ME 200 L31: Review for Examination 3 ME 200 L31: Review for Examination 3 Thu 4/10/14 Examination 3 (L22 – L30) 6:30 – 7:30 PM WTHR 200, CL50 224, PHY.

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ME 200 L31: ME 200 L31: Review for Examination 3Review for Examination 3Thu 4/10/14 Examination 3 (L22 – L30) 6:30 – 7:30 PM

WTHR 200, CL50 224, PHY 112

Kim See’s Office ME Gatewood Wing Room 2172 Please check HW and Examination Grades on BlackboardPlease pick up all graded Home Work and Examinations

from Class or Room 2172 after class today!

https://engineering.purdue.edu/ME200/ThermoMentor© Program

Spring 2014 MWF 1030-1120 AMJ. P. Gore

[email protected] Wing 3166, 765 494 0061

Office Hours: MWF 1130-1230TAs: Robert Kapaku [email protected]

Dong Han [email protected]

Examination 3

• Extra Time Students 6-9 pm ME2063• Civil Engineering Banquet Students 6-7 pm

ME1051 Can accommodate ~50 students. If you are not a CE banquet student go to your regular room

• Division 4 (Gore Students) WTHR 200

2

Isentropic Processes for Ideal Gases (Air) with Constant Specific HeatIsentropic Processes for Ideal Gases (Air) with Constant Specific Heat

3

T

s

2

1

p2v2

T2s2

T1

p1

v1

s1

State 1 and State 2 are defined in the Figureby intersection of any two curves or linespassing through the points on the graph. Thelines and curves passing through these pointsdefine additional states where one of the properties is constant. For example the verticalline is the isentropic line along which s is Constant.

22 1 2 1

1

2 22 1

1 1

2 2

1 1

0o o

oo o

o

r

r

ps s s s R ln

p

p exp( s / R )exp ( s s ) / R

p exp( s / R )

p p;

p p

Tabulated as a function of T in Table A22 for air.for air.rp

222 2

1 1 1 1

rr

r r

( RT / p )v v

v v ( RT / p )

Tabulated as a function of T in Table A22 for air.for air.rv

Isentropic Processes for Ideal Gases with Constant Specific HeatIsentropic Processes for Ideal Gases with Constant Specific Heat

4

Constant specific heat assumption allows further simplification of the p-vrelation for the isentropic process for ideal gases:

2 22 1

1 1

2 2

1 1

1 1

2 2 2 2 2 22 2 1 1

1 1 1 1 1 1

0

1 1 1p

p

R / c

p p v p

k

k kk k

T ps s c ln R ln

T p

T p; R / c ( c c ) / c / k ( k ) / k

T p

T p p v v pp v p v

T p p v v p

Isentropic processes for ideal gases can be analyzed using pv=mRT and pvk = constant if specific heat is independent of temperature.

Isentropic Processes for Ideal Gases with Constant Specific HeatIsentropic Processes for Ideal Gases with Constant Specific Heat

5

T

s

n = k

n = -1; p and v both increase and decrease together

n = 1

v = Constant; n = ±∞

p = const; n = 0

2 2 1 1n np v p v

2 2 1 1k kp v p v

pv RT Constant 1 T Constant 2

Polytropic Processes for Ideal Gases with Constant Specific HeatPolytropic Processes for Ideal Gases with Constant Specific Heat

6

p

v

n = kn = 1

2 2 1 1n np v p v

n = -1

n = -1; p and v both increase and decrease together

v = Constant; n = ±∞

p = const; n = 0

Maximum performance measures (reversible processes)

• Power cycles

• Refrigeration cycles

• Heat pump cycles

7

H

C

H

CT

T

Q

Q 11max

maxC C

H C H C

Q T

Q Q T T

CH

H

CH

HTT

T

QQ

Q

max

Carnot efficiency

Example

8

A refrigeration cycle operating between two reservoirs receives energy QC from the one at TC = 280 K and rejects energy QH to one at TH = 320 K. For each of the following cases determine whether the cycle operates reversibly, irreversibly, or is impossible:

– QC = 1500 kJ, Wcycle = 150 kJ. Impossible– QC = 1400 kJ, QH = 1600 kJ. Reversible– QH = 1600 kJ, Wcycle = 400 kJ. Irreversible– β = 5. Irreversible (βmax= 280/40 =7)

Example: Use of Clausius Inequality

QH=1000 kJ, TH=500 K and QC=600 kJ at TC:

(a) 200 K, (b) 300 K, (c) 400 K. Find if each cycle is reversible, irreversible or ideal.

Solution: Use the given QH, QC values to find

work and ensure that the work produced does not result in a negative value for cycle

CHcycle

H C

QQ

T T

b

T

Q∫

Example: Use of Clausius Inequality

(b) kJ/K 0K 300

kJ 600

K 500

kJ 1000cycle cycle = 0 kJ/K = 0

(a) kJ/K 1K 200

kJ 600

K 500

kJ 1000cycle cycle = +1 kJ/K > 0

Irreversibilities present within system

No irreversibilities present within system

(c) kJ/K 5.0K 400

kJ 600

K 500

kJ 1000cycle cycle = –0.5 kJ/K < 0

Impossible

Entropy is a Property►Entropy is easier to understand if thought of as a

property analogous to specific volume. It is defined as:

Int .Rev.

Int .Re

Int .Rev.

Rev

Re

v

v

.

Q Wds analogous to dv

T p

QQds ;The second RHS

not emphasized when de

is clearer.T T

Subscript "

fining work to av

Rev" relevant

oid confu

to path functions.

T is a p

si

oint

on

fu

.

ncti

Rev

Rev

on. So T not written after integration

Tds Q

Entropy Change Calculations

QH=1000 kJ, TH=500 K and QC=600 kJ at TC:

(a) 200 K, (b) 300 K, (c) 400 K. Find if each heat transfer is reversible find entropy change for (the material that makes up) the reservoir.

Solution: Use the given QH, QC and TH, TC values

to find the entropy change for the reservoirs:

CHH C

H C

QQS and S

T T

RevQS

T

Examples: Entropy Change for the Reservoirs (Property of materials making up the reservoirs)

(a) 1000 600

2 3500 200H C

kJ kJS kJ/K; S kJ/K

K K

Entropy can decrease and increase!

Net entropy change can be zero!

(b) 1000 600

2 2500 300H C

kJ kJS kJ/K; S kJ/K

K K

(c) 1000 6002 1 5

500 400H C

kJ kJS kJ/K; S . kJ/K

K K

Net entropy change can be negative!However, can’t continue that process in a cyclic manner!Since by definition reservoir properties are fixed, these changes must be compensated by reverse actions!

Entropy Change using Tables: Example 3Entropy Change using Tables: Example 3

14

Given: 0.5 kg/s of steam at 280oC, 20 bar is expanded in a turbine to 1 bar in a constant entropy process.

Find: Find the work produced by the steam in kW and show the process on a T-s diagram. If the process was not a constant entropy process and resulted in saturated steam at 1 bar, find the decrease in work and increase in entropy in kW/K.

Assumptions: Change in PE neglected, No heat transfer, work done onTurbine shaft, Steady state, Steady flow, Mass is conserved.

2

2

2

2

CVCV CV i i

H ,L B,S I

CVe e i e

E I E

CVS i i es CVA i i e

dE VQ W m ( h gZ )

dt

dmVm ( h gZ ) ; m m

dt

W / m h h ,W / m h h

St P,T h s x

i 20,280 2976.4 6.6828 -

es 1,99.63 2423.2 - 0.8883

e 1,99.63 2675.5 7.3594 -

x=(s-sf)/(sg-sf)=(6.6828-1.3026)/(7.3594-1.3026)=5.3802/6.0568=0.8883hes = 417.46+0.8883(2258)=2423.22 = 2976.4-2423.2=553.18 kJ/kg; =2976.4-2675.5=300.9 kJ/kg

= 150.45 kW 276.59 kW

CVS iW / m CVa iW / m

CVaW CVsW

State 2a: Saturated

State 1: 20 bar, 280 C

On the T-s diagram drawn to scale State 1 and State 2 are close to eachother as illustrated below.

State 2s: Mixture

Entropy Change using Tables: Example 4Entropy Change using Tables: Example 4

16

Given: Consider R134 throttled from p3 =120 lbf/in2 to p4 =40 lbf/in2.

Find: Find the change in entropy of R134.

Assumptions: Change in KE, PE neglected, No heat transfer, No work doneother than flow work, Steady state, Steady flow, Mass is conserved.

3 4

4

23

24 4

4

40 91

40 91

40 91 20 57 85 31 0 2384

120 0 0839 10

40 0 0452 0 2197

1

f4 4 g4 f4 4 f4 fg4

of

o of f g

h h . Btu / lbmTable A10E

40.91=h +x (h -h ); x ( . h ) h

x ( . . ) / . .

s s ( lbf / in ) . Btu / lbm R(Table A E )

s s ( lbf / in ) . Btu / lbm R ;s . Btu / lbm R

s (

0 2384 0 0452 0 2384 0 2197 0 03442 0 05238

0 0868 o

. ) . . ( . ) . .

. Btu / lbm R

Adiabatic throttle with a pressure loss and phase change lead to increase inEntropy while keeping Enthalpy constant.

17

State 4

State 3

T-s Diagram and Demonstration of Throttle Action; h-s diagram

State 3

State 4

Entropy Change for Solids: Example

2 ft3 of sand is heated from 32oF to 70oF find heat added and change in entropy.Assume: Sand is incompressible and has constant specific heat given in Table A-19E ρ=94.9 lbm/ft3, cp = c=

0.191Btu/lbm-oFSolution:

70

32

2

1

2 94 9 0 191 2 94 9 0 191 70 32 1377 57

70 459 672 94 9 0 191

32 459 67

529 672 94 9 0 191 2 6971

491 67

Int .ReInt .Rev.

Int .Rev.

v.Q W

TdS pdV mcdT

Q ( . )( . )dT ( . )( . )( ) . Btu

TdT ( . )dS mc S mcln ( . )( . )ln

T T ( . )

( . )( . )( . )ln

( . )

U

.

d

B

otu / R

Entropy Change for Liquids: Example

0.15 m3 of water at 350 K is heated to 400 K find heat added and change in entropy considering constant specific heat at 375 K from Table A-19.Assume: Constant specific heat (and constant density) at 375 K from Table A-19.

Solution:

400

350

2

1

0 15 956 8 4 22 0 15 956 8 4 22 400 350

30282 72

400143 52 4 22

350

80 874

InInt .Rev.

Int .Rev.

t .Rev.Q W

TdS pdV mcdT

Q . ( . ) . dT . ( . ) . ( )

. kJ

TdT ( )dS mc S mcln . ( . )ln

T T ( )

. kJ / K

dU

Specific Entropy “s” (kJ/kg-K or BTU/lbm-oR) Change for Ideal Gases, Constant cp, cv

► Remember Chapter 3 (Property Relations). Divide by “m” kg

2

1

2

1

2 2 2

1 1 1v

T

v v

T c const .

T

p

v p

p v

v v

p p

T

v T vdTs c

Tds du pdv dh pdv vdp pdv

Tds c dT pdv c dT vdp

For Ideal Gases :c c R; pv RT

dT dv dT dvds c p c R

T T T v

dT dp d

Rln c ln R lnT v T v

T dpalso,ds c v c R

T T

dTs c

T

T p

2 2 2

1 1 1p

p

c const .

p T pRln c ln R ln

p T p

Specific Entropy “s” (kJ/kg-K or BTU/lbm-oR) Change for Ideal Gases, Variable cp, cv

► Remember Chapter 3 (Property Relations). Divide by “m” kg

2

1

2

1T

p

T

p

p

p

Tds c dT vdp

For Ideal Gases : pv RT

cdT dp dpd

c ps

s c v dT RT T T

dT RlnT p

p

2

1

2

1

22 22 23 23

Tpo o

T

o o

o o

cpS m s Rln ; s dT

p T

Air s inTables A ,A E and Idea l gas s A ,A E

s M s ;M ( kg / kmol ) MolecularWeight

Entropy Change for Air: Example0.15 kg of Air at 700 K, 10 atm pressure is heated to 1200 K in a piston cylinder device at constant pressure, find heat added and change in entropy considering variable specific heat and properties from Table A-22.Solution:

1200

2 1

700

22 1

1

0 15 933 33 512 33

1000 0 287 1200 900 1000

63 15 12 91 76 06

0 15 3 17888 2 57277 0

0 0

o

Int .RevInt .Rev.

T

Int .Rev.

T

Int .Rev

.

.

o

Q W

Q m du p( v v )

Q . (( . . )

( . ( ) / )

. . . kJ

pS m ( s s ) R ln

p

. (( )

.

dU

. . )

909 Int .Rev.kJ / K;Sameorder as( Q / T )

T, K u, kJ/kg So

kJ/kg-K

700 512.33 2.57277

1200 933.33 3.17888

Entropy Change for Ideal Gases: Example

2 lbm of N2 is heated at constant volume from 32oF to 70oF find

heat added and change in entropy.Assume: Ideal Gas and variable specific heat with properties given in Table A23E

Solution:

2

529 67

529 67 491 67

491 67

2 22 1 2 1

1 1

70 529 67 32 491 67

23 2 2636 6 2437 89 28 14 193

o o o o

o o o oInt .Rev.

T .

. .

T .

Int .Rev.

Int .Rev.Q W T F . R;T F . R

mTdS pdV m dU u u

M

Interp.in A E : Q ( . . ) / . Btu

p Tm mS ( s s ) R ln ( s s ) R ln

M p M

dU

T

I

,

2 529 6723 45 699 45 128 1 986

28 491 67

0 0302 0

0 02783

o

oInt .Rev.

.nterp.in A E : ( . . ) . ln

.

S . Btu / R as it should be.

ComparabletoQ / T . Btu / R as it should be.

Entropy Generation using Tables: Example 1Entropy Generation using Tables: Example 1

24

Given: Steam at 120oC, 0.7 bar is pressurized through a diffuser to 1 bar, 160oC and negligible velocity. Find: Find the change in entropy of steam in kJ/kg-K and comment on whether the diffuser can be adiabatic and the resulting impact.Assumptions: Change in PE neglected, No heat transfer, No work doneother than flow work, Steady state, Steady flow, Mass is conserved.

2 2

2

2 2

2000 2796 2 2719 6 390 892

4 120 0 7 7 6395

160 1

CVi e CVj

CVCV CV i i e e

H ,L B,S I E

CV ii e e i i

I E

oI

oE

dSm( s s ) Q T

dt

dE V VQ W m ( h gZ ) m ( h gZ )

dt

dm Vm m ;h h ;V ( . . ) . m / s

dt

Table A : s s( C, . bar ) . kJ / kg K

s s( C, b

7 6597

7 6597 7 6395 0 0202CV e i

ar ) . kJ / kg K

/ m ( s s ) . . . kJ / kg K

25

State 1: 0.7bar, 100 C

State 2: 1 bar, 160 C

T-s Diagram and Diffuser Action

State 2: 1 bar, h2>h1, s2>S1

State 1: 0., 7bar, 100 C

On the T-s diagram drawn to scale State 1 and State 2

2

1

Entropy Generation Calculation: Example 2Entropy Generation Calculation: Example 2

27

Given: 0.5 kg/s of steam at 280oC, 20 bar is expanded in a turbine to 1 bar in a constant entropy process. If the process was not a constant entropy process and resulted in saturated steam at 1 bar, find the decrease in work and increase in entropy in kW/K. Find: Find the work produced by the steam in kW and show the processes on a T-s diagram. If the process was not a constant entropy process and resulted in saturated steam at 1 bar, find the decrease in work and increase in entropy in kW/K.Assumptions: Change in PE neglected, No heat transfer, work done on Turbine shaft, Steady state, Steady flow, Mass is conserved.

2

2

2

2

CVCV CV i i

H ,L B,S I

CVe e i e

E I E

CVS i i es CVA i i e

CVCV i i e e CV

H ,L I I

dE VQ W m ( h gZ )

dt

dmVm ( h gZ ) ; m m

dt

W / m h h ,W / m h h

dSQ / T m s m s

dt

St P,T h s x

i 20,280 2976.4 6.6828 -

es 1,99.63 2423.2 6.6828 0.8883

e 1,99.63 2675.5 7.3594 1.0000

x=(s-sf)/(sg-sf)=(6.6828-1.3026)/(7.3594-1.3026)=5.3802/6.0568=0.8883; hes= 417.46 +0.8883 (2258) = 2423.22

CVS iW / m 2976.4 2423.2 553.18 kJ / kg 00CVa iW / m 2976.4 2675.5 3 .9 kJ / kg

150 150 276 59 54 23CVa TW kW ; / . . % ( low ) 276 59CVsW . kW ;

0 5 7 3594 6 6828 0 3383CV e e i iI I

m s m s . ( . . ) . kW / K

State 2a: Saturated

State 1: 20 bar, 280 C

On the T-s diagram drawn to scale State 1 and State 2 are close to eachother as illustrated below.

State 2s: Mixture

Entropy Generation: Example 3Entropy Generation: Example 3

29

Given: Consider R134 throttled from p3 =120 lbf/in2 to p4 =40 lbf/in2.Find: Find the change in entropy of R134.Assumptions: Change in KE, PE neglected, No heat transfer, No work doneother than flow work, Steady state, Steady flow, Mass is conserved.

3 4

4

23

24 4

4

40 91

40 91

40 91 20 57 85 31 0 2384

120 0 0839 10

40 0 0452 0 2197

1

f4 4 g4 f4 4 f4 fg4

of

o of f g

h h . Btu / lbmTable A10E

40.91=h +x (h -h ); x ( . h ) h

x ( . . ) / . .

s s ( lbf / in ) . Btu / lbm R(Table A E )

s s ( lbf / in ) . Btu / lbm R ;s . Btu / lbm R

s (

0 2384 0 0452 0 2384 0 2197 0 03442 0 05238

0 08

0 0868 0 0839 0

68

0029

o

oCV

. ) . . ( . ) . .

. Btu

/ m . . . Bt

/ lb

u / lbm

m

R

R

Adiabatic throttle with a pressure loss and phase change lead to increase inEntropy while keeping Enthalpy constant. Entropy is generated by fluid frictionOr viscosity in this case, in spite of the process being (externally) adiabatic.

30

State 4

State 3

T-s Diagram and Demonstration of Throttle Action; h-s diagram

State 3

State 4

Entropy Generation: Example 4Entropy Generation: Example 4

31

Given: Consider R134 condensed from saturated vapor (state 2) to saturated liquid at p3 =120 lbf/in2.Find: Find the change in entropy generation rate in the process of condensing R134.Assumptions: Change in KE, PE neglected, Heat transfer to a sink at 90.54oF-δ and heat transfer to sink at 80.54oF. No work done other than flow work, Steady state, Steady flow, Mass is conserved.

3

23 2

2 3

3 2

10 40 91

120 0 0839 0 2165

113 82 40 91 72 91

72 91 72 83 459 67 0 0839 0 2165

0 1369 0 1326 0

f g

of

CV

CV

CV

Table A E : h h . ; h Btu / lbm

s s ( lbf / in ) . ;s . Btu / lbm R

q h h . . . Btu / lbm

q / T ( s s )

. / ( . . ) ( . . )

. .

/ m

.I

2h = =113.82

3 2

62 8

72 83

72 91 62 83 459 67 0 1326

0 148

3

0

o

CV

o

CV

o

f receiving heat sin k is at . F

q / T ( s s ) . / ( . . ) .

. Btu / lbm Entropy generationbecauseof heat transfer

through finite delta T if receiving heat sin k is

R.

/ m

at . F.

32

State 2State 3

T-s Diagram and Demonstration of Condenser Action; h-s diagram

State 2

State 3

Sink Temperature

33

Conservation LawsConservation Laws

Property Relations and Efficiency DefinitionsProperty Relations and Efficiency Definitions

Cycle Efficiency DefinitionsCycle Efficiency Definitions