Repeat Problem 12.62 with inlet temperature of 1400 K for the carbon dioxide and 300 K for the n

step 1 Hi, everybody. So 63 is a second part two of 62. And this one is wanting us to find estimate exo

Repeat Problem 12.62 with inlet temperature of 1400 K for...

Repeat Problem 12.62 with inlet temperature of $1400 \mathrm{K}$ for the carbon dioxide and $300 \mathrm{K}$ for the nitrogen. First estimate the exit temperature with the specific heats from Table $A .5$ and use this to start iterations with values from A.9.

Key Concepts

Iterative Estimation Techniques

Iterative techniques involve making an initial estimate and then refining that estimate through successive calculations until convergence is achieved. This method is particularly useful when dealing with nonlinear equations or when properties such as specific heats vary with temperature, allowing for improved accuracy in final predictions like exit temperature.

Energy Balance in Flow Systems

An energy balance is a principle statement reflecting conservation of energy in a system, equating the energy entering with the energy leaving plus any accumulation or work interactions. In flow systems, this balance is critical for predicting exit temperatures when working with streams of gases or liquids, based on inlet conditions and thermodynamic properties.

Specific Heat Capacities

Specific heat capacities indicate the amount of energy needed to raise the temperature of a unit mass of a substance by one degree. This concept is fundamental in thermodynamics and heat transfer as it helps in calculating changes in internal energy and enthalpy during heating or cooling processes.

Temperature-Dependent Properties

Many thermodynamic properties, including specific heats, vary with temperature. This variation necessitates the use of tabulated data or temperature-dependent equations, especially when dealing with large temperature differences or high-precision applications.

Transcript

00:01 Hi, everybody.

00:02 So 63 is a second part two of 62.

00:08 And this one is wanting us to find estimate exot temperature with specific heats.

00:13 And we're using the charts.

00:14 Okay.

00:15 And so for this, i'm going to find the cp of this.

00:24 And that's just so i don't have to break everything up later on for you guys.

00:28 So we have two thirds of the cp.

00:31 P of co2 and that's going to be 0 .842 times 44 .01.

00:42 Okay.

00:42 Plus and it looks like this is going to be one third times 1 .042 times 28 .013.

00:55 Okay.

00:55 And this is going to give us 34 .4 .4 .3.

00:56 Okay.

00:57 And this is going to give us 34 .4 .4.

01:01 4, 3, 3, 8 kilojoules per k mole times k, okay? and so we have that information.

01:13 So we can do the energy equation, okay? and from this energy equation, we know that h equals h 300 plus cp times t minus 300, okay? and you should have already done this in the last section.

01:37 So i'm just going to write in n2 times, i'm going to write without the h.

01:44 Okay, the cp of n2 times n2 minus 300 plus n of the co2 times c0 2 minus 300 plus n of the co2 times c2.

01:59 2 times t co2 minus 300 equals 3n2 times t mix minus 300 of the cp mix.

02:18 And now this is going to be in n2 of cpn2 of tn2 plus two and n2 of c .p .n2 times t -o2 equals 3n2 t mix times c -o2.

02:43 And we're just going to scroll down.

02:48 And now we have n -n2 times 1 .042 times 28 .013.

02:59 Times 300, okay, plus 2 in 2 times 0 .0, i mean 0 .8, 42 times 44 .01 times 1 ,400 equals 3n2 and 2 times 2 times t ,000 ,000, equals 3 and 2.

03:33 Mix equals 34 .4338.

03:40 And so we have 8756.

03:46 0 .86 plus 10375757.

03:52 0 .976 equals 103 .301 t mix.

04:01 And t -mix is going to equal to 1089 .194 kelvin.