Repeat Problem 12.60 with inlet temperatures of 1400 K for the carbon dioxide and 300 K for the

Repeat Problem 12.60 with inlet temperatures of 1400 K for...

Key Concepts

Energy Balance

The energy balance is a fundamental principle that involves equating the energy entering and leaving a system. In thermodynamics, especially for adiabatic processes where no heat is exchanged with the surroundings, it is used to determine unknown properties like exit temperature by ensuring that the total energy remains constant throughout the process.

Specific Heat (Cp)

Specific heat is the property that quantifies the amount of energy required to raise the temperature of a substance by one degree. It is a crucial variable in energy calculations and can be treated as either constant or variable with temperature. In many engineering problems, tables with specific heat values are used to estimate energy requirements or temperature changes in the system.

Iteration Methods

Iteration methods are mathematical techniques used to find solutions to equations where direct methods may not be feasible, particularly for non-linear problems. In thermodynamic calculations, such as finding the exit temperature when specific heats vary with temperature, an initial estimate is refined repeatedly until convergence is achieved.

Temperature-Dependent Properties

Temperature-dependent properties refer to the behavior of thermodynamic variables, like specific heat, that change with temperature. Accurate engineering analyses require these properties to be evaluated at the correct operational conditions, often necessitating the use of detailed property tables and iterative methods to account for their variation with temperature.

Transcript

0:00 Hi, everybody.

00:01 So for this one, we need to first estimate the exit temperature with specific heats from table 8 .5 and use this to start the erasions and values of table 8 .8.

00:15 Okay.

00:18 So for this, we can do the continuous equation, continuous equation, which is m -c -o -2 plus m -nitrogen 1 equals m -m -m -m -m -m -mix.

00:31 Okay.

00:34 And here we got 2m and 2 plus m and 2 equals the m mix.

00:43 And this is going to equal to 3m and 2 equals m mix.

00:51 And now we can do the energy equation.

00:58 And so now we have m, c .o2, h, c, o2, plus m.

01:05 Mn 2, h, n2, h, and 2 equals m -m -mix times h -mix.

01:14 And this is going to equal to 2m -n2 times h -c -o2 plus m -n2 times h -n1 equals h -n -1, equals 3m -n2 equals h -mix.

01:30 And from this, we get 2mn2 plus hn2 equals 3h mix.

01:48 And your h is going to equal to h 300 plus cp of t minus 300.

02:00 Okay.

02:00 So we're just going to find this.

02:02 And because of this, let me do that so that way you guys can see it a lot better.

02:10 So now we're going to plug the end.

02:11 So we've got two c .p.

02:14 Co .2 times tco2 minus 300 plus cpn2 times tn1 minus 300 equals three cp mix times three hundred.

02:41 And now we just need to find the cp really quick, okay? so i'm just going to do this so we can find the cp mix.

02:53 And that is going to be two thirds times cp co2 plus one -third of the cp of nitrogen 2, which is going to be two -thirds times 0 .842.

03:12 Plus one -third times 1 .042.

03:18 Okay? this is going to equal to 0 .9086 kilojoules per kilogram times kelvin.

03:30 And now we're going to go back to our original, this one right here.

03:34 So we have two times 0 .842.

03:38 We got all these numbers from the charts.

03:40 1 .40 plus 1 .402 times 300 equals 3 .906 times t mix.

04:00 And so we get a t mix that is equal to 979 .6 .02 kelvin's, okay? and now what i want to do is we need to find the t -mix at 1 ,100 ,000...

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