A tank has two sides initially separated by a diaphragm,...

A tank has two sides initially separated by a diaphragm, shown in Fig. P12.72. Side $A$ contains $1 \mathrm{kg}$ of water and side $B$ contains $1.2 \mathrm{kg}$ of air, both at $20^{\circ} \mathrm{C}$ and 10.0 kPa. The diaphragm is now broken, and the whole tank is heated to $600^{\circ} \mathrm{C}$ by a $700^{\circ} \mathrm{C}$ reservoir. Find the final total pressure, heat transfer, and total en. tropy generation.

Key Concepts

Heat Transfer with a Reservoir

Heat transfer is the process by which energy moves from a higher temperature source (the reservoir) to the system until a thermal equilibrium is reached. In this problem, the presence of a reservoir at a higher temperature provides the driving force for heat transfer into the combined system, which is crucial for reaching the final state at the desired temperature.

Entropy Generation in Irreversible Processes

Entropy generation is associated with the irreversibility of real processes such as mixing and non-equilibrium heat transfer. This concept stems from the second law of thermodynamics and quantifies the increase in disorder during the process. The calculated entropy generation reflects the deviation from reversible behavior and provides insight into the efficiency and irreversibility of the combined heating and mixing process.

Closed System Energy Balance

This concept involves applying the first law of thermodynamics to a system in which mass remains constant but energy can change due to work or heat transfer. In the context of the problem, the overall energy in the combined tank is conserved while it interacts with an external heat reservoir, leading to changes in internal energy and temperature until a new equilibrium is reached.

Equilibrium State Determination

Determining the final equilibrium state involves understanding how the state properties (pressure, temperature, and volume) adjust after internal mixing and external heating. This concept includes applying constraints such as mass conservation and energy balance to characterize the final uniform state of the system after the diaphragm is removed.

Transcript

0:00 Hi, everybody.

00:01 So for this one, we need to find the total pressure heat transfer and total entropy generation.

00:09 Okay? so we have the specific volume, you found a specific volume of a1, and that's going to be 0 .287 times 20, and that's temperature, so it's 20 celsius, adding to 273.

00:30 Point one five get it to two kelvin's divided by a hundred and get 0 .84 m 1 .4 kilogram and now we need to find the volume of the mix so volume of the mix is going to be m a v a 1 plus m w b w and here we get 1 .2 .84 plus 1 .2 .8 4 plus 1 times 0 .002 and you get 1 .1 meter cute.

01:10 Okay? and now we're going to look at the v of the w2 and you get v over mw, which is going to be to be 1 .01 over 1, which is 1 .01 meter cube.

01:32 And then we get a v -a2 which is going to be v over m -a and it's going to be 1 .01 divided by 1 .2 which is 0 .8 -417 meter and i just like messed up my there we go and let me just like rewrite that 8 so we'll get nicer.

02:00 There we go.

02:02 A lot nicer.

02:04 And then we need to do the final pressure so the final pressure i'll put pf and that's going to be m a r t2 divided by the a 2 so we got 0 .287 okay times 600 plus 273 .15 okay and that's just because we got to convert it and now we have 0 .84 17 down here and we get 297 .7k pascal's okay and the final pressure using the relation so the final pressure using relation got pw t plus p a 2 and that's going to be 400 and so this is p82 let me just rewrite that nicer so 400 plus 297 .7 equals 697.

03:19 Equals 697.

03:21 Kastcles, okay? and now for the heat transfer, we have take u2 equals m .a2 minus w .a1 plus m1.

03:37 Plus mw2 plus w1.

03:46 So from here we get 1 .2 times 6 .552 .2 .52.

03:52 2 .58 minus 209 .45, plus 1 times 3 ,0 .2 .2 minus 83 .9 .9 .4.

04:07 And povetic...

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