Chapter 4
Engines and Refrigerators COP is
largest When Th and To aren't
very different
·
: .
= operated in reverse
Heat Carnot
4 7 .
engines You can use the cycle to find the maximum
possible COP
Heat engine Any device into work
that absorbs heat and converts part of that energy tomakehetfowintheopsiedirection theworkingsubstancemusbe slightly notter than Ta while het is
·
:
,
↳ Not all the absorbed converted to work ,
as neat can be
energy
because the heat brings along entropy as it flows in ,
which must
4 3 Real Heat Engines
be disposed of before the cycle starts again
.
Heat cump some waste heat into its environment Internal Combustion
engine must
Engines
Work the is the between the heat absorbed and the Gasoline
produced by engine difference engine Otto
Cycle
·
>
·
:
waste heat expelled 1
. Mixture injected into a
cylinder and
adiabatically compressed by piston
Heat absorbed by the engine : Qh > comes from
hot reservoir (Th) .
2 Spark plug ignites the mixture , raising its temperature and
pressure while the volume doesn't change
·
-
·
Waste heat : Qc >
-
dumped in cold reservoir (Tc) .
3 High pressure gas pushes piston outward < adiabatic expansion ;
produces mechanical work
Th andTc are assumed fixed 4
. Hot exhaust expelled and replaced by a new mixture at lower temperature and pressure
gases are
WQQ Scannotbegreater e
tan
benefit Pa
efficiency
- -
< e =
·
cost Qh
1st
&
law :
energy is conserved > state of the engine must be unchanged at the end of a * No < thermal
not reservoir
energy produced internally by bur
·
.
the energy it absorbs 1Qh) must be
cycle :
precisely ning fuel
↳ Qh Qc + W it ignition
=
equal to the
energy expels ,
Power
ladiabatic
&
expansions
and Total
: exhaust
·
law :
entropy of the engine plus its surroundings can increase but not decrease adiabatic
[
compression &
<
Cycle : Entropy it expels must be at least as much the entropy it absorbs .
>
·
Entropy extracted from theh ot reservoir :
Shothyentropy absorbed ·
Efficiency : Net work produced during the
cycle divided
by the "heat" absorbed during ignition step .
= Jentropy 4e of these temperature ratios in
3
either
·
Entropy expelled to the cold reservoir : Scold expelled = 1 -
(v)U- = 1
. greater than the ratio of the extreme
sinvoke and law :
& Oh h
or
QC
Qh
To
Th
adiabatic process : TVU-1 is constant
temperatures T/z that appears in the
Carnot formula
,
> Otto
,
is less
engine
-
efficient than Carnot engine
This
gives : es1-
I the smaller the ratio TyTh ,
the more efficient your engine Steam
engine
Steam Steam does turbine , while heat is provided by burning fossil fuel
engine : work by pushing a
piston or a
·
The Carnot Water pumped to
Cycle <maximum possible efficiency 1.
high pressure and then flows into a boiler
In boiler heat is added under
Working substance : material that absorbs heat expels waste heat and does work .
2 constant pressure
·
,
↳ a
gas
Steam hits turbine where it expands
3
. ,
adiabatically
1. Gas absorbs heat Qh from
h ot reservoir Entropy of reservoir decreases by Qh/Th while . Cools and ends up at the
original low pressure and cooled further in a "condenser"
4
.
,
entropy of gas increases by Qh/Tgas P
Water
M
Steam
↑ &
. avoid
28 13 Not
2
making new entropy- need to make Th1 Tgas t isn't possible because heat won't flow Boiler &
* an ideal
gas !
temperature (
↑
between objects of the same &
↑
↳
pump a - turbine
↑
-
Rankine
.
3
Keep the gas at this temperature as is absorbs heat > isothermal
expansion # Condenser -
Cycle
1104 -
Gas (Water + Steam)
. To avoid
dumps waste heat into Tgas
4
. cold reservoir creating we want
-
new
entropy -
To. -
infinitesimally greater than >
V
H U + PV
> How to get gas from one temperature to the other and back without heat flow > -
adiabatic
Enthalpy : = <
energy plus work needed to make room for it in a constant
pressure environment
↳ in enthalpy heat absorbed under constant pressure conditions
Pa change :
Rankine Heat is absorbed under constant
pressure in the boiler in the
Cycle
&
: and expelled at constant
pressure
·
condenser
isothermal which
gives for the
efficiency
ee
:
adia -
expansion
a S
batic
&
compres adiabatic ↳ we can make the
assumption Hi He because the pump adds little to the water
very
=
Sion
energy .
expansion
a
I
E
*
isothermal
compression >
V
4. Real Refrigerators
P liquid
1. Fluid/Gas is compressed adiabatically raising temperature and pressure
M condenser
zi
-
&
· Gas
2
. Gives up in the condenser &
on
heat and
slowly liquefies ..
4 2 .
Refrigerators 3
. Passes
through throttling value Inarrow opening/porous plug) > lower temperature
4
. Absorbs heat and turns back into gas in evaporator
and pressure
-a compressor
Refrigerator : Heat engine operated in reverse ↳ constant pressure
·
, -
1 Heat absorbed : QC =
H1-Hy /Change in
enohalpy :
&
benefit o =Q ! > -
Cop Heat expelled : Qh = H2 H3
Efficiency Coefficient of performance 7
-
<
-
=
, :
evaporator
·
Q Hy
&
-
H4
CO4
,
-
=
Hy He in refrigerator cycle (04 (Liquid + Gas)
Q
> =
than 1 :
-
-
Qh
Ha Hz Het Hy
-
>
-
-
Throttling process /Joule -
Thomson process)
·
and law : Entropy dumped into the not reservoir must be at least as much as the
entropy absorbed
from the cold reservoir Qh QC Throttle Pretend that fluid is being pushed through the plug by piston exerting pressure 4i
Oh
: ·
: a while a second piston ,
- or , ,
Ta To /reverse of heat engine because exerting pressure Pf moves backward on the other side to make room .
entropy is
flowing in opposite
direction) We Wright PiVi-PeVe
Thy-1 Te
T - No heat flow during this process ,
thus
energy of the fluid : Uf-Ui =
Q+ W =
of + =
↳ Uf + P7VF is constant
This gives : COP : To =
UitPiVi -Hi =
He lenthalpy during throttling process)
·
Purpose of throttling : Cool the fluid to below the temperature of the cold reservoir ,
so it can absorb heat as required
* If COP is 5 .
9 then ,
for each joule of electrical
energy ,
the coolant can suck as much as 5 .
9] of In a dense
gas or liquid ,
the
energy U contains also a
potential energy term due to the forces between the molecules :
# distances
Upotential + Ukinetic + PV ↳ wearly attractive at long
at short distances
strongly repulsive
heat from the inside of the
refrigerator Waste heat dumped into kitchen is 6 .
9J
↳MostofTimesAttraction
is becomes less
.
dominates :Upotential negative negative if pressure drops - distance bee e
molecules increases)
, Chapter 5 : Free energy and Chemical Thermodynamics . 2:
5 Free energy as a force toward
equilibrium
. 7:
5 Free energy as available work For isolated the entropytends to increase the is what of spontaneous
an
system ,
System's entropy governs the direction
·
,
change
·
Enthalpy (H) : Energy plus the work needed to make room for it in an environment with constant What if a system is not isolated ? Now energy can pass between the
system and the environment, so what tends to
pressure P .
increase is the total entropy of the system plus environment
H = u + PV u+ w Assume that the environment acts as a "reservoir" of energy large enough that it can absorb or release unlimited amount?
=
,
↳ the total energy you would need to create the system out of nothing and put of energy without changing its temperature .
it in such an environment (Or if you could annihilate the that V and N for the reservoir are
,
completely system ,
assume
in and out of system
fixed ,
the work done by energy travels
I is the energy you could recover -> the System's energy plus ↑ only
collapsing the atmosphere ( ·
Total entropy of the universe :
DStotal-SS reservoir
1dS = +
-N) =D
>
If the environment is of constant temperature the can extract heat from this environ 3) The temperature of the reservoir is the same as the temperature of the
system , while one
-
one
system
.
,
is minus the
ment for free >
-
all we need to provide to create a
System from nothing is
any additional change dup in the reservoir's energy change du in the system's energy .
work needed. aStotal = dS -
FdU =
-
f(du -
TaS) =
-
Edf - underi
These conditions ,
an increase in
total entropy of the universe is the
because we
↳ if we annihilate the system generally can't recover all its energy as work
we ,
same
thing as a decrease in the
minimize its Helmholth
have to dispose of its
entropy by dumping some heat into the environment The
system will do whatever it can to free energy Helmholtz free energy of the system .
given by Tas 4) AFSW- if no work is done the F
on
system ,
can
only decrease
Helmholtz free
energy : F= -TS heat you
·
minus the can
↳ total energy needed to create the system ,
get for free from If instead
T
environment at temperature .
an we let the volume of the
system change ,
but keep it at constant pressure ,
then this
gives :
that must be provided as work , if you're creating
system
> energy
aS-Fan-FaV F1dU-TdS PdV) #66 it is the Gibbs free energy that tends to decrease
-
+ = -
>
-
aStotal
-
of nothing
=
out
.
=
lor if you annihilate the system, the energy that comes out as work is F ,
Since you dump heat (TS) into environment to get rid of entropy .
S
* The more entropy a
system has , the more of its
energy can enter as heat At constant and volume , Stends to increase
energy
·
allowed to enter leave the
·
Gibbs free energy : G = U-TS + PV ·
at constant
temperature and volume , Ftends to decrease noparticles or
system
7) Work need to do to create a
system or the work
you recover
you ,
you destroy
When it ·
at constant temperature and pressure ,
G tends to decrease
I is the system's energy , minus the heat term that's in F , plus the at
TS work term that's in H.
mospheric
-
>
S
>
-
We want to look at the
changes in these
quantities In a constant temperature environment
saying that F tends to decrease is the Jame as
saying that Utends to
F
-
U
, ,
decrease while Stends to increase
Thermodynamic
.
+ PV
Potentials The decrease , but only because when the loses its environment
spontaneously system
> can
system's energy energy
-
-6
H ,
that and therefore the entropy of the environment increases
gains energy
.
↳ at low
temperature more
entropy gained ,
since it scales with T .
For in the place Two
any change System that takes under constant
temperature T the change in F is :
,
ways to increase entropy of environment :
Q + W-TAS if is created during the process then Q TAS It
3
A AU-TAS > no new
entropy acquire energy from the system
can
=
The System's U and V want to decrease while S wants to increase
-
= = , ,
·
, ,
↓ ↳work so the the work done on It can acquire volume from the system
done on the
system change in F is
equal to lall in the interest of maximizing the total entropy of the universe
·
heat the system
added
↳ if new entropy is created , then TAS will be than Q , So AF will be less than W Extensive and Intensive Quantities
greater .
* F SW at constant T
) includes
↳ all work done on the work done Extensive quantities : Quantities that double if you of stuff
system including any automatically by its double the amount
·
,
collapsing environment V, N, S U H F G mass
expanding or . , , , , ,
Intensive quantities : Quantities that remain
unchanged when the amount of souff doubles
·
T P M, density , ,
* Constant pressure and temperature >
-
use G ·
extensive Intensive = extensive
use F extensive
* Constant temperature >
-
· - intensive
extensive
made
·
extensive extensive - neither (you've probably a mistake
Any change in the System that takes
place under constant temperature and pressure :
*G = AU-TAS + PaV = Q+ W-TAS + PAV
meby
Gibbs free
↳ Q-TAS is always zero or
negative energy and Chemical Potential
↳ Wincludes all the work environment--PAV plus Other "work
do
the
any
M
,
W PAV + Wother if you add (T, Pfixed) , the Gibbs free energy
one particle to
System increases
by M ladd more particles the e
·
a
-
=
,
,
each particle adds M)
This leaves G Wother at constant T P 6= Gibbs free energy per particle
: a >
WM >
-
, .
There G the easiest measuringAl for the reaction by . 3 :
5 Phase transformations of pure substances
are
many ways to measure ,
one is meas. .
the heat absorbed when the reaction takes place at constant and no "other" work
uring pressure
. Then calculate the
entropies and AG Alt-TaS Phase transformation : Discontinuous in the properties of its environment is
is done use =
change a substance , as
changed only infinitesim
·
ally .
Phases : The different forms of
Electrolysis fuel ,
cells and batteries ·
a substance .
H20(HatOn >
-
electrolysis of liquid water into
hydrogen and oxygen gas
·
Phase-diagram : A
graph showing the equilibrium phases as a function of temperature and pressure
↳ for this reaction DH = 206 (the amount of heat you would
k] get out if you burned a
mole of hydrogen , running the reaction in reverse Vapor-pressure Pressure at which a can coexist with its solid or
liquid phase
:
gas
·
This means will need
you 206k] of heat to form hydrogen and
oxygen out of water
Plbarin
To determine what this heat can be used as free
·
Triple point
part of work , we check the
change in the
sys. 221 - critical point : All 3 phases coexist
entropy
: S ad water at lower pressures liquid
ice "sublimates" directly into vapor
,
water cannot exist (in equilibrium) ,
free
: 3free
ice
energy
work Carbon dioxide dry ice frozen carbon dioxide
selectricity)
·
>
- =
Steam
0 006-------
ThermodynamicIdentities Most substances are like carbon dioxide
.
> :
applying more
-
triple point pressure raises its melting temperature
dU = TaS -
PAV + MAN * From these
derivatives
one can derive a
variety of formulas for partial -23 do 374 'T( Tce is unusual
T
applying pressure lowers its
,
↳ Because ice is less dense than water .
melting temperature
dH = TdS + VaP +
MAN
Of
af = SaT-PAV +
Man , S =
-1 wip--NiMON ·
Liquid-Gas phase boundary always has a positive slope :
If you have liquid and gas in equilibrium ,
and
you raise
the temperature ,
you must apply more pressure to keep the
liquid from vaporizing
(84Niv M=
46 SaT + VaP + MaN S as the pressure increases the
gas becomes more dense , so the difference between liquid and gas grows less
= , = =
: , .
N
↳ Point where there is no longer any discontinuous change from liquid to gas > Critical point (substance is a fluid)
Maxwell Relations ·
Liquid-Crystal phase : Molecules more around
randomly as in a
liquid ,
but still tend to be oriented parallel to each other
·
Example : Start from UIS V) , ,
then we find : of 4 Diamonds and Graphite
Then use the
Thermodynamic identity for U: dU-TdS-PdV :
OU-T O- : ·
Diamond and Graphite are the two familiar phases of elemental carbon > both solids but with different
crystal struc
tures
(ov
.
Substitute 04/05 and /OV : 01T) g (0) =
= =
~ Stable
phase is always the one with lower Gibbs free energy .
Pressure dependence of the Gibbs free energy is determined by the volume of the substance N :
volume than a mole of graphice thus its Gibbs free energy will grow more
rapidly
(s -(8)r
↳ mole of diamond has smaller
=
OP a a ,
=
o+ v as the pressure is raised
(85)s ( =
o 04T =
* Natural diamonds must form at
very great depths <the higher the temperature , the more pressure is required before
diamonds becomes the stable phase
