Problem 4 two tanks are connected by a pipe containing a...

Problem 4. Two tanks are connected by a pipe containing a valve. Initially, the valve is closed, one tank is evacuated, and the other tank contains 0.3 kg of water that is a saturated vapor ($x = 1$) at 10 bars. The evacuated tank is 60% the volume of the other tank. The valve is opened and the steam flows into the evacuated tank until equilibrium of $P$ and $T$ prevails. The steam pressure at the end of the process is 5 bars, Treating the two-tank system as a closed system, determine: a) the final temperature (°C). b) the heat transfer during this process (kJ).

Transcript

00:01 So here we have two tanks that are connected by a valve.

00:08 One of them, and let me see here, i may maybe draw something here to indicate that.

00:15 One of them is insulated.

00:16 So this one has insulation around it.

00:19 So it is not losing any heat.

00:24 So a is antibiotic.

00:27 B, there is heat loss.

00:29 And then there's a valve between them.

00:31 So we have steam in each of these.

00:37 And we're given the initial, the volume of this tank is 0 .2 cubic meters.

00:46 The initial pressure is 400 kilopascals and it has a quality factor of 80 % initially in here.

00:53 Now initially in b, we have 3 kilograms of steam, a pressure of 200 kilopascals, and a temperature of 250 degrees c.

01:04 Now, we open the valve and steam flows from tank a to tank b.

01:15 And let's see here, steam flows from tank a to tank b and the pressure drop.

01:20 So the pressure now in tank a after we open the valve is 300 kilopascals.

01:26 We're told that 900 kilojoules of energy, of heat is transferred out of this tank.

01:35 During after we open the valve.

01:39 Surrounding temperature we're told to take a zero degree c and let's see here we know then that the because we have no work here and then no heat loss we know and again there is this second here.

02:01 Why we have the entropy doesn't change in the in the in the first time.

02:09 Tank.

02:12 So it is no heat transfer there.

02:17 And so again, there's no work.

02:22 And so we have, again, this condition here because there's no entropy generated in this tank here because it's because of the adiabatic condition.

02:39 Let's see here.

02:45 Now we can get property.

02:48 At the initial properties in tank a, specific volume, specific internal energy, specific entropy.

02:57 We can get thermodynamic properties in there at the end in tank a because we have this condition.

03:07 So we have now the ending pressure and now we have the ending entropy.

03:16 So it turns out that the temperature then is 133 .5 degrees c.

03:23 And the quality factor after we open the valve in tank a is about 79%.

03:31 We can get the specific volume and the specific internal energy also, which we'll need those later.

03:38 So now turning our attention to tank b, we have two thermodynamic properties there initially.

03:46 And we also know the mass.

03:48 So we can get the specific volume, the specific internal energy, and the specific entropy, all in si units, standard si units, kilojoules per kilogram calvin here.

04:01 Now, since we know the volume, we know the mass in here.

04:07 We don't know the mass in here, but we can get the mass in here because we know the volume and we know the specific volume.

04:14 So dividing the volume by the specific volume, we figure out that initially, this tank had about 0 .54 kilograms in it.

04:25 So we had much less stuff in here than we had in here to start with.

04:31 But we had a higher pressure here, so that's why this is flowing this way.

04:35 Now, let's see here, where are we going now? so afterwards, the volume of tank a didn't change, but the specific volume does...

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