Consider Problem 12.41 and find the value for the mixture...

Consider Problem 12.41 and find the value for the mixture heat capacity, mole basis, and the mixture ratio of specific heats, $k_{\operatorname{mix}},$ both estimated at $850 \mathrm{K}$ from values (differences) of $h$ in Table A.9. With these values make an estimate for the reversible adiabatic exit temperature of the turbine at $100 \mathrm{kPa}.$

Key Concepts

Mixture Heat Capacity

This concept involves calculating the effective heat capacity of a gas mixture based on the mole fractions and individual heat capacities of the component gases. It is essential for determining how a gas mixture absorbs energy when its temperature changes, and it forms the basis for analyzing energy balances in thermodynamic processes such as those occurring in turbines.

Ratio of Specific Heats

Commonly denoted as k or ?, the ratio of specific heats (cp/cv) for a gas mixture is a critical parameter in characterizing the thermodynamic behavior of the mixture. It is used in analysis of adiabatic and isentropic processes, helping to relate changes in temperature and pressure, which is fundamental in evaluating turbine performance and other flow processes.

Reversible Adiabatic (Isentropic) Process

This concept describes an idealized process in which a system undergoes energy conversion without heat transfer and without any entropy increase. In such processes, the relationships among pressure, temperature, and volume can be defined precisely using the ratio of specific heats, making it a key tool in predicting outlet conditions such as the exit temperature from turbines.

Thermodynamic Property Estimation

Estimating thermodynamic properties such as enthalpy, heat capacity, and specific heat ratios from tabulated data (for example, through differences in enthalpy values at specific temperatures) is fundamental in practical engineering calculations. This approach enables engineers to determine the state and behavior of mixtures under different operating conditions, which is imperative for efficient design and analysis of thermodynamic devices.

Transcript

0:00 Hi everybody.

00:01 So this one is asking us, with these values, make an estimate the reversible atiabiotic exit temperature of the turbine at 100k pastels.

00:12 Okay.

00:13 So for this, we can find the cp, and that cp looks really bad.

00:20 Okay.

00:21 C .p over co2.

00:25 And we get 28030 minus 2 .2.

00:30 806 and it's going to be 900 minus 800 so it's going to be a hundred okay and this is going to be 52 .24 kilojoules per kilogram mole times kelvin and now we got our c -p -h -2 -o and again we can find all this information in the the chart that they gave us, which is 8 .9.

01:05 8 .9, okay? so this is going to be 21937 minus 1802, divided by 100 equals 39 .35 times kilojoules divided by kilograms per mole.

01:30 Kelvin, okay? and the next one is cp times nitrogen 2 equals 182, 2, 2, 3 minus 1, 5 ,0, 4, 6 divided by 100 again, equals 31 .77 kilojoules times kilograms times mole for kelvin, okay? and now we're just going to do the cp mix, which is going to equal to 0 .1 times 52 .24, which is we're just using these numbers here.

02:15 And these are, this is percentage, okay, plus 0 .1 times 39 .35 plus 0 .8.

02:28 And that was 8%.

02:30 80%.

02:31 At times my 39 looks like a 1 does not...

02:38 There we go.

02:39 39.

02:40 A lot better.

02:41 31 .77.

02:44 And this is going to equal to 34 .575.

02:50 This is kilojoules per kilogram, merle, kelvin...