Chapter 6 Homework For the wet vapor the entropy is given by

(c) Tc647.1 K⋅:= Pc220.55 bar⋅:= ω 0.345:=

Tr1

T1

Tc

:= Pr1

P1

Pc

:= Tr2

T2

Tc

:= Pr2

P2

Pc

:=

Tr1 1.11752=Pr1 0.13602=Tr2 0.63846=Pr2 0.01066=

The generalized virial-coefficient correlation is suitable here

∆H∆Hig

RT

c

⋅HRB Tr2 Pr2

,ω,

()

HRB Tr1 Pr1

,ω,

()

−

()

⋅

molwt

+:=

(b) T1450 273.15+()K⋅:= T2140 273.15+()K⋅:=

T1723.15 K=T2413.15 K=

P13000 kPa⋅:= P2235 kPa⋅:=

Eqs. (6.95) & (6.96) for an ideal gas: molwt 18 gm

mol

:=

∆Hig

R ICPH T1T2

,3.470,1.450 10 3−

⋅, 0.0,0.121 105

⋅,

()

⋅

molwt

:=

∆Sig

R ICPS T1T2

,3.470,1.450 10 3−

⋅, 0.0,0.121 105

⋅,

()

ln P2

P1

⎛

⎜

⎝

⎞

⎠

−

⎛

⎜

⎝

⎞

⎠

⋅

molwt

:=

164

6.42 Table F.4, sat.vapor, 300(psi):

T1417.35 459.67+( ) rankine⋅:= H11202.9 BTU

lbm

⋅:=

Superheated steam at 300(psi) & 900 degF

H21473.6 BTU

lbm

⋅:= S21.7591 BTU

lbmrankine⋅

⋅:= S3S2

:=

6.41 Data, Table F.2 superheated steam at 550 kPa and 200 degC:

Step 1–2: Const.-V heating to 800 kPa. At the initial specific volume

and this P, interpolation gives t = 401.74 degC, and

U22963.1 J

gm

⋅:= S27.5782 J

gm K⋅

⋅:= Q12 U2U1

−:=

Q12 322.5 J

gm

=

Step 2–3: Isentropic expansion to initial T.

Q23 0=S3S2

=S37.5782 J

gm K⋅

⋅:=

Step 3–1: Constant-T compression to initial P.

T 473.15 K⋅:= Q31 TS

1S3

−

()

⋅:= Q31 268.465−J

gm

=

For the cycle, the internal energy change = 0.

165

x 0.95:= S2Sliq xS

vap Sliq

−

()

⋅+=

So we must find the presure for which this equation is satisfied. This

occurs at a pressure just above 250 kPa. At 250 kPa:

Sliq 1.6071 J

gm K⋅

⋅:= Svap 7.0520 J

gm K⋅

⋅:=

6.44 (a) Table F.2 at the final conditions of saturated vapor at 50 kPa:

S27.5947 kJ

kg K⋅

⋅:= H22646.0 kJ

kg

⋅:= S1S2

:=

For the cycle, the internal energy change = 0.

6.43 Data, Table F.2, superheated steam at 4000 kPa and 400 degC:

S16.7733 J

gm K⋅

⋅:= For both parts of the problem: S2S1

:=

(a) So we are looking for the pressure at which saturated vapor has the given

(b) For the wet vapor the entropy is given by

Sliq 0.6493 kJ

kg K⋅

⋅:= Svap 8.1511 kJ

kg K⋅

⋅:=

Hliq 191.832 kJ

kg

⋅:= Hvap 2584.8 kJ

kg

⋅:=

6.46 Table F.2 for superheated vapor at the initial conditions, 1300 kPa and 400

degC, and for the final condition of 40 kPa and 100 degC:

H13259.7 kJ

kg

⋅:= S17.3404 kJ

kg K⋅

⋅:= H22683.8 kJ

kg

⋅:=

6.45 Table F.2 for superheated vapor at the initial conditions, 1350 kPa and 375

degC, and for the final condition of sat. vapor at 10 kPa:

H13205.4 kJ

kg

⋅:= S17.2410 kJ

kg K⋅

⋅:= H22584.8 kJ

kg

⋅:=

If the turbine were to operate isentropically, the final entropy would be

S2S1

:=

Table F.2 for sat. liquid and vapor at 10 kPa:

VRVR

molwt

T

P

⋅−:=

The enthalpy of an ideal gas is independent of pressure, but the entropy

DOES depend on P:

The generalized virial-coefficient correlation is suitable here

Sliq 1.0261 kJ

kg K⋅

⋅:= Svap 7.6709 kJ

kg K⋅

⋅:=

Hliq 317.16 kJ

kg

⋅:= Hvap 2636.9 kJ

kg

⋅:=

6.47 Table F.2 at 1600 kPa and 225 degC: P 1600 kPa⋅:=

V 132.85 cm3

gm

⋅:= H 2856.3 J

gm

⋅:= S 6.5503 J

gm K⋅

⋅:=

Table F.2 (ideal-gas values, 1 kPa and 225 degC)

Hig 2928.7 J

gm

⋅:= Sig 10.0681 J

gm K⋅

⋅:= P01 kPa⋅:=

T 225 273.15+()K⋅:= T 498.15 K=

168

∆Hlv HvHl

−:=

Sl2.1382 J

gm K⋅

⋅:= Sv6.5828 J

gm K⋅

⋅:= ∆Slv SvSl

−:=

∆Vlv 193.163 cm3

gm

= ∆Hlv 2.014 103

×J

gm

= ∆Slv 4.445 J

gm K⋅

=

By Eqs. (3.61) + (3.62) & (3.63) along with Eq. (6.40)

6.48 P 1000 kPa⋅:= T 179.88 273.15+()K⋅:= T 453.03 K=

(Table F.2) molwt 18.015 gm

mol

:=

Vl1.127 cm3

gm

⋅:= Vv194.29 cm3

gm

⋅:= ∆Vlv VvVl

−:=

Hl762.605 J

gm

⋅:= Hv2776.2 J

gm

⋅:=

169

dPdT P−

T2Slope⋅K⋅:= dPdT 22.984 kPa

K

=

The generalized virial-coefficient correlation is suitable here

By Eqs. (3.61) + (3.62) & (3.63) along with Eq. (6.40)

Hig 2841.1 J

gm

⋅:= Sig 9.8834 J

gm K⋅

⋅:= P01 kPa⋅:=

The enthalpy of an ideal gas is independent of pressure; the entropy

DOES depend on P:

(d) Assume ln P vs. 1/T linear and fit three data pts @ 975, 1000, & 1050 kPa.

Data: pp

975

1000

1050

⎛

⎜

⎝

⎞

⎟

⎠

kPa⋅:= t

178.79

179.88

182.02

⎛

⎜

⎝

⎞

⎟

⎠

:= (degC) i13..:=

xi

1

ti273.15+

:= yiln ppi

kPa

⎛

⎜

⎝

⎞

⎠

:= Slope slope x y,():= Slope 4717−=

170

∆Slv SvSl

−:=

∆Vlv 2.996 ft3

lbm

= ∆Hlv 863.45 BTU

lbm

=

(a) GlHlTS

l

⋅−:= GvHvTS

v

⋅−:=

For enthalpy and entropy, assume that steam at 358.43 degF and 1 psi is

an ideal gas. By interpolation in Table F.4 at 1 psi:

6.49 T 358.43 459.67+( ) rankine⋅:= T 818.1 rankine=P 150 psi⋅:=

(Table F.4) molwt 18.015 gm

mol

:=

Vl0.0181 ft3

lbm

⋅:= Vv3.014 ft3

lbm

⋅:= ∆Vlv VvVl

−:=

Hl330.65 BTU

lbm

⋅:= Hv1194.1 BTU

lbm

⋅:= ∆Hlv HvHl

−:=

Sl0.5141 BTU

lbmrankine⋅

⋅:= Sv1.5695 BTU

lbmrankine⋅

⋅:=

171

Reduced conditions: ω0.345:= Tc647.1 K⋅:= Pc220.55 bar⋅:=

Tr

T

Tc

:= Tr0.7024=Pr

P

Pc

:= Pr0.0469=

The enthalpy of an ideal gas is independent of pressure; the entropy DOES

depend on P:

(d) Assume ln P vs. 1/T linear and fit threedata points (@ 145, 150, & 155

psia)

Data: pp

145

150

155

⎛

⎜

⎝

⎞

⎟

⎠

psi⋅:= t

355.77

358.43

361.02

⎛

⎜

⎝

⎞

⎟

⎠

:= (degF) i13..:=

172

HR0 2.496−R⋅Tc

⋅:= HR1 0.586−R⋅Tc

⋅:=

HR0 7.674−103

×J

mol

=HR1 1.802−103

×J

mol

=

By Eqs. (3.61) + (3.62) & (3.63) along with Eq. (6.40)

6.50 For propane: Tc369.8 K⋅:= Pc42.48 bar⋅:= ω 0.152:=

T 195 273.15+()K⋅:= T 468.15 K=P 135 bar⋅:= P01 bar⋅:=

173

6.52 For propane: ω0.152:=

Tc369.8 K⋅:= Pc42.48 bar⋅:= Zc0.276:= Vc200.0 cm3

mol

⋅:=

If the final state is a two-phase mixture, it must exist at its saturation

temperature at 1 bar. This temperature is found from the vapor pressure

equation:

6.51 For propane: Tc369.8 K⋅:= Pc42.48 bar⋅:= ω 0.152:=

T 70 273.15+()K⋅:= T 343.15 K=P0101.33 kPa⋅:= P 1500 kPa⋅:=

Assume propane an ideal gas at the initial conditions. Use

generalized virial correlation at final conditions.

174

r1

HR

RT

c

⋅

⎛

⎜

⎝

⎞

⎠

1

=

and

r0

HR

RT

c

⋅

⎛

⎜

⎝

⎞

⎠

0

=

For Step (1), use the generalized correlation of Tables E.7 & E.8, and let

The sum of the enthalpy changes for these steps is set equal to zero, and

the resulting equation is solved for the fraction of the stream that is liquid.

ENERGY BALANCE: For the throttling process there is no enthalpy

change. The calculational path from the initial state to the final is made up

of the following steps:

(1) Transform the initial gas into an ideal gas at the initial T & P.

(2) Carry out the temperature and pressure changes to the final T & P in

the ideal-gas state.

(3) Transform the ideal gas into a real gas at the final T & P.

(4) Partially condense the gas at the final T & P.

PT() P

cexp AτT()⋅BτT()

()

1.5

⋅+ CτT()

()

3

⋅+ DτT()

()

6

⋅+

1τT()−

⎡

⎢

⎣

⎤

⎥

⎦

⋅:=

The latent heat of vaporization at the final conditions will be needed for an

energy balance. It is found by the Clapeyron equation. We proceed

exactly as in Pb. 6.17.

175

6.53 For 1,3-butadiene: ω0.190:= Tc425.2 K⋅:=

Pc42.77 bar⋅:= Zc0.267:= Vc220.4 cm3

mol

⋅:= Tn268.7 K⋅:=

T1370 K⋅:= P1200 bar⋅:=

Tr

T1

Tc

:= Tr1.001=Pr

P1

Pc

:= Pr4.708=

For Step (2) the enthalpy change is given by Eq. (6.95), for which

∆H2R ICPH T1T,1.213,28.785 10 3−

⋅, 8.824−10 6−

⋅, 0.0,

()()

⋅:=

For Step (3) the enthalpy change is given by Eq. (6.87), for which

∆H3RT

c

⋅HRB TrPr

,ω,

()

⋅:= For Step (4), ∆H4x−∆Hlv

⋅=

Hvap R ICPH T0T,2.734,26.786 10 3−

⋅, 8.882−10 6−

⋅, 0.0,

()

⋅HR

+:=

HR1 3.153−103

×J

mol

=HR0 2.436−103

×J

mol

=

HR1 0.892−R⋅Tc

⋅:=HR0 0.689−R⋅Tc

⋅:=

Z 0.718=ZZ

0ωZ1

⋅+:=Z10.1366−:=Z00.7442:=

Use Lee/Kesler correlation. HOWEVER, the values for a saturated vapor lie

on the very edge of the vapor region, and some adjacent numbers are for the

liquid phase. These must NOT be used for interpolation. Rather,

EXTRAPOLATIONS must be made from the vapor side. There may be

some choice in how this is done, but the following values are as good as any:

177

P 1435 kPa⋅:= T0273.15 K⋅:= P0101.33 kPa⋅:=

Tr

T

Tc

:= Tr0.87=Pr

P

Pc

:= Pr0.378=

Use Lee/Kesler correlation. HOWEVER, the values for a saturated vapor lie

on the very edge of the vapor region, and some adjacent numbers are for the

liquid phase. These must NOT be used for interpolation. Rather,

EXTRAPOLATIONS must be made from the vapor side. There may be

some choice in how this is done, but the following values are as good as any:

∆HnRT

n

⋅

1.092 ln Pc

bar

⎛

⎜

⎝

⎞

⎠1.013−

⎛

⎜

⎝

⎞

⎠

⋅

0.930 Tn

Tc

−

⎡

⎢

⎣

⎤

⎥

⎦

⋅:= ∆Hn22449 J

mol

=

⎜

6.54 For n-butane: ω0.200:= Tc425.1 K⋅:=

Pc37.96 bar⋅:= Zc0.274:= Vc255 cm3

mol

⋅:= Tn272.7 K⋅:=

T 370 K⋅:=

178

∆H 15295.2 J

mol

=∆H∆Hn

1T

r

−

1Tn

Tc

−

⎛

⎜

⎞

⎟

0.38

⋅:=

By Eq. (4.13)

⎢

⎥

Svap R ICPS T0T,1.935,36.915 10 3−

⋅, 11.402−10 6−

⋅, 0.0,

()

ln P

P0

⎛

⎜

⎝

⎞

⎠

−

⎛

⎜

⎝

⎞

⎠

⋅SR

+:=

Hvap R ICPH T0T,1.935,36.915 10 3−

⋅, 11.402−10 6−

⋅, 0.0,

()

⋅HR

+:=

SR1 0.835−R⋅:=SR0 0.485−R⋅:=

179

We can replace Vtank by m2V2, and rearrange to get

m2

m1Hprime H1

−

()

⋅Vtank P2P1

−Hfg2

Vfg2

−

⎛

⎜

⎝

⎞

⎠

⋅−

Hprime Hf2

−Vf2

Hfg2

Vfg2

⋅+

=

We consider this storage leg, and for this process of steam addition to

a tank the equation developed in Problem 6-74 is applicable:

mprime 6000 kg

hr 4000 kg

hr

−

⎛

⎜

⎝

⎞

⎠θ⋅:=

The steam stored during this leg is:

This situation is also represented by the equation:

Demand

(kg/hr)

6,000

2/3 hr 1/3 hr

1 hr

4,000 kg/hr

10,000 kg/hr

net storage

of steam

net depletion

of steam

time

Under the stated conditions the worst possible cycling of demand can be

represented as follows:

6.55

180

We find from the steam tables

P1700kPa:=

Initial state in accumulator is wet steam at 700 kPa.

Now we need property values:

x2

Vf2

19Vg2 Vf2

+

=

or

19 1x

2

−

()

Vf2

⋅

x2Vg2

⋅

=

therefore

From the given information we can write:

In this equation only x1 is unknown and we can solve for it as follows. First

we need V2:

Therefore our equation becomes (with Hprime = Hg2)

In this equation we can determine from the given information everything

except Hprime and Vprime. These quantities are expressed by

Making this substitution and rearranging we get

181

Given

Hg2 Hf2

−

()

Vf2

Hfg2

Vfg2

⎛

⎜

⎝

⎞

⎠

⋅+

V2

P2

+P1

−Hfg2

Vfg2

−Hg2 Hf1

−x1Hfg1

⋅−

Vf1 x1Vfg1

⋅+

=

Solve for m1 and m2 using a Mathcad Solve Block:

Guess: m1

mprime

2

:= m2m1

:=

Final state in accumulator is wet steam at 1000 kPa. P21000kPa:=

From the steam tables

Solve Eq. (C) for V2

Next solve Eq. (B) for x1Guess: x10.1:=

182

The throttling process, occurring at constant enthalpy, may be split into two

steps:

(1) Transform into an ideal gas at the initial conditions, evaluating property

changes from a generalized correlation.

(2) Change T and P in the ideal-gas state to the

final conditions, evaluating property changes by equations for an ideal gas.

Property changes for the two steps sum to the property change for the

process. For the initial conditions:

P01 bar⋅:=P 38 bar⋅:=T 400.15 K⋅:=

Pc46.65 bar⋅:=Tc365.6 K⋅:=ω 0.140:=

Propylene:6.56

m1

m2

⎛

⎜

⎝

⎞

⎠Find m1m2

,

()

:=m2m1

−2667lb=

m2

m1

Hprime Hf1

−

Hprime Hf2

−

=

Given

m2

m1

Hprime U1

−

Hprime U2

−

=Hprime Uf1

−

Hprime Uf2

−

=Hprime Hf1

−

Hprime Hf2

−

=

One can work this problem very simply and almost correctly by ignoring the

vapor present. By first equation of problem 3-15

Note that just to store 1333.3 kg of saturated vapor at 1000 kPa would

require a volume of:

183