Chapter 20: The second law f th d iof thermodynamics

1. Directions of thermodynamic processes2. Heat engines3 Internal combustion engines3. Internal combustion engines4. Refrigerators5 Th d l f th d i5. The second law of thermodynamics6. The Carnot cycle7. Entropy8. Microscopic interpretation of entropy8. Microscopic interpretation of entropy

Nature dictates a preferred direction f h d ifor thermodynamics processes

• Which one of these processes seem more pnatural? Do they violate the first law of thermodynamics? Why some of them don’t happen naturally?happen naturally? – Converting thermal energy to mechanical energy. What

is the maximum efficiency?– Converting mechanical energy to thermal energy. What

is the maximum efficiency?– Heat flowing from hot object to cold object– Heat flowing from hot object to cold object– Heat flowing from cold object to hot object– An egg cooking– An egg uncooking

Directions of thermodynamics processes and heatprocesses and heat

• Thermodynamics processes have a preferred irreversible direction. – A book slides down an inclined

surface and heats op the table– A book starts moving up an inclined

table and the table cools downtable and the table cools down. • The only way to make a process

reversible is to have it happen in small steps and at equilibrium. p q

• Reversible and equilibrium are idealizations that never happen in nature. A reversible process phappens in quasi-equilibrium state.

• Irreversible processes: – Heat flow with finite temperature p

difference– Free expansion of a gas– Conversion of work to heat by friction

Directions of thermodynamics d d

• Why organizing your room is more difficult

processes and randomnessWhy organizing your room is more difficult than messing it up?

• What is the natures direction regarding• What is the natures direction regarding order? Wh h i l i t d t• When mechanical energy is converted to heat, what happens to the order of the

t ?system?

Summary: directions of h d ithermodynamics processes

Heat engines• To take heat from a hot source

and invert as much of it as possible to mechanical energy ispossible to mechanical energy is essence our technology. Name few examples.

• Is there an upper limit to the efficiency of this process?

• Heat engine: any device thatHeat engine: any device that transforms heat partially into work or mechanical energy.

• Working substance: is a quantity of material that undergoes some change during the process g g p(expansion/contraction, phase change, inflow/outflow)

Thermal efficiency of a heat engineF li i hi h• For a cyclic process in which internal energy of the working substance is the same at beginning and end of the cycle:beginning and end of the cycle:

U2-U1=0=Q-W or Q=W• For a heat engine with cyclic

process we have:process we have:Q=QH+QC=|QH|-|QC|W=Q =|QH|-|QC||

• Thermal efficiency of an engine is the fraction of QH that is converted to work:

1H C C

H H H

Q Q QWeQ Q Q

−= = = −

• Attention to the units of all Qs and Ws

Example: heat enginesA i t k i 10000 J f h t d d li 2000 J fAn engine takes in 10000 J of heat and delivers 2000 J of

mechanical work per cycle. The heat is provided by burning gas with heat of combustion LC=5.0x104J/ga) What is he thermal efficiency of this engine? (20%) b) How much heat is discarded in each cycle? (8000 J)c) How much gasoline is burned I n each cycle? ( 0 20 g)c) How much gasoline is burned I n each cycle? ( 0.20 g)d) If the engine goes through 25 cycles per second, what is

its power output in watts? In horsepower? (1hp=746 W; 67 hp)67 hp)

e) How much gasoline is burned per second? Per hour? (5.0 g/s, 18 kg/h)

Identify, set up, execute (signs of Qs and Ws, power is not energy, it is energy/time) evaluate

About $100 000 lost energy from diving 1000 trucks in a city inAbout $100,000 lost energy from diving 1000 trucks in a city in 24 hours. How important it is to make more efficient engines?

Summary: Heat engines

Internal-combustion enginesV to rV

V to rVr is the

rV to VAdi b ti

V to rVAdiabatic expansion

rV to VReady for

r is the compression ratio 8-10

Adiabatic compression

Pushing on the piston and doing work

next stroke

g

The Otto cycle: PV diagram of idealized th d i i b ti ithermodynamics processes in a combustion engine

• Is it a good approximation if we consider it cyclic?• Is the process cyclic?

Calculating efficiency of an Otto cycleb c and d a are constant0volume proceses (no work)→ →

H V

C V

b c and d a are constant0volume proceses (no work)Q ( ) 0Q ( ) 0

c b

a d

nC T TnC T T

→ →= − >= − <C V

H

Q ( )The thermal efficiency is:

Qe=

a d

C c b a dQ T T T T+ − + −=

-1

e

For an adiabatic proces in an ideal gas e ha e: TV constant

H c bQ T T

γ

−

1we have: TV constant Th

γ =

-1 -1 -1 -1b d

en we write:T (rV) T (V) and T (rV) T (V)γ γ γ γ= =a b d c

1 1 1

1 1 1

T (rV) T (V) and T (rV) T (V)

( )( 1)e=( )

d a a d d a

d a d a

T r T r T T T T rT r T r T T r

γ γ γ

γ γ γ

− − −

− − −

− + − − −=

− −

1

11 thermal efficiency in Otto cycle always less than unity.erγ −

= −

ExampleExample

• Calculate efficiency of an engine that usesCalculate efficiency of an engine that uses air and gasoline and compression ratio of the engine is 8the engine is 8.

Diesel CycleDiesel Cycle

Typical r is 15 to 20 T i l i 0 65 t 0 75Typical e is 0.65 to 0.75

Refrigerator: a heat engine operating in reverseC C HIt takes heat (Q ) from a cold reservoir (T ) and dumps heat (Q ) C C H

H

(Q ) ( ) p (Q )into a hot reservoir (T ) with help of some work (W) done on the working substance. Coefficient of performance: g p

C CQ QK

W Q= = or

Q−W QH=rate of heat removed from cold reservoir

W

H CQ

WP= input powertHt H

[ ]

Ht HK= = dimensionlesPt P

BTU/hE Effi i R ti EER[ ]Eenergy Efficiency Rating EER =W

EER is the comercial unit of K. 1 EER=3.413 K

Two principles used in refrigerationd i di i iand air conditioning

• Temperature of a non-ideal gas dropsTemperature of a non-ideal gas drops during an adiabatic expansion

• Temperature of a non ideal gas increases• Temperature of a non-ideal gas increases during an adiabatic compression

Mechanical refrigeration cycle(adiabatic)

Tgas/liquid<TH

TCTH

Tgas/liquid>TH

(adiabatic)(adiabatic)Gas and liquid states co-exist during the cycle

A practical refrigeratorp g

Air Conditioner: an extended refrigerator

The second law of thermodynamics• Essence of the second law: Conversion of heat to

work with 100% efficiency is impossible.y• Two forms of the second law:

– Engine statement (Kelvin-Planck): It is impossible for any system to undergo a process in which it absorbs heat from a reservoir at a single temperature and converts the heat completely into mechanical work, with the system ending incompletely into mechanical work, with the system ending in the same state in which it began.

– Refrigerator statement (Clausius): It is impossible for any f f fprocess to have as its sole result the transfer of heat from a

cooler to a hotter body

The Carnot CycleThe Carnot Cycle• Sadi Carnot (1796-1832): What is the maximum possible ( ) p

efficiency for an engine?• In order to achieve maximum efficiency machine we

id th i ibl havoid the irreversible processes as much as we can. • Carnot cycle is built based on avoiding irreversible

processes.processes.• Only isothermal and adiabatic processes happening

at thermal and mechanical equilibrium are i iblirreversible.

• Carnot imagined a cycle consisting of two reversible isothermal and two reversible adiabatic processes.isothermal and two reversible adiabatic processes.

The Carnot Cycle

Summary: Internal-combustion iengines

Typical r is 8 to 10T i l i 35%Typical e is 35%

Summary: Refrigerators

Why we can’t cool our house by Leaving the frig door open?

The second law of thermodynamicsThe second law of thermodynamics

The Carnot Cycle

heat transfer in a Carot engineQQ T T or CC C C

H H H H

QQ T TQ T Q T

= − =

Ex 20 2: Carnot engineEx. 20.2: Carnot engine

• A Carnot engine takes 2000 J of heat fromA Carnot engine takes 2000 J of heat froma reservoir at 500K, does some work, anddiscards some heat to a reservoir at 350discards some heat to a reservoir at 350K. How much work does it do, how muchheat is discarded and what is theheat is discarded, and what is theefficiency? (e=0.30, QC=-1400J, W=600J)

Ex 20 3: Carnot engine IIEx. 20.3: Carnot engine II Suppose 0.200 mole of an ideal gas (γ=1.40) undergoes a Carnot cycle with temperatures 2270Cundergoes a Carnot cycle with temperatures 2270C and 270C. The initial pressure is pa=1.0x105 Pa, and during the isothermal expansion at the higherand during the isothermal expansion at the higher temperature the volume doubles.

a) Find the pressure and volume at each of the points a, b, c, and d in the pV-diagram of fig 20.11

b) Find Q, W, and ΔU for each step and for the entire cycle.) D t i th ffi i di tl f th lt f tc) Determine the efficiency directly from the results of part

(b) and compare it with the result from Eq. 20.14

The Carnot Refrigerator Garage

Every step in Carnot cycle is a reversible one. So if we reverse the entire cycle, we will have

C C C H

a refrigerator. Coefficient of performance would be: Q Q Q / Q

K C C C H

H C C H

K=W Q Q 1 Q / Q

And with Q / Q / weT T

= =− −

= getC HAnd with Q / Q / we C HT T=

Carnot

get

K CTT T

=

When is small is large and frige has better H C

H C

T TT T K

−

−performance. When is large, like putting the frige in the garage, the performance suffers.

H CT T−

Ex. 20.4: Analyzing a Carnot f irefrigerator

• If the cycle of the Caront engine in Ex. 20.3 is runIf the cycle of the Caront engine in Ex. 20.3 is run backwards, what is the coefficient of performance?

• Carnot engines and refrigerators have the highest g g gpossible efficiency among the machines that operating at similar temperatures.

• The engineers want to increase TH-TC to maximize the efficiency of engines. What are the limitations?

• The engineers want to minimize TH-TC to increase the coefficient of performance of the refrigerators. Wh t th li it ti ?What are the limitations?

Entropy IE t id tit ti f di dEntropy provides a quantitative measure for disorder. Consider isothermal infinitesmal expansion of an ideal gas:

nRT nRTdQ=dW=pdV and pV=nRT p=V V

dQ dV→ → =

1 dV dQV nR T

→ =

dQ represents order of T

dVV

∝ the system

dQdS= is defined as ENTROPY T

Unit of entropy: energy over temperature (J/K).

Entropy II

2

Generalizing to any reversible rocess: 2

2 11

S=S entropy change in a reversible processdQST

Δ − = ∫The pathe 1 to 2 is a process consisiting of all reversible processes. Like internal energy: 1) the abso

1 2

lute value of the entropy can't be determined 2) the path from S to S does not matter in SΔ1 2) pWhen S is positive, the system becomes more disordered When S is negative the system is more ordered tha

ΔΔ t beforeWhen S is negative, the system is more ordered thaΔ t before

Entropy change in meltingEntropy change in melting• One kg of ice at 00C is

melted and converted to water at 00C. Compute its change inCompute its change in entropy, assuming that melting is donemelting is done reversibly. The heat of fusion of water is Lf=3.34x105 J/Kg.

Entropy changeEntropy change

• 20 6: One kg of water at 00C is heated to20.6: One kg of water at 0 C is heated to 1000C. Compute its change in entropy.

• 20.7: A gas expands adiabatically and reversibly. What is the change in its entropy?

Entropy change in a free expansion process: