Thermodynamics

THERMODYNAMICSCOURSE OUTLINE:

1 Introduction2&3 Some concepts and definitions4&5 Properties of a pure substance6 Work and heat 7 The first law of thermodynamics 8,9 First law analysis for a control volume 10 The second law of thermodynamics 11,12 Entropy13,14 Second law analysis for a control volume15,16 Cycles

ENGR. MANSAF ALI ABRO

1 Introduction

2,3 Some concepts and definitions

4,5 Properties of a pure substance

6 Work and heat

7 The first law of thermodynamics

8,9 First law analysis for a control volume

10 The second law of thermodynamics

11,12

Entropy

13,14 Second law analysis for a control volume

15,16

Cycles

For Notes and Self Study -Books

Lecture:1Introduction

What is Thermodynamics

Historical perspective

Philosophy of science note/Application area of thermodynamics

Thermodynamics

The name thermodynamics stems from the Greek words therme (heat) and dynamis (motion), which is heat in motion.

Thermodynamics is an engineering science, which deals with the science of “motion” (dynamics) and the transformation of “heat” (therme) into various other energy forms.

Examples:

1 The chemical energy of combustion of fossil fuels (oil,coal,gas), is used to produce heat which in turn is used to provide mechanical energy in reciprocating engines.

Thermodynamics

2 Uranium atoms are bombarded asunder and the nuclear energy released is used as heat. This heat is converted in nuclear power plants.

3 Hydro Electric power plant ( figure )

4 Wind energy power plants.

Historical perspective

Although the principles of thermodynamics have been in existence since the creation of the universe, It emerged as a science with construction of steam engines in England by Thomas Savery in 1697 and Thomas Newcomen in 1712.

The term thermodynamics was first used in a publication by Lord Kelvin in 1849. The first thermodynamic textbook was written in 1859 by William Rankine, a professor at the University of Glasgow.

Application area of thermodynamics

The heart is constantly pumping blood to all parts of the human body.

Various energy conversions occur in trillions of body cells. Body’s heat generated is constantly rejected to the environment. We try to

control this heat transfer rate by adjusting of our clothing to the environmental conditions.

Other applications of thermodynamics are :An ordinary house, household utensils such as air-conditioning systems, the refrigerator, the pressure cooker, the water heater, the shower, the iron, and even the computer and the TV.

On a larger scale, thermodynamics plays a major part in the design and analysis of automotive engines, rockets, jet engines, and conventional or nuclear power plants.

Application area of thermodynamics

Power plants

The human body

Air-conditioning

systems

Airplanes

Car radiators Refrigeration systems

LECTURE:2,3Some concepts and definitions

Macroscopic versus microscopic Fundamental/Primary and Secondary/derived

dimensions and units Thermodynamic system and control volume Properties and state of a substance Processes and cycles Zeroth law of thermodynamics

Macroscopic versus microscopic approach to study thermodynamics

How Gas is compressed inside cylinder?

What is Biology of Molecules?

Behavior of the gas particles inside the cylinder.

It would be sufficient to attach a pressure gauge to the container. Pressure gauge attached outside the container gives macroscopic approach/Classical thermodynamics that does not require a knowledge of the behavior of inside individual particles.

A more elaborate approach, based on the average behavior of large groups of inside individual particles, is called microscopic approach/Statistical thermodynamics.

Fundamental/Primary and Secondary/Derived dimensions and units

Some basic dimensions such as mass m, length L, time t, and temperature T are selected as primary or fundamental dimensions, while others such as velocity V, energy E, and volume V are derived from the primary dimensions and are called secondary dimensions, or derived dimensions.TABLE 1–1The seven fundamental (or primary)dimensions and their units in SIDimension (Unit)Length meter (m)Mass kilogram (kg)Time second (s)Temperature kelvin (K)Electric current ampere (A)Amount of light candela (cd)Amount of matter mole (mol)

Dimension Unit

Length Meter(m)

Mass Kilogram(kg)

Time Second(s)

Temperature Kelvin(K)

Electric current ampere(A)

Amount of light Candela(cd)

Amount of matter Mole(mol)

Thermodynamic systems and control volume

System: Any thing under consideration or study. Surrounding: The mass or region outside the system is called the surroundings. Boundary: The real or imaginary surface that separates the system from its surroundings is called the boundary.

i. Close System/Control mass:ii. Open System /Control Volume: Mass flow in and out of a system are modeled as control volumes(open systems).iii. Isolated system:ExampleContainer on fire with and withOut lid , and a thermo flask

Thermodynamic systems and control volume

Examples: Control Volume A water heater, a car radiator, a turbine, and a compressor all involve mass

flow and should be analyzed as control volumes (open systems) instead of as control masses (closed systems).

Properties and state of a substance

Any characteristic of a system is called a property. Some familiar properties are pressure P, temperature T, volume V, and mass m.

Intensive properties are those that are independent of the mass of a system, such as temperature, pressure, and density.

Extensive properties are those that are dependent of the mass of a system. Total mass, total volume, and total momentum are some examples of extensive properties.

Properties and state of a substance

Extensive properties per unit mass are called specific properties. Some examples of specific properties are specific volume (v V/m) and specific total energy (e E/m).

NOTE:An easy way to determine whether a property is intensive or extensive is to divide the system into two equal parts with an imaginary partition, as shown in Fig. 1–20. Each part will have the same value of intensive properties but half the value of the extensive properties.

STATE AND EQUILIBRIUM

State: A system not undergoing any change is state.

What is your state at the moment?

State properties have fixed values. If the value of even one property changes, the state will change to a different one. In Fig. 1–23 a system is shown at two different states.

Thermodynamics deals with equilibrium states.


Thermal equilibrium :If the temperature is the same throughout the entire system, as shown in Fig.1–24(b)

Mechanical equilibrium: If there is no change in pressure at any point of the system with time.

Equilibrium:


The State Postulate:

The state of a simple compressible system is completely specified by two independent, intensive properties.

Temperature and specific volume, are always independent properties, and together they can fix the state of a simple compressible system (Fig. 1–25).

Processes and cycles

Process: Any change that a system undergoes from one equilibrium state to another is called a process.

Path of Process: series of states through which a system passes during a process is called the path of the process (Fig. 1–26).


Quasistatic or Quasi-equilibrium process:

When a process proceeds in such a manner that the system remains infinitesimally close to an equilibrium state at all times, it is called a quasistatic, or quasi-equilibrium, process.

It is a slow process so that elements inside may adjust.


isothermal process: T remains constant.

Isobaric process: pressure P remains constant.

isochoric (or isometric) process: is a process during which volume v remains constant.

Cycle: A system is said to have undergone a cycle if it returns to its initial state at the end of the process. That is, for a cycle the initial and final states are identical.

Zeroth law of thermodynamics

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

Assignment : 1

Enlist as many as intensive , extensive and specific properties.

Enlist examples of thermodynamics in daily life , Each student should submit unique example.

Explain why the color of the sky is blue?

Assignment must be hand written.

Note:Submission Date:14-10-2015 (IC LAB)

Assignment : 2

What is Nuclear Power Plant? Explain its working principle with a schematic diagram.

What is Hydro Electric Power Plant? Explain its working principle with a schematic diagram.

What is Wind Energy Power Plant? Explain its working principle with a schematic diagram.

Assignment must be hand written.

Note:

Submission Date: 19-10-2015 (IC LAB)

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Quiz Number : 1

Q 1 What is Thermodynamics?Q 2 What is the applications of thermodynamics?Q 3 What is difference between classical and

statistical thermodynamics?Q 4 What is Zeroth law of thermodynamics?

Time Allowed 10 minutes.

Lecture 4,5Properties of a pure substance

The pure substance Vapor-liquid-solid phase equilibrium Independent properties Thermal equations of state

Pure Substance

A substance that has a fixed chemical composition throughout is called a pure substance.

It is not necessary that a pure substance is made up of a single chemical element(O2) or compound, however, A mixture of various chemical elements or compounds also qualifies as a pure substance as long as the mixture is homogeneous. (Air) , (N2) , (H20)

What is mixture of Oil and Water?

Is it a pure substance?

Pure Substance

Phases of A Pure Substance

The substances exist in different phases, e.g. at room temperature and pressure, copper is solid and water is a liquid.

There are 3 Principal phases• solid• Liquid• gas

Each with different molecular structures.

Pure Substance

Phase-change Processes of Pure Substances There are many practical situations where two phases of a pure

substances coexist in equilibrium.

Solid: strong intermolecular bond Liquid: intermediate intermolecular bonds Gas: weak intermolecular bond

Solid Liquid Gas

E.g. water exists as a mixture of liquid and vapor in the boiler and etc.

Phase-change Processes

Compressed liquid or sub cooled liquidIt is not about to vaporize

Saturated liquidIt is about to vaporize

Midway about the vaporization line

State 4 is a saturated vapor state

A substance at states between 2 and 4 is referred to as a saturated liquid–vapor mixture since the liquid and vapor phases coexist in equilibrium at these states.

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Once the phase-change process is completed, we are back to a single-phase region again (this time vapor), and further transfer of heat results inan increase in both the temperature and the specific volume .

as long as the temperature remains above 100°C no condensation will occur

A vapor that is not about to condense (i.e., not a saturatedvapor) is called a superheated vapor.


This constant-pressure phase-change process is illustrated on a T-v diagram in Fig. 3–11.


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It probably came as no surprise to you that water started to boil at 100°C. Strictly speaking, the statement “water boils at 100°C” is incorrect. The correct statement is “water boils at 100°C at 1 atm pressure.” The only reason water started boiling at 100°C was because we held the pressure constant at 1 atm (101.325 kPa). If the pressure inside the cylinder were raised to 500 kPa by adding weights on top of the piston, water would start boiling at 151.8°C.

Saturation

Saturation Temperature and Saturation Pressure

Saturation is defined as a condition in which a mixture of vapor and liquid can co-exist together at a given temperature and pressure.

Saturation pressure it’s the pressure at which the pure substance changes its phase, at given temperature.

Saturation temperature : it’s the temperature at which the pure substance changes its phase, at given pressure.

For a pure substance there is a definite relationship between saturation pressure and saturation temperature. The higher the pressure, the higher the saturation temperature

The graphical representation of this relationship between saturation temperature and saturation pressure at saturated conditions is called the Liquid Vapor saturation curve

Saturation

Latent Heat

Latent heat: The amount of energy absorbed or

released during a phase-change process.

Latent heat of fusion: The amount of energy absorbed

during melting. It is equivalent to the amount of energy

released during freezing.

Latent heat of vaporization: The amount of energy

absorbed during vaporization and it is equivalent to the

energy released during condensation.

At 1 atm pressure, the latent heat of fusion of water

is 333.7 kJ/kg and the latent heat of vaporization is

2256.5 kJ/kg.

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Why it takes shorter time for a pressure cooker to cook meal than at STP?

It is clear from Figure that Temperature increases with Pressure. Thus, a substance at higher pressures boils at higher temperatures. In the kitchen, higher boiling temperatures mean shorter cooking times and energy savings. A beef , for example, may take 1 to 2 h to cook in a regular pan that operates at 1 atm pressure, but only 20 min in a pressure cooker operating at 3 atm absolute pressure (corresponding boiling temperature: 134°C).

The atmospheric pressure, and thus the boiling temperature of water,decreases with elevation. Therefore, it takes longer to cook at higher altitudes than it does at sea level (unless a pressure cooker is used).

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Enthalpy and Entropy

Entropy: It’s the measure of dis orderness. To improve or beat entropy work is needed.

Example: a Maintained room, a fatty man , Global warming etc Enthalpy : In the analysis of processes such as power generation and

refrigeration , we come across to face a new property that is enthalpy. Sum of internal energy and product of pressure and volume is termed as

ENTHALPY

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Saturated Liquid and Saturated Vapor States

Tables A–4 and A–5 Use of A-4

The subscript f is used to denote properties of a saturated liquid, and thesubscript g to denote the properties of saturated vapor.

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During a vaporization process, a substance exists as part liquid and partvapor. That is, it is a mixture of saturated liquid and saturated vapor(Figure)

So, we need to know the proportions of saturated liquid and saturated vapor. For this a new property is defined as Quality.

Quality

Mixture of liquid and vapor

Quality

Quality (x) is defined as the ratio of the mass of the vapor to the total mass of both vapor and liquid(mixture)

The quality is zero for the saturated liquid and one for the saturated vapor (0 ≤ x ≤ 1)

For example, if the mass of vapor is 0.2 g and the mass of the liquid is 0.8 g, then the quality is 0.2 or 20%.

xmass

mass

m

m msaturated vapor

total

g

f g

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Consider a tank that contains a saturated liquid–vapor mixture. The volume occupied by saturated liquid is Vf, and the volume occupied by saturated vapor is Vg. The total volume V is the sum of the two:

Quality

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Quality

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Quality

INTERNAL ENERGY OF SYSTEM The internal energy U of a system is the

total of all kinds of energy possessed by the particles that make up the system.

Usually the internal energy consists of the sum of the potential and kinetic energies of the working gas molecules.

TWO WAYS TO INCREASE THE INTERNAL ENERGY, U.

HEAT PUT INTO A SYSTEM

(Positive)

+U

WORK DONE ON A GAS (Positive)

WORK DONE BY EXPANDING GAS: W is

positive

WORK DONE BY EXPANDING GAS: W is

positive

-UDecreas

e

-UDecreas

e

TWO WAYS TO DECREASE THE INTERNAL ENERGY, U.

HEAT LEAVES A SYSTEM Q is negative

Qout

hot

WoutWout

hot

Moisture Content

The moisture content of a substance is opposite of its quality.

Moisture is defined as the ratio of the mass of the liquid to the

total mass of both liquid and vapor

Recall the definition of quality x

Then

xm

m

m

m mg g

f g

m

m

m m

mxf g

1

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Moisture Content

Take specific volume as an example. The

specific volume of the saturated mixture

becomes

Pro

pert

y T

ab

le

Important Definition

o Critical point - the temperature and pressure above which there

is no distinction between the liquid and vapor phases.

o Triple point - the temperature and pressure at which all three

phases can exist in equilibrium.

o Sublimation - change of phase from solid to vapor.

o Vaporization - change of phase from liquid to vapor.

o Condensation - change of phase from vapor to liquid.

o Fusion or melting - change of phase from solid to liquid.

Ideal Gas Law Robert Boyle formulates a well-known law that states the pressure of a

gas expanding at constant temperature varies inversely to the volume,

or

constant2211 VPVP

As the result of experimentation, Charles concluded that the pressure of

a gas varies directly with temperature when the volume is held

constant, and the volume varies directly with temperature when the

pressure is held constant, or

2

1

2

1

2

1

2

1

T

T

P

Por

T

T

V

V

By combining the results of

Charles' and Boyle's

experiments, the following

relationship can be obtained

The constant in the above

equation is called the ideal gas

constant and is designated by

R; thus the ideal gas equation

becomes

In order to make the equation

applicable to all ideal gas, a

universal gas constant RU is

introduced

constantT

Pv

mRTPVorRTPv

M

RR U

For example the ideal gas constant for air, Rair

KkgkJM

RR

air

airUair ./2871.0

96.28

3144.8

)(

)(

The amount of energy needed to raise the temperature of a unit of

mass of a substance by one degree is called the specific heat at

constant volume Cv for a constant-volume process and the specific

heat at constant pressure Cp for a constant pressure process. They

are defined as

PP

vv T

hCand

T

uC

Using the definition of enthalpy (h = u + Pv) and writing the

differential of enthalpy, the relationship between the specific heats

for ideal gases is

The specific heat ratio, k is defined as

v

P

C

Ck

P V

P V

h u Pv

dh du RT

C dt C dt RdT

C C R

For ideal gases u, h, Cv, and Cp are functions of temperature alone.

The Δu and Δh of ideal gases can be expressed as

)( 1212 TTCuuu v

)( 1212 TTChhh P

Example 2.6

An ideal gas is contained in a closed assembly with an initial pressure and temperature of 220 kPa and 700C respectively. If the volume of the system is increased 1.5 times and the temperature drops to 150C, determine the final pressure of the gas.

Solution:given

1

1

2

2 1

1

220

70 273 343

2

15 273 288

1.5

state

P kPa

T K K

state

T K

V V

From ideal-gas law:

1 1 2 2

1 2

312

1

288220 10

1.5 343

123.15

PV PV

T T

VP

V

kPa

Example 2.7A closed assembly contains 2 kg of air at an initial pressure and temperature of 140 kPa and 2100C respectively. If the volume of the system is doubled and temperature drops to 370C, determine the final pressure of the air. Air can be modeled as an ideal gas.

Solution:given

1

1

2

2 1

1

140

210 273 483

2

37 273 310

2

state

P kPa

T K K

state

T K

V V

From ideal-gas law:

1 1 2 2

1 2

312

1

310140 10

2 483

44.93

PV PV

T T

VP

V

kPa

Example 2.8

An automobile tire with a volume of 0.6 m3 is inflated to a gage pressure of 200 kPa. Calculate the mass of air in the tire if the temperature is 20°C.

Solution:given

1

200 100

20 273 293

state

P kPa

T K K

From ideal-gas law:

3

3 2

.

300 10 0.6

287 293

2.14

Nm

Nmkg K

PVm

RT

m

K

kg

THE FIRST LAW OF THERMODYAMICS:

• The net heat put into a system is equal to the change in internal energy of the system plus the work done BY the system.

Q = U + W final - initial)

• Conversely, the work done ON a system is equal to the change in internal energy plus the heat lost in the process.

SIGN CONVENTIONS FOR FIRST LAW

• Heat Q input is positive

Q = U + W final - initial)

• Heat OUT is negative

• Work BY a gas is positive• Work ON a gas is negative

+Qin

+Wout

U

-Win

-Qout

U

APPLICATION OF FIRST LAW OF THERMODYNAMICS

Example 1: In the figure, the gas absorbs 400 J of heat and at the same time does 120 J of work on the piston. What is the change in internal energy of the system?

Q = U + W

Apply First Law:

Qin

400 J

Wout =120 J

Example 1 (Cont.): Apply First Law

U = +280 J

Qin

400 J

Wout =120 J

U = Q - W = (+400 J) - (+120 J) = +280 J

DW is positive: +120 J (Work OUT)

Q = U + W

U = Q - W

DQ is positive: +400 J (Heat IN)

Example 1 (Cont.): Apply First Law

U = +280 J

The 400 J of input thermal energy is used to perform 120 J of external work, increasing the internal energy of the system by 280 J Qin

400 J

Wout =120 J

The increase in internal energy is:

Energy is conserved:

FOUR THERMODYNAMIC PROCESSES:

Isochoric Process: V = 0, W = 0 Isobaric Process: P = 0 Isothermal Process: T = 0, U = 0 Adiabatic Process: Q = 0

Q = U + W

Q = U + W so that Q = U

ISOCHORIC PROCESS: CONSTANT VOLUME, V = 0, W = 0

0

+U -U

QIN QOUT

HEAT IN = INCREASE IN INTERNAL ENERGY

HEAT OUT = DECREASE IN INTERNAL ENERGY

No Work Done

ISOCHORIC EXAMPLE:

Heat input increases P with const. V

400 J heat input increases internal energy by 400 J and zero work is done.

B

A

P2

V1= V2

P1

PA P B

TA T B

=

400 J

No Change in volume:

Q = U + W But W = P V

ISOBARIC PROCESS: CONSTANT PRESSURE, P = 0

+U -U

QIN QOUT

HEAT IN = Wout + INCREASE IN INTERNAL ENERGY

Work Out

Work In

HEAT OUT = Wout + DECREASE IN INTERNAL ENERGY

ISOBARIC EXAMPLE (Constant Pressure):

Heat input increases V with const. P

400 J heat does 120 J of work, increasing the internal energy by 280 J.

400 J

BAP

V1 V2

VA VB

TA T B

=

ISOBARIC WORK

400 J

Work = Area under PV curve

Work P V

BAP

V1 V2

VA VB

TA T B

=

PA = PB

ISOTHERMAL PROCESS: CONST. TEMPERATURE, T = 0, U = 0

NET HEAT INPUT = WORK OUTPUT

Q = U + W AND Q = W

U = 0 U = 0

QOUT

Work In

Work Out

QIN

WORK INPUT = NET HEAT OUT

ISOTHERMAL EXAMPLE (Constant T):

PAVA =

PBVB

Slow compression at constant temperature: ----- No change in U.

U = T = 0

B

APA

V2 V1

PB

ISOTHERMAL EXPANSION (Constant T):

400 J of energy is absorbed by gas as 400 J of work is done on gas.

T = U = 0

U = T = 0

B

APA

VA VB

PB

PAVA = PBVB

TA = TB

ln B

A

VW nRT

V

Isothermal Work

Q = U + W ; W = -U or U = -W

ADIABATIC PROCESS: NO HEAT EXCHANGE, Q = 0

Work done at EXPENSE of internal energy

INPUT Work INCREASES internal energy

Work Out

Work In

U +U

Q = 0

W = -U U = -W

ADIABATIC EXAMPLE:

Insulated Walls: Q = 0

B

APA

V1 V2

PB

Expanding gas does work with zero heat loss. Work = -DU

Work and heat

Work Heat

The first law of thermodynamics Representations of the first law Specific internal energy for general

materials Specific enthalpy for general materials Specific heat capacity Caloric equations of state Time-dependency Final comments on conservation

First law analysis for a control volume

Detailed derivations of control volume equations

Mass conservation in brief Energy conservation in brief Some devices Introduction to the Rankine cycle Preview: equations of continuum

mechanics Problems

The second law of thermodynamics Statements of the second law Reversible and irreversible processes Analysis of Carnot heat engines The absolute temperature scale Analysis of Carnot refrigerators and heat

pumps Rejected thermal energy on a national

scale Problems

Entropy

Second law in terms of entropy Entropy for ideal gases Isentropic relations for an ideal gas Two cycles Entropy of mixing Summary statement of thermodynamics Problems

Second law analysis for a control volume

Rankine Rankine v/s Carnot Cycle efficiency

Cycles

Brayton Otto Diesel

Documents

Thermodynamics