Consider Problem 12.39 and find the value for the mixture...

Consider Problem 12.39 and find the value for the mixture heat capacity, mass basis, and the mixture ratio of specific heats, $k_{\text {mix }},$ both estimated at $850 \mathrm{K}$ from values (differences) of $h$ in Table A.8. With these values make an estimate for the reversible adiabatic exit temperature of the turbine at $100 \mathrm{kPa}.$

Key Concepts

Specific Heat Capacity (Mass Basis)

This concept refers to the amount of energy per unit mass required to raise the temperature of a substance by one degree. In thermodynamic analyses, particularly for gas mixtures, it is essential to determine an effective specific heat capacity at a given temperature to accurately predict how the mixture will change its thermodynamic state when energy is added or removed.

Ratio of Specific Heats (k)

The ratio of specific heats, commonly denoted as k (or gamma), is defined as the ratio of the specific heat capacity at constant pressure (Cp) to that at constant volume (Cv). This parameter is pivotal in characterizing the behavior of compressible fluids, especially in adiabatic processes, where it helps relate temperature and pressure changes under isentropic conditions.

Reversible Adiabatic (Isentropic) Process

A reversible adiabatic process is one in which there is no heat transfer with the surroundings and the process occurs without generating entropy. Such processes are termed isentropic and are fundamental in the analysis of turbines and compressors, as they allow the use of idealized relations to determine exit temperatures and pressures based solely on energy conservation and ideal gas relations.

Mixture Property Estimation

When dealing with a gas mixture, its overall thermodynamic behavior is characterized by weighted averages of the individual species properties. Estimating the mixture heat capacity and ratio of specific heats involves combining the properties of the constituents based on their mass or mole fractions. This approach simplifies the analysis of processes, such as those occurring in turbines, by allowing the use of single-property evaluations for the entire mixture.

Transcript

0:00 Hi everybody.

00:01 So this one is asking us, with these values, make an estimate for the reversible antibiotic exit temperature of the turbine at 100k pascal's.

00:12 Okay.

00:14 And so we have your cp.

00:17 And let me make that cp look nice.

00:21 So cp is going to equal to the delta 8 over delta t for temperature.

00:27 And this is going to be h -900 minus 8 -800 divided by 900 minus 800 and going to give you 8 so now we can do the cp of your co2 and that is going to be 849 72 minus 7302 minus 731 2 divided by a hundred and you clearly see this hundred here okay and this is going to equal to 1 .187 kilojoules per kilogram times kelvin and now we got cp of the h2o and again we're finding all this information in the charts that they gave us 1768 .60 minus 155 .0 .1 .00 0 .13 divided by 100.

01:47 And in here we got equals 2 .1847 kilojoules per kilograms times kelvin.

01:58 Okay.

02:00 And now we can do the cp of in and that's going to be 960 .25.

02:11 There we go.

02:13 Minus 846 .885 divided by 100 equals 1 .134 .4 kilo.

02:27 Killed per kilograms times kelvin.

02:32 Okay.

02:35 And now what we can do is find the cp mix.

02:45 And this is going to equal to, so what we have is the c .1, and we're going to ties it by the c.

02:54 Co2 that we just found.

02:56 So it's going to be 1 .187 plus 0 .1 times the cp we just found above for the h2.

03:04 Which is 2 .1847 plus 0 .8 8 times 1 .134.