Entropy Print

Advanced Thermodynamics Note 4 The Second Law of Thermodynamics Lecturer: 郭修伯

Differences between heat and work • A conversion efficiency of 100% by elimination of friction, a dissipative process that transforms work into heat. • All efforts to devise a process for the continuous conversion of heat completely into work or into mechanical or electrical energy has failed. Conversion efficiencies do not exceed about 40%. • The flow of heat between two bodies always takes place from the hotter to the cooler body, and never in the reverse direction.

The second law • No apparatus can operate in such a way that its only effect is to convert heat absorbed by a system completely into work done by the system. – It is impossible by a cyclic process to convert the heat absorbed by a system completely into work done by the system.

• No process is possible which consists solely into the transfer of heat from one temperature level to a higher one.

• Heat engine: devices or machines that produce work from heat in a cyclic process. – Essential to all heat-engine cycles are absorption of heat into the system at a high temperature, rejection of heat to the surroundings at a lower temperature, and production of work. – In operation, the working fluid of a heat engine absorbs heat |QH| from a hot reservoir, produces a net amount of work |W|, discard heat |QC| to a cold reservoir, and returns to its initial state.

The first law:

W = QH − QC

The thermal efficiency of the engine: η ≡

W QH

=

QH − QC QH

= 1−

QC QH

• If a thermal efficiency of 100% is not possible for heat engines, what determines the upper limits? • A heat engine operating in a completely reversible manner is called a Carnot engine (N.L.S. Carnot, 1824): – Step 1: A system at the temperature of a cold reservoir TC undergoes a reversible adiabatic process that causes its temperature to rise to that of a hot reservoir at TH. – Step 2: The system maintains contact with the hot reservoir at TH, and undergoes a reversible isothermal process during which heat |QH| is absorbed from the heat reservoir. – Step 3: The system undergoes a reversible adiabatic process in the opposite direction of step 1 that brings its temperature back to that of the cold reservoir at TC. – Step 4: The system maintains contacts with the reservoir at TC, and undergoes a reversible isothermal process in the opposite direction of step 2 that returns to its initial state with rejection of heat |QC| to the cold reservoir.

The cycle traversed by an ideal gas serving as the working fluid in a Carnot engine is shown by a PV diagram. a → b adiabatic compression b → c isothermal expansion TH c → d adiabatic expansion d → a isothermal compression P b |QH| For the isothermal steps b → c and d → a c

QH = RTH ln

TC

Vc Vb

and QC = RTC ln

Vd Va

For adiabatic processes a → b and c → d a |QC|

d

∫

TC

V

QH QC

TH ln(Vc / Vb ) TH = = TC ln(Vd / Va ) TC

η=

TH

W QH

CV dT Va = ln and R T Vb

= 1−

QC QH

TC = 1− TH

∫

TH

TC

CV dT Vd = ln R T Vc

300 η = 1− = 0.5 600

Rough practical limits for ηof a Carnot engine; actual heat engines are irreversible and ηrarely exceed 0.35.

A central power plant, rated at 800,000 kW, generates steam at 585K and discards heat to a river at 295 K. If the thermal efficiency of the plant is 70% of the maximum possible value, how much heat is discarded to the river at rated power?

η max = 1 −

295 = 0.4957 585

η = 0.7 × 0.4957 = 0.347

η=

W QH

= 1−

η = 0.347

QC QH

= 1−

TC TH

W = 800000 kW

QC = 1505500 kW

• A Carnot engine:

QH TH

=

QC TC

QH QC + =0 TH TC dQH dQC + =0 TH TC

dQrev dQrev dQrev dS t = =0 ∫ called entropy S,T which is an intrinsic • There exists a property T T property of a system, ∆S t = ∫

functionally related to the measurable coordinates which characterize the system.

t – If a process is reversible and adiabatic, dQrev = 0, dS = 0. The entropy of a system is constant during a reversible adiabatic process and the process is isentropic. – When a system undergoes an irreversible process between two equilibrium states, the entropy change of the system ΔSt is evaluated by an arbitrarily chosen reversible process. Since entropy is a state function, the entropy changes of the irreversible and reversible processes are identical.

Entropy changes of an ideal gas • For one mole of fluid undergoing a mechanically reversible process in a closed system: dU = dQrev − PdV dH = dU + PdV + VdP

dQrev = dH − VdP dH = C Pig dT V = RT / P dS =

dQrev dT dP = C Pig −R T T P

ig T C ∆S P P dT =∫ − ln T 0 R R T P0

For an ideal gas with constant heat capacities undergoing a reversible adiabatic process: ig T C ∆S dT P =∫ P − ln T0 R T R P0

∆S = 0

C Pig T2 P 0= ln − ln 2 R T1 P1

T2  P2  =   T1  P1 

R / C Pig

ig C C =C +R γ = P ig P

T2  P2  =   T1  P1 

ig V

( γ −1) / γ

CVig

The 2nd law mathematical statement • Let a quantity of heat |Q| be transferred from the hotter (TH) to the cooler (TC) reservoir. The entropy changes of the two reservoirs are: −|Q| −|Q| t t ∆S H =

∆Stotal

∆SC =

TH

TC

 TH − TC = ∆S + ∆S =| Q |   TH TC t H

t H

  

• For the process of irreversible heat transfer, ΔStotal is always positive, approaching zero as the process becomes reversible.

∆Stotal ≥ 0

A 40 kg steel casting (CP= 0.5 kJ/kgK) at a temperature of 450°C is quenched in 150 kg of oil (CP= 2.5 kJ/kgK) at 25°C. If there are no heat losses, what is the change in entropy of (a) the casting, (b) the oil, and (c) both considered together? No heat losses: ( 40)(0.5)(T − 450) + (150)(2.5)(T − 25) = 0 t (a) the casting ∆S = ∫

(b) the oil

∆S t = ∫

dQ C dT T kJ = m∫ P = mC P ln 2 = −16.33 T T T1 K dQ C dT T kJ = m∫ P = mCP ln 2 = 26.13 T T T1 K

t t (c) total entropy change ∆Stotal = ∆S cast + ∆S oil = 9.80

kJ K

T = 46.25 C

Entropy balance for open systems Net rate of change in entropy of flowing streams

Time rate of change entropy in control volume

Time rate of change entropy in surroundings

t d (mS ) cv dS surr ∆ ( Sm ) fs + + = S G ≥ 0 dt dt

Q j d (mS ) cv ∆( Sm ) fs + −∑ = S G ≥ 0 dt j Tσ , j

Total rate of entropy generation

In a steady-state flow process, 1 mol/s of air at 600 K and 1 atm is continuously mixed with 2 mol/s of air at 450K and 1 atm. The product stream is at 400K and 1 atm. Determine the rate of heat transfer and the rate of entropy generation for the process. Assume that air is an ideal gas with CP = (7/2)R, that the surroundings are at 300K, and that kinetic- and potential-energy changes are negligible. mol n = 3 Q  s S G = n S − n A S A − n B S B − mol Tσ n A = 1 T = 400 K s Q = n A ( S − S A ) + n B ( S − S B ) − TA = 600 K Tσ

C.V.

Q

mol n B = 2 s TB = 450 K

T T Q = n AC P ln + n B C P ln − TA TB Tσ 7 400 400  8729.7  = ( )(8.314) (1) ln + (2) ln +  2 600 500 300   J = 10.446 Ks

Q = n H − n A H A − n B H B = n AC P (T − TA ) + n B C P (T − TB ) = −8729.7 J / s

An inventor claims to have devised a process which takes in only saturated steam at 100 °C and which by a complicated series of steps makes heat continuously available at a temperature level of 200°C. The inventor claims further that, for every kilogram of steam taken into the process, 2000 kJ of energy as heat is liberated at the temperature level of 200°C. Show whether or not this is possible. In order to give the inventor the benefit of any doubt, assume cooling water available in unlimited quantity at a temperature of 0°C.

∆H = Q + Ws

T’ = 200°C

Q = Q′ + Qσ = −2000 + Qσ

Q′ = −2000kJ Saturated steam at 100 °C kJ H 1 = 2676 kg kJ S1 = 7.3554 kgK

apparatus

Liquid water at 0 °C H 2 = 0.0 S 2 = 0.0

Qσ Tσ = 0°C

∆H = 0.0 − 2676 = −2000 + Qσ Qσ = −676.0 kJ kJ K 2000 kJ ∆S t = = 4.227 200 + 273.15 K 676.0 kJ ∆S t = = 2.4748 0 + 273.15 K ∆S = 0.0 − 7.3554 = −7.3554

∆S total = −7.3554 + 4.227 + 2.4748 = −0.6536

kJ K

Calculation of ideal work • In a process producing work, there is an absolute maximum amount which may be accomplished as the result of a given change of state of the fluid flowing through the control volume. • The limiting value obtains when change of state associated with the process is accomplished completely reversibly. • The entropy generation is zero: Q ∆( Sm ) fs − =0 Tσ Wideal

Q = Tσ ∆( Sm ) fs

= ∆H − Tσ ∆S W ideal = ∆( Hm ) fs − Tσ ∆( Sm ) fs

 1   ∆  H + u 2 + zg m  = Tσ ∆( Sm ) fs + W s (rev) 2   fs   1   W ideal = ∆  H + u 2 + zg m  − Tσ ∆( Sm ) fs 2   fs 

W ideal W s ηt ( work required ) =  ηt ( work produced ) =  Ws Wideal

What is the maximum work that can be obtained in a steady-state flow process from 1 mol of nitrogen (assumed an ideal gas) at 800 K and 50 bar? Take the temperature and pressure of the surroundings as 300 K and 1.0133 bar? For an ideal gas, enthalpy is independent of pressure, and its change is given: T

∆H = ∫ C Pig dT = 8.314 × ICPH (800,300;3.280,0.593E − 3,0.0,0.040 E + 5) = −15060 T0

T2

∆S = ∫ C Pig T1

dT P 1.0133 J − R ln 2 = −15060 − 8.314 ln = 3.042 T P1 50 mol

Wideal = ∆H − Tσ ∆S = −15060 − (300)(3.042) = −15973

J mol

Or, Step 1, reversible, adiabatic expansion from initial state to 1.0133 bar, T’ Step 2, cooling to the final temperature Step 1

Q + WS = ∆H

WS = ∆H = ( H ′ − H1 )

T − Tσ = (dQ ) T

WCarnot = Q − Tσ ∫

Step 2

dWCarnot

Total

Wideal = WS + WCarnot = ∆H − Tσ ∆S

T2

T'

dQ = ( H 2 − H ' ) − Tσ ∆S T

J mol

An inventor claims to have devised a process which takes in only saturated steam at 100 °C and which by a complicated series of steps makes heat continuously available at a temperature level of 200°C. The inventor claims further that, for every kilogram of steam taken into the process, 2000 kJ of energy as heat is liberated at the temperature level of 200°C. Show whether or not this is possible. In order to give the inventor the benefit of any doubt, assume cooling water available in unlimited quantity at a temperature of 0°C.

∆H = 0.0 − 2676 = −2676

T’ = 200°C

Q′ = −2000kJ Saturated steam at 100 °C kJ H 1 = 2676 kg kJ S1 = 7.3554 kgK

apparatus

Liquid water at 0 °C H 2 = 0.0

∆S = 0.0 − 7.3554 = −7.3554

Wideal = ∆H − Tσ ∆S = −2676 − (273.15)(−7.3554) = −666.9

J mol

| Q |=| W |

T Tσ − T

S 2 = 0.0

Qσ Tσ = 0°C

kJ K

= 666.9

200 + 273.15 = 1577.7 kJ 200 − 0

Lost work • Work that is wasted as the result of irreversibilities in a process is called lost work: Wlost ≡ WS − Wideal  1   W S = ∆  H + u 2 + zg m  = Q 2   fs 

 1   W ideal = ∆  H + u 2 + zg m  − Tσ ∆( Sm ) fs 2   fs 

W lost = Tσ ∆( Sm ) fs − Q Surrounding temperature Tσ

W lost = Tσ S G ≥ 0 Wlost = Tσ SG ≥ 0

 S = ∆( Sm ) − Q G fs Tσ

The two basic types of steady-flow heat exchanger are characterized by their flow patterns: cocurrent and countercurrent. Consider the two cases, for each of which the following specifications apply:

TH 1 = 400 K TH 2 = 350 K TC1 = 300 K

n H = 1 mol

s

The minimum temperature difference between the flowing streams is 10K. Assume that both streams are ideal gases with CP = (7/2)R. Find the lost work for both cases. Take Tσ = 300 K.

Fig 5.9

 1   W S = ∆  H + u 2 + zg m  = Q 2   fs  ∆( Sn ) fs = n H (∆S ) H + n C (∆S ) C

n H (∆H ) H + n C (∆H ) C = 0

n H C P (TH 2 − TH 1 ) + n C C P (TC 2 − TC1 ) = 0

Negligible pressure change

 TH 2 n C TC 2     ∆( Sn) fs = nH C P  ln + ln  TH 1 n H TC1 

W lost = Tσ ∆( Sn ) fs − Q

340  J 7  350 Case I, cocurrent: ∆( Sn ) fs = (1) (8.314) ln + 1.25 ln  = 0.667 300  Ks 2  400 n C 400 − 350 = = 1.25 J n H 340 − 300 W lost = Tσ ∆( Sn ) fs = 300 × 0.667 = 200.1 s

Case II, countercurrent:

390  J 7  350 ∆( Sn ) fs = (1) (8.314) ln + 0.5556 ln  = 0.356 300  Ks 2  400

n C 400 − 350 J = = 0.5556   W = T ∆ ( S n ) = 300 × 0 . 356 = 106 . 7 n H 390 − 300 lost σ fs s

From thermodynamic point of view, the countercurrent case is much more efficient.

The third law of thermodynamics • The absolute entropy is zero for all perfect crystalline substances at absolute zero temperature. – When the form is noncrystalline, e.g., amorphous or glassy, calculations show that the entropy of the more random form is greater than that of the crystalline form. – The absolute entropy of a gas at temperature T: S=∫

Tf

0

∆H f Tv (C ) T (C P ) g (C P ) S ∆H v P l dT + +∫ dT + +∫ dT T T f v T Tf T Tv T