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19. 2 nd Law of Thermodynamics. Reversibility & Irreversibility The 2 nd Law of Thermodynamics Applications of the 2 nd Law Entropy & Energy Quality. Most energy extracted from the fuel in power plants is dumped to the environment as waste heat, here using a large cooling tower.

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19. 2 nd Law of Thermodynamics


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19 2 nd law of thermodynamics

19. 2nd Law of Thermodynamics

Reversibility & Irreversibility

The 2nd Law of Thermodynamics

Applications of the 2nd Law

Entropy & Energy Quality

slide2

Most energy extracted from the fuel in power plants is dumped to the environment as waste heat,

here using a large cooling tower.

Why is so much energy wasted?

2nd law: no Q  W with 100% efficiency

19 1 reversibility irreversibility
19.1. Reversibility & Irreversibility

Block slowed down by friction:

irreversible

Bouncing ball:

reversible

Examples of irreversible processes:

  • Beating an egg, blending yolk & white
  • Cups of cold & hot water in contact

Spontaneous process:

order  disorder

( statistically more probable )

got it 19 1
GOT IT? 19.1.
  • Which of these processes is irreversible:
  • stirring sugar into coffee.
  • building a house.
  • demolishing a house with a wrecking ball,
  • demolishing a house by taking it apart piece by piece,
  • harnessing the energy of falling water to drive machinery,
  • harnessing the energy of falling water to heat a house?
19 2 the 2 nd law of thermodynamics
19.2. The 2nd Law of Thermodynamics

Heat engine extracts work from heat reservoirs.

  • gasoline & diesel engines
  • fossil-fueled & nuclear power plants
  • jet engines

Perfect heat engine: coverts heat to work directly.

2nd law of thermodynamics ( Kelvin-Planck version ):

There is no perfect heat engine.

No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.

Heat dumped

slide7

cylinder compressed adiabatically

( T rises to Th adiabatically)

  • gas in contact with Th , expands isothermally to do work; heat Qh = Wh absorbed
  • cylinder expands adiabatically

( T drops to Tc )

  • cylinder in contact with Tc , gas compressed isothermally heatQc = Wc dumped

simple heat engine

(any engine)

Efficiency

(any cycle)

(Simple engine)

carnot engine
Carnot Engine

Ideal gas:

AB: Heat abs.

B  C: Work done

C  D:

Heat rejected:

D  A: Work done

Adiabatic processes:

  • isothermal expansion: T = Th , W1 = Qh > 0
  • Adiabatic expansion: Th Tc, W2 > 0
  • isothermal compression: T = Tc , W3 =  Qc < 0

Adiabatic compression : Tc Th , W4 =  W2 < 0

example 19 1 carnot engine
Example 19.1. Carnot Engine

A Carnot engine extracts 240 J from its high T reservoir during each cycle,

& rejects 100 J to the environment at 15C.

How much work does the engine do in each cycle?

What’s its efficiency?

What’s the T of the hot reservoir?

work done

efficiency

engines refrigerators the 2 nd law
Engines, Refrigerators, & the 2nd Law
  • Carnot’s theorem:
  • All Carnot engines operating between temperatures Th & Tc have the same efficiency.
  • No other engine operating between Th & Tc can have a greater efficiency.

Refrigerator: extracts heat from cool reservoir into a hot one.

work required

slide11

perfect refrigerator: moves heat from cool to hot reservoir without work being done on it.

2nd law of thermodynamics ( Clausius version ):

There is no perfect refrigerator.

No process is possible whose sole result is the transfer of heat from a body of lower temperature to a body of higher temperature.

slide13

Hypothetical engine,

e = 70%

Carnot refrigerator, e = 60%

 Carnot engine is most efficient

eCarnot = thermodynamic efficiency

eCarnot erev > eirrev

19 3 applications of the 2 nd law
19.3. Applications of the 2nd Law

Power plant

Turbine

Generator

Electricity

Boiler

fossil-fuel : Th = 650 K

Nuclear : Th = 570 K

Tc = 310 K

Condenser

Heat source

Cooling water

Waste water

Actual values:

efossil ~ 40 % enuclear~ 34 % ecar ~ 20 %

Prob 54 & 55

application combined cycle power plant
Application: Combined-Cycle Power Plant

Turbine engines: high Th ( 1000K  2000K ) & Tc ( 800 K ) … not efficient.

Steam engines : Tc ~ ambient 300K.

Combined-cycle : Th ( 1000K  2000K ) & Tc ( 300 K ) … e ~ 60%

example 19 2 combined cycle power plant
Example 19.2. Combined-Cycle Power Plant

The gas turbine in a combined-cycle power plant operates at 1450 C.

Its waste heat at 500 C is the input for a conventional steam cycle, with its condenser at 8 C.

Find e of the combined-cycle, & compare it with those of the individual components.

refrigerators
Refrigerators

Coefficient of performance (COP) for refrigerators :

1st law

Max. theoretical value (Carnot)

COP is high if Th Tc .

W = 0 ( COP =  ) for moving Q when Th= Tc .

example 19 3 home freezer
Example 19.3. Home Freezer

A typical home freezer operates between Tc =  18C to Th = 30 C.

What’s its maximum possible COP?

With this COP, how much electrical energy would it take to freeze 500 g of water initially at 0 C?

Table 17.1

2nd law: only a fraction of Q can become W in heat engines.

a little W can move a lot of Q in refrigerators.

heat pumps
Heat Pumps

Heat pump: moves heat from Tc to Th .

Heat pump as AC :

Heat pump as heater :

Ground temp ~ 10C year round (US)

got it 19 2
GOT IT? 19.2.
  • A clever engineer decides to increase the efficiency of a Carnot engine by cooling the low-T reservoir using a refrigerator with the maximum possible COP.
  • Will the overall efficiency of this system
  • exceed.
  • be less than.
  • equal that of
  • the original engine alone?

see Prob 32 for proof

19 4 entropy energy quality
19.4. Entropy & Energy Quality

2nd law:

Energy of higher quality can be converted completely into lower quality form.

But not vice versa.

Energy quality Q measures the versatility of different energy forms.

conceptual example 19 1 energy quality end use cogeneration
Conceptual Example 19.1. Energy Quality, End Use, & Cogeneration

You need a new water heater, & you’re trying to decide between gas & electric.

The gas heater is 85% efficient, meaning 85% of the fuel energy goes into heating water.

The electric heater is essentially 100% efficient.

Thermodynamically, which heater makes the most sense?

Ans.

Only 1/3 of fuel energy is converted to electricity at a power plant.

With this in mind, the gas heater is a better choice.

Cogeneration:

Waste heat from electricity generation used for low Q needs.

making the connection
Making the Connection

If the electricity comes from a more efficient gas-fired power plant with e = 48%,

compare the gas consumption of your two heater choices.

Gas heater: 1 unit of fuel energy becomes 0.85 unit of heat.

Electric heater: 1 unit of fuel energy becomes 0.48 unit of electric energy,

then becomes 0.48 unit of heat.

 Electric heater consumes 0.85/0.48 = 1.8 times the fuel consumed by gas heater.

entropy
Entropy

Carnot cycle (reversible processes):

Qh = heat absorbed

Qc = heat rejected

Qh , Qc = heat absorbed

lukewarm: can’t do W, Q

C = Carnot cycle

C = any closed path

Irreversible processes can’t be represented by a path.

S = entropy

[ S ] = J / K

Contour = sum of Carnot cycles.

slide25

S = 0 over any closed path

 S21+S12= 0

 S21=S21

Entropy change is path-independent.

( S is a thermodynamic variable )

entropy in carnot cycle
Entropy in Carnot Cycle

Ideal gas:

Heat absorbed:

Heat rejected:

Adiabatic processes:

irreversible heat transfer
Irreversible Heat Transfer

Cold & hot water can be mixed reversibly using extra heat baths.

T1 = some medium T.

reversible processes

T2 = some medium T.

Actual mixing, irreversible processes

adiabatic free expansion
Adiabatic Free Expansion

Adiabatic  Qad.exp. = 0

S can be calculated by any reversible process between the same states.

isothermal

p = const.

Can’t do work

Q degraded.

entropy availability of work
Entropy & Availability of Work

Before adiabatic expansion, gas can do work isothermally

After adiabatic expansion, gas cannot do work, while its entropy increases by

In a general irreversible process

Coolest T in system

example 19 4 loss of q
Example 19.4. Loss of Q

A 2.0 L cylinder contains 5.0 mol of compressed gas at 300 K.

If the cylinder is discharged into a 150 L vacuum chamber & its temperature remains at 300 K,

how much energy becomes unavailable to do work?

a statistical interpretation of entropy
A Statistical Interpretation of Entropy

Gas of 2 distinguishable molecules occupying 2 sides of a box

1/4

2 ¼ = ½

1/4

slide32

Gas of 4 distinguishable molecules occupying 2 sides of a box

1/16 = 0.06

4 1/16 = ¼ =0.25

6 1/16 = 3/8 = 0.38

4 1/16 = ¼ =0.25

1/16 = 0.06

slide33

Gas of 1023 molecules

Gas of 100 molecules

Equal distribution of molecules

Statistical definition of entropy :

  •  # of micro states
entropy the 2 nd law of thermodynamics
Entropy & the 2nd Law of Thermodynamics

2nd Law of Thermodynamics :

in any closed system

S can decrease in an open system by outside work on it.

However, S  0 for combined system.

 S  0 in the universe

Universe tends to disorder

Life ?

got it 19 3
GOT IT? 19.3.
  • In each of the following processes, does the entropy of the named system increase, decrease, or stay the same?
  • a balloon inflates
  • cells differentiate in a growing embryo, forming different organs
  • an animal dies, its remains gradually decays
  • an earthquake demolishes a building
  • a plant utilize sunlight, CO2 , & water to manufacture sugar
  • a power plant burns coal & produces electrical energy
  • a car’s friction based brakes stop the car.