Carbon dioxide gas at 320 K is mixed with ni trogen at 280 K in an insulated mixing chamber. Bot

Carbon dioxide gas at 320 K is mixed with ni trogen at 280 K...

Key Concepts

Ideal Gas Behavior and Specific Heats

Many mixing problems assume that the gases involved behave ideally, meaning that their properties can be described using ideal gas laws and constant specific heats. Specific heats, particularly at constant pressure or volume, enable one to express the enthalpy or internal energy changes as linear functions of temperature. This assumption simplifies the calculation of exit conditions such as temperature and also the evaluation of entropy changes associated with the heating or cooling of the gas streams.

Irreversible Processes and Entropy Generation

When streams at different temperatures are mixed, the process is inherently irreversible because it involves spontaneous mixing and thermal equilibration. This irreversibility leads to the generation of entropy. The entropy generation in such processes quantifies the loss of thermodynamic quality and can be calculated by comparing the actual entropy change of each stream with the change that would occur in a reversible process. It reflects the internal production of disorder due to mixing of gases with different initial temperatures.

Energy Conservation in Mixing Processes

When two streams are mixed in an insulated (adiabatic) system, the total energy is conserved. This means that the sum of the individual enthalpies (or internal energies, depending on process assumptions and pressure work) of the incoming streams must equal the total enthalpy of the exit mixture. This principle allows us to determine properties such as the exit temperature by setting up an energy balance equation based on the mass flow rates and specific heats of the gas streams.

Adiabatic Mixing

Adiabatic mixing occurs in a system that does not exchange heat with its surroundings. In such a process, while energy is conserved, the mixing of streams at different temperatures leads to an unequal redistribution of energy among the constituents, resulting in a uniform mixture temperature that can be found from an energy balance. The adiabatic condition ensures that any change in entropy is due solely to the irreversibilities within the process rather than external heat interactions.

Transcript

0:00 Hi everybody.

00:01 So here we need to find the exit temperature of the total entropy generation per kilogram of the exit mixture.

00:09 So if we found the ratio of the mass of co2 over the mass of n2 we have a 2 to 1 ratio and for the co2 over little m of n02 it's a 2 to 1 ratio as well.

00:30 So this means that your m -c -o -2 is equal to your 2m -n -2.

00:39 And from the continuity equation, we have an m -m -m -m -m -m -2.

00:45 That is equal to 3m -n -2.

00:47 Okay, so from the energy equation, let's just skip to energy equation.

00:51 So i will put energy equation here.

00:55 Here we have m c o two h c o two plus m in two h n2 equals m mix h mix and so now we have h equals h 300 plus cp of t minus 300 okay and and that's for a reference here so that's just for reference and i'll box it off the reference so now going back to the energy equation we have m c o2 h oops this becomes um c o2 temperature of c o2 minus 300 plus mn2 cn2 t of the nitrogen to minus 300 equals m mix which is times cp mix times t mix minus 300 okay so i'm scroll down and so the 300 we don't have to worry about so now we have m c -o -2 c -p -o -2 t -c -o -2 plus m -n -2 2, cn2, t, and, oops, it's supposed to be t.

02:38 Tn2 equals m mix, c -p -mix, and t -mix.

02:47 Okay, and what we can do is add in what we know.

02:51 So we have 2m -n -2, 0 .842, 320, plus m -2, um 1 .042 280 equals 3m and 2 cp mix and mix and um so from this um you get 8 30 64 equals 3 cp mix mix t mix and our, so from this, and our cp is, so from 830 .64 equals 3, and we get 0 .9086.

03:49 The c .p mix, t mix, and we get a t mix of 304 .732k.

04:02 Okay.

04:03 And now we need to find the mole fractions of co2.

04:08 And we have the c -i -mi, so 0 .666 -4 .01, plus 0 .3333, 28 .0 .56, 4401 plus 0 .333, 28 .013, equals 0 .56.

04:31 And then we have into 0 .333 divided by 28 .0133, 0 .666, 44 .01, plus 0 .333, 28 .014 .014.

04:57 Okay...

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