UNIT-1 [Thermodynamics (R16) JNTU Kakinada]

Contents

System, boundary, Surrounding, control volume, Universe,

Types of Systems, Macroscopic and Microscopic viewpoints, Concept of Continuum,

Thermodynamic Equilibrium, State, Property, Process, Cycle – Reversibility – Quasi – static

Process, Irreversible Process, Causes of Irreversibility – Energy in State and in Transition,

Types, Work and Heat, Point and Path function. Zeroth Law of Thermodynamics –Concept of

Temperature – Principles of Thermometry –Reference Points – Const. Volume gas Thermometer

LECTURE CONTENT

Thermodynamic system

It can be defined as the quantity of matter which is focused under certain thermodynamic processes.

Surroundings:

The space around the TD system is known as surroundings of TDS. We can simple say that the external presented out of system is known as surroundings.

Boundary:

The envelop which it is covered the TD system is known as boundary of TD system.

Universe:

The combination of Thermodynamic system, Boundary and Surroundings is treated as Universe of Thermodynamics. 

Universe = System + Boundary + Surroundings

Thermodynamic Systems: The systems which are undergone for TD process are known as TD systems.
Thermodynamic systems are classified as below 3 types.
1) Closed system
2) Open system
3) Isolated system

Closed TD system:
In which system the energy crosses the boundary and mass cannot crosses  the boundary is known as closed thermodynamic system.
Example:
Water boiling in side of vessel with lid close.
In this case the heat energy can transfer from water to out of the surroundings due to convection phenomenon. But the matter presented in side the vessel doesn't crosses.

Open TD system:
In which system both energy and mass crosses the boundary is known as open TD system.
Example:
Water boiling inside a vessel with out lid. Energy crosses the boundary as discussed in closed TD system. Mass transfer occurs as water converts in to steam / vapours.

Isolated TD system:
The system which it does not allows mass & energy cross over the boundary is known as isolated TD system.
Example:
*There is no perfect Isolated system in universe. Ice box insulated with thermocoal can be assumed as isolated TD system.

Control volume:
Take an open system i.e, it has both mass and energy flowing through the system. If you have to analyze this system you cant take a fixed amount of mass as you do in a closed system because this mass wont be fixed in the space and is constantly flowing. Therefore you have to take a volume through which the mass flows over a period of time and then determine the properties by observing this fixed volume. This volume is known as a control volume and the boundary of the control volume is known as a control surface.

Macroscopic and Microscopic Approaches

Thermodynamics studies are undertaken by the following two different approaches:

• Macroscopic approach (Macro mean big or total)

• Microscopic approach (Micro means small)

The state  or condition of the system can be completely described by measured values of pressure, temperature and volume which are called macroscopic or time-averaged variables. In the classical thermodynamics, macroscopic approach is followed. The results obtained are of sufficient accuracy and validity. Statistical thermodynamics adopts microscopic approach. It is based on kinetic theory. The matter consists of a large number of molecules, which move, randomly in chaotic fashion. At a particular moment, each molecule has a definite position, velocity and energy. The characteristics change very frequently due to collision between molecules. The overall behavior of the matter is predicted by statistically averaging the behavior of individual molecules.Microscopic view helps to gain deeper understanding of the laws of thermodynamics. However, it is rather complex, cumbersome and time consuming. Engineering thermodynamics analysis is macroscopic and most of the analysis is made by it.
These approaches are discussed below:

Macroscopic Approach:

• In this approach a certain quantity of matter is considered without taking into account the events occurring at molecular level. In other words this approach to thermodynamics is concerned with gross or overall behavior. This is known as classical thermodynamics.

• The analysis of macroscopic system requires simple mathematical formula.

• The value of the properties of the system are their average values. For examples consider a sample of gas in a closed container. The pressure of the gas is the average value of the pressure exerted by millions of individual molecules.

• In order to describe a system only a few properties are needed.

Microscopic Approach:

• The approach considers that the system is made up of a very large number of discrete particles known as molecules. These molecules have different velocities are energies. The values of these energies are constantly changing with time. This approach to thermodynamics, which is concerned  directly with the structure of the matter, is known as statistical thermodynamics.

• The behavior of the system is found by using statistical methods, as the number of molecules is very large. So advanced statistical and mathematical methods are needed to explain the changes in the system.

• The properties like velocity, momentum, impulse, kinetic energy and instruments cannot easily measure force of impact etc. that describe the molecule.

• Large numbers of variables are needed to describe a system. So the approach is complicated.

First Law of Thermodynamics N.F.E.E  Applications (Problems with solutions)

Q1) In a nonflow process there is heat transfer loss of 1055 kJ and an internal energy increase of 210 kJ. Determine the work transfer and state whether the process is an expansion or compression.
Solution: 

Closed system for which the first law of Thermodynamics applies,

Q - W = 🔺U

1055 – W = 210

Hence the work done can be found as: W = -1265 kJ

Since negative, it must be work input, i.e compression.

Q2) Ina non flow process carried out on 5.4 kg of a substance, there was a specific internal energy    decrease of 50 kJ/kg and a work transfer from the substance of 85 kJ/kg. Determine the   heat transfer and state whether it is gain or loss.

Solution:

Closed system for which the first law of Thermodynamics applies, 
Q - W = 🔺U

Q         = DU + W

=[ 5.4*(-50)] +[5.4*85]

= + 189 kJ,

Since Q is positive, it implies heat is entering the control volume, i.e Gain.

Q3)  During the working stroke of an engine the heat transferred out of the system was 150 kJ/kg of the working substance. If the work done by the engine is 250 kJ/kg, determine the change in internal energy and state whether it is decrease or increase.

Solution:

Closed system for which the first law of Thermodynamics applies, 
Q - W = 🔺U

Hence

 🔺U     = Q – W

= (-150) – 250

= -400 kJ/kg

Since the sign is negative, there is a decrease in internal energy.

Q4)     Steam enters in to cylinder fitted with a piston at a pressure of 20 MN/m2 and a temperature of 500 deg C. The steam expands to a pressure of 200 kN/m2 and a temperature of 200 deg C. During the expansion there is a net heat loss from the steam through the walls of the cylinder and piston of 120 kJ/kg. Determine the displacement work done by one kg of steam during this expansion.

Solution:

State 1

at 20 MPa, 500 C: u = 2942.9 kJ/kg

State 2

at 200 kPa, 200C: u = 2654.4 kJ/kg

Closed system for which the first law of Thermodynamics applies, 

Q - W = 🔺U

Rearranging to determine the work done:

W = Q - 🔺U = (-120) – (2654.4 -2942.9) = 168.5 kJ/kg