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8/18/2019 termodinamika sol

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Problem 7 10.213

a) Estimate the residual properties using Lee/Kessler generalized correlations

T=200 °C = 473.15 KP = 1400 kPa = 1400 * bar/100 kPa = 14 bar

The critical conditions of water from Table B.1Tc=647.1 KPc=220.55 barω=0.345M=18.015

The reduced temperature and pressureTr= 473.15/647.1 = 0.73Pr= 14/220.55 = 0.0635

Use Tables in Appendix to get Z 0, Z1 by interpolation (see solution of problem 5 for how to do linearinterpolation)

Note: The values in the table are not italic which means that the uid at the conditions given is in thevapor phase. We already know from the reduced variables that we are not in the supercritical region.

Z0=0.9431Z1=-0.0542Z=0.9431+0.345*(-0.0542)=0.9244

Repeat the above procedure for the enthalpy

Do the same for the entropy

( )

( ) molcm

4.212bar14

K15.473*Kmol

barcm14.83

*19244.0

PRT

1ZP

RTP

RTZVVV

3

3

igR

−=

−=

−=−=−=

( )

( ) 2106.0RTH

1207.0RTH

c

1R

c

0R

−=

−=

1934.0)2106.0(*345.01207.0RT

H

c

R

−=−+−=

molJ

2.1040K1.647*Kmol

J314.8*1934.0H R −=−=


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b) Repeat part (a) using the steam tables

Here, we will use the denition of the residual property

There are two types of steam tables: saturated and superheated. The saturated steam tables give infor-mation about the two-phase region while the superheated tables give information about the vapor phase.Since we know from part (a) that our point lies in the vapor phase, we will use the superheated steamtables. If we looked rst at the saturated steam table, we would have found that the saturated pressure is1554.5 kPa (p. 671). Since we are interested in a lower pressure, we make out that our point lies in thevapor phase.

The values given in the steam tables are the real values, which can be used for the rst term of the resid-ual property denition. So let us get the three real values from the steam table at T=200 °C and P=1400kPa.

V=142.94 cm 3 g-1

H=2801.4 kJ kg -1

S=6.4941 kJ kg -1 K -1

(Units are given in the rst table p.668)

Now we need to get the hypothetical ideal values of the properties at the same temperature and pressure.But the tables only give real values and not ideal ones. However, we can get the ideal values at the sametemperature and very low pressure because we know that the ideal gas is a very good approximation atlow pressures. So from the steam table, we get those properties at P=1 kPa and the same temperature.

Vig(1 kPa) = 218350 cm 3 g -1

H ig(1 kPa) =2880.1 kJ kg -1

Sig(1 kPa) =9.9679 kJ kg -1 K -1

But we need the ideal properties at P=1400 kPa and not at 1 kPa. For the volume, we can use the ideal

gas low for an isothermal process.

P1V1 = P 2V2

Vig(1400 kPa) = 218350 * 1 / 1400 = 155.96 cm 3 g-1

For the enthalpy, we know that

dH ig = Cp dT (i.e. enthalpy is not a function of pressure for an ideal gas)

( )

( )

KmolJ

6098.1Kmol

J314.8*1936.0S

1936.0)2395.0(*345.01110.0R

S

2395.0R

S

1110.0R

S

R

R

1R

0R

−=−=

−=−+−=

−=

−=

( ) ( ) ( )P,TMP,TMP,TM igR −=


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Therefore,

H ig(1 kPa) = H ig(1400 kPa) =2880.1 kJ kg -1

For the entropy, we know that

For an isothermal process, we can write

Therefore

Now we can get the residual properties

VR=142.94-155.96=-13.02 cm 3 g -1

HR=2801.4-2880.1=-78.7 kJ kg -1

SR=6.4941-6.6247=-0.1306 kJ kg -1 K -1

PdP

RT

dTCdS igp

ig −=

∫

−=−=∆

2

1

P

P 1

2ig

PP

lnRP

dPRS

( ) ( ) ( )

−=∆+=

11400

lnKmol

J314.8kPa1SSkPa1SkPa1400S igigigig

( ) ( )

KkgkJ

6247.6

1400lnkgg1000

g015.18mol

J1000kJ

KmolJ

314.8KkgkJ

9679.9kPa1400Sig

=

−=

molcm

36.234mol

g18g

cm02.13

33

−=−=

molJ

6.1416kJ

J1000g1000

kgmol

g18kgkJ

7.78 −=−=

KmolJ

3508.2kJ

J1000g1000

kgmol

g18Kkg

kJ1306.0 −=−=


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c)

Ln(P) vs. H. diagram

Constant temperature lineFor the liquid phase and the ideal gas, the line is a vertical line because H is not a function of pressurefor these two cases (see lecture notes)Inside the dome, there is only one degree of freedom (2 phases). Hence for a dened T, P is set. Theconstant temperature line becomes a horizontal line.

Constant entropy lineFrom the denition of dHdH = TdS + VdPFor constant entropy, dS=0Therefore,

That means that the constant entropy lines in a ln P vs. H diagram always have positive slopes that areequal to 1/PV.

V1

HP

S

=

∂∂

PV1

HP

P1

HPln

SS

=

∂∂=

∂

∂


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d) Slope of an isotherm

We also start from dH

For constant T,

We have

Then,

e) The slope for an ideal gas

For an ideal gas we have

Substituting in the slope equation, we get

The slope the constant temperature lines of an ideal gas in a ln P vs. H diagram are always vertical lines.

dPTV

TVdTCdHP

p

∂∂−+=

PT TV

TVPH

∂∂−=

∂∂

TT HP

P1

HPln

∂∂=

∂

∂

∂∂

−

=

∂∂=

∂

∂

PT

T

TV

TVP

1

PH1

P1

HPln

PRT

V =

PR

TV

P

=

∂∂

( ) ∞==−=

−

= ∂∂ 01

VVP1

PR

TVP

1H

Pln

T

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