Chapter 3

The First Law of Thermodynamics:

Closed System

1

Energy Transfer by Heat,Work and Mass

Thermodynamics –Sixth Edition (Chapter 2 & 4)

FIRST LAW OF THERMODYNAMICS (CLOSED SYSTEM)

OBJECTIVES:

At the end of this chapter students should be able to:

Define heat and work and relationship betweenthem.

State the concept of the first law of thermodynamics.

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State the concept of the first law of thermodynamics.

Develop equations related to the first law ofthermodynamics.

Solve the problems related to first law ofthermodynamics for closed system.

Energy is neither created nordestroyed

• It can change forms

• During an interaction between a systemand its surroundings, the amount ofenergy gained by the system must be

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energy gained by the system must beexactly equal to the amount of energy lostby the surroundings

How does energy cross asystem boundary?

• Heat

• Work

4

Q WSystem

Heat transfer

• Heat is defined as the form of energy thatis transferred between two systems byvirtue of a temperature difference

• A process with no heat transfer is

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• A process with no heat transfer isadiabatic

– Greek – not to be passed

Symbols

• Q

– Total heat transferred

– kJ or BTU

• q

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• q

– Heat/mass

– kJ/kg or BTU/lbm

•

– Rate of heat transfer

– kJ/sec = kW

Q

Heat transferred to asystem is positive

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Caloric Theory

• Antoine Lavoisier

– Heat is a fluid like substance called caloric

– It can be poured from one container (system)to another

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to another

– This model didn’t last very long

Modes of Heat Transfer

• Conduction

– Transfer of heat as a result of interactionsbetween particles

• Convection

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• Convection

– Heat transfer between a solid surface and agas or liquid that is in motion

• Radiation

– Does not require an intervening medium

Conduction

Q kAT

xcond

ThermalConductivity

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Q A kdT

dxcond t

Conductivity

Thermal Conductivities

Material Thermal Conductivity

W/(m K)

Diamond 2300

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Diamond 2300

Copper 401

Iron 80

Glass 1.4

Water 0.613

Air 0.026

Convection

12

13

( )Q h A T Tconv s f

Convective Heat Transfer Coefficient

Convective Heat TransferCoefficients

Typical h W/(m2 K)

Free convection – gas 2 - 25

Free convection – liquid 50 - 1000

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Free convection – liquid 50 - 1000

Forced convection – gas 25 – 250

Forced convection –liquid

50 – 20,000

Boiling andCondensation

2500 – 100,000

Radiation

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A lot more complicated thanconduction and convection

All bodies at a temperature aboveabsolute 0 emit thermal radiation

4sATQ

Stephan Boltzman Constant(5.67x 10-8 W/(m2.K4)

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(5.67x 10-8 W/(m2.K4)

This is the maximum -Called black body radiation

Real materials emit less – so we need a factor,

called the emissivity, e

e4sATQ

incidentQQ absorbed

absorptivity

Calculating the net radiative heat

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Calculating the net radiative heattransfer is complicated, but

44surrsrad TTAQ

For the special case where the surface issmall and completely enclosed

Work

• Work is the energy transferred with forceacting through a distance

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Work

• W = Fs

– N m = J usually we’ll use kJ

• w = W/m

– kJ/kg

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– kJ/kg

•

– kJ/sec

– kW

W

Work done by asystem is positive

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Similarities between heat and work,

• Both heat and work are dynamic forms ofenergy

– They are recognized as they cross aboundary

• Systems possess energy, but not heat orwork

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work

• Both are associated with a process, not astate

• Both are path functions (i.e., their magnitudesdepend on the path followed during a processas well as the end states).

Path Functions• Have inexact differentials

• Properties are point functions – they have

Q not dQ

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• Properties are point functions – they haveexact differentials

• If we want to know how much work wasdone, we need to know the path

Properties are point functionshave exact differentials (d ).

Path functions have inexactdifferentials ( )

D

D

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Example 3-1 (page 64)Burning of a candle in an insulated room

A candle is burning in a well-insulated room. Taking the room (the air

plus the candle) as the system, determine

(a) If there is any heat transfer during this burning process and

(b) If there is any change in the internal energy of the system

Solution:

(a) The interior surfaces of the room form the system boundary, as indicatedby the dashed lines. As pointed out earlier, heat is recognized as it

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by the dashed lines. As pointed out earlier, heat is recognized as itcrosses the boundaries. Since the room is well insulated, we have anadiabatic system and no heat will pass through the boundaries.Therefore, Q = 0 for this process.

(b) The internal energy involves energies that exist in various forms(sensible, latent, chemical, nuclear). During the process described above,part of the chemical energy is converted to sensible energy. Since thereis no increase or decrease in the total internal energy of the system,

U = 0 for this process.

Example 3-2 (page 64) Heating of a potato in an oven

A potato initially at room temperature (25ºC) is being baked in an oventhat is maintained at 200ºC. Is there any heat transfer during thisbaking process?

Solution

This is not a well-defined problem since the

system is not specified.

Heat

Potato

oven

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system is not specified.

Let us assume that we are observing the

Potato, which will be our system. Then the skin

Of the potato may be viewed as the system boundary.

Part of the energy in the oven will pass through the skin to the potato.

Since the driving force for this energy transfer is a temperaturedifference, this is a heat transfer process.

Types of Work

• Electrical Work

VIW electrical

Electrical power in terms of resistance

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• Mechanical Work

W = Fs

Electrical power in terms of resistanceR, current I, and potential difference V.

When force is not constant

The work done is proportional to theforce applied (F) and the distancetraveled (s).

Types of Work

• Spring Work

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Elongation of a spring underthe influence of a force.

Mechanical Work

• There are many kinds of mechanical work

• The most important for us will be movingboundary work

• W

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• Wb

If there is nomovement, no work isdone.

Moving boundary work or Boundary workis associated with the expansion orcompression of a gas in a piston-cylinderdevice.

Moving Boundary Work

The moving boundary

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Moving boundary work is the primaryform of work involved in automobileengines.

During their expansion, the combustiongases force the piston to move, which inturn forces the crankshaft to rotate.

Gas

Piston in automobile engine

30Crankshaft (red), pistons (gray) in their cylinders (blue), and flywheel (black)

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2

1

2

1

2

1

2

1

Pdv

AdsA

FFdsWW bb

Theareaunderthecurve

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curveis thework

The workdependson theprocess

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processpath

Consider some special cases

• Constant volume

• Constant pressure

• Isothermal

• Polytropic

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• Polytropic

Constant Volume

P 1

2

10PdVWb

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V

2

Example 3-3

Boundary work during a constant-volume process

A rigid tank contains air at 500 kPa and 150ºC. As aresult of heat transfer to the surroundings, thetemperature and pressure inside the tank drop to65ºC and 400 kPa, respectively. Determine theboundary work done during this process.

Solution:

The boundary work can be determined to be

0

500

400

P,kPa

1

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This is expected since a rigid tank has a constantvolume and dV = 0 in the above equation. Therefore,there is no boundary work done during this process.That is, the boundary work done during a constant-volume process is always zero. This is also evidentfrom the P-V diagram of the process (the area underthe process curve is zero).

2

10PdVWb

0 400

V

2

Constant Pressure

P 1 2

W

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12

2

1

2

1VVPdVPPdVWb

V

W

Example 3-4

A mass of 5 kg of saturated water vapour at 200 kPa isheated at constant pressure until the temperature reaches300ºC. Calculate the work done by the steam during thisprocess.

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Isothermal

PmRT

V

2

1

2

1dV

V

mRTPdVWb

P 1

2

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1

2

11

lnV

VmRT

dVV

PdVWb

V

2

Example 3-5

A piston-cylinder device initially contains 0.4 m3 of air at100 kPa and 80ºC. The air is now compressed to 0.1 m3 insuch a way that the temperature inside the cylinderremains constant. Determine the work done during thisprocess.

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Polytropic

P 1

2

PV n constant

2

1

2

1 V

dVConstPdVW

nb

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V

2

1,ln

1,1

1

2

1122

11

nV

VPV

nn

VPVP

V

Lets go through that step-wise

CPV n

nV

CP nCV

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V

2

1

2

1dVCVPdVW n

b

1

1

n

VCW

n

b

2

1

1

11

12

n

VVC

nn

But…C

P2 And….C

P

1

11

12

n

VVC

nn

Wb

So…

nn V

CV

V

CV

1

1

2

2

1 n

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nVP

2

2 And…. nVP

1

1 So…

1122 VPVPWb

1 n

So far, we have not assumed an idealgas in this derivation, if we do, then….

mRTPV mRTmRT

W

12

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n

mRTmRTWb

112

n

TTmRWb

112

First Law of Thermodynamics

• Energy can be neither be created nordestroyed during a process.

• It can only change forms.

• Therefore, every bit of energy should be

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• Therefore, every bit of energy should beaccounted during a process.

Energy Balances

Energy balance for any system undergoingany process

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Energy balance in the rate form

Let’s Look at Closed Systems• There is no mass transfer into a closed

system

• The only way energy can get into or out ofa closed system is by heat transfer or work

• It is common practice to assume heat to

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• It is common practice to assume heat tobe transferred into the system, Qnet,in andwork to be done by the system, Wnet,outEWQ outnet,innet,

What is DE equal to?

• Since

PEKEUE

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PEKEUE

Usually we don’t need to worry about kineticenergy or potential energy

PEKEUWQ outnetinnet ,,

How to solve problems

• Draw a picture

• List the given data

• Identify the goal (What do you want tosolve for)

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solve for)

• List the equations you know

• Draw a Process Diagram (if necessary)

• Solve for the unknowns

Let’s do some of the example problems

Example :

A rigid tank is divided into two equal parts by apartition. Initially, one side of the tank contains 5kg of water at 200 kPa and 250C, and the otherside is evacuated. The partition is then removed,and the water expands into the entire tank. The

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and the water expands into the entire tank. Thewater is allowed to exchange heat with itssurroundings until the temperature in the tankreturns to the initial value of 250C. Determine (a)the volume of the tank (b) the final pressure and(c) the heat transfer for this process

Specific Heats

• Also called the heat capacity

• Specific heat at constant volume, cv:The energy required to raise thetemperature of the unit mass of asubstance by one degree as the volume is

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substance by one degree as the volume ismaintained constant.

• Specific heat at constant pressure, cp:The energy required to raise thetemperature of the unit mass of asubstance by one degree as the pressureis maintained constant.

• Units => kJ/(kg 0C) or kJ/(kg K)

Consider a constant volumesystem

E=U+KE +PEE=U+KE +PE

v

vT

uC

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E=U+KE +PEE=U+KE +PE

ddE =E = dUdU

DE = DDE = DUU

== mCmCvvdTdT Constant-volume specificheats cv (values are forhelium gas).

•Cv is related to the changes in internal energy.

Consider a constant pressuresystem

• It takes more energy to warm up aconstant pressure system, because thesystem boundaries expand

• You need to provide the energy to

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• You need to provide the energy to

– increase the internal energy

– do the work required to move the systemboundary

Constant pressure system

p

pT

hC

h includes theinternal energy andthe work required toexpand the systemboundaries

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CCpp is always bigger than Cis always bigger than Cvv

CCpp = C= Cvv + R+ R constant-pressurespecific heats cp

(values are for heliumgas).

•Cp is related to the changes inenthalpy.

Cp and Cv are properties

• Both are expressed in terms of u or h, andT, which are properties

• Because they are properties, they areindependent of the process!!

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independent of the process!!

• The constant volume or constant pressureprocess defines how they are measured,but they can be used in lots of applications

Ideal Gases

RTPv

Joule determined that internal energy for an

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Tuu

Joule determined that internal energy for anideal gas is only a function of temperature

Pvuh

RTPv

RTuh

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RTuh

Which means that h is also only afunction of temperature for idealgases!!

For non ideal gases both h and u varywith the state

Specific Heats varywith temperature –but only withtemperature

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Note that the Noblegases have constantspecific heats

1. By using the tabulated u and h data. This is the easiest andmost accurate way when tables are readily available.

2. By using the cv or cp relations (Table A-2c) as a function oftemperature and performing the integrations. This is veryinconvenient for hand calculations but quite desirable forcomputerized calculations. The results obtained are veryaccurate.

Three ways of calculating u and h

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accurate.

3. By using average specific heats. This is very simple andcertainly very convenient when property tables are notavailable. The results obtained are reasonably accurate ifthe temperature interval is not very large.

If Cv and Cp are functions of temperature – how canwe integrate to find DU and DH?

Use an average value, and let the heat capacity bea constant

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2

112,12 TTCdTTCuuu avevv

2

112,12 TTCdTTChhh avepp

That only works, if thevalue of heat capacitychanges linearly in therange you areinterested in.More Accurate

Not Accurate

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The specific heat valuesfor some commongases are listed as afunction of temperaturein Table A2-b(page912)

How to get betterapproximation?

All of these functionshave been modeled usingthe form

Cp = a + bT + cT2 + dT3

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Cp = a + bT + cT2 + dT3

The values of theconstants are in theappendix of our book –Table A2-c (page 913)

2

1

322

1)( dTdTcTbTadTCh p

432

41

42

31

32

21

22 TTdTTcTTb

aTh

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This is a cumbersome!! Only do if you really need to bevery accurate!!

Most of the cases we only use the simple one(table, average)

Use the Ideal Gas Tables

• Table A-17 (page 936)

• Both u and h are only functions of T – notpressure

• Relative values have been tabulated for

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• Relative values have been tabulated formany ideal gases

• If the gas isn’t ideal – then it’s a function ofboth T and P and these tables don’t work!!

Three Ways to Calculate Du

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v

p

C

Ck

Specific Heat Ratio

k does not vary as strongly withk does not vary as strongly withtemperature as the heat capacitytemperature as the heat capacity

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temperature as the heat capacitytemperature as the heat capacity

k = 1.4 for diatomic gases (like air)k = 1.4 for diatomic gases (like air)

k = 1.667 for noble gasesk = 1.667 for noble gases

Solids and Liquids

• Incompressible substance: A substancewhose specific volume (or density) isconstant. Solids and liquids areincompressible substances.

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incompressible substances.

Cp = Cv = C

The specific volumes ofincompressible substancesremain constant during aprocess.

Let we proof the statement

du C dT CdTV

u C T C T T ( )2 1

h u Pv

Internal Energy Change

Enthalpy Change

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dh du Pdv vdP

h u v P C T v P

But dv is 0 if the system isincompressible0

small

END OF CHAPTER 3

THANK YOU…….

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