1. Chapter 2 Basic Concepts of Thermodynamics 1

2. Thermodynamic System (System)
System is defined as a quantity of matter or a region in space chosen for study.
surroundings: The mass or region outside the
system.
Boundary: The real or imaginary surface that
separates the system from its surroundings 2

3. Closed System
Closed system (also known as a control mass)
Consists of a fixed amount of mass, and no mass can cross the boundary, i.e. no mass can enter or leave the closed system
Energy can cross the boundary
The volume of a closed system does not have to be fixed 3

4. Closed System

5. In isolated system, even energy is not allowed to cross the boundary5

6. Open System In Open system (also called control volume)
Both mass and energy can cross the boundary of
a control volume.
Boundaries of a control volume are called a control surface
Control surface can be real or imaginary
A control volume can be fixed in size and shape or it may involve a moving boundary
This involves mass flow such as a compressor, turbine, or nozzle. 6

8. System Properties1/2
Any characteristic of a system is called a property.
Some familiar properties are
Pressure, P,
Temperature, T,
Volume, V,
Mass, m. 8

9. System Properties2/2
Density is defined as mass per unit volume, ρ. Specific gravity (SG), or relative density, and is defined as the ratio of the density of a substance to the density of some standard substance at a specified temperature (usually water density at 4oC is (1000kg/m3), and thus, the SG =1 (ie density/ density water). For gases, SG= density/ density of Air). Specific volume defined as the volume per unit mass: ν=V/m = 1/ ρ

10. Types of Properties: Intensive properties are those that are independent of the size of a system, such as temperature, pressure, and density. Extensive properties are those whose values depend on the size—or extent—of the system such as : Mass m, Volume V, and total Energy E. Specific properties. Some examples of specific properties are specific volume (v=V/m) and specific total energy (e=E/m).

11. How To Differentiate Between Intensive and Extensive Properties?
• Divide the system into two equal parts with a partition
• If each part has the same property value as the
original system, the property is intensive
• If each part has only half the property value of the
original system, the property is extensive

12. State and Equilibrium1/2
The condition of a system at which a set of properties can
completely describe it
• At a given state, all the properties of a system have fixed
values
• If the value of one property changes, the state will change
Thermodynamics deals with
equilibrium states.
Equilibrium
• In an equilibrium state, there are no unbalanced potentials
(or driving forces) within the system

13. State and Equilibrium2/2
Thermodynamic Equilibrium:
A system is said to be in thermodynamic equilibrium if it maintains thermal, mechanical, phase, and chemical equilibrium.
Thermal Equilibrium: The temperature throughout the
entire system is uniform
Mechanical Equilibrium: The pressure throughout the
Phase Equilibrium: The mass of each phase is in
equilibrium
Chemical Equilibrium: Chemical composition does not
change with time, i.e.no chemical reactions take place 13

14. Processes and Cycles
Process: Any change that a system undergoes from one equilibrium state to
another
Path: The series of states through which a system passes during a process.
How to describe a process completely? one should specify the initial and
final states AND the path the process follows
Cycle: The system returns to its initial state at the end of the process,

15. Types of Processes
In most of the processes we study, one property of thermodynamic is held constant, also, the prefix iso- is often used to designate a process for which a particular property remains constant.
isothermal process, is a process during which the temperature T remains constant;
isobaric process is a process during which the pressure P remains constant; and
isochoric (or isometric) process is a process during which the specific volume v remains constant.
Isentropic process: Entropy held constant 15

16. Steady and Unsteady States
Steady state implies no change with time. The term uniform, however, implies no change with location over a specified region.
Unsteady state or transient, the opposite of steady
The steady-flow process:
a process during which a fluid flows through a control volume steadily. That is, the fluid properties can change from point to point within the control volume, but at any fixed point they remain the same during
the entire process (V, m & E), ie not change with time. 16

17. Forms of Energy1/4

18. Forms of Energy2/4
The total energy of a system is divided into two forms:
The macroscopic forms of energy are those which a system possesses as a whole with respect to some outside reference frame, such as kinetic and potential energies .
2. The microscopic forms of energy are those which are related to the molecular structure of a system and they are independent of outside reference frame. The sum of all microscopic forms of energy is called the internal energy of a system and is denoted by U. 18

19. Forms of Energy3/4
kinetic energy: The energy that a system possesses as a result of its motion relative to some reference
KE = ½ m v2 (J),
potential energy: The energy that a system possesses as a result of its elevation in a gravitational field and thus,
PE = mgz (J)
Internal Energy
The sum of all the microscopic forms of energy
that is denoted by U 19

20. Forms of Energy4/4
The total energy of a system consists of the kinetic, potential, and internal energies (ie the sum of all form of energy):
E = U + KE + PE
= U + ½ m v2 + mgz
Furthermore, when changes in kinetic and potential energy are negligible (eg Most closed systems remain stationary) , the change in total energy E becomes identical to the change in its internal energy U, and thus,
ΔE=ΔU 20

21. Types of associated energies with internal U
Portion of the internal energy of a system associated with the kinetic energies of the molecules is called the sensible energy.
The internal energy associated with the phase of a system is called the latent energy
The internal energy associated with the atomic bonds in a molecule is called chemical energy,
The energy associated with the strong bonds within the nucleus of the atom itself is called nuclear energy.
The only two forms of energy interactions associated
with a closed system are heat transfer (Q) and work
(W). An energy interaction is heat transfer if its driving
force is a temperature difference. Otherwise it is work. 21

22. Types of Energy

23. Temperature and the Zeroth Law of Thermodynamics:
A measurement of the overall microscopic kinetic energy
of a body or a system,
fast-moving molecules have high kinetic energy → they
have high temperature.
when an object is in contact with another object that is at a
different temperature, heat is transferred from the object
at higher temperature to the one at lower temperature
until both objects attain the same temperature
(ie Thermal Equilibrium).

24. Temperature and the Zeroth Law of Thermodynamics:
If two bodies are in thermal equilibrium with a third body, they are also in thermal equilibrium with each other
If the third body is a thermometer and it is in thermal equilibrium with two different bodies, then the two bodies must also be in equilibrium and at the same temperature 24

26. Example 1
Solution 26

27. Example 2
Convert 18 oC to oF
Convert 550 R to oC
R=460+F 27

28. Pressure
1 bar = 105 N/m2

29. Gage and Vacuum Pressure
Pgage = Pabs - Patm (for pressures above Patm)
Pvac = Patm Pabs (for pressures below Patm)
P gage
P atm
P vac
P atm
P abs
P atm
P abs
P abs = 0
Absolute vacuum 29

30. Pressure Conversion Factors
Pressure has units of kPa, psi, atm, inches of Hg, mm of Hg, ft of H2O, and in of H2O, etc.
29.92 in. of Hg = 1 atm
1 atm = 760 mmHg 30

31. The pressure P will denote absolute pressure unless specified otherwise. Often the letters “a” (for absolute pressure) and “g” (for gauge pressure) are added to pressure units (such as psia and psig) Example 1: A vacuum gauge connected to a chamber reads 5.8 psi at a location where the atmospheric pressure is 14.5 psi. Determine the absolute pressure in the chamber. Solution: Pabs = Patm - Pvac = = 8.7 psi

32. Example 2 What is the absolute pressure in psi of a vacuum of
15 inch of Hg? (1 atm=29.92 in Hg) 32

33. Example 3
Consider a diver swimming at 50 ft below the surface of the ocean. Determine the gauge and absolute pressure on diver in psi. 33

34. Pressure in Depth
pressure in a fluid does not change in the horizontal direction. Pressure in a fluid increases with depth because more fluid rests on deeper layers, and the effect of this “extra weight” on a deeper layer is balanced by an increase in pressure
P=Patm + ρgh and Pgage =∆P = P1 – P2 = ρgh (h = Pressure head) 34

35. The Manometer
The manometer is an instrument used to measure the pressure differences of liquids and/or gases.
Small and moderate pressure differences can be measured by the manometer.
P1 = P2 (horizontally)
and
P2 = Patm + ρgh 35

36. Solution: Example: SG = Density substance / water density
A manometer is used to measure the pressure in a tank. The fluid
used has a specific gravity of 0.85, and the manometer column
height is 55 cm. If the local atmospheric pressure is 96 kPa,
determine the absolute pressure within
the tank.
Patm=96 atm
P ??
gas
h=55cm
Solution:
SG = Density substance / water density
Density of substance = 0.85 (1000kg/m3) = 850 kg/m3
P = Patm + ρgh = 96 kPa kg/m3 * 9.81 m2/s *
0.55m (1 kPa/1000N/m2) * ( 1 N / 1kg. m/s2)
= kPa

37. P1=Patm + ρ1gh1 + ρ2gh2 + ρ3gh3 or
P1-Patm= ρ1gh1 + ρ2gh2 + ρ3gh3

38. P1+ ρ1g(a + h) = P2+ ρ2gh + ρ1ga or
P1–P2= ρ2gh + ρ1ga – ρ1g(a + h)

39. Determine the pressure (psi) at the bottom of a 10 inch
Example :
Determine the pressure (psi) at the bottom of a 10 inch
column of Hg (sp. g.=13.55).
Solution:
P= ρgh+ Patm 39

41. P1 + ρwater gh1 + ρoil gh2 = Patm + ρmercury gh3 or ρwater=1000 kg/m3,
Solution
P1 + ρwater gh1 + ρoil gh2 = Patm + ρmercury gh3 or
ρwater=1000 kg/m3,
ρmercury=13600 kg/m3 ,
ρoil=850 kg/m3
h1=0.1 m, h2=0.2 m , h3=0.35 m, Patm=85.6 kPa
Solving for P1 and substituting,
85.6 kPa + (9.81 m/s2)[(13,600 kg/m3) (0.35 m) -(1000 kg/m3) (0.1 m)
– (850 kg/m3) (0.2 m)] [(1 N/1 kg · m/s2)( 1 kPa/1000 N/m2)]
= kPa h3 h2 h1 1 2
Water
Oil
Mercury
ρwater= 1000 kg/m3
ρmercury= kg/m3
ρoil= 850 kg/m3 41

42. Example
Freshwater and seawater flowing in parallel horizontal pipelines are connected to each other by a double U-tube manometer, as shown in the figure. Determine the pressure difference between the two pipelines. Take the density of seawater at that location to be ρ=1035 kg/m3. The density of freshwater= 1000 kg/m3 and for mercury=13600 kg/m3 .Can the air (ρ=1.22 kg/m3) column be ignored in the analysis?

43. Solution
The densities of seawater and mercury are given to be sea = 1035 kg/m3 and Hg = 13,600 kg/m3. We take the density of water to be  w =1000 kg/m3.
Analysis Starting with the pressure in the fresh water pipe (point 1) and moving along the tube by adding (as we go down) or subtracting (as we go up) the terms until we reach the sea water pipe (point 2), and setting the result equal to P2 gives

44. Rearranging and neglecting the effect of air column on pressure,
Substituting,
Therefore, the pressure in the fresh water pipe is 3.39 kPa higher than the pressure in the sea water pipe.
Discussion: A m high air column with a density of 1.2 kg/m3 corresponds to a pressure difference of kPa. Therefore, its effect on the pressure difference between the two pipes is negligible.

45. Barometer and the Atmospheric Pressure
The atmospheric pressure (barometric pressure) is measured by a device called a barometer. The atmospheric pressure is measured by inverting a mercury -filled tube into a mercury container that is open to the atmosphere. Note that the cross- sectional area of the tube have no effect on the height of the fluid column of a barometer ,ie Patm = ρgh 45

46. and (g = 9.807 m/s2). Called 760mmHg. 1 mmHg = 1 torr = 133.3 Pa
standard atmosphere, which is defined as the pressure produced by a column of mercury 760 mm in height at 0 °C (ρ= 13,595 kg/m3)
and (g = m/s2). Called 760mmHg.
1 mmHg = 1 torr = Pa
1 atm = 760 mm Hg = 760 mm torr
If water instead of mercury were used to measure the standard atmospheric pressure, a water column will be about 10.3 m. How?
As we go up atmospheric pressure drops. Why? 46