1

Second Law of Thermodynamics

The law that entropy always increases -- the second law of thermodynamics -- holds I think, the supreme position among the laws of Nature. If someone points out to you that your pet theory of the universe is in disagreement with Maxwell's equations - then so much worse for Maxwell equations. If it is found to be contradicted by observation - well these experimentalists do bungle things sometimes. But if your theory is found to be

Week 6 & 7 1

g y yagainst the second law of Thermodynamics, I can give you no hope; there is nothing for it but to collapse in deepest humiliation.

Sir Arthur Stanley Eddington, in The Nature of the Physical World. Maxmillan, New York, 1948, p. 74.

Laws of Thermodynamics (Sort of….)

• First Law: You can't get anything without working for it.

• Second Law: The most you can accomplish by• Second Law: The most you can accomplish by work is to break even.

• Third Law: You can't break even.

• For Sanitation Engineers:

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– 0th: There is sh*#. – 1st: You can't get rid of it. – 2nd: It gets deeper. – 3rd: A nice, empty trashcan is wishful thinking.

http://www.xs4all.nl/~jcdverha/scijokes/2_18.html#subindex

The Benoit/Blamey Theory of Thermo-Sock-Dynamics:

Why bother to do laundry, when the inevitable loss of a sock will just increase entropy and contribute to the eventual heat

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death of the universe anyway?

Second Law of Thermodynamics

• Why:– First Law does not restrict the

_____________ of a process– Some processes will only occur in

___________________ (ie: liquid cooling. You cannot cool a liquid by placing it in an

i t th t i h t )

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environment that is hot.)• For a process to occur, it must satisfy both

the first and second law.

2

2nd Law of Thermodynamics

• Energy spontaneously tends to flow only f b i t t d i lfrom being concentrated in one place to becoming diffused or dispersed and spread out.

• i.e.: A hot frying pan cools down when it is taken off the kitchen stove. Its thermal

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taken off the kitchen stove. Its thermal energy ("heat") flows out to the cooler room air. The opposite never happens.

http://www.secondlaw.com/

1st Law vs. 2nd Law

• 1st Law:of energ– _____________ of energy

– Energy Transformations• 2nd Law:

– “____________” of energy (but not “x”)• High temperature energy ⇒ high quality

E d d ti d i

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– Energy degradation during a process– Limit of performance– Degree of completion of chemical reactions

2nd Law of Thermodynamics

• Heat cannot of itself pass from a _________________________________. (Rudolph Clausius)

• Conclusions of this statement:1. The entropy of an isolated system not at

equilibrium will tend to increase over time,

Week 6 & 7 7

q ,approaching a maximum value.

2. Any process occurring on its own is thermodynamically irreversible.

http://en.wikipedia.org/wiki/Second_law_of_thermodynamics

Some Definitions

• Thermal Energy Reservoir: body with a large thermal capacity that can supply orlarge thermal capacity that can supply or absorb heat without changing its temperature (ie: ___________________________________________________________________) (heat reservoir)

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(heat reservoir)– Source: type of reservoir that _______ energy– Sink: type of reservoir that _________ energy

3

Heat Engine• Device to convert heat to work

1. __________________ heat from a high-temperature source

2. ______________ part of the heat to work3. __________________ waste heat to a

low temperature source4. Operates on a cycle

• Typically, moves a fluid from one place to another (working fluid)

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Tea Kettle- Pinwheel ExampleHigh Temp Source (Thermal Energy Reservoir): Stove BurnerLow Temp Sink: Surrounding AirHeat Engine: Pinwheel

Steam Power Plant• Type of heat engine• External Combustion Engineg

– Combustion occurs _____________ engine– Thermal energy is _____________ to the steam(heat supplied to steam in boiler from high-temp source)

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(work delivered by steam as it expands in turbine)

(work required to compress water to boiler pressure)

(heat rejected from steam in condenser to a low-temp sink)

Thermal Efficiency

• Some heat is lost (wasted) when(wasted) when completing any cycle

• Performance:

• Efficiency:

Desired OutputPerformance =Required Input

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• Efficiency:

η out outth

in in

Thermal Efficiency =

W Q= =1-Q Q

Cyclic Devices• Heat Engines• Refrigerators• Refrigerators• Heat Pumps• Operate between a high temperature (TH) and

low temperature reservoir (TL)• Heat Terms:

QH = of heat transfer between

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QH ________________ of heat transfer between device and high temperature reservoir

QL = ________________ of heat transfer between device and low temperature reservoir

4

Heat Engine

Work and Efficiency:

η

out H L

th

W = Q -Q = =1-

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The 2nd Law of Thermodynamics: Kelvin-Planck Statement

• It is impossible for any device that operates l h f l on a cycle to receive heat from a single

reservoir and produce a net amount of work.

• OR: No cyclic process is possible that does nothing but convert heat into work (must be some waste heat)

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some waste heat)

• (___________________________________________________________________________.)

Refrigerators

• Heat transfer from a temperature_______-temperature

reservoir to a _______-temperature reservoir

• Objective: Maintain a low-temperature by A = Evaporator

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low temperature by removing heat from it

pB = CompressorC = Expansion ValveD = Condenser (Back of fridge)

D

Refrigerators

• Coefficient of Performance: ffi i f f i tefficiency of a refrigerator

and by definition, it can exceed 1.– Amount of heat removed can

____________ the work input.

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=−

out H L

L LR

in H L

W = Q -QQ QDesired Output COP = = =

Required Input W Q Q

5

Heat Pumps• Heat transfer from a low-temperature

reservoir to a high-temperature reservoir• Objective: maintain a heated space at a• Objective: maintain a heated space at a

______________ temperature

____________=

−

out H L

H HHP

in H L

W = Q -QQ QDesired Output 1COP = = =

Required Input W Q Q

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Cool Env. @ TL

Warm Heated Space @ TH > TL

The 2nd Law of Thermodynamics: Clausius Statement

• It is impossible to construct a device that operates in a cycle and produces no effect other p y pthan the transfer of heat from a lower-temperature body to a higher temperature body.

• OR: No cyclic process is possible that does nothing but transfer heat from a cold body to a hot one

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• Refrigerator ________________ operate without a net work input from an external source.

Reversible Process

• Idealized processA th t b ith t• A process that can be _____________ without leaving any trace on the surroundings

• Irreversible process: some loss to the surroundings occurs

• Can assume a reversible process for some d i ( )

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devices (_______________________________)• Can be used to calculate the “Theoretical Limit”

of a process

Irreversibilities

• Friction• Unrestrained Expansion of Gas• Heat Transfer

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6

Reversible Processes

• Internally Reversible: no irreversibilities th t b d i________ the system boundaries

• Externally Reversible: no irreversibilities ________ the system boundaries

• Totally Reversible (Reversible): no irreversibilities

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irreversibilities

Carnot Cycle

• Reversible Process• Heat Engine

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Carnot Power Cycle in a Gas Piston-Cylinder Assembly(Reversible Process)

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1 → 2 2 → 3 3 → 4 4 → 1________work during Expansion

________Work of Compression is done on the sys

________Compression occurs

________Boundary Work Expansion

Carnot Power Cycle in a Gas Piston-Cylinder Assembly

QH

QL

TH

TL

(Reversible Process)QL

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1 → 2 2 → 3 3 → 4 4 → 1

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Carnot Cycle:Heat Engine

• Reversible Cycle4 P

QH

QL

TH

TL

• 4 Processes:• 1 to 2: Adiabatic expansion: Reversible adiabatic expansion

during which the system does work as the working fluid temperature decreases from TH to TL.

• 2 to 3: Isothermal heat rejection: The system is brought in contact with a heat reservoir at TL<TH and a reversible isothermal heat exchange occurs while work of compression is done on the system

• 3 to 4: Isentropic compression: A reversible adiabatic

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p pcompression process increases the working fluid temperature from TL to TH.

• 4 to 1: Isothermal heat supply: Reversible isothermal heat addition at high temperature (TH>TL) to the working fluid in a piston-cylinder device that does some boundary work.

• Carnot cycle involving two phases—

• Still two adiabatic processes and two isothermal processes.

• Always reversible by definition.

TL

Week 6 & 7 26TL

TL Cool Animation Web Site:http://www.corrosion-doctors.org/Biographies/carnotcycle.htm

Carnot Cycle: Refrigeration

• Reversal of Heat Engine (heat arrowsEngine (heat arrows are in __________ direction) (what we just discussed)

• The heat transfer is in the

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the ____________ direction; the rest is the same.

Carnot Principles

1. The efficiency of an irreversible heat i i l th thengine is always _______ than the

efficiency of a reversible one operating between the same two reservoirs.

2. The efficiencies of all reversible heat engines operating between the same two

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engines operating between the same two reservoirs are the _________.

8

REVIEW from 1st Week:Thermodynamic Temperature

Scale

• Independent of substance properties• Look at derivation on Pg. 251-252• Kelvin scale

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Carnot Heat Engine Efficiency

η L LQ T=1- =1-ηth,revH H

=1- =1-Q T

__________For any Heat Engine, Reversible or Irreversible

__________For Carnot Heat Engine, __________

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• Maximum Efficiency:TL (or QL) is very low or TH (or QH) is very high

Carnot Heat Engine• ηth,rev = Highest efficiency a heat engine

operating between two heat reservoirs hcan have.

η Lth,rev

H

T=1-T

Lth

H

Q=1-Q

η

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η ηη ηη η

<

=

>

th th,rev

th th,rev

th th,rev

______________ heat engine ______________ heat engine ______________ heat engine

Carnot Heat Engine Example 1

• An inventor claims to have an engine that i 100 BTU f h t d dreceives 100 BTU of heat and produces

25 BTU of work when operating between a source of 140oF and 0oF. Is the claim valid?

Week 6 & 7 32http://www.engineersedge.com/thermodynamics/carnot_cycle.htm

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Actual vs. Ideal (Carnot) Efficiency for a Heat Engine

• The actual efficiency of a steam cycle is 18% Th f ilit t f t18%. The facility operates from a steam source at 340oF and rejects heat to the atmosphere (60oF). Compare the Carnot and actual efficiency.

Don’t worry about ”s” right

Week 6 & 7 33http://www.engineersedge.com/thermodynamics/carnot_cycle.htm

TL

Don’t worry about ”s” right now.

The important thing is the “shape” of cycle diagram.

Carnot Heat Engine Example 2

• A Carnot Engine operates between TH= 850 K and TL = 300 K. L

• For each cycle, the engine performs 1,200 J of work.

a) What is the engine efficiency?b) How much heat is extracted from the

high temperature reservoir during each cycle?

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yc) How much heat is rejected to the low

temperature reservoir during each cycle?

Carnot Refrigerator and Heat Pump

1 1− −R,rev

H H

1 1COP = =Q T1 1

1 1− −

<

L L

HP,revL L

H H

R R,rev

Q T1 1COP = =Q TQ T

COP COPCOP COP

irreversible refrigeratorreversible refrigerator

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=

>R R,rev

R R,rev

COP COPCOP COP

reversible refrigerator impossible refrigerator

(The same can be said for Heat Pumps)

Carnot Refrigerator Example 1

• A Carnot Refrigerator operates between TH = 850 K and TL = 300 K. H L

• For each cycle, the engine needs 1,200 J of work.

a) What is the Coefficient of Performance for this refrigerator?

b) How much heat is extracted from the low temperature reservoir during

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p geach cycle?

c) How much heat is rejected to the high temperature reservoir during each cycle?

10

Limitations of Carnot Cycle• The conceptual value of the Carnot cycle is that

it establishes the possibleit establishes the ___________ possible efficiency for an engine cycle operating between TH and TC.

• It is not a practical engine cycle because the ____________________________________________________________________________. A S h d t it "d 't b th i t lli

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• As Schroeder puts it: "don't bother installing a Carnot engine in your car; while it would increase your gas mileage, you would be passed on the highway by pedestrians."

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/clausius.html

Second Law of Thermodynamics

• Energy of all kinds in our material world ________________________________ out if it is not hindered from doing so.

• Entropy is the __________ measure of that kind of spontaneous process: how much energy has flowed from being

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much energy has flowed from being localized to becoming more widely spread out (at a specific temperature).

http://www.entropysimple.com/content.htm

Review: The 2nd Law of Thermodynamics:

Kelvin Planck Statement: It is impossible for any• Kelvin-Planck Statement: It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work.– OR: No cyclic process is possible that does nothing but convert

heat into work (must be some waste heat)• Clausius Statement: It is impossible to construct a

device that operates in a cycle and produces no effect th th th t f f h t f l t t

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other than the transfer of heat from a lower-temperature body to a higher temperature body.– OR: No cyclic process is possible that does nothing but transfer

heat from a cold body to a hot one

Clausius Inequality

• Another corollary of the 2nd Law.• Deals with increments of heat and work, δQ and δW, not Q and W.

• Will use the symbol: , which means to integrate over all the parts of the cycle.Used to define entropy a measure of

∫

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• Used to define entropy, a measure of system disorder

11

Clausius InequalityThe cyclic integral of δQ/T for a closed system is always equal to or less than zerosystem is always equal to or less than zero.

δ≤∫

cycle

Q 0T

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Where : T = absolute temperature at the boundary

Clausius Inequality

Week 6 & 7 42http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/clausius.html

• Applies to any _______________ cycle • Implies a ____________ change in entropy on the cycle.

– The entropy to the environment during the cycle is

Clausius Inequality

– The entropy __________ to the environment during the cycle is larger than the entropy ___________ to the engine by heat from the hot reservoir.

– In the simplified heat engine where the heat, QH, is added at temperature TH:

• To complete the cycle, the amount of entropy: dS = QH/TH– is added to the system– must be removed to the environment.

In general:

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– In general:• The engine temperature will be ______________________ when

heat is being added• Any temperature difference implies an ________________ process.

– Excess entropy is created in any irreversible process, and therefore more heat must be dumped to the cold reservoir to get rid of this entropy. This leaves less energy to do work.

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/clausius.html

For a Carnot Cycle• Manipulate ηth,revEquation:

L L

H H

L L

H H

Q T1- =1-Q T

Q T= or __________________ = 0Q T

If Q = heat added to system, then heat removed from the system

Week 6 & 7 44

η L Lth,rev

H H

Q T=1- =1-Q T

∑ i

i i

then heat removed from the system will be negative.

QOR : = 0T

12

For a Carnot Cycle

• Or in a more general expression:

dQ = 0T

Which is the Clausius Theorem

∫

Week 6 & 7 45

for a reversible heat engine (Carnot Cycle)

What’s Entropy?

• Extensive property• Found in Tables in back of book• Symbol: S or s• Units:

S: kJ/K/

Week 6 & 7 46

s: kJ/K.kg

What’s Entropy?• measure of disorder, nature tends toward

maximum entropy for any isolated systempy y y• What’s “disorder” with respect to entropy?

– A measure of the “_____________" associated with the state of the objects.

– If a given state can be accomplished in many ways, then it is more probable than one which can be accomplished in only a few ways.

• When “throwing dice“:

Week 6 & 7 47

When throwing dice :– Throwing a seven is more probable than a two because you

can produce seven in six different ways and there is only one way to produce a two.

– Seven has a higher multiplicity than a two, or a seven represents higher "disorder" or higher entropy.

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/clausius.html

For a Heat Engine:• Entropy exists.• Must define the change in entropyMust define the change in entropy.• For a process going from state 1 to state 2:

Δ 2 1S = S -S =

O i diff ti l f d t i t i th l

Week 6 & 7 48

Or in differential form and at any point in the cycle :dQdS = T

13

For a “Real” Heat Engine• Irreversibilities occur.• Efficiency is than for a Carnot Cycle• Efficiency is ________ than for a Carnot Cycle

– Less heat flow to the system and/or more heat flow out of the system.

• Leads to Clausius Inequality:

≤∫dQ 0T

Due to __________________

Week 6 & 7 49

• Any real engine cycle will result in more entropy going to the environment than was taken from it (or a net increase in entropy).

∫ T Due to __________________

Review: Second Law of Thermodynamics

• Why:– First Law does not restrict the direction of a

process– Some processes will only occur in one

direction (ie: liquid cooling. You cannot cool a liquid by placing it in an environment that is h t )

Week 6 & 7 50

hot.)• For a process to occur, it must satisfy both

the first and second law.

Review: 1st Law vs. 2nd Law

• 1st Law:Q antit of Energ– Quantity of Energy

– Energy Transformations• 2nd Law:

– “Quality” of energy (but not “x”)• High temperature energy ⇒ high quality

E d d ti d i

Week 6 & 7 51

– Energy degradation during a process– Limit of performance– Degree of completion of chemical reactions

Change in Entropy

• Like energy, changes in entropy can be quantifiedquantified.

• Entropy is a ______________ (fixed values at fixed states)

• Good for an internally, reversible process (not an irreversible path)

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• For an irreversible path, must choose to integrate along a convenient internally, reversible path

14

Special Case: Internally Irreversible, Isothermal Heat

Transfer ProcessI th l h t t f i i t ll• Isothermal heat transfer process is internally reversible

• Because isothermal, only have one temperature (T0).

• Entropy change reduces to:

Week 6 & 7 53

( )δ δ⎛ ⎞⎜ ⎟⎝ ⎠∫ ∫

2 2

01 1int rev int rev

Q 1ΔS = = Q =T T

For any heat engine

Why does Entropy Increase?

• Why does ice melt in a warm room? – ____________________________________

________________________________________________________________________________________________________________________________________________ Thi f ll th d l d th f it i

Week 6 & 7 54

– This follows the second law and therefore it is a spontaneous process involving an increase in entropy in the ice as it melts to form water.

http://www.entropysimple.com/content.htm

Why does Entropy increase?δ

δ δ

≤

⎛ ⎞

∫2 1

Q 0TQ Q

(Clausius Inequality)

δ δ

δ

δ

⎛ ⎞+ ≤⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

≤

∫ ∫

∫

∫

2 1

int rev1 21

2 1int rev2

2

1 21

Q Q 0T T

QBut : S -S =T

QOR : + S - S 0T

1

int rev2

QTδ⎛ ⎞⎜ ⎟⎝ ⎠∫

2

1

QTδ∫

Week 6 & 7 55

δ

δ

δ

≥

=

1

QRearrange and in differential form : dST

QdSTQ dS >T

Where : for a reversible process and

for an irreversible process

2

Increase of Entropy Principle

• The entropy of an isolated system (ie: di b ti l d t ) fadiabatic, closed system) for a process

always increases, or, for a reversible process, remains constant.

• Entropy _______________ decreases.

δQ

Week 6 & 7 56

δ≥

QdST

15

Entropy Changes

• For an isolated system, the change in entropy is the sum ofchange in entropy is the sum of the entropy changes for each subsystem.

• But for an open system, the system and the surroundings can be considered as two

Week 6 & 7 57

be considered as two subsystems. OR:Sgen = ΔStotal = ____ + _____ ≥ 0

Entropy Principle Summary

• Entropy change can be negative within a b t tprocess, but entropy

_________________________________.• Increase of Entropy:

Sgen > 0 Irreversible ProcessS = 0 Reversible Process

Week 6 & 7 58

Sgen 0 Reversible ProcessSgen < 0 Impossible Process

Entropy Conclusions

1. Processes occur in a certain direction ______. A process must comply with: S ≥ 0A process must comply with: Sgen ≥ 0.

2. Entropy is a ______________ property. Entropy is conserved in reversible, ideal processes (not in real, irreversible ones).

3. Entropy is a

Week 6 & 7 59

____________________________________of a process. The greater the irreversibility, the greater the entropy.

Prob. 7-25, Pg. 335• Air is compressed by a 12

kW compressor from P1 to P2p 1

P2. The air temperature is maintained at a constant 25oC during this process and, as a result of heat transfer to the surrounding medium at 10oC

AIRT =

const.

P

&outQ

&inW =12kW

Week 6 & 7 60

10 C. 1. Calculate the entropy

change of the air. 2. State your assumptions.

P1

16

Entropy and Pure Substances

• Table values are like ν and h.• Note: under vapor dome, quality (x)

applies.

Week 6 & 7 61http://www.eh.doe.gov/techstds/standard/hdbk1012/h1012v1.pdf

T-s Diagram for water

• Vapor dome curve and constantand constant pressure lines look like what you’re used to.

• Constant ν lines are steeper than

Week 6 & 7 62

are steeper than constant pressure lines

Isentropic Process

• “Isoentropic”• Constant entropy

– ______________________– ______________________

• Examples: pumps, nozzles, turbines, diffusers

Week 6 & 7 63

diffusers• Reversible, adiabatic process is isentropic

Prob. 7-39, Pg. 336

• An insulated piston-cylinder device t i 0 05 3 f t t d f i tcontains 0.05 m3 of saturated refrigerant-

134a vapor at 0.8 MPa pressure. The refrigerant is now allowed to expand in a reversible manner until the pressure drops to 0.4 MPa. Calculate:

Week 6 & 7 64

a) The final temperature in the cylinderb) The work done by the refrigerant

17

Heat Transfer & T-S Curve

• Area under T-S curve is the heat transfer Qheat transfer, QδQint rev = TdS

OR:

∫2

Week 6 & 7 65

∫2

int rev 1Q =

T-s Diagram

• Isentropic process is a __________ line T dion a T-s diagram.

– No heat transfer, so area under curve must be zero.

2T

Iso-Pressure Lines

Week 6 & 7 66

1

ss1=s2

T-s Diagram

• Fig. A-9, Pg. 1098: T-s diagram for water. • Example 7-6: T-S diagram of Carnot

Cycle

2 3

TQH = TΔS23

W

Week 6 & 7 67

http://www.personal.psu.edu/faculty/d/h/dhj1/classes/thermo/chapter7.html

1 4

S

QL = TΔS14

WoutWin

T-s & P-ν Diagram for a Carnot Cycle

2 3

TQH = TΔS23

WoutWinWin

Week 6 & 7 68

1 4

S

QL = TΔS14

Wout

18

h-s Diagram• Δh = change in enthalpy

between entrance and exitbetween entrance and exit• Δs = change in entropy or

irreversibilities in process• Used mostly for steam

power plant design• Figure A-10, Pg. 1099

Week 6 & 7 69

g g(Mollier Diagram)– Can use instead of tables

(but you probably don’t want to….)

Not on Handouts…

Mollier (h-s) Diagram

• Relates:– Outside Vapor

Dome: h, s, T and pressure

– Within vapor dome, includes

lit f

Week 6 & 7 70

quality of mixture

http://commons.wikimedia.org/wiki/Image:HS-Wasserdampf_engl.png

Why do we care about h-s Diagram?

U d h d i i f id l• Used to show deviation from ideal behavior

• Ideal: – used for quick calculations or to see if feasible

Sometimes calculate and then use efficiency

Week 6 & 7 71

– Sometimes calculate and then use efficiency to correct for non-ideal (real) behavior

• Real: used to size equipment, determine operation costs….

Real vs. Ideal (ie: Reversible) h-s & T-s Diagram

• Reversible processes will have “vertical” transitions from one iso-pressure line to another

• Real processes will have losses and transitions from one iso-pressure line to another will have a “slope”

Week 6 & 7 72

19

Tds

• What happens when temperature changes i ?in a process?

• We have defined equations for isothermal processes but not where temperature changes.

Week 6 & 7 73

For an Internally, reversible process for a Closed System, Simple Compressible

Substanceδ δint,rev int,rev,outQ - W = dU

δ δint,rev int,rev,outBut : Q = TdS and W =Pd VTherefore : TdS = dU + Pd V

νOR : Tds = du + Pd (per mass basis)

This is Gibbs Equation.

Week 6 & 7 74

Note: For a simple compressible system, only boundary work can be done for an internally reversible process.

But what about enthalpy?

νTds = du + Pd (Gibbs Equation from previous slide)

νν ν

ν ν

h = u +Pafter differentiating : dh = du + Pd + dP

Rearranging: du = dh - Pd - dP

Substitute into Gibbs Equation :

(definition of enthalpy)

Week 6 & 7 75

Why are they important?νTds = du + Pd

• Relate entropy to other properties• Allow use of tables to solve problems

νTds = dh - dP

Week 6 & 7 76

20

3rd Law of Thermodynamics(Nernst Heat Theorem)

• If one could reach absolute zero, all bodies would have the same entropy. py– In other words, a body at absolute zero could exist in

only one possible state, which would possess a definite energy, called the zero-point energy. This state is defined as having zero entropy.

http://www.infoplease.com/ce6/sci/A0861526.html

• The entropy of a pure crystalline substance at ________________________________ is zero.

Week 6 & 7 77

• Provides an absolute reference point for entropy.

• Useful in the analysis of chemical reactions.

Review of Laws of Thermo

• Zeroth Law: defines temperature scale• First Law: energy is conserved • Second Law: everything moves toward

equilibrium because of something called entropyThird Law: there is a lowest temperature

Week 6 & 7 78

• Third Law: there is a lowest temperature, called absolute zero, where this entropy stuff is zero

Laws 1-3: http://www.madsci.org/posts/archives/dec96/835000890.Ph.r.html

Entropy Change in Liquids & Solids

Approximate by assuming incompressible.ν

≈∫

P V

2av

1

du dds = +PT T

du dTOR : ds = = C , Where : C = C = CT T

TIntegrate : C lnT

2

2 1 1

dTs - s = C(T)T

Week 6 & 7 79

≈

1

2av 2 1

1

TTIsentropic : C ln = 0 OR T = TT2 1

T

s - s

NOTE: For an incompressible substance, the temperature is constant for an isentropic process.

Prob. 7-50, Pg. 337

• A 12 kg iron block initially at 350oC is h d i i l t d t k th tquenched in an insulated tank that

contains 100 kg of water at 22oC. Assuming the water that vaporizes during the process condenses back in the tank, calculate the total entropy change for this

Week 6 & 7 80

process.

21

Entropy Change of Ideal Gases νTds = du + Pd

Td dh dP• Need a way to find the change in entropy

using these two expressions– Why:

νTds = dh - dP

Week 6 & 7 81

• Need specific internal energy (u) and specific enthalpy (h) for some calculations

Entropy Change of Ideal Gases –Internal Energy

du dd +P ν

V

V

ds = +PT T

RTWhere : du = C dT, and P =

dT dds = C +R

νν

Note: Typically, assume constant specific heats for

Week 6 & 7 82

2

1

Vds C RT

Integrate : Rln2

2 1 V1

dTs - s = C (T)T

ννν

+∫

specific heats for Cv and CP.

Entropy Change of Ideal Gases -Enthalpy

d d Note: Typically,

P

du dds = +PT T

RTSimilarly : dh = C dT, and = P

ν

ν

Note: Typically, assume constant specific heats for Cv and CP.

Week 6 & 7 83

2

1

PIntegrate : RlnP

2

2 1 P1

dTs - s = C (T)T

+∫

Assuming Constant Specific Heats

νT 2

1

νν

+

+

2v,ave

1

2 2P,ave

1 1

TC ln RlnTT PC ln RlnT P

2 1

2 1

s - s =

s - s =

Week 6 & 7 84

kJWith units : kg K

22

Isentropic Processes of Ideal Gases with Constant Specific

Heats

2 2

1 1

2 2v,ave

1 1 v,ave

2 2 2 2P,ave

1 1 1 P ave 1

T T RC ln Rln ln lnT T CT P T PRC ln Rln ln lnT P T C P

2 1

2 1

s - s = 0 = OR :

s - s = 0 = OR :

ν νν ν

+ = −

+ = −

Week 6 & 7 85

1 1 1 P,ave 1

p vWhere : R = C -C

Isentropic Processes of Ideal Gases with Constant Specific Heats – More Common Form

These expressions simplify to :

2

1

1

k-1k-1k

2 2

1 1s=constantk

2

These expressions simplify to :

T PT P

P

νν

ν

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

Week 6 & 7 86

1

2

,

2

1 s=constant

p

v v

P

C RWhere : k = = k -1C C

ν=⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

General Energy Equation

in o t in o t i e S sQ -Q + W - W +E -E = ΔE Form we’ve been using in classin out in out i e Sys

in out in out i i e e 2 2 1 1

21

1 1 1

22

2 2 2

Q Q W W E E ΔE

Q -Q + W - W +mθ -m θ = m e - m e

ve = u + +gz = energy due to initial mass within control volume (State 1)2ve = u + +gz = energy due to final mass within control v2

∑ ∑

2i

olume (State 2)

vθ h d t fl i i t t l l

g

Week 6 & 7 87

ii i i

2e

e e e

θ = h + +gz = energy due to mass flowing into control volume2vθ = h + +gz = energy due to mass flowing out of control volume2

Why h or u?

• h = u + Pνθ fl

21v• θ = flow energy

Pν = flow work = energy due to fluid flowing into and/or out of control volume (entrances/exits)

• e = energy of the non-

11 1 1

22

2 2 2

2i

i i i

2

ve = u + +gz2ve = u + + gz2

vθ = h + +gz2v

Week 6 & 7 88

• e = energy of the non-flowing fluidh = u + Pν reduces to h = u or

just “u”

ee e e

vθ = h + + gz2

23

General Energy Equation

in out in out i e SysQ -Q + W - W +E -E = ΔE

in out in out i i e e 2 2 1 1

Sys 2 2 1 1

Q -Q + W - W +mθ -m θ = m e - m e

Steady State : ΔE = m e - m e = 0Special Cases:

∑ ∑

∑ ∑

Week 6 & 7 89

i e i i e e 1 2

in

Closed System: E -E mθ -m θ = 0 & m = m = m

Adiabatic (or Isentropic) : Q

=

out-Q = 0

Handout:• Flowchart of

System &System & Typical Assumptions

Week 6 & 7 90