A vertical cylinder fitted with a piston contains 10 lbm of R-22 at 50 F, shown in Fig. P 5.64 H

A vertical cylinder fitted with a piston contains 10 lbm of...

A vertical cylinder fitted with a piston contains 10 lbm of $\mathrm{R}-22$ at $50 \mathrm{F}$, shown in Fig. $\mathrm{P} 5.64$ Heat is transferred to the system, causing the piston to rise until it reaches a set of stops at which point the volume has doubled. Additional heat is transferred until the temperature inside reaches $120 \mathrm{F}$, at which point the pressure inside the cylinder is $2001 b f /$ in $^{2}$ a. What is the quality at the initial state? b. Calculate the heat transfer for the overall processs

A vertical cylinder fitted with a piston contains 10 lbm of $\mathrm{R}-22$ at $50 \mathrm{F}$, shown in Fig. $\mathrm{P} 5.64$ Heat is transferred to the system, causing the piston to rise until it reaches a set of stops at which point the volume has doubled. Additional heat is transferred until the temperature inside reaches $120 \mathrm{F}$, at which point the pressure inside the cylinder is $2001 b f /$ in $^{2}$
a. What is the quality at the initial state?
b. Calculate the heat transfer for the overall processs

Key Concepts

Heat Transfer Calculation

Heat transfer calculation involves quantifying the energy transferred into or out of a system during a process. This requires an understanding of both the specific thermodynamic properties of the fluid and the work interactions, and it often relies on applying the energy balance over multiple process stages. Accurate heat transfer computation is essential for predicting process outcomes in thermodynamic cycles.

Energy Balance for Closed Systems

The first law of thermodynamics for a closed system is used to relate the energy added or removed (via heat transfer and work) to the changes in internal energy of the system. This analysis is central to determining how different stages of a process, such as heating or expansion, affect temperature, pressure, and phase-relative properties.

Thermodynamic State Properties

This refers to the properties that define the state of a substance, such as temperature, pressure, and specific volume. In thermodynamic analyses, these properties allow you to determine the phase of the substance (liquid, vapor, or mixture) and are often obtained from property tables. Recognizing the state properties is crucial for characterizing the conditions of a substance at different stages of a process.

Phase Change and Quality

This concept involves the two?phase region where a fluid exists simultaneously as both liquid and vapor. The quality is defined as the mass fraction of the vapor within the mixture. Determining the quality of a substance is essential in problems involving phase change because it indicates the proportion of the substance that has transitioned between phases under given thermodynamic conditions.

Piston–Cylinder Process Analysis

In piston–cylinder devices, the working fluid is contained in a variable volume system where the piston movement is associated with work interactions. Understanding the behavior of the system as the piston moves—whether it is due to heating, cooling, or phase change—is key to analyzing how changes in volume and pressure relate to energy transfer and state changes.

Transcript

00:02 Refligent r22 is contained in a given state within a piston cylinder device.

00:08 Heat is then transferred to the system and this causes the piston to rise until it reaches a set of stops at which point the volume has now doubled.

00:18 Additional heat is then transferred until the temperature reaches a given value and the pressure reaches a different value.

00:25 We want to calculate first the quality of the initial state as well as the total heat transfer for the process.

00:32 So if we take the r20 as our control volume, we have a control mass going through three processes, one to two and two to three.

00:41 So three states and two processes another.

00:45 Now, we have that also that v3 is equal to v2, and this is two times the initial volume, the volume doubles, two times v1.

01:01 And if we go to our tables, we get state 3 has a specific volume v3 of 0 .2959 cubic feet per kg.

01:25 So having v3 essentially gives us v2 and v1, these specific volumes.

01:33 Now the specific internal energy u3 is equal to the enthalpy at the state, minus the pressure time specific volume.

01:44 So again, from our tables, this is 117 minus 200 times 0 .2959, and then conversion factor 144 over 778, which gives us a specific internal energy of 106 .1 btu per pound.

02:15 So now that we have state three, we can actually determine state one and two.

02:22 So firstly, state one has specific volume v1 to be 0 .14795.

02:39 We've calculated v3, but we know this is equal to 0 .0 -1 -282.

02:47 So we write this in terms of the quality...

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