A piston/cylinder assembly contains 1 kg of liquid water at 20^∘ C and 300 kPa. Initially the p

step 1 So guys, let us start with the answer of this question. First of all, draw the PV diagram. We've

A piston/cylinder assembly contains 1 kg of liquid water at...

A piston/cylinder assembly contains $1 \mathrm{~kg}$ of liquid water at $20^{\circ} \mathrm{C}$ and $300 \mathrm{kPa}$. Initially the piston floats, similar to the setup in Problem $3.173,$ with a maximum enclosed volume of $0.002 \mathrm{~m}^{3}$ if the piston touches the stops. Now heat is added so that a final pressure of $600 \mathrm{kPa}$ is reached. Find the final volume and the heat transfer in the process.

Key Concepts

Process Constraints and Boundary Conditions

The physical limits imposed by system design—such as maximum allowable volume from piston stops—play a key role in defining the process path and the associated work interactions. Recognizing these constraints helps in determining whether the system’s behavior is limited by mechanical boundaries or by the thermodynamic conditions imposed (like specified final pressures).

Control Volume Energy Analysis (First Law of Thermodynamics for Closed Systems)

In a closed system, where no mass enters or leaves, the first law of thermodynamics is applied to relate heat transfer, work done by the system, and changes in internal energy. In piston-cylinder problems, this energy balance is used to connect the added heat to the physical work (volume change) performed by or on the fluid.

Piston-Cylinder Assemblies

This concept involves systems in which a fluid is contained within a cylinder whose volume can change due to the movement of a piston. The dynamics of the piston—whether it moves freely or is limited by stops—affect the work interaction between the system and its surroundings, and must be considered when analyzing processes such as heating or compression.

Thermodynamic Properties of Water

A thorough understanding of water’s properties, including specific volume, saturated liquid-vapor states, and compressed liquid behavior, is crucial when evaluating state changes under different temperatures and pressures. These properties are used to determine the volume and energy state of water through processes such as heating.

Transcript

00:01 So guys, let us start with the answer of this question.

00:03 First of all, draw the pv diagram.

00:05 We've got p on the y axis and v on the x axis.

00:10 So you've got p3, then we have b2.

00:15 Here we have v1 and v3.

00:18 So these are the three intersection points.

00:22 Stage 1, 2, and 3.

00:28 So we are given the falling data for water, that is t1 is equal to 20 degrees centigrates, m is equal to 1 kg, p1 is equal to 300 khal, and then we've got p3 is equal to 600 kepascal, we've got v3 is equal to 0 .002 cubic meters.

00:56 So in this problem we got two stages.

00:58 One is constant volume process and the other is constant pressure process.

01:02 So to find total work and heat transfer we have to calculate it in each process and then we summit.

01:09 To make it easier, we have to look at the pv diagram.

01:12 So for this problem, during process two to three volume doesn't change.

01:17 So the final volume is the same as in state two.

01:21 So v3 is equal to v2 and that is equal to 0 .002 cubic meters.

01:31 So from saturated water table corresponding to temperature, t20 degrees centigrade, t1, 20 degrees centigrade, and you are 20 degrees we can obtain saturation pressure, specific volume, and internal energy.

01:48 So p saturation is equal to 2 .339 kilo -pascal.

01:55 Then we've got v1 is equal to vf, and which is 0 .00122 cubic meters per kg.

02:07 And we've got u1 is equal to uf and that is equal to 83 .94 kilojoules per kg and then we've got ufg is equal to 2318 .98 kilojoules per kg so the specific volume at stage 2 is equal to v2 is equal to capital b2 over m and that is 0 .002 cubic meters per kg...

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