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.
Reversibility & Irreversibility
The 2nd Law of Thermodynamics
Applications of the 2nd 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.
Why is so much energy wasted?
2nd law: no Q W with 100% efficiency
Block slowed down by friction:
Examples of irreversible processes:
( statistically more probable )
Heat engine extracts work from heat reservoirs.
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.
( T rises to Th adiabatically)
( T drops to Tc )
simple heat engine
AB: Heat abs.
B C: Work done
D A: Work done
Adiabatic compression : Tc Th , W4 = W2 < 0
A Carnot engine extracts 240 J from its high T reservoir during each cycle,
& rejects 100 J to the environment at 15C.
How much work does the engine do in each cycle?
What’s its efficiency?
What’s the T of the hot reservoir?
Refrigerator: extracts heat from cool reservoir into a hot one.
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.
e = 70%
Carnot refrigerator, e = 60%
Carnot engine is most efficient
eCarnot = thermodynamic efficiency
eCarnot erev > eirrev
fossil-fuel : Th = 650 K
Nuclear : Th = 570 K
Tc = 310 K
efossil ~ 40 % enuclear~ 34 % ecar ~ 20 %
Prob 54 & 55
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%
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.
Coefficient of performance (COP) for refrigerators :
Max. theoretical value (Carnot)
COP is high if Th Tc .
W = 0 ( COP = ) for moving Q when Th= Tc .
A typical home freezer operates between Tc = 18C 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?
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 pump: moves heat from Tc to Th .
Heat pump as AC :
Heat pump as heater :
Ground temp ~ 10C year round (US)
see Prob 32 for proof
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.
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?
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.
Waste heat from electricity generation used for low Q needs.
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.
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.
Entropy change is path-independent.
( S is a thermodynamic variable )
Cold & hot water can be mixed reversibly using extra heat baths.
T1 = some medium T.
T2 = some medium T.
Actual mixing, irreversible processes
Adiabatic Qad.exp. = 0
S can be calculated by any reversible process between the same states.
p = const.
Can’t do 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
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?
Gas of 2 distinguishable molecules occupying 2 sides of a box
2 ¼ = ½
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
Gas of 100 molecules
Equal distribution of molecules
Statistical definition of entropy :
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