Water at 1000 kPa, 250^∘ C is brought to saturated vapor in a piston/cylinder with an isothermal

Water at 1000 kPa, 250^∘ C is brought to saturated vapor in...

Key Concepts

First Law of Thermodynamics

The First Law of Thermodynamics is a statement of energy conservation that relates the change in internal energy of a system to the net heat transfer into the system and the work done by the system. This fundamental principle is used to analyze energy exchanges during thermodynamic processes, ensuring that the energy input, output, and changes within the system are properly balanced.

Saturated Vapor

The concept of saturated vapor refers to a state where a substance is at the boundary between its gaseous and liquid phases. This state is characterized by specific pressure and temperature conditions known as saturation conditions. Understanding saturated vapor is essential in phase change processes and in determining precise thermodynamic properties at the point of condensation or evaporation.

P-v Diagram Analysis

The Pressure–Volume (P-v) diagram is a graphical representation of the thermodynamic process, where the area under the process curve represents the work done by or on the system. Estimating work from the area on a P-v diagram provides a visual and approximate method for analyzing the work interaction, which is especially useful in processes with complex paths.

Heat Transfer

Heat transfer in thermodynamic processes involves the exchange of energy between a system and its surroundings. In an isothermal process, the heat transfer is directly correlated with the work done, since the internal energy remains constant. Accurately evaluating heat transfer is vital for energy balance and efficiency analysis in many practical applications.

Specific Work

Specific work refers to the work done per unit mass of a substance. It is calculated by integrating the pressure with respect to specific volume over the process path. Understanding specific work is essential in identifying performance parameters of systems where energy changes per mass unit are critical, such as in piston-cylinder arrangements.

Isothermal Process

An isothermal process is one in which the temperature remains constant throughout the entire process. In thermodynamics, this implies that any work done by or on the system is exactly balanced by heat transfer to maintain the constant temperature, making it a key concept for understanding energy interactions in processes that occur at a steady thermal state.

Transcript

00:01 In this question it is given that during an isothermal process, water is brought to saturated vapor state in a piston cylinder arrangement.

00:10 The initial pressure of the water was 100 kilo -pascal and the initial temperature was 250 degrees centigrade.

00:21 We are required to evaluate the specific work and heat transfer.

00:26 After that, we need to calculate specific work from the area under the pv.

00:32 Diagram and then we need to compare this specific work with the work obtained earlier so let's see how to solve this question refer table b .1 .3 the superheated vapor water so the value of specific volume b1 corresponding to pressure thousand kilo pascal and temperature 250 degrees centigrade is equals to 0 .23268 meter cube per kilogram.

01:17 Similarly from that table we get value of specific internal energy u1 is equal to 2 ,7009 .91 kilojoules per kilogram and entropy from this table s1 is equals to 6 .9246 kilojoule per kilogram kelvin and now refer table b .1 .1 the properties of saturated water.

01:58 So the pressure p2 corresponding to temperature 250 degree centigrade is equals to 3 ,9703 kilopascal.

02:14 Specific volume corresponding to temperature 250 degrees centigrade and this pressure is 0 .05013 meter per kilogram and from the same table the value of specific internal energy u2 is equals to 2 ,602 .37 kilojoule per kilogram and entropy s2 is equals to 6 .0729 kilojoules per kilogram kelvin.

02:59 Now let's find the heat transfer from entropy equation.

03:05 So let's write heat transfer q during process 1 to 2, is equals to t into s2 minus s1 now substitute all the values so we will have heat transferred during process 1 to 2 and this will be equals to temperature t that means 250 degrees centigrade plus we are adding 273 to convert this temperature into kelvin and it is multiplied by s2 that means 6 .0729 minus s 1 that means 6 .07 9 to 46.

03:46 So when we further calculate we get specific heat transfer q during process 1 to 2 is equals to minus 445 .6 kilojoules per kilogram...

Please give Ace some feedback