Solved Problems in Thermodynamic Property Relations

1.) Verify the validity of the Maxwell relation IV (Eq. 3.20) for steam at 250°C and 300 kPa.

Solution:

( ∂ v∂ T )

p

=−( ∂ s∂ p )

T

( ∆ v∆ T )

p=300 kPa

≅−( ∆ s∆ p )

T =250℃

[ v300℃−v200℃

(300−200 )℃ ]p=300 kPa

≅−[ s400 kPa−s300 kPa

(400−200 ) kPa ]T=250℃

from steam tables:

(0.8753−0.7163 ) m3/kg(300−200 ) ° K

≅−(7.3789−7.7086 ) kJ / kg∙°K

( 400−200 ) kPa0.00159 m3 /kg ∙° K≅ 0.00165 m3/kg ∙ ° K

2.) Using the Clausius-Clapeyron equation, estimate the value of the enthalpy of vaporization of refrigerant-134a at 20°C.

Solution:

h fg=T v fg( dpdT )

from refrigerant tables:

v fg=( vg−v f )T=20℃=(0.0358−0.0008157 )m3 /kg=0.035 m3/kg

( dpdT )≅( ∆ p

∆T )= psat,24℃−psat ,16℃

(24−16 )° K=

(645.66−504.16 ) kPa(24−16 )° K

=17.6875 kPa/° K

substituting these values to the equation for hfg:

h fg=(20+273 ) ° K (0.035 m3/kg ) (17.6875 kPa /°K )h fg=181.385 kJ /kg

The actual value of hfg is 181.09 kJ/kg.

3.) Estimate the saturation pressure of refrigerant-134a at-50°F, using the data available in the refrigerant tables.

Solution:Using the Clausius-Clapeyron equation the saturation pressure can be extrapolated in the absence of a compete table.

dpp

=( hfg

R )( dTT2 )

integrating the Clausius-Clapeyron equation:

ln ( p2

p1)

sat

≅ ( hfg

R )( 1T 1

− 1T 2

)sat

let T1 = -40°F and T2 = -50°F

ln ( p2

7.49 psia )≅ ( 95.82 Btu / lbm0.01946 Btu / lbm ∙ °R )( 1

420°R− 1

410°R )p2=5.627 psia

The actual value of the saturation pressure p2 at -50°F is 5.505 psia.

4.) Show that c p−cv=R for an ideal gas.

Solution:This relation is proved by showing that the right-hand side of (5.31) is equivalent to the gas constant R of the ideal gas:

c p−cv=−T ( ∂ v∂T )

p

2

( ∂ p∂ v )

T

p= RTv

( ∂ p∂ v )

T

=−RT

v2=−p

v

v=RTp

( ∂ v∂ T )

p

2

=( Rp )

2

subtituting:

c p−cv=−T ( ∂ v∂T )

p

2

( ∂ p∂ v )

T

=−T (Rp )

2

(−pv )=R

5.) Show that the Joule-Thompson coefficient of an ideal gas is zero.

Solution:

μJT=( ∂ T∂ p )

h

=−1c p

[v−T ( ∂ v∂T )

p]

For an ideal gas v=RT / p, and thus

( ∂ v∂ T )

p

=Rp

substituting:

μJT=−1cp

[v−T ( ∂ v∂ T )

p]=−1

c p[v−T

Rp ]=−1

cp

(v−v )=0

6.) Determine the enthalpy change and entropy change of oxygen per unit mole as it undergoes a change of state from 220°K and 5 MPa to 300°K and 10 MPa (a) by assuming ideal-gas behavior and (b) by accounting for the deviation from ideal-gas behavior.

Solution:(a) The enthalpy values at the initial and the final temperatures can be determined from

the ideal-gas table at the specified temperatures:(h2−h1 )ideal=(8736−6404 ) kJ /kmol

(h2−h1 )ideal=2332 kJ /kmol

The entropy depends on both temperature and pressure even for ideal gases. Under the ideal-gas assumption, the entropy change of oxygen is determined from

( s2−s1 )ideal=s2∘−s1

∘−R lnp2

p1

where:

s∘=∫0

T

c p (T ) dTT

( s2−s1 )ideal=(205.213−196.171 ) kJ /kmol ∙ °K−(8.314 kJ /kmol ∙ °K ) ln 10 MPa5 MPa

( s2−s1 )ideal=3.279 kJ/ kmol ∙ °K

(b) The deviation from the ideal-gas behavior can be accounted for by determining the enthalpy and entropy from the generalized charts at each state:

T R1=T1

T cr

= 220154.8

=1.42

pR1=p1

pcr

=5

5.08=0.98 }Zh1=0.53 , Zs 1=0.25

and

T R2=T2

T cr

= 300154.8

=1.94

pR2=p2

pcr

=10

5.08=1.97 }Zh 2=0.48 , Zs 2=0.20

Then,h2−h2=( h2−h1 ) ideal−R T cr (Zh 2−Zh 1 )

¿2332 kJ /kmol−(8.314 kJ /kmol ∙ °K ) [154.8 ° K (0.48−0.53 ) ]h2−h2=2396.35 kJ/ kmol

and

s2−s1=( s2−s1) ideal−R ( Z s 2−Z s 1)¿3.279 kJ /kmol ∙ ° K−( 8.314 kJ /kmol ∙ °K ) (0.20−0.25 )

s2−s1=3.695 kJ /kmol ∙ °K

7.) Using p-v-T data for saturated water, calculate the (a) latent heat of vaporization at 100°C and (b) sg – sf.

Solution:(a)

( dpdT )= hg−h f

T (v¿¿ g−v f )=hfg

T v fg

¿

h fg=hg−h f=T (v¿¿g−v f )( dpdT )¿

from steam tables:

h fg=(373 °K ) (1.678−0.0010435 ) m3/ kg[ (120.82−84.55 ) kPa(105−95 )° K ]

h fg=2261.941 kJ /kg

This agrees very closely with the value read from the steam tables, hfg = 2257 kJ/kg.

(b)

sg−sf =hg−h f

T

sg−sf =2261.941 kJ /kg

373° Ksg−sf =6.064 kJ /kg ∙ ° K

8.) Determine the ug – uf of water at 100°C.

Solution:h2−h1=u2−u1+ ( p2 v2−p1 v1 )

since vaporization occurs at constant pressure:

ug−u f=hg−hf −psat (v g−v f )

from steam tables:

ug−u f=2257 kJ /kg−101.4 kPa (1.678−0.0010435 ) m3/kgug−u f=2086.957 kJ /kg

9.) For liquid water at 1 atm and 20°C, estimate the percent error in cv that would result if it were assumed that cp = cv.

Solution:

c p−cv=vT β2

α

c p−cv=(0.001001 m3/kg ) (293° K ) (206.6 ×10−6° K−1 )2

4.59 ×10−7 kPa−1

c p−cv=0.027 kJ/ kg ∙° K

cv=c p−0.027 kJ/ kg ∙° Kcv= (4.187−0.027 ) kJ /kg ∙° K

cv=4.16 kJ / kg ∙° K

% error=( c p−cv

cv)(100 )=( 0.027 kJ /kg ∙ °K

4.16 kJ /kg ∙ ° K ) (100 )

% error=0.65 %

10.) Using hfg, vfg and psat at 10°F from the refrigerant-134a tables, estimate the saturation pressure at 20°F.

Solution:

( dpdT )= h fg

T v fg

integrating the above equation yields:

p2=p1+h fg

v fg

lnT 2

T 1

p2=26.651 psi+88.53 Btu/ lb (778 )1.7131 ft3/ lb (144 )

ln480 ° R470 ° R

p2=32.529 psi:

Reference: Y. A. Yunus and M. A. Boles. Thermodynamics: An Engineering Approach, 4th Ed. New

York: McGraw-Hill, 2002.

M. J. Moran and H. N. Shapiro. Fundamentals of Engineering Thermodynamics, 5th Ed. John Wiley & Sons, Inc., 2004.