Thermo Chap III

7/28/2019 Thermo Chap III

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FIRST LAW OF THERMODYNAMICS

One of the basic needs of Thermodynamics is to understand

Energy. The law which governs Conservation of Energy is known

as FIRST LAW OF THERMODYNAMICS. This law is the

cornerstone of Thermodynamics and we need to understand it

fully. However before we look at the Law itself, we need to define

two major components of the First Law which is

HEAT INTERACTION and WORK INTERACTION

HEAT INTERACTION

HEAT is the form of energy which is transferred from a system to

its surroundings , because of the Temperature Difference between

the system and the surroundings.

The form of energy which is transferred is known as

THERMAL ENERGY.

Heat Interaction is the actual transfer of Thermal Energy between

the system and its surroundings due to the temperature difference

between them. It flows from High Temp to low Temp

The Thermal Energy must cross the boundary of the system

If thermal energy crosses the boundary then the state of the

system will change , hence heat Interaction is a process. The units

of Heat Interaction is JOULES

System at T1

Surroundings at T2

Q = Heat flow or heat interaction as T1

> T2

The interaction is ve for the system and+ve for the surroundings

Q is +ve when it is given to the systemQ is ve when it is given by the system


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Since Heat interaction is a process so the change of state can be

shown as a path on a property diagram. This can be shown on a

P-T diagram

Now we can also have a process in which there is no Heat transfer

at all . Such a process is called ADIABATIC PROCESS.

Since Heat transfer is a process so we use the symbol

Q1-2 meaning that heat transfer during process 1 to 2.

We can also have 1 21 2

/Q

q Heat transfer unit massmass

= =

The rate of Heat transfer is*

1 2Q QTime

=

[ ]

2* *

1 2

1* *

1 2

varies

constant

Now if Q with time then Q Q dt

Now if Q is then Q Q time

=

=

T2

T1

T

PThe system undergoes thepath as shown and as a resultof the process the state of thesystem changes. The processwill stop when systemachieves temperature of thesurroundings.


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MODES OF HEAT TRANSFER : There are three methods by

which Heat Transfer takes place. These are Conduction,

Convection and Radiation.

CONDUCTION is the transfer of energy from high energy particles

to low energy particles. It takes place in Solids,Liquids, and Gases.

It is predominant in Solids and less dominant in Liquids and

Gases. It is governed by FOURIER LAW

*

cross sectional area

Cond

dTQ kA where k Thermal Conductivity

dxA

dTTemperature Gradient in flow direction

dx

= =

=

=

CONVECTION is the transfer of energy , due to physical

movement of mass and conduction.

The mass has to have a velocity. The velocity can be generated by

a. External Forces ( Forced Convection)

b. Density Changes (Free convection)

It is governed by NEWTONS LAW OF COOLING

( )*

Convection s f s f Q hA T T with T T = >

Tf

h

Tsurface

A = surface areah = Heat transfer

co-efficient dueto convection

A = surface Area


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RADIATION : Thermal Energy transmitted in the form of electro-

magnetic waves. It does not need a medium. It is governed by the

STEPHAN-BOLTZMANN LAW . The energy is emitted due to the

fact that a body has temperature above absolute Zero.

It also depends upon the type of body emitting Radiation

For a perfect Radiator which is called a black body

( )*

4

emitted SurfaceQ A T= and if the body is not a perfect radiator

then it is called a Real Body and here

( ) ( )*

4

1

emitted SurfaceQ A T where emmissivity of body

For a black body

= =

=

Bodies also have the capability to absorb Radiation. This is

governed by

RadiationAbsorbed

Radiation Recieved

1

Absorbitivity

For a black body

= =

=

Black Body absorbs all Radiation Received so thus

* *

Absorbed Incident or recievedQ Q=Thus is a body is emitting as well as absorbing .

The net Heat Transfer by Radiation is thus dependent of the rates

of emission and absorbing. If a body is completely enclosed by

another body then

12

( )( )

*4 4

1 21For Body

Q A T T =


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We will later see how these are evaluated in Semester V

WORK is also an energy transfer between a system and its

surroundings. It is not due to difference of temperature but it is due

to a FORCE ACTING IN A SPECIFIC MANNER.

We willsee that work is generated in several ways

The units of Work is also Joules =N-m

*

W Joulesw Work per unit mass in

mass kg

W JoulesPower W in units of Watts

Time kg

=

= = =

Surr

System

Pistonmoving

Moving BoundaryWork

System

Shaft Work

RI

Electrical Work

System

Q= -ve

Q= +veW = -ve

W = +ve Heat and workinteractions imply achange of state ofsystem


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Heat and Work Interactions

They are

a. Interactions which cross boundaries

b. They are not properties.

c. They are associated with a process and occur because

of change of state.

d. Both are path functions , and thus the value of W and

Q depend upon the path taken.

2

1 2

1

2 1

det minW W and we have to er e the Inexact Differential and

and it is not equal to W W as Work is not a property

=

2

2 1

1

2 1

V dV V V This is an exact differential because

V and V are properties

= =

1 2 1 2Q and W are never written as W We will always have to specify the path.

A

B

C

2

1

A B C

A B C

W W W

Q Q Q


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Lets look at a few examples of how work and heat can be

assessed considering the system and its surroundings


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FORMS OF WORK

System is room which is completelyinsulated.Candle located in room.

Q=0 and W=0Isolated system

200 oCSystem is potato at 25 oC which islocated in an oven having T=200 oCQ= +ve and W=0

Potato

Room

Oven

-ve+ve

System is oven and element. Thesystem is completely insulated.Q= 0 and W=-veWork is Electrical work

Electronscrossboundary

Element

HeatGiven

Oven

Oven

System is oven and is not insulated.Q= +ve0 and W= 0Heat given by fire.


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Electrical Work: When electrons cross the from the system

boundary then electrical work is done by the system. If electrons

cross into the system then electrical work is done on system

We = V(N) where V=voltage and N = Coulombs of electron

Now*

( )eN

Current I soW V I Power Time

= = = =

and thus power is expressed in Watts.

( ) ( )

( ) ( ) ( )

2

1

e

e

If V and I are functions of time then W V I dT

If V and I are not functions of time then W V I t

=

=

MECHANICAL WORKFrom Physics we know that W = Force X Distance = F(s)

Insulated

Ice Waterat 0oC

Oil at 0oC10 kJ Initially Oil ( System A) and

Ice Water (System B) are at0oC.

When 10 kJ of electrical work is given to system A , it converts toheat and increases temperature of Oil and it rises.Now as oil gets heated it gives heat to ice water and some icemelts.Eventually both systems go to 0oC.System A gets 10 kJ of Work and gives 10 kJ of Heat to ice waterand comes back to its original state.System B gets 10 kJ of Heat from system A

For the combined System A+BWin

= 10kJ and Qout

= 0

We should be sure about our system definition and theinteractions


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If force is not constant then

2

1

W Fds=

This form of work is called Mechanical Work

Now in Thermodynamics we have system and surroundings.Here

Work Interaction requires that we have

a. Force Acting on system boundary

b. Movement of boundary due to force

If we have forces on boundary and no movement of boundary

W =0

If boundary moves without any force acting on it then again

W =0

So in Thermodynamics Mechanical Work is only possible due to

Moving Boundary work due to a force

Moving whole system work due to a force.

We will now look at them in details

MOVING BOUNDARY WORK


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Let us look at a cylinder with gas in it. The cylinder has a piston

which can move.

Generally if the process is very fast , then properties inside the

system do not change uniformly.

THERMODYNAMICALLY PROPERTY CHANGE SHOULD BE

UNIFORM IN THE SYSTEM.

Thus to analyse processes we use the concept of QUASI-STATIC

or QUASI-EQUILIBRIUM PROCESS . This requires that the

process is so slow that the property change is uniform in the

system.

NON QUASI-STATIC PROCESS CAUSE LOSES AND EFFECT

EFFICIENCY

For our course we will assume that all processes are QUASI-

STATIC.

P

AreaThe pressure of the gas will force thepiston up, and it moves a distanceds. The Pressure is constant and isabsolute Pressure

so moving boundary work is

orPdV work

ds


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Now when the boundaries expand then dV is positive. So PdV

work will be positive when the boundaries expand. So expansion

gives positive work.

When boundaries are compressed then dV is negative. So here

PdV work will be negative. Hence Compression gives negative

work.

2

1

Now PdV work is W PdV

so we plot the processs on a P V diagram

=

The area under the path of process 1-2 is the PdV work as2

1

W PdV Moving Boundary Work = =

so if we have three different processes between the same states

Although the change of state is same but the path decides the

value of Work. Here Process A is the best.

1

2

P

V

1 2

1

2

P

V

A

B

C

2

1

2

1

2

1

10

8

5

A

B

C

W PdV kJ

W PdV kJ

W PdV kJ

= =

= =

= =

Area is different undereach curve


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Now supposing we have a process which undergoes a cycle as

shown in the next diagram

The process starts from State 1 goes to State 2 and again comes

back to 1.

When initial state is equal to the final state we have a cycle.

2 1

1 2

1 2 2 1

2 1

Net for cycle

Compression Work PdV ve Expansion Work PdV ve

W W W ve

because area under curve is greater

= = = = +

= + = +

This is how most engines work by operating in a cycle and getting

positive work.

We can also have shaft work in that we use the net work of the

cycle to rotate a shaft ( Car Engine)

The boundary can also be made to expand by heating it to get

positive work. ( Baloon rising)

Basically Mechanical work is mostly due to PdV work and Shaft

work. We represent shaft work by WX .We will later learn how to

evaluate the value of shaft work as it requires information about

Torque and RPM.

We shall now look at a few examples.

2

1

P

V

2 1


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PROCESSES IN THERMODYNAMICS AND WORK

EVALUATION

Ideal Gas Polytropic Process:- For Ideal gases generally we have

a process where P,V and T change together. This is called a

POLYTROPIC PROCESS. It is represented by the equation

( )

2 2 1 1

2 1b

1 1

1 1 2 2

2 2 2

b

Constant=C where n is an index depending

upon the type of polytropic process. But n 1

So for such a process the PdV work can be

W1

W

n

n n

n

n n

n

PV

V VPdV CV dV C n

but C PV P V

P V V

+ +

=

= = = + = =

=

( ) ( ) ( )

( )

1 1

1 1 1 2 2 1 1

1 1 1 1 2 2 2 2

2 2 1 1b

2 1

b

1 1

W tan1

W1

n n nPV V P V PV

n n

But PV m RT and P V m RT so

m RT m RT and if m cons t

n

mR T T

n

+ + =

+

= =

= =

=

This is a very important process as we shall see later.

We can also have other processes of Ideal gases. These are


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( )

2

1

2

1

2

2

1

11

2 1

tan

Constant

ln

tan

tan

0 0

V

b

V

V

b

V

b

Isothermal with cons t mass

where PV mRT

VmRTW PdV dV mRT

V V

Isobaric with cons t mass

W PdV P V V

Isochoric with cons t mass

W as V

= =

= = =

= =

= =

GRAVITATIONAL WORK

Work done by or against a Gravitational Filed. In such work the

force acting on a body is its weight.

F = mg where g = acceleration due to gravity so

2

2 1

1

( )gW Fdz mg z z = = where z is measured from a reference

point usually the Sea level.

ACCELERATIONAL WORK

Due to change of velocity. A body moves when it is subject to a

Force

( )2 2

2 2

2 1

1 1

1

2a

dV dV F ma where a so F m

dt dt

dsbut V so ds Vdt hencedt

dVW Fds m Vdt m V V

dt

= = =

= =

= = =

SHAFT WORK


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The energy transmitted by a rotating shaft due to a Torque.

The shaft rotates so distance traveled in one rotation is 2r

( )

( )( )*

/ sec 2

2 2 Shaft

sIf shaft rotates at n revolutions then r n

t

Fs workso r n n W Shaft Power

t Time r

=

= = = = =

SPRING WORK

Spring Work is produced by change in length of Spring.

For any spring its force is given by F = k (x) where

k = spring constant and x is the distance length changes

( ) ( )2 2

1 1

2 2

2 1

1

2

x x

Spring

x x

W Fdx kx dx k x x = = =

WORK OF ELASTIC SOLID BAR behaves like a spring and hence

2

1

/

x

Elastic n n

x

FW Adx where A c s area and A

= = =

WORK OF A LIQUID FILM DUE TO SURFACE TENSION

2

1

.

A

ST s s

A

FW dA where Surface Tension Unit Length

and dA increase in length of film area

= =

=

We can also have non-mechanical work such as

a. Electrical = Volts X Amps. , b. Magnetic due to dipole

movement and c. Electrical Polarization work caused by dipole

movement due to electricity

Having defined Work and Heat Interaction we now look at the

ShaftTorque = =F rr = radius of shaft


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FIRST LAW OF THERMODYNAMICS

This law deals with Conservation of energy. It reaffirms that

Energy can neither be destroyed nor created. It can change forms.

This then points towards the Concept of Energy Balance

Total Energy going

into system-

Total Energy

coming out of system=

Changes in total

energy of system

Ein - Eout = Esystem

P.E. = MK.E. = 0

P.E. = M- NK.E. = N

If we have a rock at a high point it hasPotential Energy and no Kinetic EnergyAs it falls the PE starts changing into KEPotential energy reduces and Kinetic Energyincreases.Increase in KE = Decrease in PE

If we heat a closed system then the heatenergy adds to the the total energy of thesystem.Here E

System= Q = 10kJ

Q = 10 kJW = 0 kJ

If we have a closed system on which we dowork then the Work energy adds to the totalenergy of the systemHere E

System= W = 10kJ

On the other hand if 10 kJ of work is takenout of the system then E

System= W = -10kJ

Q = 0 kJW = 10 kJ


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E is the summation of all the different energies ( Internal, KE,PE,

Electrical , Magnetic, Surface Tension etc. )

For our simple systems we will neglect Electrical , Magnetic,

Surface Tension etc. and so for us

( )( ) ( ) ( )( ) ( )2 2

2 12 1 2 1 2 1

1

2

0

E U KE PE or

E E m u u m v v mg z z

For Stationary systems KE PE so E U

= + + = + + = = =

MECHANISM OF ENERGY TRANSFER

Energy transfer can take place by the following methods

a. Heat Transfer = Q

b. Work Transfer = W

c. Mass Flow

Now if a system is closed then mass flow is zero. So for closed

systems . Q-W = E is the FIRST LAW OF THERMODYNAMICS

tanQ W E or q w e as mass is cons t or

General Q W E

Stationary System Q W U

Closed system q w e u per unit mass

Differential form for closed system q w de

= =

= =

= =

=

If we have a process from State 1 to State 2 then

Q1-2 W1-2 = E2 E1

If we have a cycle then E2 = E1 and so Q1-2 = W1-2

E

V

1

2

If we have a process where energytransfer takes place thenE

System= E

Final State E

Initial State

= E2 E1


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The first Law can be represented by a postulate called PMM1 or

Perpetual Motion Machine 1

A machine which operates in a cycle , and keeps on giving work

without any form of Heat Transfer

We now look at a few examples.

W


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SPECIFIC HEATS

It is defined as the ENERGY REQUIRED TO RAISE

TEMPERATURE OF A UNIT MASS OF A SUBSTANCE BY ONE

DEGREE .

To achieve this we can either keep Volume constant or Pressure

constant and hence we have two types of Specific Heats in

Thermodynamics

CV = Specific Heat at Constant volume

CP = Specific heat at Constant Pressure and they are

exp

v P

v P

o o

v P

v P

o o

u hC and C T T

kJ kJ with units or

kg K kg C

They can also be ressed on molar basis

u hC and C

T T

kJ kJ with units or kmol K kmol C

= =

= =

CP and Cv are expressed in terms of other properties so they are

also properties. So the specific heats are to be defined by two

independent intensive properties. They are thus not dependent

upon the process but on the state of the system.

For most substances Cp and Cv are functions of Temperature.

and we see this in figure 3-76

CP

oK

CO2 H

2O

Air


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For ideal gases

( ) ( )

( ) ( )v P

u f T and h f T so

C f T and C f T

= =

= =

So for Ideal gases they are not constants. However for simplicity

and for ease of calculation they are at times taken as constant.

Such that

2 1 2 1

2 1 2 1

( )

( )

v v

P P

duC so u u C T T and

dT

dh

C so h h C T T dT

= =

= =

If they are functions of temperature then

2 2

2 1 2 1

1 1

v Pu u C dT and h h C dT = =

Sometimes we take average values of CP and Cv between the

temperature ranges and then we evaluate

( ) ( )2 1 2 1 2 1 2 1( ) ( )v PAvg Avgu u C T T and h h C T T = =

We can thus find values of the Specific Heats from

a. Tables

b. Integratung h,u with respect to T

c. By using Average values or constant values. This method is

mostly preferred.


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SPECIFIC HEAT RELATIONS OF IDEAL GASES

( )

introduce

vary

1.4

P v P v

P v u

PP v

v

Air

We have h u Pv and for an ideal gas Pv Rt so

dh duh u RT or dh du RdT or R

dT dT so C C R or C C R for an ideal gas

If we use molar basis C C R

Also we a ratio of Specific Heats

Ck or f T as both C and C with T

C

F

= + =

= + = + = +

= + =

=

= =

=2 2

1 1

P vor Ideal gases h C dt and u C dt = =

U,H AND CP AND CV FOR SOLIDS AND LIQUIDS

Solids and Liquids are in-compressible and their density does not

change significantly with Pressure or Temperature. For them

CP = Cv = C. Values are given in Table A-3

[ ]

[ ] [ ]

2 1 2 1

2 1 2 1 2 1

2 1 2 1 2 1

0

0

. .. .

v av

av

here du C dt so u u C T T

and for enthalpy changes h u Pv

so dh du Pdv vdP du vdP as dv

so h h C T T v P P

For solids P P so h h u u

For Liquids a if pressure is const then h ub If const temperature h v P

= =

= +

= + + = + =

= +

= =

=

=

[ ]

( )@ , @

@

2 1 2 1

@ ,

@ ,

0

1 2

P T T

T

P T f f Sat at T

P T f

as u

so for this case if we have a process from

h h v P P

so to find h h v P P

or h nearly equal to h

=

=

= +

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Thermo Chap III