A flow of 1 kg / s carbon dioxide at 1600 K, 100 kPa is mixed with a flow of 2 kg / s water at 8

A flow of 1 kg / s carbon dioxide at 1600 K, 100 kPa is...

A flow of $1 \mathrm{kg} / \mathrm{s}$ carbon dioxide at $1600 \mathrm{K}, 100$ $\mathrm{kPa}$ is mixed with a flow of $2 \mathrm{kg} / \mathrm{s}$ water at 800 $\mathrm{K}, 100 \mathrm{kPa},$ and after the mixing it goes through a heat exchanger where it is cooled to $500 \mathrm{K}$ by a $400 \mathrm{K}$ ambient. How much heat transfer is taken out in the heat exchanger? What is the entropy generation rate for the whole process?

Key Concepts

Mixing of Fluid Streams

Mixing of fluid streams is a process where flows with differing properties such as temperature or composition combine. The mixing process, often assumed to be adiabatic when no heat is lost to the surroundings, requires an energy balance to calculate the resulting thermodynamic state, and it inherently involves irreversibilities due to the spontaneous nature of mixing.

Entropy Generation

Entropy generation is used to quantify the irreversibility of a thermodynamic process. In real processes such as mixing of streams at different temperatures and subsequent heat transfer in a heat exchanger, entropy is generated due to factors like heat transfer across finite temperature differences and mixing irreversibility. Evaluating the rate of entropy generation helps in assessing the efficiency and performance loss of the overall process.

Energy Balance Analysis

This concept involves the application of the first law of thermodynamics to a system, where the energy entering the system in the form of mass flow, work, or heat must equal the energy leaving the system plus the change in stored energy. In processes that involve mixing and heat transfer, energy balances are used to determine quantities like the final temperature of a mixture and the magnitude of heat transfer required in a heat exchanger.

Heat Exchanger Analysis

Heat exchanger analysis focuses on understanding how energy is removed or added to a fluid stream through a heat exchange process with an external medium. This involves using parameters such as the mass flow rate, specific heat capacity, and temperature change of the fluid to determine the rate of heat transfer, and it considers the thermodynamic interactions between the process stream and the ambient environment.

Transcript

00:01 Hi, everybody.

00:01 So how much heat transfer is taken out in the heat exchange? what is entropy generation rate for the whole process? okay, so we find the n of the co2, which is going to be the mco2 over the mass total, which is going to be 1 over 4401 equals 0 .027.

00:30 Kilomoles per second and then we're going to do the same thing with h2o which is going to be m h2 o over the m 2 over 18015 equals 0 .1 .1 .1 .0 kilo 1 .01 kilo 1 kilo 1 kilo 1 kilo 1 kilo 1 kilo 1 kilo 1 kilo kilo per second okay so now we need to find the mole fraction so we need to find the mole fraction so we need to find the y co2.

01:02 Okay, so the whole fraction here of the co2.

01:06 And the, oops, i did a double equal, and i didn't even realize it.

01:10 So it's going to be 0 .0227 divided by 0 .11111 plus 0 .027, which equals 0 .1697.

01:29 And now we need to find the mole fraction of h.

01:33 2 -0 and that's going to be one minus the more fraction of the co2 okay and that's going to be 1 minus 0 .1697 which equals 0 .8303 okay and now we're going to apply the energy equation to the whole system and work done on the system okay so that is going to look like q plus so we have energy equation we separate that equation here.

02:09 Equation to the whole system and work done on the system.

02:24 And i'm just going to down a little bit.

02:31 So we got q plus m1, h1, plus m2, h2, h2, equals m4, h4, okay.

02:42 And so now we have q plus mh2o, h2o plus mco2, hco2 equals mh2o, h4, hco2, plus.

03:17 Plus m, co2, h4, co2.

03:22 And let's be two here, and two here.

03:27 There we go.

03:29 Now i'm just going to keep solving.

03:32 So we got q equals mh2o times h4h20, minus h2h2o, plus mco2c22, plus mco2c2.

03:47 C .o .2.

03:49 And h4, c .o .2 minus h2, c .o .2.

03:56 Okay.

03:57 And now if we look at, i try to think that they give us in book, we have two.

04:06 And this is nine at three five point twelve minus one, five, zero point thirteen plus, and and it looks like this is one.

04:21 But i'm just going to not put one out in front, but you can if you want.

04:26 I just don't like putting one out in the front because it always just looks that's pointless, my opinion, but you can if you want.

04:34 Okay...

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