A cylinder fitted with a piston contains ammonia at 120 F, 20 % quality with a volume of 60 in

A cylinder fitted with a piston contains ammonia at 120 F,...

Key Concepts

Quality (Thermodynamic Property)

Quality is a fundamental property in thermodynamics that defines the mass fraction of vapor in a liquid-vapor mixture, and it is used to characterize the state of a substance when it is undergoing a phase change. It provides a means to interpolate between the properties of the saturated liquid and the saturated vapor in the mixture, which is essential in calculations involving energy and mass balances during phase transitions.

Phase Change in Saturated Mixtures

The concept of phase change in saturated mixtures refers to the situation where a substance exists partly as liquid and partly as vapor at a constant temperature and pressure, known as the saturation conditions. This is a critical state in many thermodynamic processes where a substance absorbs or releases latent heat, and understanding the properties at these saturation points is fundamental to analyzing energy transfer during phase changes.

Isothermal Process

An isothermal process is one in which the temperature remains constant throughout the entire process. In the context of thermodynamics, this implies that any work done by or on the system must be balanced by heat transfer to or from the surroundings to maintain a constant temperature. Such processes are pivotal in analyzing systems like pistons or cylinders where temperature control is maintained despite volume changes.

First Law of Thermodynamics for Closed Systems

The First Law of Thermodynamics relates the change in internal energy of a closed system to the net heat transfer into the system and the work done by the system. It is essentially a statement of energy conservation and is applied to calculate energy exchanges during processes such as expansion or compression. In problems involving constant temperature processes and phase change, it helps in equating the added heat to the work done, given that the change in internal energy may be zero for an ideal isothermal process.

Work Calculation in Quasi-static Processes

Work in quasi-static processes, especially in systems with a moving piston, is typically calculated using the integral of pressure with respect to volume. In an isothermal expansion or compression, this often involves integrating pressure, which may be a function of volume according to state equations or saturation properties, to determine the work done by or on the system. This concept is critical for understanding energy interactions in thermodynamic cycles.

Heat Transfer and Energy Balance

Heat transfer in thermodynamic processes is the means by which energy is exchanged between a system and its surroundings. In processes where work is performed, such as the expansion of a gas, the first law of thermodynamics is used to link the work done to the heat transferred, particularly in isothermal processes where the internal energy remains constant. Understanding this energy balance is essential for determining the performance and efficiency of thermal systems.

Transcript

00:01 In this question it is given that ammonia is contained in a piston cylinder arrangement at a temperature of 120 fernheite and the dryness fraction x is 20%.

00:13 The volume is 60 inch cube.

00:21 Now, during a constant pressure process, ammonia expands slowly.

00:27 Heat transfer takes place during this process and the process continues until all the liquid is gone.

00:34 We are required to find the work transfer and heat transfer during this process.

00:39 So let's see how to solve this question.

00:42 So we have initial temperature t1 is equal to 120 pernheid.

00:49 Plus let's add 450 .67 to convert this into ranking and this will be equals to 5779 .67 rennkine.

01:03 Also we have initial drainage section x1 and initial volume v1.

01:11 Since the process is isothermal therefore final temperature t2 will also be equals to 120 fahrenheit or we can write 579 .67 rennkine and during the process all the liquid is gone therefore final dryness section x2 is equal to 1 now refer table f .8 .1 so from there pressure p2 corresponding to temperature t2 is equal to 286 .5 lbf per inch square the corresponding b2 is is equal to 1 .04 56 feet cube per albm and corresponding entropy as 2 is equals to 1 .14 btu per albm are now let's apply the formula to calculate specific volume for state 1 so we can write v1 is equal to vf1 plus x1 into vfg 1 now substitute all the values so we will have v1 is equal to 0 .2836 plus x1 that means 0 .2 multiplied by vfg and from the table its value is 1 .0171.

02:54 When we further calculate we get 0 .2318 feed cube per albm.

03:03 Now let's apply the formula to calculate and try to calculate.

03:07 At state 1, so this will be equal to s1 is equal to sf1 plus x1 into sfg1.

03:19 Now substitute all the values, so we will have s1 is equal to 0 .3571 plus 0 .2 multiplied by 0 .7829...

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