Thermodynamics Heat Content and the First Law (Kinetic ...

Heat Content and the First LawThermodynamics

(Kinetic Theory of Gases)

Dh

Agenda

• Kinetic theory of ideal gases,

• Work and heat: First Law of TD

• Equation of State

• Reversible processes (expansion/Compression)

• Carnot cycles, entropy

• Steady flow energetics

• TD of real gases

• Steam engines

• Carnot and Rankine cyclic engines

• Otto motor cycle

• Turbines, combined cycle thermal plants

ESTS 486 2019 Jane Doe

Sol

ar P

roj

2

Land lease/rental in NWNY: 1 acre @ $11,000/a

Mechanical Energy: Work and Heat

Ideal Gas: Diluted ensemble of N structure-less, independent particles with macroscopic, observable properties described by Equation of State.

→ (Sole interactions: Elastic scattering)

Enclosed in a volume V, IG exerts macroscopic (kinetic) pressure p on container walls.

→ Compression or decompression of (N=const.) changes its (kinetic-) energy content

U = capacity to perform work w. (Additional work types wel , rxns).

Heating or cooling = transfer of disorganized energy = “heat” q (motion of particles in container walls).

|0 :

( )0 :

ext

V Gas Expansionp Force Area p F h

V GasComw p

pres oV

si n

D = = D + D =

D = − D

Sign Convention: Work and heat are counted positive (w, q > 0), when they increasethe internal energy U of the gas, and negative when done or emitted by the gas.

Dh

Vacuum Pump

1. Law of Thermodynamics(Conservation of Energy in isolated system):

...dU dq dw dq p dV= + = − +

w p h= − D

=

,

,

extensive variables V Np V N R T

intensive variables p T

real gases

“inexact” differentials

F

W. Udo Schröder, 2019

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Random (IG) Velocity/Kinetic-Energy Spectra

( ) ( )

( )

− −= =

→→ = =

3 3/2

2

24

22

3

BTk

B

B

Maxwell

Boltz

dP dP uue

d m d u k T

Mean thermal energy U Nm

u

ann

k

m

T

( ) ( )

( )

= −

→ = =

3 2 2

2

0

2( ): 4 exp

2

8

B B

B

dP u m umIG particles mass m u

du k T k T

dP u kMean thermal speed du u

du

Tu

m

Plausibility of EOS (Gas Law) :N =# of particles colliding with a wall <u> → momentum transfer to wall per particle <u>→ pressure on wall p <u>2 T (see EOS)

dP(u

)/du

(a.u

.)dP(

)/d

(a

.u.)

Bp V N k T =

Bound Lattice low T Lattice→Fluid/Real Gas

Unbound Gas, T=300K

Lattice

Real Gas

Ideal Gas i

(t)/

kB

i (t)/kB (i=1,..,1000)

Velocity u (km/s)

Kinetic Energy (10-20 J)

Transition


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Heat Transfer

Heat (Q) conduction, flux=current density through area A

( )

= − = − + +

' :

( )

q

T T TFourier s Law j T r i j k

x y z

Thermal conductivity W mK

Heat convection: Heat transfer via mass flow

( )= − −

2

'

( )

ambient

dQNewton s Law of cooling A T T

dt

Heat transfer coefficient W mh K

h

Heat radiation: Heat transfer via elm. photons (like light)

( )

−

−

= −

=

− =

4 4

8 2 4

( 1)

5.6703 10

SB

SB

Q ambient

Stefan Boltzmann Law

Radiated thermal flux j T T

Emissivity often

Stefan Boltzmann constant W m K

( ) /qj dQ dt A=

Th Tl

Thermometers

6

The Ideal-Gas Equation of State

{p, V, T}

A

State “functions:” {p, V, T} (N=const.). Molar {p, V, T} hyper-plane (monotonic) contains all possible gas states A. Set of “independent coordinates.” There are no other states of the gas. → All state functions can be expressed as {p, V, T}.

Ideal Gas Constant R

R = 0.0821 liter·atm/mol·K

R = 8.3145 J/mol·K

R = 8.2057 m3·atm/mol·K

R = 62.3637 L·Torr/mol·K or

L·mmHg/mol·

Boltzmann Constant kB

kB= 1.381·10-23 J/K


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p(V,T)= n R T/V

p·V = n·R·T; n=# moles; equivalent: p·V = N·kB·T (N=# particles)

Interacting only via elastic scattering, no bonding →→ Only gas phase!

,rando

U Cm

m iT

nC N

ot o

7

Transitions Between States

AB

State functions p, V, T,… describe the system states but not the processes connecting states. Two states A, B can be reached by different processes representing different pathways on the {p,V,T} hyperplane. The two processes and Energy transfers A→B differ by relative magnitudes of heat absorbed vs. work done.

Process 1 A→ B

along Path 1

Process 2 A→ B

along Path 2


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Two different states (A, B), same gas.

1A B⎯⎯→ 2A B⎯⎯→

1 1

2 2

B A

q wU U U

q w

D + D D = − =

D + D

,U C T C N

8

Of interest for cyclic machines.“Slow,” equilibriumprocesses A → B,

subject to boundary conditions of:

1. Dp = 0 (isobaric)

2. DV = 0 (isochoric)

3. DT = 0 (isothermal)

4. q = 0 (adiabatic)

follow well-defined, constrained routes in the {p, V, T} hyper-plane of states. Can easily be inverted →reversible

processes.

Reversible Processes

T

q=0

Reversibility is not guaranteed for all processes involving an ideal gas.


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Examples of reversible IG processes

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w = - area under curve p(V)Total work (1 →2, T = const.) :

Reversible Isothermal Expansion/Compression


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2 2

1 1

1

2

1

( )

ln

0 ( )

0

0

0

V V

V V

p V R T

w

w implies system loses internal energy

by doing work on surr

But

Use for expanding mole

dVp V d

q absorbs heat

compensates energy loss by he

V R TV

VR T

ound

at ab

U

s

T

o

ing

V

Us

rptio

= − = − =

=

D

=

→

→

D

=

.Needs contact to heat bath T

n

const→ Intersection of {p,V,T} hyper-plane with plane T=const.

1

0 2

ln1. 0:Law of Thermodynamics Isotherm expansV

q U w wion R TV=

=D − = − = −

higher T more

w

r

T

wo k

→

10

Compress 1 mole at p=const.

Heat transfer

Total heat transfer (1 →2)

Reversible Isobaric Compression

1

2

1

2

2 1

( ) :

( )

0: ,

50

2

( ) :

5

[ ] 0

(

.

.

0

2p

V

p

VV

p

V

system has to be coole

p VT

R

Work done on system

p V dV

p V Shaded Area

T emitting

w

Enthalpy change

emit

w p d

ted he

R

a

d

q C

C T C

V

H q

f

p

o

>0R T

r

t i

p

const

T T

const

= − =

= − D =

D

= = −

=

D =

−

D = − =

−

=

D

=

D

D

2 1

( )

[ 0

)

]

( )p

V

U q

nternal energy

C R T

T T

w H

C

−D = + = D

= −

D

Internal energy change

DV


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Inverse process: heating at constant p, e.g., p=patm , leads to expansion, V2 → V1>V2 → drives piston out of its cylinder.

EOS

11

Isochoric (V = const.) decompression → of 1 mole w =-pDV=0

Heat transfer

Total energy change (1 →2)

Reversible Decompression

2 1

1

2 1

2

0

0,

1.

0

( ) ( )

[ ]

[

( )

( )

[ ]

]

:

V

V

V

p

V

p

Work done on system w

But U system emits heat

Law of Thermodynamics :

q w

pV R

q C T C T T heat ba

qU C T T

U C

th

since p

Enthalpy change

H T

con

C T T always H C

stB

<

U

0

T

T

H q

=

D →

= + = =

D = + D = + D

=

= D =

D

− D

−

=

D

D

D

→

−


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0Vq C T= D

Inverse process: heating at constant V, leads to increased temperature and pressure.

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1) Isothermal expansion at T1=const.

2) Isochoric decompression at V2=const.,

3) Isothermal compression at T2 =const.

4) Isochoric compression V1=const.,

Energy balance:

1) gas does work w1 = - q1; DU = 0

2) gas emits heat q < 0; DU < 0

3) gas receives work w2 = - q2; DU = 0

4) gas absorbs heat q > 0; DU > 0

Total energy change: DU = 0 (cyclic)

Total work done: w = w1+ w2 < 0

Total heat absorbed: q = q1+ q2=-w > 0

Expansion-Compression Cycles

In one cycle the gas absorbs net heat energy and does net work,

w = w1 + w2 = -q = CV∙[T2-T1]

Not all absorbed heat is converted, some has to be dumped as waste heat.


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V1 V2

work dw=-p·dV

1

2

3

4T1

T2

Observation: IG systems absorbing external (random) heat can produce mechanical work

on surroundings (engine). Continuous operation requires cyclic process (in p-V-T space).

Needs contact to 2 heat baths T1 and T2.

13

Make an arbitrary cyclic process out of

elementary isothermal and isochoric

processes →

Heat energy q1 is absorbed at a high

temperature(s) T1, and partially

dumped, |q2| < |q1|, at a lower

temperature(s) T2.

The difference (q1 + q2)= q1-|q2| is

converted into useful work w < 0 done

on surroundings by the gas.

Thermal Engines: Principle of Operation

p

V

Net work done by gas

T2

T1

Random heat energy is converted into orderly collective energy (work, pushing a piston, turning a wheel) !!!!!!! → Practical use


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q1

q2

Horizontal paths traveled in both directions do not contribute net work → Area within closed p-V paths = total work done in cyclic process.

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Isothermal expansion at Th=T1

Adiabatic expansion Th→ Tc=T2

Isothermal compression at Tc <Th

Adiabatic compression Tc→ Th,

Carnot Cycles

1

1 2

2 12

1 2

:

" "

:

0

(

.

)

.

A BA B

Entropy S S

For any process

sign for reve

Entropy is conserved i

S state function descripto

n reversible

cycli

q

c processe

rsible

q

T T

qS

T

A B o

r

s

nly

S S

→→

D + D =

= −D = D

= →

→ =

= −

D Reversible adiabatic expansion or compression: DS = q/T= 0 since q= 0.Irreversible adiabatic exp./compr.: DS 0.

V

p

(p2, V2)

(p3, V3)

(p4, V4) work

(p1, V1)

T1

T2

q=0

q=0


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−=

=

=1( 0)

p V

T V conA sdiabatic q Eo

c

tS

c

2

1

1

4

32

2

1

4

3

2

1

2

1 1 ln 0

ln 0

V

VRT V

V

V

V

w p dVq T

q

R

w p dV RV

T

V

V=

=− = =

=−

= =

Energy balance: w = q1 + q2 > 0on isothermal portions:

Adiabatic expansion/compression →V4/V1= V3/V2 → V4/V3= V1/V2

→ Adiabatic works cancel

17

Example: Adiabatic (q=0) expansion A’ → B’

into partial vacuum (both A’, B’ are legitimate states)

→ Free expansion A’→ B’, p → pext=0, wirrev = 0

occurs spontaneously. Actually, it is a non-equilibrium transport process eliminating p-differential, more probable configuration.

Equilibration A’→B’ → wirrev =0 .This work is smaller in magnitude than reversible

work wirrev ≤ 0, | wirrev | < | wrev | done by gas in equilibrium with environmentWhy can process A’→B not be reversed?

>>> Reversal would de-randomize thermal motion of gas @ equilibrium.A “free contraction” Vin → Vfin< Vin with fewer

occupied particle states would require a correlated motion of many particles towards a point in space.

Free expansion occurs only Vin → Vfin > Vin

Irreversible/Spontaneous Processes


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AA

DT=0rev expansion

V

pgas

irrev

decom

pre

sio

n

A

B

A’

B’

pext

pext

pext

irrev

rev

B

A

A’Opening valve → Process.

X

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Efficiency of Carnot Engines

Theoretical Carnot efficiency

Tc

Th

-w = qh+qc=DS·(Th- Tc)

qh= DS·Th

qc= -DS·Tc

11 1h

cC

h

h cC

h h

c

T

h

q qw

q q

q

Tq

T

→

+−= =

+ = − ⎯⎯⎯→=


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In practice, Th depends on fuel heating value (max temperature Tad). Transfer from fuel to hot reservoir: ( ) ( )F ad h ad cT T T T = − −

1 c ad hC F

h ad c

T T T

T T T

−= = −

− → Effective Carnot efficiency:

( )

( )

, . U const

h ch c h

h

h ch

h

Process at p T const q H

T Tw T T S T S

T

T TH T

T

== ⎯⎯⎯→ = D →

−=− − D = − D

−= − D

DS = const

Heat Bath

Cold Sink

19

Entropy Flow in Carnot Engines

Entropy DS from the hot reservoir enters the engine with a heat energy of DS·Th,

produces work and leaves it again with a heat energy of DS·Tc, which is dumped into the cold sink.

Tc

Th

-w = qh+qc== DS·(Th -Tc)

qh= DS·Th

qc= -DS·Tc

Analog: Stream of water DM from a reservoir carries energy DM∙g∙h1 , enters a hydro-turbine, produces work, and leaves with an energy DM∙g∙h2 , which is dumped into the river.

Hydrodynamic Power Plant

DM∙g∙h1

DM∙g∙h2


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Turbine

Inlet

Outlet

Reservoir

.M

S

Mass flow j dM dt

Entropy flow j dS dt

Ideal Otto Cycle


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1) Intake stroke ( 1→2 ), gasoline vapor and air drawn into engine.

2) Compression stroke (2→3) . p,T increase.

3) Combustion (spark) (3→4), short time, V= constant. Heat

absorbed from high-T “reservoir”.

4) Power stroke: expansion (4→5).

5) Valve exhaust: Valve opens, gas can escape.

6) Emission of heat (5→6) to low-T reservoir.

7) Exhaust stroke (2→1), piston evacuates cylinder.

crank shaft

cams

fly wheel

fuel intake

Energetics of Otto Cycle


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1

( ) ( )

( )

( )

( )

4 3 6 53 4 5 6

3 4 3 4 4 3

5 6

4 3

:

1

. 1V VR c

V V

V

R c

c T T c T Tq qwEfficiency

q q c T T

T T

T T

Adiabatic EoS T V co rnst

→ →

−

→ →

− + −+= = =

−

−

= −

= −−

= → Effic

iency

Compression Ratio r

=Cp/Cv


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Work in Steady-Flow Processes

1. Law of Thermodynamics (Conservation of total energy in isolated system):

21

2= + + potE U M V

( ) ( ) = = =

:

A Ai oi o

Mass conservation

dM dt dM dt

, ,i i i

u dVi

dQ≈0

dW<0

M= mass, = flow velocity, Vpot =potential

energy (often ≈0)Mass density = m (kg/m3), assume homogeneous = M/VInternal energy density u (J/m3)Enthalpy density H/V =: h = u + pDifferentials

( )

= =

=

=

=

=

2

;

(A dx )

dV A dx

1 2 (A dx )

(A dx )

i i i i

i i i

i i i i i

i i i i

i input o output

Internal energy dU u

with

Kinetic energy dK

Mechanical work dW p

dxi

dxo


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Work in Steady-Flow Processes

1. Law of Thermodynamics (Conservation of total energy in isolated system):

21

2= + + potE U M V

= +

=

+

+ +

/

2

2

0

1(A dx ) (A dx )

2(A dx )

(A dx )1

(A dx ) (A dx )

,

2

Exp comp workFlow energyInternal

i i i i i i i i

o o o o o o o

i i i

o o o

incom

Carried by mass flow through

Additional

sys

mec

p

h

ing

dE u

outg

anical work

oing

dE

outpu

u p

t m

t dW

e

( )− = +

: in out

d E E dW dQ

dtSteady sta

h

tedt d

eat out d

t

put Q

= + − +

+ 2 21 1A A

2 2i i i i i o o o o o

dQ dWh h

dtd d

dW

t t

( ) ( ) = = =

:

A Ai oi o

Mass conservation

dM dt dM dt

+ − +

22

2 2o oi i

i o

hhdW dM

dt dt

, ,i i i

u

, ,o o o

u

Power Fuel Flow

dVi

dVo

if well insulated

dQ≈0

dW<0


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Thermodynamics Heat Content and the First Law (Kinetic ...

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