The substance R-410a (see Problem 12.48) is at 100 kPa and 290 K . It is now brought to 250 kPa

The substance R-410a (see Problem 12.48) is at 100 kPa and...

Key Concepts

Polytropic Process

A polytropic process is a thermodynamic process that follows the relation PV^n = constant, where n is the polytropic exponent. This process encompasses a variety of practical processes, including isothermal and adiabatic behaviors as special cases. Understanding the polytropic relation is crucial for calculating how properties such as volume and pressure change during the process when it is subject to work or heat transfer.

Thermodynamic States and Properties

Thermodynamic states are characterized by properties such as pressure, temperature, and specific volume that define the condition of a system at equilibrium. These properties are essential for describing and analyzing the energy and mass transfers during a process. In the context of any thermodynamic problem, understanding the state points provides the necessary basis for computing changes in other properties like enthalpy and entropy.

Reversible Process

A reversible process is an idealized thermodynamic process that occurs infinitely slowly, ensuring the system remains in equilibrium at every stage. Because it does not generate any internal irreversibilities, the path of the process can be reversed without any net change in the system and its surroundings. This concept is integral when analyzing the exact balance of energy and entropy changes in a process.

Specific Volume

Specific volume is defined as the volume per unit mass of a substance. It is an important property in thermodynamics, especially in processes involving phase change or compressible fluids, as it directly relates to the density of the material. Changes in specific volume during a process can influence the work interactions and efficiency of thermodynamic systems.

Specific Enthalpy

Specific enthalpy is a measure of the total heat content of a substance per unit mass and is defined as the sum of its internal energy and the product of its pressure and specific volume (h = u + Pv). It is a critical property for energy analysis in processes that include heating or work interaction under constant pressure conditions. Enthalpy changes are often used to evaluate the energy required or released during a process.

Specific Entropy

Specific entropy quantifies the degree of randomness or disorder per unit mass in a system and is a key indicator of the energy dispersal in thermodynamic processes. It is used to assess the feasibility and efficiency of a process, particularly in the context of the second law of thermodynamics. Changes in specific entropy help in evaluating the irreversibility of real processes even when idealized reversible processes are assumed.

Transcript

00:01 Hi, everybody.

00:01 So this one's question is asking us, find the change of specific volume, specific entropy, and specific entropy for the process.

00:09 Okay, so for this, we are going to find the r mix.

00:14 And again, these values you can find in the charts in your book, okay? so 0 .5 times 0 .1598 plus 0 .5 times 0 .1598 plus 0 .5 times 0 .5.

00:30 0 .06927.

00:35 And this is going to equal to 0 .0114535 kilyloules per kilograms times kelvin.

00:50 Okay.

00:52 And now we have cp mix.

00:57 And that is going to be equal to 0 .5 .5.

01:01 Times 0 .822 plus 0 .5 times 0 .791 equals 0 .8065 kilojoules per kilogram times kelvin.

01:21 Okay.

01:23 And now what we can do is find the volume.

01:30 And so we have p1 v1 equals r mix times temperature and so we have volume one equals r mix times temperature 1 of pressure 1 and we go 0 .114 -5335 times 290 divided by 100 the pressure for pressure 1 and this is going to be 0 .333.

02:02 2 .15 meter cube divided by a kilogram.

02:07 Now we can find volume 2, which is going to be the exact same thing as this, except for everything is replaced with 2.

02:16 So temperature 2 and pressure 2, okay? so your r -mix is still the same.

02:21 1 -1 -4.

02:24 So it looks like i added a 0.

02:30 There we go.

02:32 1 -1 -4 -5.

02:39 35300 divided by and it's going to be 250.

02:43 Okay.

02:44 And this is going to equal to 0 .183256 times meter cube divided by a kilogram.

02:53 Okay.

02:53 So now we can do the overall volume, which is what we need.