} \text { ) } & 426.48 & 426.12 & 425.76 \end{array}\) (The specific constant-pressure heat capacity is \(\left.0.7649 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~kg}^{-1}\right)\) (b) Computer \(\mu\) at \(1.00 \mathrm{MPa}\) and \(350 \mathrm{~K}\) from the following data (all referring to \(350 \mathrm{~K}\) ): \(\begin{array}{llll} \text { PIMPa } & 0.80 & 1.00 & 1.2 \\ \text { Specific enthalpy/(k] } \mathrm{kg}^{-1} \text { ) } & 461.93 & 459.12 & 456.15 \end{array}\) (The specific constant-pressure heat capacity is \(1.0392 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~kg}^{-1}\).)

Short Answer

Expert verified
The Joule-Thomson coefficient \(\mu\) at 0.100 MPa and 300 K is \(-23.53 \text{ K/MPa}\), and at 1.00 MPa and 350 K is \(-13.54 \text{ K/MPa}\).

Step by step solution

01

Calculate the Joule-Thomson Coefficient at 0.100 MPa and 300 K

The Joule-Thomson coefficient \(\mu\) is given by the relation \(\mu = \left(\frac{\partial T}{\partial P}\right)_H = \frac{1}{C_P} \cdot \left(\frac{\partial H}{\partial P}\right)_T\), where \(C_P\) is the specific heat capacity at constant pressure, and \(\frac{\partial H}{\partial P}\right)_T\) is the rate of change of specific enthalpy with pressure at constant temperature. Approximate \(\frac{\partial H}{\partial P}\right)_T\) by the slope of the enthalpy between the closest given pressures at constant temperature of 300 K: \(\frac{\Delta H}{\Delta P} = \frac{426.12 \text{ kJ/kg} - 426.48 \text{ kJ/kg}}{0.100 \text{ MPa} - 0.080 \text{ MPa}} = \frac{-0.36 \text{ kJ/kg}}{0.02 \text{ MPa}}\).
02

Plug Values into the Joule-Thomson Coefficient Formula for 300 K

Now, calculate \(\mu\) at 300 K using the specific constant-pressure heat capacity of \(0.7649 \text{ kJ} \text{ K}^{-1} \text{ kg}^{-1}\) and the rate of change of specific enthalpy. \(\mu = \frac{1}{0.7649} \cdot \frac{-0.36}{0.02} = -23.53 \text{ K/MPa}\).
03

Calculate the Joule-Thomson Coefficient at 1.00 MPa and 350 K

For the condition at 1.00 MPa and 350 K, approximate the enthalpy change with pressure similarly: \(\frac{\Delta H}{\Delta P} = \frac{459.12 \text{ kJ/kg} - 461.93 \text{ kJ/kg}}{1.00 \text{ MPa} - 0.80 \text{ MPa}} = \frac{-2.81 \text{ kJ/kg}}{0.20 \text{ MPa}}\).
04

Plug Values into the Joule-Thomson Coefficient Formula for 350 K

Calculate \(\mu\) at 350 K using the specific constant-pressure heat capacity of \(1.0392 \text{ kJ} \text{ K}^{-1} \text{ kg}^{-1}\) and the calculated enthalpy change rate. \(\mu = \frac{1}{1.0392} \cdot \frac{-2.81}{0.20} = -13.54 \text{ K/MPa}\).

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Key Concepts

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

Thermophysical Properties
Thermophysical properties are important characteristics of substances that describe how they change under different temperatures, pressures, and other physical conditions. These properties include specific enthalpy, density, viscosity, thermal conductivity, specific heat capacities, and the Joule-Thomson coefficient, among others. They play a critical role in fields such as thermodynamics, fluid mechanics, and heat transfer, influencing the design and operation of numerous engineering systems. For instance, when dealing with refrigerants like 1,1,1,2 tetrafluoroethane (HFC-134a), it's essential to accurately understand and calculate these properties to optimize the performance of refrigeration cycles.

Students aiming to master the calculation and application of thermophysical properties should focus on understanding the principles underlying each property, how they relate to one another, and the method of their calculation. This approach will not only assist them in solving textbook problems but also empower them with a practical understanding that is applicable in laboratory experiments and real-world engineering challenges.
Specific Enthalpy
Specific enthalpy, denoted as 'H', is a thermodynamic property that measures the total heat content per unit mass of a substance. It's a combination of the internal energy of the material and the work needed to make room for it by displacing its environment. This property is fundamental in understanding and analyzing processes like heating, cooling, and phase changes. In the context of the given exercise, specific enthalpy values at different pressures but constant temperature aid in calculating the Joule-Thomson coefficient, which is pivotal for designing efficient refrigeration and air conditioning systems.

To improve the grasp of this concept, students should prioritize practical comprehension over rote memorization. They can do this by engaging with exercises that require them to calculate specific enthalpy changes under various conditions, and by applying these calculations to understand real-world systems, such as how the specific enthalpy of a refrigerant impacts the cooling process in a refrigerator.
Constant-Pressure Heat Capacity
Constant-pressure heat capacity (\( C_P \)) is a measure of the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin) at constant pressure. This thermophysical property is particularly important in processes that occur at constant atmospheric pressure, such as those happening on Earth's surface. It's a critical factor when it comes to understanding heat exchange in gases and liquids and is extensively used in calculations concerning energy transfer and thermodynamics.

In teaching this concept, educators should emphasize the role of constant-pressure heat capacity in everyday phenomena, such as the warming of air in the atmosphere or the cooling effects of a refrigerant in air conditioning systems. By linking these abstract concepts to tangible examples, students are more likely to comprehend and retain the knowledge. Exercises that demonstrate the practical application of these calculations can further reinforce the concepts. For example, by calculating the heat capacity of a refrigerant, students can predict how efficient it will be at removing heat from an enclosed space, thus deepening their understanding of the thermodynamics behind refrigeration cycles.

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Most popular questions from this chapter

The thermodynamic equation of state \((\partial U / \partial V) \mathrm{T}=T(\partial p / \partial T)_{v}-p\) was quoted in the chapter. Derive its partner from it and the general relations between partial differentials.

Show that the following functions have exact differentials: (a) \(x^{2} y+3 y^{2}\), (b) \(x \cos x y\), (c) \(x^{3} y^{2}\), (d) \(t\left(t+e^{3}\right)+s\).

Concerns over the harmful effects of chlorofluorocarbons on stratospheric ozone have motivated a search for new refrigerants. One such alternative is 2,2 -dichloro-1,1,1-trifluoroethane (refrigerant 123). Younglove and McLinden published a compendium of thermophysical properties of this substance (B.A. Younglove and M. McLinden, \(J\). Phys. Chem. Ref Data 23,7 (1994)), from which properties such as the Joule-Thomson coetficient \(\mu\) can be computed. (a) Compute \(\mu\) at \(1.00\) bar and \(50^{\circ} \mathrm{C}\) given that \((\partial H / \partial p)_{T},=-3.29\) \(\times 10^{3} \mathrm{~J} \mathrm{MPa}^{-1} \mathrm{~mol}^{-1}\) and \(C_{p \cdot m}=110.0 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\). (b) Compute the temperature change that would accompany adiabatic expansion of \(2.0 \mathrm{~mol}\) of this refrigerant from \(1.5\) bar to \(0.5\) bar at \(50^{\circ} \mathrm{C}\).

A gas obeys the equation of state \(V_{m}=R T / p+a T^{2}\) and its constantpressure heat capacity is given by \(C_{p^{\prime} m}=A+B T+C p\), where \(a, A, B\), and Care constants independent of \(T\) and \(p .\) Obtain expressions for (a) the JouleThomson coefficient and (b) its constant-volume heat capacity.

An average human produces about \(10 \mathrm{M} \mathrm{J}\) of heat each day througn metabolic activity. If a human body were an isolated system of mass \(65 \mathrm{~kg}\) with the heat capacity of water, what temperature rise would the body experience? Human bodies are actually open systems, and the main mechanism of heat loss is through the evaporation of water. What mass of water should be evaporated each day to maintain constant temperature?

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