Chapter 6 Homework Property changes by equations for an ideal gas

A 1.637:= B22.706 10 3−

⋅

K

:= C6.915−10 6−

⋅

K2

:=

Solve energy balance for final T. See Eq. (4.7).

τ1:= (guess) Given

HRRAT⋅τ1−

()

⋅B

2T2

⋅τ

21−

()

⋅+

⎡

⎢

⎣

⎤

⎥

⎦

C

3T3

⋅τ

31−

()

⋅+

⎡

⎢

⎣

⎤

⎥

⎦

⋅=

Step (1): Use the Lee/Kesler correlation, interpolate.

H00.863−R⋅Tc

⋅:= H10.534−R⋅Tc

⋅:= HRH0ωH1

⋅+:=

Step (2): For the heat capacity of propylene,

184

A 1.213:= B28.785 10 3−

⋅

K

:= C8.824−10 6−

⋅

K2

:=

Solve energy balance for final T. See Eq. (4.7).

τ1:= (guess) Given

HRRAT⋅τ1−

()

⋅B

2T2

⋅τ

21−

()

⋅+

⎡

⎢

⎣

⎤

⎥

⎦

C

3T3

⋅τ

31−

()

⋅+

⎡

⎢

⎣

⎤

⎥

⎦

⋅=

6.57 Propane: ω0.152:= Tc369.8 K⋅:= Pc42.48 bar⋅:=

T 423 K⋅:= P 22 bar⋅:= P01 bar⋅:=

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:

Tr

T

Tc

:= Tr1.144=Pr

P

Pc

:= Pr0.518=

Step (1): Use the generalized virial correlation

Step (2): For the heat capacity of propane,

185

ω0.094:= Tc373.5 K⋅:= Pc89.63 bar⋅:=

T1400 K⋅:= P15 bar⋅:= T2600 K⋅:= P225 bar⋅:=

Tr1

T1

Tc

:= Pr1

P1

Pc

:= Tr2

T2

Tc

:= Pr2

P2

Pc

:=

Tr1 1.071=Pr1 0.056=Tr2 1.606=Pr2 0.279=

Use generalized virial-coefficient correlation for both sets of conditions.

Eqs. (6.91) & (6.92) are written

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

T 100 273.15+()K⋅:= T 373.15 K=P01 bar⋅:= P 10 bar⋅:=

Tr

T

Tc

:= Tr1.009=Pr

P

Pc

:= Pr0.235=

Assume ideal gas at initial conditions. Use virial correlation at final conditions.

6.59 H2S:

186

A 5.457:= B1.045 10 3−

⋅

K

:= D 1.157−105

⋅K2

⋅:=

Solve energy balance for final T. See Eq. (4.7).

τ1:= (guess) Given

6.60 Carbon dioxide: ω0.224:= Tc304.2 K⋅:= Pc73.83 bar⋅:=

T 318.15 K⋅:= P 1600 kPa⋅:= P0101.33 kPa⋅:=

Throttling process, constant enthalpy, may be split into two steps:

(1) Transform to ideal gas at initial conditions, generalized correlation

for property changes.

(2) Change T and P of ideal gas to final T & P.

Property changes by equations for an ideal gas.

Assume ideal gas at final T & P. Sum property changes for the process.

For the initial T & P:

Step (2): For the heat capacity of carbon dioxide,

187

(b) Ethylene: ω0.087:= Tc282.3 K⋅:= Pc50.40 bar⋅:=

Tr0

T0

Tc

:= Tr0 1.85317=Pr0

P0

Pc

:= Pr0 0.75397=

At final conditions as calculated in (a)

Tr

T

Tc

:= Tr1.12699=Pr

P

Pc

:= Pr0.02381=

Use virial-coefficient correlation.

The entropy change is now given by Eq. (6.92):

τ0.5:= (guess) Given

6.61 T0523.15 K⋅:= P03800 kPa⋅:= P 120 kPa⋅:=

∆S0J

mol K⋅

⋅:= For the heat capacity of ethylene:

A 1.424:= B14.394 10 3−

⋅

K

:= C4.392−10 6−

⋅

K2

:=

(a) For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15)

with D = 0:

τ0.4:= (guess) Given

∆SRAlnτ

()

⋅BT

0

⋅CT

02

⋅τ1+

2

⎛

⎜

⎝

⎞

⎠

⋅+

⎡

⎢

⎣

⎤

⎥

⎦τ1−

()

⋅+ ln P

P0

⎛

⎜

⎝

⎞

⎠

−

⎡

⎢

⎣

⎤

⎥

⎦

⋅=

188

∆S0J

mol K⋅

⋅:= For the heat capacity of ethane:

A 1.131:= B19.225 10 3−

⋅

K

:= C5.561−10 6−

⋅

K2

:=

(a) For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15)

with D = 0:

τ0.4:= (guess) Given

∆SRAlnτ

()

⋅BT

0

⋅CT

02

⋅τ1+

2

⎛

⎜

⎝

⎞

⎠

⋅+

⎡

⎢

⎣

⎤

⎥

⎦τ1−

()

⋅+ ln P

P0

⎛

⎜

⎝

⎞

⎠

−

SRB τT0

⋅

Tc

Pr

,ω,

⎛

⎜

⎝

⎞

⎠SRB Tr0 Pr0

,ω,

()

−+

…

⎡

⎢

⎣

⎤

⎥

⎦

⋅=

The work is given by Eq. (6.91):

∆Hig R ICPH T0T,1.424,14.394 10 3−

⋅, 4.392−10 6−

⋅, 0.0,

()

⋅:=

6.62 T0493.15 K⋅:= P030 bar⋅:= P 2.6 bar⋅:=

189

⎜

⎢

⎥

⎜

Use virial-coefficient correlation.

The entropy change is now given by Eq. (6.83):

τ0.5:= (guess) Given

∆SRAlnτ

()

⋅BT

0

⋅CT

02

⋅τ1+

2

⎛

⎜

⎝

⎞

⎠

⋅+

⎡

⎢

⎣

⎤

⎥

⎦τ1−

()

⋅+ ln P

P0

⎛

⎜

⎝

⎞

⎠

−

SRB τT0

⋅

Pr

,ω,

⎛

⎜

⎞

,ω,

()

−+

…

⎡

⎢

⎤

⎥

⋅=

The work is given by Eq. (6.91):

∆Hig R ICPH T0T,1.131,19.225 10 3−

⋅, 5.561−10 6−

⋅, 0.0,

()

⋅:=

(b) Ethane: ω0.100:= Tc305.3 K⋅:= Pc48.72 bar⋅:=

Tr0

T0

Tc

:= Tr0 1.6153=Pr0

P0

Pc

:= Pr0 0.61576=

At final conditions as calculated in (a)

190

HRB00.05679−:=

The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0:

(guess) τ0.4:=

Given ∆SRAlnτ

()

⋅BT

0

⋅CT

02

⋅τ1+

2

⎛

⎜

⎝

⎞

⎠

⋅+

⎡

⎢

⎣

⎤

⎥

⎦τ1−

()

⋅+ ln P

P0

⎛

⎜

⎝

⎞

⎠

−

SRB T0τ⋅

Tc

Pr

,ω,

⎛

⎜

⎝

⎞

⎠SRB Tr0 Pr0

,ω,

()

−+

…

⎡

⎢

⎣

⎤

⎥

⎦

⋅=

The work is given by Eq. (6.91):

∆Hig R ICPH T0T,1.935,36.915 10 3−

⋅, 11.402−10 6−

⋅, 0.0,

()

⋅:=

6.63 n-Butane: ω0.200:= Tc425.1 K⋅:= Pc37.96 bar⋅:=

T0323.15 K⋅:= P01 bar⋅:= P 7.8 bar⋅:=

∆S0J

mol K⋅

⋅:= For the heat capacity of n-butane:

A 1.935:= B36.915 10 3−

⋅

K

:= C11.402−10 6−

⋅

K2

:=

Tr0

T0

Tc

:= Tr0 0.76017=Pr0

P0

Pc

:= Pr0 0.02634=

Pr

P

Pc

:= Pr0.205=

HRB Tr0 Pr0

,ω,

()

0.05679−=

191

For the compressed liquid at

325 K and 8000 kPa, apply

Eqs. (6.28) and (6.29) with

P18000 kPa⋅:=

T 325 K⋅:= β460 10 6−

⋅K1−

⋅:=

For sat. vapor at 8000 kPa, from Table F.2:

6.64 The maximum work results when the 1 kg of steam is reduced in a

completely reversible process to the conditions of the surroundings,

where it is liquid at 300 K (26.85 degC). This is the ideal work.

From Table F.2 for the initial state of superheated steam:

H13344.6 kJ

kg

⋅:= S17.0854 kJ

kg K⋅

⋅:=

From Table F.1, the state of sat. liquid at 300 K is essentially correct:

6.65 Sat. liquid at 325 K (51.85 degC), Table F.1:

Hliq 217.0 kJ

kg

⋅:= Sliq 0.7274 kJ

kg K⋅

⋅:= Vliq 1.013 cm3

gm

⋅:=

Psat 12.87 kPa⋅:=

192

6.67 For sat. liquid water at 20 degC, Table F.1:

H183.86 kJ

kg

⋅:= S10.2963 kJ

kg K⋅

⋅:=

For sat. liquid water at 0 degC, Table F.1:

H00.04−kJ

kg

⋅:= S00.0000 kJ

kg K⋅

⋅:=

For ice at at 0 degC:

H2H0333.4 kJ

kg

⋅−:= S2S0

333.4

273.15

kJ

kg K⋅

⋅−:=

Work as a fraction of heat added:

The heat not converted to work ends up in the surroundings.

6.66 Treat the furnace as a heat reservoir, for which

Qdot 2536−kJ

kg

⋅10⋅kg

sec

⋅:= T 600 273.15+()K⋅:= T 873.15 K=

193

S20.0 kJ

kg K⋅

⋅:= Q’ 2000−kJ

kg

⋅:= Tσ273.15 K⋅:=

The system consists of two parts: the apparatus and the heat reservoir at

elevated temperature, and in the equation for ideal work, terms must be

included for each part.

Wideal ∆Happaratus.reservoir Tσ∆Sapparatus.reservoir

⋅−=

H2333.44−kJ

kg

=S21.221−kJ

kg K⋅

=

Tσ293.15 K⋅:= mdot 0.5 kg

sec

⋅:= ηt0.32:=

By Eqs. (5.26) and (5.28):

6.68 This is a variation on Example 5.6., pp. 175-177, where all property values

are given. We approach it here from the point of view that if the process

is completely reversible then the ideal work is zero. We use the notation of

Example 5.6:

H12676.0 kJ

kg

⋅:= S17.3554 kJ

kg K⋅

⋅:= H20.0 kJ

kg

⋅:=

194

6.70 From Table F.3 at 430 degF (sat. liq. and vapor):

Vliq 0.01909 ft3

lbm

⋅:= Vvap 1.3496 ft3

lbm

⋅:= Vtank 80 ft3

⋅:=

Uliq 406.70 BTU

lbm

⋅:= Uvap 1118.0 BTU

lbm

⋅:= mliq 4180 lbm

⋅:=

VOLliq mliq Vliq

⋅:= VOLliq 79.796 ft3

=

6.69 From Table F.4 at 200(psi):

H11222.6 BTU

lbm

⋅:= S11.5737 BTU

lbmrankine⋅

⋅:= (at 420 degF)

(Sat. liq.

and vapor)

Hliq 355.51 BTU

lbm

⋅:= Hvap 1198.3 BTU

lbm

⋅:=

Sliq 0.5438 BTU

lbmrankine⋅

⋅:= Svap 1.5454 BTU

lbmrankine⋅

⋅:= x 0.96:=

H2Hliq xH

vap Hliq

−

()

⋅+:= S2Sliq xS

vap Sliq

−

()

⋅+:=

H21.165 103

×BTU

lbm

=S21.505 BTU

lbmrankine⋅

=

Neglecting kinetic- and potential-energy changes,

on the basis of 1 pound mass of steam after mixing, Eq. (2.30) yields for

195

Property values below are for sat. liq. and vap. at 420 degF

From Table F.3 we see that the enthalpy of saturated vapor changes

from 1203.9 to 1203.1(Btu/lb) as the temperature drops from 430 to 420

degF. This change is so small that use of an average value for H of

1203.5(Btu/lb) is fully justified. Then

m2U2

⋅m1U1

⋅−

0

m

mH

⌠

⎮

⌡d+0=

Integration gives:

dm

tUt

⋅

()

Hdm⋅+ 0=

(Subscript t denotes the contents of the tank.

H and m refer to the exit stream.)

By Eq. (2.29) multiplied through by dt, we can write,

196

6.72 This problem is similar to Example 6.8, where it is shown that

Q∆mtHt

⋅

()

H∆mt

⋅−=

Here, the symbols with subscript t refer to the contents of the tank,

whereas H refers to the entering stream.

We illustrate here development of a simple expression for the first term on

the right. The1500 kg of liquid initially in the tank is unchanged during the

process. Similarly, the vapor initially in the tank that does NOT condense

is unchanged. The only two enthalpy changes within the tank result from:

1. Addition of 1000 kg of liquid water. This contributes an enthalpy change of

2. Condensation of y kg of sat. vapor to sat. liq.

This contributes an enthalpy change of

6.71 The steam remaining in the tank is assumed to have expanded isentropically.

Data from Table F.2 at 4500 kPa and 400 degC:

S16.7093 J

gm K⋅

⋅:= V164.721 cm3

gm

⋅:= Vtank 50 m3

⋅:=

197

0.1640

0.06628

0.03126

⎛

⎜

⎞

⎟

kg

1.396

3.600

7.775

⎛

⎜

⎞

⎟

80

90

100

⎛

⎜

⎞

⎟

Data for saturated nitrogen vapor:

mtank 30 kg⋅:=T1295 K⋅:=C 0.43 kJ

kg K⋅

⋅:=

Hin 120.8−kJ

kg

⋅:=Vtank 0.5 m3

⋅:=

Given:6.73

QH

liq ∆mt

⋅y∆Hlv

⋅− H∆mt

⋅−=

Whence

∆mtVt

⋅

()

Vliq ∆mt

⋅y∆Vlv

⋅−=0=

Similarly,

198

Vs lspline T V,():=Us lspline T U,():=

Fit tabulated data with cubic spline:

U

56.006

62.579

63.325

⎛

⎜

⎞

⎟

kJ

kg

=

Calculate internal-energy values for saturated vapor nitrogen at the given

values of T:

(A)

mvap Uvap

⋅Hin mvap

⋅− Q=mtank C⋅Tvap T1

−

()

⋅=

Subscript t denotes the contents of the tank; H and m refer to the inlet

stream. Since the tank is initially evacuated, integration gives

dn

tUt

⋅

()

Hdm⋅− dQ=

By Eq. (2.29) multiplied through by dt,

78.9

⎛

⎞

199

which is later solved for m2

m2HU

liq.2

−

Vtank

m2

Vliq.2

−

∆Vlv.2

∆Ulv.2

⋅−

⎛

⎜

⎝

⎞

⎟

⎠

⋅m1HU

1

−

()

⋅=

Eliminating x2 from these equations gives

V2

Vtank

m2

=V2Vliq.2 x2∆Vlv.2

⋅+=

U2Uliq.2 x2∆Ulv.2

⋅+=

Also

m2HU

2

−

()

⋅m1HU

1

−

()

⋅=

Whence

m2U2H−

()

⋅m1U1H−

()

⋅− Q=0=

The result of Part (a) of Pb. 3.15 applies, with m replacing n:6.74

Uvap Tvap

()

Hin

−mtank C⋅T1Tvap

−

()

⋅Vvap Tvap

()

⋅

Vtank

=

Given

200

From Table F.2 at 400 kPa and 240 degC

U2H=

Whence

n1Q=0=

The result of Part (a) of Pb. 3.15 applies, with 6.75

Data from Table F.2 @ 800 kPa:

U1Uliq.1 x1∆Ulv.1

⋅+:=x1

V1Vliq.1

−

∆Vlv.1

:=

201