Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Saturated R-134a vapor at \(100^{\circ} \mathrm{F}\) is condensed at constant pressure to a saturated liquid in a closed pistoncylinder system. Calculate the heat transfer and work done during this process, in Btu/lbm.

Short Answer

Expert verified
Answer: The heat transfer, \(Q_{12}\), during this process is approximately \(-61.235\, \mathrm{Btu/lbm}\), and the work done, \(W_{12}\), is approximately \(-1.805\, \mathrm{Btu/lbm}\).

Step by step solution

01

Find the initial state properties

We are given that the initial state of the system is a saturated R-134a vapor at \(100^{\circ} \mathrm{F}\). Convert the temperature to Kelvin by adding \(273.15\) to the Fahrenheit temperature, then use the Refrigerant-134a tables to locate the initial state properties. \(T_{1} = 100^{\circ}\mathrm{F} \times \frac{5}{9} + 273.15 \approx 310.93\,\mathrm{K}\) From R-134a tables, initial properties at \(T_1 \approx 310.93\,\mathrm{K}\): \(h_1 = 98.91\,\mathrm{Btu/lbm}\) \(v_1 = 4.038 \times 10^{-2}\,\mathrm{ft}^3\mathrm{/lbm}\) \(P_1 \approx 136.26\,\mathrm{psia}\)
02

Find the final state properties

The final state of the system is a saturated R-134a liquid at the same pressure as the initial state. We will use the R-134a tables again to locate the final state properties. From R-134a tables, saturated liquid properties at \(P_1 \approx 136.26\,\mathrm{psia}\): \(h_2 = 39.48\,\mathrm{Btu/lbm}\)
03

Calculate the heat transfer

In a piston-cylinder system, heat transfer \(Q_{12}\) can be calculated from the enthalpy changes between the initial and final states, as well as the work done during the process according to first law of thermodynamics: \(Q_{12} = \Delta h + W_{12}\) Enthalpy change: \(\Delta h = h_2 - h_1\) Calculate \(\Delta h\): \(\Delta h = 39.48\,\mathrm{Btu/lbm} - 98.91\,\mathrm{Btu/lbm} \approx -59.43\,\mathrm{Btu/lbm}\)
04

Calculate the work done

In a closed piston-cylinder system undergoing a constant pressure process, work done \(W_{12}\) is given by: \(W_{12} = P \times{\Delta V}\) Since the volume can be expressed as \(V=m\times v\), we can rewrite the work done expression as: \(W_{12} = P \times \Delta(mv)\) Dividing by \(m\), we get the work done per unit mass: \(W_{12} = P \times{\Delta v}\) Calculate the final specific volume \(v_2\) from R-134a tables using the final state pressure: \(v_2 = 0.02933 \,\mathrm{ft}^3\mathrm{/lbm}\) Then, calculate the specific work done: \(\Delta v = v_2 - v_1 \approx -0.01038\,\mathrm{ft}^3\mathrm{/lbm}\) \(W_{12} = 136.26\,\mathrm{psia} \times (-0.01038\,\mathrm{ft}^3\mathrm{/lbm}) \times \frac{1}{778.17} \approx -1.805\,\mathrm{Btu/lbm}\)
05

Calculate the heat transfer

Now, we can calculate the heat transfer during the process using the enthalpy change and the work done: \(Q_{12} = \Delta h + W_{12}\) \(Q_{12} = -59.43\,\mathrm{Btu/lbm} - 1.805\,\mathrm{Btu/lbm} \approx -61.235\,\mathrm{Btu/lbm}\) During this condensation process, both heat transfer and work done are negative, which means the system loses heat and work is done on the system. The heat transfer, \(Q_{12}\), during this process is approximately \(-61.235\,\mathrm{Btu/lbm}\) and the work done, \(W_{12}\), is approximately \(-1.805\,\mathrm{Btu/lbm}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Saturated Vapor
Understanding the concept of a saturated vapor is crucial when dealing with refrigeration cycles or any thermodynamic problem that involves phase changes. A saturated vapor is a state where a substance exists simultaneously in its liquid and vapor (gaseous) phases at an equilibrium condition. This occurs at a specific combination of pressure and temperature unique to each substance, known as the saturation point.

For example, in the context of the R-134a refrigerant used in the exercise, the saturated vapor state is where the refrigerant is just about to condense into a liquid. Any additional removal of energy (heat transfer out of the system) will lead to condensation. This is what happens inside many refrigeration systems where the refrigerant absorbs heat at low pressure and releases it at a higher pressure, often passing through the saturated vapor state during this cycle.
R-134a Refrigerant Properties
R-134a, also known as Tetrafluoroethane, is a hydrofluorocarbon (HFC) refrigerant used in automobile and domestic refrigeration applications. It's known for its low ozone depletion potential and has been used as a replacement for more environmentally harmful refrigerants, such as R-12.

The properties of R-134a are critical for engineers and technicians when designing and analyzing refrigeration systems. These properties include saturation temperature and pressure, specific volume, specific enthalpy, and specific entropy. Thermodynamic tables or property software allow us to find these properties at various temperatures and pressures. For instance, when the refrigerant is a saturated vapor at 100°F, we can find its associated enthalpy and specific volume, as shown in the given exercise.

Importance in Calculations

Knowing the R-134a properties allows us to determine the heat transfer and work done in thermodynamic processes, which are pivotal in ensuring the efficiency and effectiveness of refrigeration systems.
Enthalpy Change
Enthalpy change, typically denoted as \( \Delta h \), is the measure of the total heat content change within a system undergoing a thermodynamic process. It is an essential concept in thermodynamics as it relates directly to the energy transfers that occur during phase changes, heating, or cooling processes.

The calculation of enthalpy change is straightforward when the initial and final states of the system are known. By referring to the specific enthalpy properties of the substance at these states, as done in the exercise for R-134a refrigerant, the change in enthalpy can be determined by subtracting the initial enthalpy from the final enthalpy of the process.

Significance in Heat Transfer

Enthalpy change is significant in the computation of heat transfer, especially in closed systems, where it is often the primary form of energy transfer. In many engineering applications, controlling and measuring enthalpy change is key to managing energy efficiency.
Piston-Cylinder System
A piston-cylinder system is an essential component of many thermodynamic devices, acting as the 'working volume' for the process fluid, often in engines and compressors. In such a system, a piston moves within a cylindrical chamber, allowing for volume change and thus causing pressure variations and work interactions with the environment.

In the exercise, we see a closed piston-cylinder system used to model the condensation process of R-134a. The work done by or on the system during a thermodynamic process like compression or expansion can be calculated using the pressure and the change in volume.

Application

The piston-cylinder system demonstrates basic thermodynamic principles like work and energy transfer, which are applicable in real-world engineering systems such as refrigeration cycles, hydraulics, and internal combustion engines. Its simplicity makes it an ideal model for students to grasp fundamental concepts of heat transfer and work interactions in thermodynamics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A piston-cylinder device initially contains \(0.35-\mathrm{kg}\) steam at \(3.5 \mathrm{MPa}\), superheated by \(7.4^{\circ} \mathrm{C}\). Now the steam loses heat to the surroundings and the piston moves down, hitting a set of stops at which point the cylinder contains saturated liquid water. The cooling continues until the cylinder contains water at \(200^{\circ} \mathrm{C}\). Determine \((a)\) the final pressure and the quality (if mixture), \((b)\) the boundary work, \((c)\) the amount of heat transfer when the piston first hits the stops, \((d)\) and the total heat transfer.

Argon is compressed in a polytropic process with \(n=\) 1.2 from \(120 \mathrm{kPa}\) and \(10^{\circ} \mathrm{C}\) to \(800 \mathrm{kPa}\) in a piston- cylinder device. Determine the work produced and heat transferred during this compression process, in \(\mathrm{kJ} / \mathrm{kg}\).

A piston-cylinder device, with a set of stops on for the top, initially contains 3 kg of air at \(200 \mathrm{kPa}\) and \(27^{\circ} \mathrm{C}\). Heat is now transferred to the air, and the piston rises until it hits the stops, at which point the volume is twice the initial volume. More heat is transferred until the pressure inside the cylinder also doubles. Determine the work done and the amount of heat transfer for this process. Also, show the process on a \(P\) -v diagram.

Design an experiment complete with instrumentation to determine the specific heats of a gas using a resistance heater. Discuss how the experiment will be conducted, what measurements need to be taken, and how the specific heats will be determined. What are the sources of error in your system? How can you minimize the experimental error?

A mass of \(1.5 \mathrm{kg}\) of air at \(120 \mathrm{kPa}\) and \(24^{\circ} \mathrm{C}\) is contained in a gas-tight, frictionless piston-cylinder device. The air is now compressed to a final pressure of 600 kPa. During the process, heat is transferred from the air such that the temperature inside the cylinder remains constant. Calculate the work input during this process.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free