Dr Ali JawarnehDepartment of Mechanical EngineeringDepartment of Mechanical Engineering

Hashemite University

1

• Examine the moving boundary work or P dV work g ycommonly encountered in reciprocating devices such as automotive engines and compressors.

• Identify the first law of thermodynamics for closed systems.• Develop the energy balance applied to closed systems.• Define the specific heat at constant volume and the specific

heat at constant pressure.• Relate the specific heats to the calc lation of the changes in• Relate the specific heats to the calculation of the changes in

internal energy and enthalpy of ideal gases.• Describe incompressible substances and determine the Describe incompressible substances and determine the

changes in their internal energy and enthalpy.• Solve energy balance problems for closed systems that

2

gy p yinvolve heat and work interactions for general pure substances, ideal gases, and incompressible substances.

3

4–1 MOVING BOUNDARY WORKWORK

• Boundary work (Wb ): work associated with a moving boundary

• Wb is one form of mechanical work frequently encountered in practice is associated with theencountered in practice is associated with the expansion or compression of a fluid in a piston–cylinder device.p y

• Quasi-equilibrium process: a process during which the system remains nearly in equilibrium at all times. It is also called a quasi-static process, is closely approximatedby real engines especially when the piston

4

by real engines, especially when the piston moves at low velocities

VdPdsPAdsFWδ b === (4-1)

∫=2

1

(kJ)dPWb V (4-2)

∫ ∫==2

1

2

1

VdPdAA (4-3)

Area under the process curve on a P V diagram is equal in magnitude

1 1

a P-V diagram is equal, in magnitude, to the work done during a quasi-equilibrium expansion orcompression process of a closed systemcompression process of a closed system.

5

• Work is a path function (i.e., it depends on the path followed as welldepends on the path followed as well as the end states).

• The cycle shown in Fig 4–5The cycle shown in Fig. 4 5produces a net work output (Wnet) because the work done by the system y yduring the expansion process (area under path A) is greater than the

k d h d i hwork done on the system during the compression part of the cycle (area under path B) and the differenceunder path B), and the difference between these two is the net work done during the cycle (the colored g y (area).

6

Wb for a Constant-Volume Process2

002

1

.dPWb == ∫ V

W for a Constant-Pressure Process (Isobaric Process)Wb for a Constant-Pressure Process (Isobaric Process)

)vm(vP)P(dPWb 1212

2

−=−== ∫ VVV1

7

Isothermal Compression or expansion of an Ideal GasCV

V CPorCmRTP o ===

22 VVC

1

211

1

2

11 VVV

VVV

VV lnPlnCdCdPWb ==== ∫∫

RTPP VV

12

2211

PP

mRTPP o

=

==

VV

VV

21 PV

8

Polytropic Process: thermodynamics process that obeys:• During actual expansion and compression processes of• During actual expansion and compression processes of

gases, pressure and volume are often related by PV n = C, where n and C are constants. -nCP V= (4-8) nn PPC 2211 VV ==( )

nPP

nCdVCdPW

-n-n-n

b −−

=+−−

===++

∫∫ 111122

11

12

2

1

2

1

VVVVVV (4-9)For Gas and

• For an ideal gas (PV =mRT),Eq. (4-9) can also be written as

)TmR(TPP VV

Liquid

111

121122 ≠−−

=−−

= nn

)TmR(TnPPWb

VV (4-10)

• For the special case of n=1 the boundary work becomes

22

12 VVVVV lnPdCdPW -

b === ∫∫For an ideal gas this result is equivalent to

9111 V

VVVV lnPdCdPWb ∫∫ result is equivalent to the isothermal process

Example: A piston–cylinder device contains 50 kg of water at 250 kPa andcontains 50 kg of water at 250 kPa and 25°C. The cross-sectional area of the piston is 0.1 m2. Heat is now transferred to the water, causing part of it to evaporate and expand. When the volume reaches 0 2 m3 the piston reaches a linearreaches 0.2 m , the piston reaches a linear spring whose spring constant is 100 kN/m. More heat is transferred to the water until the piston rises 20 cm more. Determine (a) the final pressure and temperature and (b) the work done duringtemperature and (b) the work done during this process. Also, show the process on a P-V diagram.

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Solution:))((xkF 20100(a) kPa

.).)((

AxkP

AFPP s 450

1020100250223 =+=+=+=

CT o25= /kgmvv 30010030≅kPaPCT

25025

1

1

== /kgm.vv Cf@251 00100300 =≅

3 3311 050001003050 m.).)((mv ===V 3

2 20 m.=V

32323 220102020 m.).)(.(.Ax p =+=+=VV

/kgm..m

v 333 00440

50220

===V

11

At 450 kPa, vf = 0.001088 m3/kg and vg = 0.41392 m3/kg. Noting that vf < v3 < vg , the final state is a saturated mixture and thus the final temperature is

C.TT okPasat@ 91474503 ==

(b) The pressure remains constant during process 1 2 and(b) The pressure remains constant during process 1-2 and changes linearly (a straight line) during process 2-3. Then the boundary work during this process is simply the total area y g p p yunder the process curve,

)(PP)(PAreaW 32 ++== VVVV

kJ.)..()..(

)()(PAreaWb

544202202

45025005020250

2 23121

=−+

+−=

−+−== VVVV

12

2

Example: A piston–cylinder device contains 0.15 kg of air initially at 2 MPa and 350°C The air is firstair initially at 2 MPa and 350 C. The air is first expanded isothermally to 500 kPa, then compressed polytropically with a polytropic exponent of 1.2 to p y p y p y p pthe initial pressure, and finally compressed at the constant pressure to the initial state. Determine the boundary work for each process and the net work of the cycle.

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Solution:k /k ( bl )Rair = 0.287 kJ/kg.K (Table A-2).

For the isothermal expansion process: 2733502870150 ))()((mRT + 3

11 013410

20002733502870150 m.))(.)(.(

PmRT

=+

==V

32733502870150 ))(.)(.(mRT + 3

22 053640

5002733502870150 m.))(.)(.(

PmRT

=+

==V

kJ).(ln))(()(lnPW 183705364001341020002 ===VV kJ.)

.(ln).)(()(lnPWb, 1837

0134100134102000

11121 ===− V

V

14

For the polytropic compression process: 32121 016902000053640500 )())((PP nn VVVV 3

321

321

3322 016902000053640500 m.)().)((PP ..nn =→=→= VVVV

kJ.).)(().)((PPWb, 8634211

0536405000169020001

223332 −=

−=

−=−

VV

For the constant pressure compression process: .nb, 211132 −−

kJ)()(PW 976016900134102000 −=−=−= VVThe net work for the cycle is the sum of the works for each

process

kJ.)..()(PWb, 97601690013410200031313 ===− VV

pkJ.).().(.WWWW b,b,b,net 65497686341837133221 −=−+−+=++= −−−

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4–2 ENERGY BALANCE FOR CLOSED SYSTEMSSYSTEMS

(4-11)(kJ)ΔEEE systemoutin =−

Net energy transfer Change in internal kineticNet energy transferby heat, work, and mass

Change in internal, kinetic, potential etc.. energies

ΔPEΔKEΔUΔE)W(W)Q(QEE systemoutinoutinoutin ++==−+−=−

systemoutnet,innet, ΔEWQ =−

outininnet, QQQ −=net heat input:Work: moving boundary, electr.

or, in the rate form, as

outininnet, QQQ

inoutoutnet, WWW −=net work output: , shaft, etc…)

or, in the rate form, as(kW)/dtdEEE systemoutin =−

••

R t f t t f Rate of change in internal

(4-12)

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Rate of net energy transferby heat, work, and mass

Rate of change in internal, kinetic, potential etc..

energies

• For constant rates, the total quantities during a time interval Δt are related to the quantities per unit time asinterval Δt are related to the quantities per unit time as

• The energy balance (Eq 4 1) can be expressed on a per(kJ)Δt(dE/dt)ΔEandΔt,WWΔt,QQ ===

••(4-13)

• The energy balance (Eq.4-1) can be expressed on a per unit mass basis as

(kJ/kg)Δeee = (4 14)• For a closed system undergoing a cycle, the initial and

final states are identical:

(kJ/kg)Δeee systemoutin =− (4-14)

final states are identical:

ii

system

EEorEE

EEΔE

==−

=−=

0

012

net,innet,outnet,innet,out

outinoutin

QWorQW

EEorEE••

==

0

(4-16)

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Taking heat transfer to the system and work done by the system to be positive quantities the energy balance for asystem to be positive quantities, the energy balance for a closed system can also be expressed as

PEΔKEΔUΔEΔWQ ++==− UWQ

Q: net heat input W: net work output

Work (W) is the sum of boundary and other forms of work (such as electrical

d h f )and shaft, etc…).bother WWW +=

For a constant pressure process:For a constant-pressure process:

ΔHΔUWb =+

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Example: A piston–cylinder device contains steam initially at 1 MPa, 450°C, and 2.5 m3. Steam is allowed to cool at constant pressure until it first starts

d i Sh h T di i hcondensing. Show the process on a T-v diagram with respect to saturation lines and determine (a) the mass of the steam (b) the final temperature and (c) theof the steam, (b) the final temperature, and (c) the amount of heat transfer.

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Solution: (kJ)ΔEEE systemoutin =−

)PEKEcesin()um(uΔUWQ outb,out 012 ==−==−−

)h(hQ )hm(hQout 12 −=−

since ΔU + Wb = ΔH during a constant pressure i ilib iquasi-equilibrium process.

The properties of water are (Tables A-4 through A-6)

CT

MPaPo450

1

1

1

=

=

kJ/kg.h/kgm.v

33371330450

1

31

==

kg..v

m 5657330450

521 ===V

20

.v 3304501

(b)MPP 1 CTT 9179 0==

vaporSat.MPaP 12 =

kJ/kg.hh

C.TT

Mpag@

Mpasat@

12777

9179

12

12

==

==

(c) Substituting the energy balance gives(c) Substituting, the energy balance gives

kJ)..(.)hm(hQout 44951277713371565721 =−=−=

21

: A piston–cylinder device initially contains 0 8 m3 of saturated water vapor at 250 kPacontains 0.8 m3 of saturated water vapor at 250 kPa. At this state, the piston is resting on a set of stops, and the mass of the piston is such that a pressure of p p300 kPa is required to move it. Heat is now slowly transferred to the steam until the volume doubles. Show the process on a P-v diagram with respect to saturation lines and determine (a) the final temperat re (b) the ork done d ring this processtemperature, (b) the work done during this process, and (c) the total heat transfer.

22

Solution: (kJ)ΔEEE systemoutin =−

system)stationary,PEKEcesin()um(uΔUWQ outb,in 013 ==−==−

outb,in W)um(uQ +−= 13

The properties of steam are (Tables A-4 through A-6) kPaP 2501 = /kgm.vv kPag@ 718730 3

2501 =≅

vaporSat. kJ/kg.uu kPag@ 825362501 =≅

kg..m 1131801 ===V kg.

.vm 1131

7187301

/kgm..v 333 43751

113161

===V g

.m3 1131

3003 kPaP = CT 662 03 =

23

437513 .v = kJ/kg.u 434113 =

(b) The work done during process 1-2 is zero (since V = const) and the work done during the constant pressure process 2-3and the work done during the constant pressure process 2 3is:

kJ)()P(dVPW 24080613003

=−=−== ∫ VV

(c)

kJ)..()P(dVPW outb, 2408061300232

==== ∫ VV

( )

kJ)..(.

W)um(uQ outb,in

12132408253643411113113

=+−=

+−=

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4–3 SPECIFIC HEATSh ill bl• Specific heat: a property that will enable us to compare

the energy storage capabilities of various substances.• The specific heat is defined as the energy required to• The specific heat is defined as the energy required to

raise the temperature of a unit mass of a substance by one degree in a specified wayg p f y

• Two kinds of specific heats:a- Specific heat at constant volume (c ):a Specific heat at constant volume (cv):can be viewed as the energy required to raise the temperature of the unit mass ofraise the temperature of the unit mass of a substance by one degree as the volume is maintainedconstant. )u(c ∂ (4 19)

25

V)T(cv ∂

= (4-19)

b- Specific heat at constant volume (cP): can be viewed as the energy required to raise the temperature of the unit mass ofenergy required to raise the temperature of the unit mass of a substance by one degree as the pressure is maintained constant. h∂ (4 20)

• cP > cv

PP )Th(c

∂∂

= (4-20)

P vbecause at constant pressure the system is ll d d dallowed to expand and

the energy for this expansion workpmust also be supplied to the system.

26cP , cv units: kJ/kg · °C or kJ/kg · K.

ΔT(°C) = Δ T(K)

cP cv units: kJ/kg · °C or kJ/kg · K.P , v g g

ΔT(°C) = Δ T(K)

The specific heats are sometimes given on a molar basis. They are then denoted by and and have the unit kJ/kmol · °C or kJ/kmol · K

Pc vckJ/kmol C or kJ/kmol K.The unit kJ/kmol is very convenient in the thermodynamic analysis of chemical reactions.

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4–4 INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF IDEAL GASESSPECIFIC HEATS OF IDEAL GASES

• State equation for ideal gas:• The internal energ for ideal gas is a f nction of the

Pv = RT• The internal energy for ideal gas is a function of the

temperature only:• The enthalpy of an ideal gas is also a function of

(4-21)u = u (T)The enthalpy of an ideal gas is also a function of temperature only:

h = u + Pv h

h = h (T) (4-22)

• Since u and h depend only on

h u + PvPv = RT

h =u + RT

temperature for an ideal gas, the specific heats cP and cv cp also v

depend, at most, on temperature only.28

• The differential changes in the internal energy and enthalpy of an ideal gas can be expressed as:enthalpy of an ideal gas can be expressed as:

• The change in internal energy or enthalpy for an ideal gasdT)T(cdu v= dT)T(cdh P= (4-24)(4-23)

The change in internal energy or enthalpy for an ideal gas during a process from state 1 to state 2 is determined by

∫=−=2

12 )kg/kJ(dT)T(cuuu vΔ (4-25) ∫=−=2

)kmol/kJ(dT)T(cuuuΔ∫1

12 )g()(v (4 25) ∫==1

12 )kmol/kJ(dT)T(cuuu vΔ

Muu ΔΔ =

Mhh Δ

Δ = M: molar mass

∫=−=2

112 )kg/kJ(dT)T(chhh PΔ (4-26) ∫=−=

2

112 )kmol/kJ(dT)T(chhh PΔ

• At low pressures, all real gases approach ideal-gas behavior, and therefore their specific heats depend on t t ltemperature only

29

• The specific heats of real gases at low pressures are called ideal-gas specific heats or zero-pressure specific heatsideal-gas specific heats, or zero pressure specific heats, and are often denoted cp0 and cv0

• The use of ideal-gas specific heat data is limited to low g ppressures, but these data can also be used at moderately high pressures with reasonable accuracy as long as the gas does

t d i t f id l b h i i ifi tlnot deviate from ideal-gas behavior significantly.• By performing the integrations in Eqs. 4–25 and 4–26

then u and h for a number of ideal gases have beenthen u and h for a number of ideal gases have been tabulated in kJ/kg for air (Table A–17) and in kJ/kmolfor other gases Table A-18 to A-25g

• The variation of specific heats with temperature is smooth and may be approximated as linear over small y pptemperature intervals

30

• Therefore the specific heat functions in Eqs. 4–25 and 4–26 can be replaced by the constant average specific heat26 can be replaced by the constant average specific heat values.

kg/kJ)TT(cuu 1212 −=− (4-27)kg/kJ)TT(cuu avg,v 1212

kg/kJ)TT(chh avg,p 1212 −=−

(4-27)

(4-28)• The specific heat values for some common gases are

listed as a function of temperature in Table A–2b.

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• The average specific heats and evaluated from Table A-2 at the average temperature (T1 + T2)/2 If the final

avg,pc avg,vcTable A 2 at the average temperature (T1 + T2)/2. If the final temperature T2 is not known, the specific heats may be evaluated at T1 or at the anticipated average temperature. Then T2 can be determined by using these specific heat values. The value of T2 can be refined, if necessary, by evaluating the specific heats at the newevaluating the specific heats at the new

average temperature. Another way of determining the average specific heatsdetermining the average specific heats is to evaluate them at T1 and T2 and thentake their average. Usually both methodstake their average. Usually both methodsgive reasonably good results, and one isnot necessarily better than the other.y

32

• The presence of the constant-volume specific heat cv in an equation should not lead one to believe that thisan equation should not lead one to believe that this equation is valid for a constant-volume process only.

• To summarize, there are three waysTo summarize, there are three waysto determine the internal energy andenthalpy changes of ideal gasespy g g

12 uuu −=Δ Table A-17, A-18 to A-25 (1)

∫=2

dT)T(cuΔ Table A 2c(2)

For Air In mole for different gases

∫=1

dT)T(cu vΔ

Tcu ΔΔ ≅ Table A-2a(at 300K or near 300 K) A-2b

Table A-2c(2)

(3)33

Tcu avg,v ΔΔ ≅ Table A 2a(at 300K or near 300 K), A 2b(3)

Specific Heat Relations of Ideal Gasesi l l i hi b d f id l• A special relationship between cp and cv for ideal gases can

be obtained by differentiating the relation: h = u + RTdh = du + R dTdh = du + R dT

replacing dh by cp dT and du by cv dT and dividing the resulting expression by dT we obtain:resulting expression by dT, we obtain:

• When the specific heats are given on a molar basis, R in the)K.kg/kJ(Rcc vP += (4-29)

When the specific heats are given on a molar basis, R in the above equation should be replaced by the universal gas constant Ru

• Specific heat ratio k, defined as:

)K.kmol/kJ(Rcc uvP += (4-30)

34v

P

cck = (4-31)

• Example: A piston–cylinder device contains 0.8 kg f it i iti ll t 100 kP d 27°C Th itof nitrogen initially at 100 kPa and 27°C. The nitrogen

is now compressed slowly in a polytropic process during which PV 1.3 constant until the volume isduring which PV constant until the volume is reduced by one-half. Determine the work done and the heat transfer for this process.

35

• Solution: k 3/k ( bl )R = 0.2968 kPa.m3/kg.K (Table A-1).

cv=0.744 kJ/kg.K (Table A-2b) @ Tavg = (369+300)/2 =335 K(T bl A 2 ) 0 743 kJ/k Kor you can use (Table A-2a) to get cv=0.743 kJ/kg.K

(kJ)ΔEEE systemoutin =−

)T(T)(ΔUQW )T(Tcm)um(uΔUQW voutinb, 1212 −=−==−

The final pressure and temperature of nitrogen are:

kPa.)(P)(PPP //.. 22641002 311

31

2

12

3111

3122 ===→=

VVVV

K...TPPT

TP

TP 336930050

1002246

11

2

1

22

2

22

1

11 =××==→=VVVV

36

The boundary work for this polytropic process:

n)TmR(T

nPPdPW inb, 11

122

1

1122

−−

−=−−

−=−= ∫VVV

kJ..

).)(.)(.( 854311

30033692968080=

−−

−=

)T(TcmWQ vinb,out 12 −−=

kJ.).)(.)(.(.

6133003369744080854

=−−=

37

A i t li d d i ith t f t th t• A piston–cylinder device, with a set of stops on the top, initially contains 3 kg of air at 200 kPa and 27°C. Heat is now transferred to the air and the piston rises until itis now transferred to the air, and the piston rises until it hits the stops, at which point the volume is twice the initial volume. More heat is transferred until the pressure inside the cylinder also doubles. Determine the work done and the amount of heat transfer for this

l h h dprocess. Also, show the process on a P-v diagram.

38

• Solution: R = 0.287 kPa.m3/kg.K (Table A-1).

(kJ)ΔEEE systemoutin =−

outb,in )um(uΔUWQ −==− 13

outb,in

outb,in

W)um(uQ +−= 13

13

The initial and the final volumes and the final temperature of air are: mRT 30028703 ××

m..

m..P

mRT

58229122

291200

30028703

313

3

1

11

=×==

=××

==

VV

V

KTPPT

TP

TP

m..

12003002200400

58229122

11

3

1

33

3

33

1

11

13

=××==→=

×

VVVV

VV

39

No work is done during process 2-3 since V2 = V3. The pressure remains constant during process 1 2 and theThe pressure remains constant during process 1-2 and the work done during this process is

kJ)()(PdPW 2582915822002

∫ VVV kJ)..()(PdPWb 258291582200231

2 =−×=−== ∫ VVV

The initial and final internal energies of air are (Table A-17) g ( )

kJ/kg.uu

kJ/kg.uu

K@

K@

33933

07214

12003

3001

==

==

gK@12003

kJ)..(Qin 241625807214339333 =+−=

Alternative solution: cv = 0.800 kJ/kg.K @Tavg=(300+1200)/2=750 K v g @ avg ( )(Table A-2b)

W)T(TcmW)um(uQ outb,voutb,in 1313 +−≅+−=

40kJ)(. 2418258300120080003 =+−××=

4–5 INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF SOLIDS AND LIQUIDSSPECIFIC HEATS OF SOLIDS AND LIQUIDS

• Liquids and solids can be approximated as incompressible substances (substance whoseincompressible substances (substance whose specific volume (or density) is constant)

• The constant-volume and constant-pressurespecific heats are identical for incompressible substancesspecific heats are identical for incompressible substances

ccc vp == (4-32)

41

Internal Energy Changesifi h f i ibl b d d• Specific heats of incompressible substances depend on

temperature onlydTc(T)dTcdu == (4 33)

• The change in internal energy between states 1 and 2 is then obtained by integration:

dTc(T)dTcdu v == (4-33)

then obtained by integration:

∫=−=2

112 (kJ/kg)dTc(T)uuΔu (4-34)

• For small temperature intervals, a c value at the average temperature can be used and treated as a constant:

1

(kJ/kg))T(TcΔu avg 12 −≅ (4-35)

42

Enthalpy Changes• Using the definition of enthalpy and noting that

v constant.Diff i h

h = u + Pv

0• Differentiate h:• Integrating:

vdPduPdvdPvdudh +=++= (4-36)PvΔTcdPvΔuΔh avg Δ+≅+= (4-37)

For solids, the term vΔP is insignificant and thus:

F li id i l l dΔTcΔuΔh avg≅=

For liquids, two special cases are commonly encountered:1. Constant-pressure processes, as in heaters(ΔP=0):

ΔTcΔuΔh ≅

2. Constant-temperature processes, as in pumps (ΔT=0):

ΔTcΔuΔh avg≅=

PΔh Δ43

PvΔh Δ=

• Comment on the last equation For a process between states 1 and 2 it can be expressed as:

PvΔh Δ=For a process between states 1 and 2 it can be expressed as:

By taking state 2 to be the compressed liquid state at a given)Pv(Phh 1212 −=−

By taking state 2 to be the compressed liquid state at a given T and P and state 1 to be the saturated liquid state at the same temperature,

The enthalpy of the compressed liquid can be expressed as:

)P(Pvhh sat@Tf@Tf@T@P T −+≅ (4-38)

As discussed in Ch. 3 the enthalpy of the compressed liquid

)( sat@Tf@Tf@T@P,T (4 38)

could be taken h@P,T hf@ T however, the contribution of the last term is often very small, and is neglected.

44

• Example: In a manufacturing facility, 5-cm-diameter brass balls (ρ 8522 kg/m3 and c 0 385 kJ/kg °C)brass balls (ρ = 8522 kg/m3 and cp = 0.385 kJ/kg · °C) initially at 120°C are quenched in a water bath at 50°C for a period of 2 min at a rate of 100 balls per minute. If the p ptemperature of the balls after quenching is 74°C, determine the rate at which heat needs to be removed from th t i d t k it t t t t t 50°Cthe water in order to keep its temperature constant at 50°C.

45

• Solution:k i l b ll h hAnalysis: We take a single ball as the system. The energy

balance for this closed system can be expressed as:

Then the rate of heat transfer from the balls to the water becomes

Therefore, heat must be removed from the water at a rate of 988 kJ/min in order to keep its temperature constant at 50°C since energy input must be equal to

t t f t h l l i t t Th t i

46

energy output for a system whose energy level remains constant. That is,