me2121 - me2121e slides chapter 3 (2014)

Mechanical Engineering,NUS

• Thermodynamic Properties of Pure Substances– A pure substance is one where its properties

remain invariant (or no change) when it undergoes state changes, from gaseous to liquid or to solid phase.

ME2121/ TM1121- Thermodynamics

Chapter 3 – 1/ 12

Water, H2O

Ice (Solid) liquid

Water, H2O

Melting (NMP)

O

H H


• State Diagrams– Water,

– Isobaric Process,(A-B-C-D)


Chapter 3 – 2/ 12

v

TCritical point

1 bar10 bar

Pcr= 221 bar

Liquid & VapourVapour region

Liquid

AB

C

Fig. 3.1 2-D State diagram

374 o C

L

L+V

V

vf = vg

g-linef-line

D

State A State B State C-D State D

Liquid phase

State C

Two-phase mixture

subcooled Saturated (T=Tsat)

Gaseousphase

L

V

Qin

100oCTA <Tsat

T>Tsat

100 bar


• 3-D state diagrams


Chapter 3 – 3/ 12

liquidice

load



Chapter 3 – 3/ 12

A thermodynamic state on the surface require two independent properties to locate it, e.g, P=P(v,T).

Constant pressure



Chapter 3 – 3/ 12

3-D and 2-D representations of a substance that contracts on freezing


• Variables in the liquid+vapour region– Quality or dryness fraction of steam

- Specific volume of atwo-phase mixture


Chapter 3 – 4/ 12

fg

g

mmm

vapourofmassliquidofmassvapourofmassx

fgv

fgf

gf

T

gg

T

ff

T

T

ggffmixture

vvxv

xvvxvorm

vmm

vmv

mV

givesmbyand

vmvmV

)(

)1(

fgfg

vtocomparedsmall

f

vv

vvv

x


• Other derived properties– enthalpy

– internal energy

- Entropy

- Quality


Chapter 3 – 5/ 12

fgfgf xhhxhhxh )1(

fgfgf xuuxuuxu )1(

fgfgf xssxssxs )1(

fg

f

fg

f

fg

f

uuu

hhh

sss

x

At a given P, T, v

Example 3.1Find the states of steam for the following given conditions:

(i) t = 34o C, v = 26.6 m3/kg, (ii) P = 0.8 bar, t = 100o C, (iii) P = 2.7 bar, x = 0.5.

Refer to the Table of saturated steam and water (pg.2), the column 1 is the temperature of mixtures in oC, and column 3 is

the specific volume at dry saturated vapour.

At v =26.6 m3/kg, note that t of steam/water is 34oC. Thus, we conclude that the state point

sits on the dry saturatedvapour line or the “g” line, shown below.

t

vV=26 m3/kg

34oCg-linef-line

Critical pt

At P =0.8 bar, t =100o C,

Look up page 3 of Table where column 1 is the saturated pressure, and at 0.8 bar, the corresponding temperature is only 93.5oC.

This temperature is found to be less than 100oC (given), thus, we conclude that the steam is super-heated (that is, temperature of state higher than the saturated temperature).

t

v=2.087m3/kg

100oC

g-linef-lineCritical pt P=0.8

bar

Tsat=93.5o

C

Superheated state

At P =2.7 bar, x=0.5,

vf = 0.00107 m3/kg (pg. 10 of Table)., vg =0.6686 m3/kg (pg.4),

Note that 2.7 bar is very much lower than Pcr =221 bar, thus, vf <<< vg., and Tsat = 130oC.

By definition, any other properties could be found;

In this region (L+V), P & T are no longer independent.

t

vNote: at low P, vf << vg,,

thus vi ~ x(vg) +(1-x)vf

g-linef-line

Critical pt P=2.7 bar

X=0.5

L+V

=0


• Interpolation technique- data in tables are sparsely tabulated. -To interpolate for v at pressure (P) when values of v1 and v2 are available from tables at P1 and P2.


Chapter 3 – 6/ 12

g-line

f-line

Critical pt P2

L+V

P1P

vv2 v1

t 12

12

11)( vv

PPPP

vPv

v (m3/kg) P1 =20 bar P=25 bar P2 =30 bar

t = 250o C 0.1115 m3/kg 0.0706 m3/kg

t =275o C =(0.1115+0.1255)/2=0.1185

=(0.1185+0.0759) / 2=0.0972

=(0.0706+0.0812)/2=0.0759

t =300o C 0.1255 0.0812


• Gaseous Phase– The equation of state (EOS),– Two forms: (i) gravimetric , (ii) molar form

or

– Compressibility factor


Chapter 3 – 7/ 12

)tan(lim

tconsgasaRTmPV

)./314.8

tan(lim

KmolJ

tconsgasuniversalRTnPV

o

gasidealforRTPvZ 0.1

Z=Pv/RT1.0

Pr = P/ Pcr

Tr = 1.0

Tr= 1.5

Tr = 10

5 10

Tr = T/Tcr


• Empirical equations for real gases Losses incurred when gas molecules can be described by Virial Equations of the form

It can represent many other forms of EOS. If written in terms of pressure,

where B2,p = B2,v /RoT, and B2,P =(B3,v –B2,v2)/(RoT)2.


Chapter 3 – 8/ 12

2,3,2

2,3,2

)()(1

,)/(

)()(1

vTB

vTB

vR

TP

vnVvolumemolaroftermsinorV

TBV

TBZ

TRPV

vvo

vv

o

2,3,21 PBPBZ

TRPV

ppo

Example

• van del Waals (1837), accounted for the intermolecularattractions or pressure ( ) and the presence or effectof gas molecules (b) on the “free volume”

Comparing with the Virial Eq.,

And expanding the van der Waals as

then we see that B2,v(T) =and B3,v (T) =

TRbvvaP o

2

P

L+VT

TC

Critical pt,

02

2

CC TT VP

VP

2

)(

2

)(,3,2

11vb

RTab

vvR

TP

TBTBvv

2b

RTab

2

,3,2 )()(1

vTB

vTB

vR

TP vvo


• Example 3.1– Derive, from first principles, the ideal gas law, PV/T = constant?– From experiments, the expansion and compressibility factors

are– and

– The volume of ideal gas in a system is V=V(P,T) and the change

–– Substitute for

Ln V - ln T+ln P = ln (constant),or

PV/T = constant. (QED)


Chapter 3– 9/ 12

TTV

V P

11

PPV

V T

11

dPPVdT

TVdV

TP

0

PdP

TdT

VdVor

PdPV

TdTV

VdPVdTdV

Example 3.3 (notes)

Calculate the van der Waals parameters (aand b) for the following gases with the experimental values at the critical point:

Write the van der Waals equation in another form as

Gases Pc (bar) Tc (K)

Benzene 49.1 562

Water 221.2 647.3

023

Pabv

Pavb

PTR

v o

And at the critical point, T=Tc, P=Pc,

023

ccc

co

Pabv

Pavb

PTRv

At the critical point, the three roots (solutions) of v, as given mathematically by a cubic equation, must converge i.e.,

03 cvv

After expanding, we have

033 3223 ccc vvvvvv

Comparing equations for the roots (solution) of v,

The root of vo:

The root of v1:

The root of v2:

Thus,

Thus, substitute for a and b, we

have

3cc

vPab

23 cc

vPa

cc

c vbP

RT 3

cc Pva 23

33 2

3c

cc

cc vPvPvb

c

coc P

TRv

83

Gases Pc (bar)or 105 Pa

Tc (K)

[Jm3 / mol2]. [m3/mol]

Benzene 49.1 562

=1.88 1.19 E-4

Water 221.2 647.3 0.552 3.04 E-5.

cPcToR

a2

6427

c

co

PTR

b8

Gas molecule effect on the free volumeEffect of intermolecular forces

5101.49

2)562(3143.86427

xxa

5101.4985623143.8

xxxb


• Steam Tables by Rogers & Mayhew


Chapter 2 – 10/ 12


• Saturated liquid properties


Chapter 3 – 11/ 12


• Superheated properties


Chapter 3 – 12/ 12

me2121 - me2121e slides chapter 3 (2014)

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