Chapter 3 PROPERTIES OF PURE SUBSTANCESmars.umhb.edu/~wgt/engr2345/old/Cengel_8th/Chapter_3...Pure...

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3-1

Solutions Manual for

Thermodynamics: An Engineering Approach

8th Edition

Yunus A. Cengel, Michael A. Boles

McGraw-Hill, 2015

Chapter 3

PROPERTIES OF PURE SUBSTANCES

PROPRIETARY AND CONFIDENTIAL

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3-2

Pure Substances, Phase Change Processes, Property Diagrams

3-1C Yes. Because it has the same chemical composition throughout.

3-2C A vapor that is about to condense is saturated vapor; otherwise it is superheated vapor.

3-3C No.

3-4C Because one cannot be varied while holding the other constant. In other words, when one changes, so does the other

one.

3-5C Yes. The saturation temperature of a pure substance depends on pressure. The higher the pressure, the higher the

saturation or boiling temperature.

3-6C At critical point the saturated liquid and the saturated vapor states are identical. At triple point the three phases of a

pure substance coexist in equilibrium.

3-7C Yes.

3-8C Case (c) when the pan is covered with a heavy lid. Because the heavier the lid, the greater the pressure in the pan, and

thus the greater the cooking temperature.

Property Tables

3-9C A given volume of water will boil at a higher temperature in a tall and narrow pot since the pressure at the bottom

(and thus the corresponding saturation pressure) will be higher in that case.

3-10C The molar mass of gasoline (C8H18) is 114 kg/kmol, which is much larger than the molar mass of air that is 29

kg/kmol. Therefore, the gasoline vapor will settle down instead of rising even if it is at a much higher temperature than the

surrounding air. As a result, the warm mixture of air and gasoline on top of an open gasoline will most likely settle down

instead of rising in a cooler environment



3-3

3-11C Yes. Otherwise we can create energy by alternately vaporizing and condensing a substance.

3-12C No. Because in the thermodynamic analysis we deal with the changes in properties; and the changes are independent

of the selected reference state.

3-13C The term hfg represents the amount of energy needed to vaporize a unit mass of saturated liquid at a specified

temperature or pressure. It can be determined from hfg = hg - hf .

3-14C Yes. It decreases with increasing pressure and becomes zero at the critical pressure.

3-15C Yes; the higher the temperature the lower the hfg value.

3-16C Quality is the fraction of vapor in a saturated liquid-vapor mixture. It has no meaning in the superheated vapor

region.

3-17C Completely vaporizing 1 kg of saturated liquid at 1 atm pressure since the higher the pressure, the lower the hfg .

3-18C The compressed liquid can be approximated as a saturated liquid at the given temperature. Thus TfPT @, vv .

3-19C Ice can be made by evacuating the air in a water tank. During evacuation, vapor is also thrown out, and thus the

vapor pressure in the tank drops, causing a difference between the vapor pressures at the water surface and in the tank. This

pressure difference is the driving force of vaporization, and forces the liquid to evaporate. But the liquid must absorb the

heat of vaporization before it can vaporize, and it absorbs it from the liquid and the air in the neighborhood, causing the

temperature in the tank to drop. The process continues until water starts freezing. The process can be made more efficient by

insulating the tank well so that the entire heat of vaporization comes essentially from the water.



3-4

3-20 Complete the following table for H2 O:

T, C P, kPa u, kJ / kg Phase description

143.61 400 1450 Saturated mixture

220 2319.6 2601.3 Saturated vapor

190 2500 805.15 Compressed liquid

466.21 4000 3040 Superheated vapor


T, C P, kPa v, m3 / kg Phase description

50 12.35 7.72 Saturated mixture

143.6 400 0.4624 Saturated vapor

250 500 0.4744 Superheated vapor




3-5

3-22 Problem 3-21 is reconsidered. The missing properties of water are to be determined using EES, and the solution

is to be repeated for refrigerant-134a, refrigerant-22, and ammonia.

Analysis The problem is solved using EES, and the solution is given below.

"Given" T[1]=50 [C] v[1]=7.72 [m^3/kg] P[2]=400 [kPa] x[2]=1 T[3]=250 [C] P[3]=500 [kPa] T[4]=110 [C] P[4]=350 [kPa] "Analysis" Fluid$='steam_iapws' "Change the Fluid to R134a, R22 and Ammonia and solve" P[1]=pressure(Fluid$, T=T[1], v=v[1]) x[1]=quality(Fluid$, T=T[1], v=v[1]) T[2]=temperature(Fluid$, P=P[2], x=x[2]) v[2]=volume(Fluid$, P=P[2], x=x[2]) v[3]=volume(Fluid$, P=P[3], T=T[3]) x[3]=quality(Fluid$, P=P[3], T=T[3]) v[4]=volume(Fluid$, P=P[4], T=T[4]) x[4]=quality(Fluid$, P=P[4], T=T[4]) "x = 100 for superheated vapor and x = -100 for compressed liquid"

SOLUTION for water

T [C] P [kPa] x v [kg/m3]

50.00 12.35 0.6419 7.72

143.61 400.00 1 0.4624

250.00 500.00 100 0.4744

110.00 350.00 -100 0.001051

SOLUTION for R-134a

T [C] P [kPa] x v [kg/m3]

50.00 3.41 100 7.72

8.91 400.00 1 0.05127

250.00 500.00 - -

110.00 350.00 100 0.08666



3-6

SOLUTION for R-22

T [C] P [kPa] X v [kg/m3]

50.00 4.02 100 7.72

-6.56 400.00 1 0.05817

250.00 500.00 100 0.09959

110.00 350.00 100 0.103

SOLUTION for Ammonia

T [C] P [kPa] X v [kg/m3]

50.00 20.40 100 7.72

-1.89 400.00 1 0.3094

250.00 500.00 100 0.5076

110.00 350.00 100 0.5269

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0

0

100

200

300

400

500

600

700

s [kJ/kg-K]

T [

C]

8600 kPa

2600 kPa

500 kPa

45 kPa

Steam

10-4 10-3 10-2 10-1 100 101 102 1030

100

200

300

400

500

600

700

v [m3/kg]

T [

C]

8600 kPa

2600 kPa

500 kPa

45 kPa

Steam



3-7

10-3 10-2 10-1 100 101 102100

101

102

103

104

105

v [m3/kg]

P [

kP

a]

250 C

170 C

110 C

75 C

Steam

0 500 1000 1500 2000 2500 3000

100

101

102

103

104

105

h [kJ/kg]

P [

kP

a]

250 C

170 C

110 C

75 C

Steam

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0

0

500

1000

1500

2000

2500

3000

3500

4000

s [kJ/kg-K]

h [

kJ/k

g]

8600 kPa

2600 kPa

500 kPa

45 kPa

Steam



3-8




155.46 550 0.001097 Saturated liquid



3-24E Complete the following table for H2 O:

T, F P, psia u, Btu / lbm Phase description

300 67.03 782 Saturated mixture

267.22 40 236.02 Saturated liquid





3-9

3-25E Problem 3-24E is reconsidered. The missing properties of water are to be determined using EES, and the

solution is to be repeated for refrigerant-134a, refrigerant-22, and ammonia.


"Given" T[1]=300 [F] u[1]=782 [Btu/lbm] P[2]=40 [psia] x[2]=0 T[3]=500 [F] P[3]=120 [psia] T[4]=400 [F] P[4]=420 [psia] "Analysis" Fluid$='steam_iapws' P[1]=pressure(Fluid$, T=T[1], u=u[1]) x[1]=quality(Fluid$, T=T[1], u=u[1]) T[2]=temperature(Fluid$, P=P[2], x=x[2]) u[2]=intenergy(Fluid$, P=P[2], x=x[2]) u[3]=intenergy(Fluid$, P=P[3], T=T[3]) x[3]=quality(Fluid$, P=P[3], T=T[3]) u[4]=intenergy(Fluid$, P=P[4], T=T[4]) x[4]=quality(Fluid$, P=P[4], T=T[4]) "x = 100 for superheated vapor and x = -100 for compressed liquid"

Solution for steam

T, ºF P, psia x u, Btu/lbm

300 67.028 0.6173 782

267.2 40 0 236

500 120 100 1174

400 400 -100 373.8

Solution for R134a


300 - - 782

29.01 40 0 21.24

500 120 100 203.5

400 400 100 173.6



3-10

Solution for R22


300 - - 782

1.534 40 0 78.08

500 120 100 240.1

400 400 100 218.7

Solution for ammonia


300 - - 782

11.67 40 0 63.47

500 120 100 785.5

400 400 100 727



3-11

3-26 Complete the following table for Refrigerant-134a:


-4 320 0.000764 Compressed liquid


33.45 850 0.02409 Saturated vapor


3-27E Complete the following table for Refrigerant-134a:

T, F P, psia h, Btu / lbm x Phase description

65.89 80 78 0.566 Saturated mixture

15 29.759 69.92 0.6 Saturated mixture

10 70 15.36 - - - Compressed liquid

160 180 129.46 - - - Superheated vapor

110 161.16 117.25 1.0 Saturated vapor

3-28 A rigid tank contains steam at a specified state. The pressure, quality, and density of steam are to be determined.

Properties At 220C vf = 0.001190 m3/kg and vg = 0.08609 m

3/kg (Table A-4).

Analysis (a) Two phases coexist in equilibrium, thus we have a saturated liquid-vapor mixture. The pressure of the steam is

the saturation pressure at the given temperature. Then the pressure in the tank must be the saturation pressure at the specified

temperature,

kPa 2320 C@220satTP

(b) The total mass and the quality are determined as

0.0269

518.1

13.94

kg 1.51894.132.504

kg 3.941/kgm 0.08609

)m (1.82/3

kg .2504/kgm 0.001190

)m (1.81/3

3

3

3

3

t

g

gft

g

gg

f

ff

m

mx

mmm

m

m

v

V

v

V

(c) The density is determined from

3mkg/ 287.8

003474.0

11

/kgm 003474.0)08609.0)(0269.0(001190.0)( 3

v

vvvv

fgf x

Steam

1.8 m3

220C



3-12

3-29 A piston-cylinder device contains R-134a at a specified state. Heat is transferred to R-134a. The final pressure, the

volume change of the cylinder, and the enthalpy change are to be determined.

Analysis (a) The final pressure is equal to the initial pressure, which is determined from

kPa 90.4

22

2

2atm12kg.m/s 1000

kN 1

/4m) (0.25

)m/s kg)(9.81 (12kPa 88

/4 D

gmPPP

p

(b) The specific volume and enthalpy of R-134a at the initial state of 90.4 kPa and -10C and at the final state of 90.4 kPa

and 15C are (from EES)

v1 = 0.2302 m3/kg h1 = 247.77 kJ/kg

v 2 = 0.2544 m3/kg h2 = 268.18 kJ/kg

The initial and the final volumes and the volume change are

3m 0.0205

1957.02162.0

m 2162.0/kg)m kg)(0.2544 85.0(

m 1957.0/kg)m kg)(0.2302 85.0(

3322

3311

V

vV

vV

m

m

(c) The total enthalpy change is determined from

kJ/kg 17.4 kJ/kg 247.77)kg)(268.18 85.0()( 12 hhmH

3-30E The temperature of R-134a at a specified state is to be determined.

Analysis Since the specified specific volume is higher than vg for 80 psia, this is a superheated vapor state. From R-134a

tables,

13E)-A (Table /lbmft 0.6243

psia 803 F80

T

P

v

3-31 A rigid container that is filled with R-134a is heated. The final temperature and initial pressure are to be determined.

Analysis This is a constant volume process. The specific volume is

/kgm 1348.0kg 10

m 348.1 33

21 m

Vvv

The initial state is determined to be a mixture, and thus the pressure is the

saturation pressure at the given temperature

11)-A (Table C40- @sat 1 kPa 51.25 PP

The final state is superheated vapor and the temperature is determined by

interpolation to be

13)-A (Table /kgm 1348.0

kPa 20023

2

2 C66.3

T

P

v

R-134a

0.85 kg

-10C

Q

R-134a

-40°C

10 kg

1.348 m3

P

v

2

1



3-13

3-32 The enthalpy of R-134a at a specified state is to be determined.

Analysis The specific volume is

/kgm 03.0kg 300

m 9 33

m

Vv

Inspection of Table A-11 indicates that this is a mixture of liquid and vapor. Using the properties at 10°C line, the quality

and the enthalpy are determined to be

kJ/kg 179.9

)80.190)(6001.0(42.65

6001.0/kgm )0007929.0049466.0(

/kgm )0007929.003.0(3

3

fgf

fg

f

xhhh

xv

vv

3-33 The specific volume of R-134a at a specified state is to be determined.

Analysis Since the given temperature is higher than the saturation temperature for 200 kPa, this is a superheated vapor state.

The specific volume is then

13)-A (Table C25

kPa 200 /kgm 0.11647 3

v

T

P

3-34 The average atmospheric pressure in Denver is 83.4 kPa. The boiling temperature of water in Denver is to be

determined.

Analysis The boiling temperature of water in Denver is the saturation temperature corresponding to the atmospheric pressure

in Denver, which is 83.4 kPa:

C94.6 kPa [email protected] (Table A-5)

3-35E The temperature in a pressure cooker during cooking at sea level is measured to be 250F. The absolute pressure

inside the cooker and the effect of elevation on the answer are to be determined.

Assumptions Properties of pure water can be used to approximate the properties of juicy water in the cooker.

Properties The saturation pressure of water at 250F is 29.84 psia (Table A-4E). The standard atmospheric pressure at sea

level is 1 atm = 14.7 psia.

Analysis The absolute pressure in the cooker is simply the saturation pressure at the cooking temperature,

psia 29.84 Fsat@250abs PP

It is equivalent to

atm 2.03

psia 14.7

atm 1psia 84.29absP

The elevation has no effect on the absolute pressure inside when the temperature is maintained constant at 250F.

H2O

250F



3-14

3-36E A spring-loaded piston-cylinder device is filled with R-134a. The water now undergoes a process until its volume

increases by 50%. The final temperature and the enthalpy are to be determined.

Analysis From Table A-11E, the initial specific volume is

/lbmft 5452.3)01143.04286.4)(80.0(01143.0 311 fgf x vvv

and the initial volume will be

3311 ft 0.4609/lbm)ft 2lbm)(3.545 13.0( vV m

With a 50% increase in the volume, the final volume will be

3312 ft 0.6913)ft 4609.0(5.14.1 VV

The distance that the piston moves between the initial and final conditions is

ft 2934.04/ft) 1(

ft)4609.06813.0(

4/ 2

3

2

12

DA

xp

VVV

As a result of the compression of the spring, the pressure difference between the initial and final states is

psia 1.152lbf/in 152.1in) 12(

in) 122934.0(lbf/in) 37(4

4/

2

22

D

xk

A

xk

A

FP

pp

The initial pressure is

11E)-A (Table psia 869.9F30- @sat 1 PP

The final pressure is then

psia 02.11152.1869.912 PPP

and the final specific volume is

/lbmft 318.5lbm 13.0

ft 0.6813 33

22

m

Vv

At this final state, the temperature and enthalpy are

EES) (from /lbmft 318.5

psia 02.11

2

2

32

2

Btu/lbm 124.7

F104.7

h

TP

v

Note that it is very difficult to get the temperature and enthalpy readings from Table A-13E accurately.

P

v

2

1



3-15

3-37E A piston-cylinder device that is filled with water is cooled. The final pressure and volume of the water are to be

determined.

Analysis The initial specific volume is

/lbmft 4264.2lbm 1

ft 4264.2 33

11

m

Vv

This is a constant-pressure process. The initial state is determined to be superheated

vapor and thus the pressure is determined to be


F600213

1

1 psia 250

PP

T

v

The saturation temperature at 250 psia is 400.1°F. Since the final temperature

is less than this temperature, the final state is compressed liquid. Using the

incompressible liquid approximation,

4E)-A (Table /lbmft 01663.0 3F200 @ 2 fvv

The final volume is then

3ft 0.01663 /lbm)ft 01663.0)(lbm 1( 322 vV m

3-38 The volume of a container that contains water at a specified state is to be determined.

Analysis The specific volume is determined from steam tables by interpolation to be

6)-A (Table/kgm 9367.1 C150

kPa 100 3

v

T

P

The volume of the container is then

3m 5.81 /kg)m 9367.1)(kg 3( 3vV m

3-39 Water is boiled at sea level (1 atm pressure) in a pan placed on top of a 3-kW electric burner that transfers 60% of the

heat generated to the water. The rate of evaporation of water is to be determined.

Properties The properties of water at 1 atm and thus at the saturation temperature of 100C are hfg = 2256.4 kJ/kg (Table A-

4).

Analysis The net rate of heat transfer to the water is

kW 8.1kW 360.0 Q&

Noting that it takes 2256.4 kJ of energy to vaporize 1 kg of saturated liquid water, the rate of

evaporation of water is determined to be

kg/h 2.872 kg/s 100.80kJ/kg 4.2256

kJ/s 8.1= 3

fgnevaporatio

h

Qm

&&

P

v

2 1

H2O

600°F

1 lbm

2.4264 ft3

Water

3 kg

100 kPa

150C

H2O

100C



3-16

3-40 Water is boiled at 1500 m (84.5 kPa pressure) in a pan placed on top of a 3-kW electric burner that transfers 60% of

the heat generated to the water. The rate of evaporation of water is to be determined.

Properties The properties of water at 84.5 kPa and thus at the saturation temperature of 95C are hfg = 2269.6 kJ/kg (Table

A-4).

Analysis The net rate of heat transfer to the water is

& . .Q 0 60 3 18 kW kW

Noting that it takes 2269.6 kJ of energy to vaporize 1 kg of saturated liquid water, the

rate of evaporation of water is determined to be

kg/h 2.855 kg/s 100.79

33kJ/kg 6.2269

kJ/s 8.1=nevaporatio

fgh

Qm

&&

3-41 A rigid container that is filled with R-134a is heated. The temperature and total enthalpy are to be determined at the

initial and final states.

Analysis This is a constant volume process. The specific volume is

/kgm 0014.0kg 10

m 014.0 33

21 m

Vvv

The initial state is determined to be a mixture, and thus the temperature is the

saturation temperature at the given pressure. From Table A-12 by interpolation

C0.61 kPa 300 @sat 1 TT

Using EES, we would get 0.65C. Then,

kJ/kg 56.54)17.198)(009351.0(71.52

009351.0/kgm )0007735.0067776.0(

/kgm )0007735.00014.0(

11

3

31

1

fgf

fg

f

hxhh

xv

vv

The total enthalpy is then

kJ 545.6 )kJ/kg 56.54)(kg 10(11 mhH

The final state is also saturated mixture. Repeating the calculations at this state,

C21.55 kPa 600 @sat 2 TT

kJ/kg 64.84)95.180)(01731.0(50.81

01731.0/kgm )0008198.0034335.0(

/kgm )0008198.00014.0(

22

3

32

2

fgf

fg

f

hxhh

xv

vv

kJ 846.4 )kJ/kg 64.84)(kg 10(22 mhH

R-134a

300 kPa

10 kg

14 L

P

v

2

1

4

H2O

95C



3-17

3-42 A piston-cylinder device that is filled with R-134a is cooled at constant pressure. The final temperature and the change

of total internal energy are to be determined.


/kgm 12322.0kg 100

m 322.12 33

1 m

Vv

The initial state is superheated and the internal energy at this state is

13)-A (TablekJ/kg 08.263 /kgm 12322.0

kPa 20013

1

1

u

P

v

The final specific volume is

/kgm 06161.02

/m 12322.0

2

33

12

kgvv

This is a constant pressure process. The final state is determined to be saturated

mixture whose temperature is

12)-A (Table kPa 200 @sat 2 C10.09TT

The internal energy at the final state is (Table A-12)

kJ/kg 52.152)25.186)(6135.0(26.38

6135.0/kgm )0007532.0099951.0(

/kgm )0007532.006161.0(

22

3

32

2

fgf

fg

f

uxuu

xv

vv

Hence, the change in the internal energy is

kJ/kg 110.6 08.26352.15212 uuu

R-134a

200 kPa

100 kg

12.322 m3

P

v

2 1



3-18

3-43 A piston-cylinder device fitted with stops contains water at a specified state. Now the water is cooled until a final

pressure. The process is to be indicated on the T-v diagram and the change in internal energy is to be determined.

Analysis The process is shown on T-v diagram. The internal

energy at the initial state is

6)-A (TablekJ/kg 8.2808 C300

kPa 2001

1

1

u

T

P

State 2 is saturated vapor at the initial pressure. Then,

5)-A (Table/kgm 8858.0 vapor)(sat. 1

kPa 200 32

2

2

v

x

P

Process 2-3 is a constant-volume process. Thus,

5)-A (TablekJ/kg 6.1508 /kgm 8858.0

kPa 10033

23

3

u

P

vv

The overall change in internal energy is

kJ/kg 1300 6.15088.280831 uuu

3-44 Saturated steam at Tsat = 40C condenses on the outer surface of a cooling tube at a rate of 130 kg/h. The rate of heat

transfer from the steam to the cooling water is to be determined.

Assumptions 1 Steady operating conditions exist. 2 The condensate leaves the condenser as a saturated liquid at 30C.

Properties The properties of water at the saturation temperature of 40C are hfg = 2406.0 kJ/kg (Table A-4).

Analysis Noting that 2406.0 kJ of heat is released as 1 kg of saturated

vapor at 40C condenses, the rate of heat transfer from the steam to the

cooling water in the tube is determined directly from

kW 86.9=

kJ/h 780,312kJ/kg) .0kg/h)(2406 130(

evap

fghmQ &&

D = 3 cm L = 35 m

40C

T

v

2

1

3

200 kPa

100 kPa 300C

Water

200 kPa

300C

Q



3-19

3-45 The boiling temperature of water in a 5-cm deep pan is given. The boiling temperature in a 40-cm deep pan is to be

determined.

Assumptions Both pans are full of water.

Properties The density of liquid water is approximately = 1000 kg/m3.

Analysis The pressure at the bottom of the 5-cm pan is the saturation

pressure corresponding to the boiling temperature of 98C:

kPa 94.39Csat@98 PP (Table A-4)

The pressure difference between the bottoms of two pans is

kPa 3.43skg/m 1000

kPa 1m) )(0.35m/s )(9.807kg/m (1000

2

23

hgP

Then the pressure at the bottom of the 40-cm deep pan is

P = 94.39 + 3.43 = 97.82 kPa

Then the boiling temperature becomes

C99.0 kPa [email protected] TT (Table A-5)

3-46 A cooking pan is filled with water and covered with a 4-kg lid. The boiling temperature of water is to be determined.

Analysis The pressure in the pan is determined from a force balance on the lid,

PA = PatmA + W

or,

kPa 102.25skg/m 1000

kPa 1

m) (0.1

)m/s kg)(9.81 (4kPa) (101

22

2

A

mgPP atm

The boiling temperature is the saturation temperature corresponding to this pressure,

C100.2kPa 102.25@sat TT (Table A-5)

40 cm

5 cm

W = mg

Patm

P



3-20

3-47 Prob. 3-46 is reconsidered. Using EES (or other) software, the effect of the mass of the lid on the boiling

temperature of water in the pan is to be investigated. The mass is to vary from 1 kg to 10 kg, and the boiling temperature is

to be plotted against the mass of the lid.


"Given data" {P_atm=101[kPa]} D_lid=20 [cm] {m_lid=4 [kg]} "Solution" "The atmospheric pressure in kPa varies with altitude in km by the approximate function:" P_atm=101.325*(1-0.02256*z)^5.256 "The local acceleration of gravity at 45 degrees latitude as a function of altitude in m is given by:" g=9.807+3.32*10^(-6)*z*convert(km,m) "At sea level:" z=0 "[km]" A_lid=pi*D_lid^2/4*convert(cm^2,m^2) W_lid=m_lid*g*convert(kg*m/s^2,N) P_lid=W_lid/A_lid*convert(N/m^2,kPa) P_water=P_lid+P_atm T_water=temperature(steam_iapws,P=P_water,x=0)

mlid [kg] Twater [C]

1 2 3 4 5 6 7 8 9 10

100.1 100.1 100.2 100.3 100.4 100.5 100.6 100.7 100.7 100.8

1 2 3 4 5 6 7 8 9 10

100

100.1

100.2

100.3

100.4

100.5

100.6

100.7

100.8

100.9

mlid [kg]

Tw

ate

r [

C]



3-21

0 1 2 3 4 5 6 7 8 9

30

40

50

60

70

80

90

100

110

z [km]

Pw

ate

r [

kP

a]

Effect of altitude on boiling pressure of water in pan with lid

mass of lid = 4 kg

0 1 2 3 4 5 6 7 8 9

70

75

80

85

90

95

100

105

z [km]

Tw

ate

r [

C]

mass of lid = 4 kg

Effect of altitude on boiling temperature of water in pan with lid



3-22

3-48 A vertical piston-cylinder device is filled with water and covered with a 40-kg piston that serves as the lid. The boiling

temperature of water is to be determined.

Analysis The pressure in the cylinder is determined from a force balance on the piston,

PA = PatmA + W

or,

kPa 126.15skg/m 1000

kPa 1

m 0.0150

)m/s kg)(9.81 (40kPa) (100

22

2

atm

A

mgPP

The boiling temperature is the saturation temperature corresponding to this pressure,

C106.2 kPa 126.15@satTT (Table A-5)

3-49 Water is boiled in a pan by supplying electrical heat. The local atmospheric pressure is to be estimated.

Assumptions 75 percent of electricity consumed by the heater is transferred to the water.

Analysis The amount of heat transfer to the water during this period is

kJ 2700s) 6030(kJ/s) 2)(75.0(timeelect fEQ

The enthalpy of vaporization is determined from

kJ/kg 2269kg 1.19

kJ 2700

boil

m

Qh fg

Using the data by a trial-error approach in saturation table of water (Table A-5) or using EES as we did, the saturation

pressure that corresponds to an enthalpy of vaporization value of 2269 kJ/kg is

Psat = 85.4 kPa

which is the local atmospheric pressure.

3-50 A rigid tank that is filled with saturated liquid-vapor mixture is heated. The temperature at which the liquid in the tank

is completely vaporized is to be determined, and the T-v diagram is to be drawn.

Analysis This is a constant volume process (v = V /m = constant), and

the specific volume is determined to be

/kgm 0.12kg 15

m 1.8 33

m

Vv

When the liquid is completely vaporized the tank will contain

saturated vapor only. Thus,

/kgm 0.12 32 gvv

The temperature at this point is the temperature that corresponds to

this vg value,

C202.9 /kgm 0.12@sat 3

gvTT (Table A-4)

H2O

90C

T

v

2

1

W = mg

Patm

P



3-23

3-51 A piston-cylinder device contains a saturated liquid-vapor mixture of water at 800 kPa pressure. The mixture is

heated at constant pressure until the temperature rises to 200°C. The initial temperature, the total mass of water, the final

volume are to be determined, and the P-v diagram is to be drawn.

Analysis (a) Initially two phases coexist in equilibrium, thus we have a saturated liquid-vapor mixture. Then the temperature

in the tank must be the saturation temperature at the specified pressure,

C158.8 kPa 600@satTT

(b) The total mass in this case can easily be determined by adding the mass of each phase,

kg 7.395

852.2543.4

kg .8522/kgm 0.3156

m 0.9

kg 543.4/kgm 0.001101

m 0.005

3

3

3

3

gft

g

g

g

f

f

f

mmm

m

m

v

V

v

V

(c) At the final state water is superheated vapor, and its specific volume is

/kgm 0.3521C020

kPa 006 32

2

2

vT

P (Table A-6)

Then,

3m 2.604 /kg)m kg)(0.3521 (7.395 322 vV tm

P

v

2 1



3-24

3-52 Prob. 3-51 is reconsidered. The effect of pressure on the total mass of water in the tank as the pressure varies

from 0.1 MPa to 1 MPa is to be investigated. The total mass of water is to be plotted against pressure, and results are to be

discussed.


P[1]=600 [kPa] P[2]=P[1] T[2]=200 [C] V_f1 = 0.005 [m^3] V_g1=0.9 [m^3] spvsat_f1=volume(Steam_iapws, P=P[1],x=0) "sat. liq. specific volume, m^3/kg" spvsat_g1=volume(Steam_iapws,P=P[1],x=1) "sat. vap. specific volume, m^3/kg" m_f1=V_f1/spvsat_f1 "sat. liq. mass, kg" m_g1=V_g1/spvsat_g1 "sat. vap. mass, kg" m_tot=m_f1+m_g1 V[1]=V_f1+V_g1 spvol[1]=V[1]/m_tot "specific volume1, m^3" T[1]=temperature(Steam_iapws, P=P[1],v=spvol[1])"C" "The final volume is calculated from the specific volume at the final T and P" spvol[2]=volume(Steam_iapws, P=P[2], T=T[2]) "specific volume2, m^3/kg" V[2]=m_tot*spvol[2]

P1 [kPa] mtot [kg]

100

200

300

400

500

600

700

800

900

1000

5.324

5.731

6.145

6.561

6.978

7.395

7.812

8.23

8.648

9.066

100 200 300 400 500 600 700 800 900 10005

5.5

6

6.5

7

7.5

8

8.5

9

9.5

P1 [kPa]

mto

t [k

g]

10-3 10-2 10-1 100 101 102100

101

102

103

104

105

106

v [m3/kg]

P [

kP

a]

200°C

SteamIAPWS

1 2

P=600 kPa



3-25

3-53E A rigid tank contains water at a specified pressure. The temperature, total enthalpy, and the mass of each phase are to

be determined.

Analysis (a) The specific volume of the water is

/lbmft 1.0lbm 5

ft 5 33

m

Vv

At 20 psia, vf = 0.01683 ft3/lbm and vg = 20.093 ft

3/lbm (Table A-12E). Thus the tank contains saturated liquid-vapor

mixture since vf < v < vg , and the temperature must be the saturation temperature at the specified pressure,

F227.92 psia 20@satTT

(b) The quality of the water and its total enthalpy are determined from

Btu/lbm 243.2893.95904897.027.196

04897.001683.0093.20

01683.00.1

fgf

fg

f

xhhh

xv

vv

H = mh = (5 lbm)(243.28 Btu/lbm) = 1216.4 Btu

(c) The mass of each phase is determined from

lbm 4.755

lbm 0.245

245.05

504897.0

gtf

tg

mmm

xmm

3-54E A rigid tank contains saturated liquid-vapor mixture of R-134a. The quality and total mass of the refrigerant are to be

determined.

Analysis At 50 psia, vf = 0.01252 ft3/lbm and vg = 0.94791 ft

3/lbm (Table A-12E). The volume occupied by the liquid and

the vapor phases are

33 ft 4andft 1 gf VV

Thus the mass of each phase is

lbm 22.4/lbmft 0.94909

ft 4

lbm 88.79/lbmft 0.01252

ft 1

3

3

3

3

g

g

g

f

f

f

m

m

v

V

v

V

Then the total mass and the quality of the refrigerant are

mt = mf + mg = 79.88 + 4.22 = 84.10 lbm

0.05018lbm 84.10

lbm 4.22

t

g

m

mx

H2O

5 lbm

20 psia

R-134a

5 ft3

50 psia



3-26

3-55E Superheated water vapor cools at constant volume until the temperature drops to 250°F. At the final state, the

pressure, the quality, and the enthalpy are to be determined.

Analysis This is a constant volume process (v = V/m = constant), and the initial specific volume is determined to be

/lbmft 3.0433F500

psia 180 31

1

1

vT

P (Table A-6E)

At 250°F, vf = 0.01700 ft3/lbm and vg = 13.816 ft

3/lbm. Thus at the

final state, the tank will contain saturated liquid-vapor mixture since

vf < v < vg , and the final pressure must be the saturation pressure at

the final temperature,

psia 29.84F250@sat PP

(b) The quality at the final state is determined from

0.219

01700.0816.13

01700.00433.32

2fg

fx

v

vv

(c) The enthalpy at the final state is determined from

Btu/lbm 426.0 41.945219.063.218fgf xhhh

T

v

2

1

H2O

180 psia

500F



3-27

3-56E Problem 3-55E is reconsidered. The effect of initial pressure on the quality of water at the final state as the

pressure varies from 100 psi to 300 psi is to be investigated. The quality is to be plotted against initial pressure, and the

results are to be discussed.


T[1]=500 [F] P[1]=180 [psia] T[2]=250 [F] v[ 1]=volume(steam_iapws,T=T[1],P=P[1]) v[2]=v[1] P[2]=pressure(steam_iapws,T=T[2],v=v[2]) h[2]=enthalpy(steam_iapws,T=T[2],v=v[2]) x[2]=quality(steam_iapws,T=T[2],v=v[2])

P1 [psia]

x2

100 122.2 144.4 166.7 188.9 211.1 233.3 255.6 277.8 300

0.4037 0.3283 0.2761 0.2378 0.2084 0.1853 0.1665 0.151 0.1379 0.1268

100 140 180 220 260 300

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

P[1] [psia]

x[2

]

10-2 10-1 100 101 102 103 1040

200

400

600

800

1000

1200

1400

v [ft3/lbm]

T [

°F]

1600 psia

780 psia

180 psia

29.82 psia

0.05 0.1 0.2 0.5

1.2 1.3 1.4 1.5 Btu/lbm-R

Steam

1

2



3-28

3-57 Superheated steam in a piston-cylinder device is cooled at constant pressure until half of the mass condenses. The final

temperature and the volume change are to be determined, and the process should be shown on a T-v diagram.

Analysis (b) At the final state the cylinder contains saturated liquid-

vapor mixture, and thus the final temperature must be the saturation

temperature at the final pressure,

C151.83 MPa [email protected] (Table A-5)

(c) The quality at the final state is specified to be x2 = 0.5. The specific

volumes at the initial and the final states are

/kgm 0.42503C200

MPa 5.0 31

1

1

vT

P (Table A-6)

/kgm .18800

)001093.037483.0(5.0001093.05.0

MPa 5.0

3

222

2

fgf x

x

Pvvv

Thus,

3m 0.14222 /kgm0.42503)kg)(0.1880 (0.6)(Δ 312 vvV m

3-58 The water in a rigid tank is cooled until the vapor starts condensing. The initial pressure in the tank is to be

determined.

Analysis This is a constant volume process (v = V /m = constant), and the

initial specific volume is equal to the final specific volume that is

/kgm 79270.0 3C124@21 gvvv (Table A-4)

since the vapor starts condensing at 150C. Then from

Table A-6,

MPa 0.30

13

1

1

/kgm 0.79270

C025P

T

v

H2O

200C

0.5 MPa

T

v

2

1

H2O

T1= 250C

P1 = ?

T

v

2

1 25

0

15

0

C



3-29

3-59 Heat is supplied to a piston-cylinder device that contains water at a specified state. The volume of the tank, the final

temperature and pressure, and the internal energy change of water are to be determined.

Properties The saturated liquid properties of water at 200C are: vf = 0.001157 m3/kg and uf = 850.46 kJ/kg (Table A-4).

Analysis (a) The cylinder initially contains saturated liquid water. The volume of the cylinder at the initial state is

3311 m 001619.0/kg)m 001157.0kg)( 4.1( vV m

The volume at the final state is

3m 0.006476 )001619.0(4V

(b) The final state properties are

kg/m 004626.0kg 1.4

m 0.006476 33

2 m

Vv

kJ/kg 5.22011

kg/m 004626.0

2

2

2

2

32

u

P

T

xkPa 21,367

C371.3v

(Table A-4 or A-5 or EES)

(c) The total internal energy change is determined from

kJ 1892 kJ/kg 850.46)-kg)(2201.5 4.1()( 12 uumU

3-60E The error involved in using the enthalpy of water by the incompressible liquid approximation is to be determined.

Analysis The state of water is compressed liquid. From the steam tables,

7E)-A (Table Btu/lbm 41.378 F400

psia 3000

h

T

P

Based upon the incompressible liquid approximation,

4E)-A (TableBtu/lbm 04.375 F400

psia 3000F400 @

fhh

T

P

The error involved is

0.89%

10041.378

04.37541.378ErrorPercent

which is quite acceptable in most engineering calculations.

Water

1.4 kg, 200°C

sat. liq.

Ethane

10 MPa

100C

Q



3-30

3-61 A piston-cylinder device that is filled with R-134a is heated. The volume change is to be determined.


13)-A (Table/kgm 33608.0 C20

kPa 60 31

1

1

v

T

P

and the initial volume is

3311 m 0.033608/kg)m 8kg)(0.3360 100.0( vV m

At the final state, we have

13)-A (Table/kgm 0.50410 C100

kPa 60 32

2

2

v

T

P

3322 m 0.050410/kg)m 0kg)(0.5041 100.0( vV m

The volume change is then

3m 0.0168 0336080050410012 ..VVV

3-62 A rigid vessel is filled with refrigerant-134a. The total volume and the total internal energy are to be determined.

Properties The properties of R-134a at the given state are (Table A-13).

/kgm 0.061687

kJ/kg .91329

C012

kPa 0053

vu

T

P

Analysis The total volume and internal energy are determined from

kJ 2639

m 0.4943

kJ/kg) kg)(329.91 (8

/kg)m 87kg)(0.0616 (8 3

muU

mvV

R-134a

60 kPa

-20C

100 g

P

v

2 1

R-134a

8 kg

500 kPa

120C



3-31

3-63 Heat is supplied to a rigid tank that contains water at a specified state. The volume of the tank, the final temperature

and pressure, and the internal energy change of water are to be determined.

Properties The saturated liquid properties of water at 200C are: vf = 0.001157 m3/kg and uf = 850.46 kJ/kg (Table A-4).

Analysis (a) The tank initially contains saturated liquid water and air. The volume occupied by water is

3311 m 001619.0/kg)m 001157.0kg)( 4.1( vV m

which is the 25 percent of total volume. Then, the total volume is determined from

3m 0.006476 )001619.0(25.0

1V

(b) Properties after the heat addition process are

kg/m 004626.0kg 1.4

m 0.006476 33

2 m

Vv

kJ/kg 5.22011

kg/m 004626.0

2

2

2

2

32

u

P

T

xkPa 21,367

C371.3v

(Table A-4 or A-5 or EES)

(c) The total internal energy change is determined from

kJ 1892 kJ/kg 850.46)-kg)(2201.5 4.1()( 12 uumU

3-64 A piston-cylinder device that is initially filled with water is heated at constant pressure until all the liquid has

vaporized. The mass of water, the final temperature, and the total enthalpy change are to be determined, and the T-v diagram

is to be drawn.

Analysis Initially the cylinder contains compressed liquid (since P > Psat@40°C) that can be approximated as a saturated liquid

at the specified temperature (Table A-4),

kJ/kg 167.53

/kgm 0.001008

Cf@401

3Cf@401

hh

vv

(a) The mass is determined from

kg 49.61/kgm 0.001008

m 0.050

3

3

1

1

v

Vm

(b) At the final state, the cylinder contains saturated vapor and thus

the final temperature must be the saturation temperature at the final

pressure,

C120.21 kPa 002@satTT

(c) The final enthalpy is h2 = hg @ 200 kPa = 2706.3 kJ/kg. Thus,

kJ 125,950 kJ/kg167.53)kg)(2706.3 (49.61)( 12 hhmH

T

v

2

1

H2O

40C

200 kPa



3-32

Ideal Gas

3-65C A gas can be treated as an ideal gas when it is at a high temperature or low pressure relative to its critical temperature

and pressure.

3-66C Ru is the universal gas constant that is the same for all gases whereas R is the specific gas constant that is different for

different gases. These two are related to each other by R = Ru / M, where M is the molar mass of the gas.

3-67C Propane (molar mass = 44.1 kg/kmol) poses a greater fire danger than methane (molar mass = 16 kg/kmol) since

propane is heavier than air (molar mass = 29 kg/kmol), and it will settle near the floor. Methane, on the other hand, is lighter

than air and thus it will rise and leak out.

3-68 A rigid tank contains air at a specified state. The gage pressure of the gas in the tank is to be determined.

Assumptions At specified conditions, air behaves as an ideal gas.

Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1).

Analysis Treating air as an ideal gas, the absolute pressure in the tank is determined from

kPa 1069.1m 0.4

K) K)(298/kgmkPa kg)(0.287 (5

3

3

V

mRTP

Thus the gage pressure is

kPa 972.1 971069.1atmPPPg

3-69E The temperature in a container that is filled with oxygen is to be determined.

Assumptions At specified conditions, oxygen behaves as an ideal gas.

Properties The gas constant of oxygen is R = 0.3353 psiaft3/lbmR (Table A-1E).

Analysis The definition of the specific volume gives

/lbmft 5.1lbm 2

ft 3 33

m

Vv

Using the ideal gas equation of state, the temperature is

R 358

R/lbmftpsia 0.3353

/lbm)ft psia)(1.5 (803

3

R

PT

v

Pg

Air

400 L

25C



3-33

3-70 The volume of a container that is filled with helium at a specified state is to be determined.

Assumptions At specified conditions, helium behaves as an ideal gas.

Properties The gas constant of helium is R = 2.0769 kJ/kgK (Table A-1).

Analysis According to the ideal gas equation of state,

3m 4.154

kPa 300

K) 273K)(27/kgmkPa kg)(2.0769 (2 3

P

mRTV

3-71 The pressure and temperature of oxygen gas in a storage tank are given. The mass of oxygen in the tank is to be

determined.

Assumptions At specified conditions, oxygen behaves as an ideal gas

Properties The gas constant of oxygen is R = 0.2598 kPa.m3/kg.K (Table A-1).

Analysis The absolute pressure of O2 is

P = Pg + Patm = 500 + 97 = 597 kPa

Treating O2 as an ideal gas, the mass of O2 in tank is determined to be

kg 19.08

K273)K)(28/kgmkPa (0.2598

)m kPa)(2.5 (597

3

3

RT

Pm

V

3-72 A balloon is filled with helium gas. The mole number and the mass of helium in the balloon are to be determined.

Assumptions At specified conditions, helium behaves as an ideal gas.

Properties The universal gas constant is Ru = 8.314 kPa.m3/kmol.K. The molar mass of helium is 4.0 kg/kmol (Table A-1).

Analysis The volume of the sphere is

333 m 7.381m) (4.5

3

4

3

4 rV

Assuming ideal gas behavior, the mole numbers of He is determined from

kmol 30.61

K) K)(300/kmolmkPa (8.314

)m kPa)(381.7 (2003

3

TR

PN

u

V

Then the mass of He can be determined from

kg 123 kg/kmol) kmol)(4.0 (30.61NMm

He

D = 9 m

27C

200 kPa

Pg = 500 kPa

O2

V = 2.5 m3

T = 28C



3-34

3-73 Problem 3-72 is to be reconsidered. The effect of the balloon diameter on the mass of helium contained in the

balloon is to be determined for the pressures of (a) 100 kPa and (b) 200 kPa as the diameter varies from 5 m to 15 m. The

mass of helium is to be plotted against the diameter for both cases.


"Given Data" {D=9 [m]} T=27 [C] P=200 [kPa] R_u=8.314 [kJ/kmol-K] "Solution" P*V=N*R_u*(T+273) V=4*pi*(D/2)^3/3 m=N*MOLARMASS(Helium)

D [m] m [kg]

5 6.111 7.222 8.333 9.444 10.56 11.67 12.78 13.89

15

21.01 38.35 63.31 97.25 141.6 197.6 266.9 350.6 450.2 567.2

P=200 kPa

5 7 9 11 13 150

100

200

300

400

500

600

D [m]

m

[kg

]

P=200 kPa

P=100 kPa



3-35

3-74E An automobile tire is under inflated with air. The amount of air that needs to be added to the tire to raise its pressure

to the recommended value is to be determined.

Assumptions 1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tire remains constant.

Properties The gas constant of air is R = 0.3704 psia.ft3/lbm.R (Table A-1E).

Analysis The initial and final absolute pressures in the tire are

P1 = Pg1 + Patm = 20 + 14.6 = 34.6 psia

P2 = Pg2 + Patm = 30 + 14.6 = 44.6 psia

Treating air as an ideal gas, the initial mass in the tire is

lbm 0.0900R) R)(550/lbmftpsia (0.3704

)ft psia)(0.53 (34.63

3

1

11

RT

Pm

V

Noting that the temperature and the volume of the tire remain constant, the final mass in the tire becomes

lbm 0.1160R) R)(550/lbmftpsia (0.3704

)ft psia)(0.53 (44.63

3

2

22

RT

Pm

V

Thus the amount of air that needs to be added is

m m m 2 1 0.1160 0.0900 0.0260 lbm

3-75 Two rigid tanks connected by a valve to each other contain air at specified conditions. The volume of the second tank

and the final equilibrium pressure when the valve is opened are to be determined.

Assumptions At specified conditions, air behaves as an ideal gas.


Analysis Let's call the first and the second tanks A and B. Treating air as an ideal gas, the volume of the second tank and the

mass of air in the first tank are determined to be

kg 309.4K) K)(283/kgmkPa (0.287

)m kPa)(1.0 (350

kPa 200

K) K)(308/kgmkPa kg)(0.287 (3

3

3

1

1

3

1

11

A

A

B

B

RT

Pm

P

RTm

V

V 3m 1.326

Thus,

kg 309.73309.4

m 326.2326.11.0 3

BA

BA

mmm

VVV

Then the final equilibrium pressure becomes

kPa 264

3

32

2m 2.326

K) K)(293/kgmkPa kg)(0.287 (7.309

V

mRTP

Air

V = 1 m3

T = 10C

P = 350 kPa

Air

m = 3 kg

T = 35C

P = 200 kPa

A B

Tire

0.53 ft3

90F

20 psig



3-36

3-76 One side of a two-sided tank contains an ideal gas while the other side is evacuated. The partition is removed and the

gas fills the entire tank. The gas is also heated to a final pressure. The final temperature is to be determined.

Assumptions The gas is specified as an ideal gas so that ideal gas relation can be used.

Analysis According to the ideal gas equation of state,

1112

12

32 VVVV

PP

Applying these,

C3327

K 3600K )273927333

11

11

1

212

2

2

1

1

2

22

1

11

11

TTTT

TT

T

P

T

P

mm

V

V

V

V

VV

VV

3-77 A piston-cylinder device containing argon undergoes an isothermal process. The final pressure is to be determined.

Assumptions At specified conditions, argon behaves as an ideal gas.

Properties The gas constant of argon is R = 0.2081 kJ/kgK (Table A-1).

Analysis Since the temperature remains constant, the ideal gas equation gives

22112211 VV

VVPP

RT

P

RT

Pm

which when solved for final pressure becomes

kPa 275 )kPa 550(5.05.02

11

11

2

112 PPPP

V

V

V

V

3-78E A rigid tank contains slightly pressurized air. The amount of air that needs to be added to the tank to raise its

pressure and temperature to the recommended values is to be determined.

Assumptions 1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tank remains constant.

Properties The gas constant of air is R = 0.3704 psia.ft3/lbm.R (Table A-1E).

Analysis Treating air as an ideal gas, the initial volume and the final mass in the tank are determined to be

lbm 33.73R) R)(550/lbmftpsia (0.3704

)ft 3psia)(196. (35

ft 196.3psia 20

R) R)(530/lbmftpsia 4lbm)(0.370 (20

3

3

2

22

33

1

11

RT

Pm

P

RTm

V

V

Thus the amount of air added is

m m m 2 1 33.73 20.0 13.73 lbm

Argon

1.5 kg

0.04 m3

550 kPa

100 g

Q

Ideal gas

927°C

V1

Evacuated

2V1

Air, 20 lbm

20 psia

70F



3-37

Compressibility Factor

3-79C All gases have the same compressibility factor Z at the same reduced temperature and pressure.

3-80C Reduced pressure is the pressure normalized with respect to the critical pressure; and reduced temperature is the

temperature normalized with respect to the critical temperature.

3-81E The temperature of R-134a is to be determined using the ideal gas relation, the compressibility chart, and the R-134a

tables.

Properties The gas constant, the critical pressure, and the critical temperature of refrigerant-134a are, from Table A-1E,

R = 0.10517 psia·ft3/lbm·R, Tcr = 673.6 R, Pcr = 588.7 psia

Analysis (a) From the ideal gas equation of state,

R 526

R)/lbmftpsia (0.10517

/lbm)ft 84psia)(0.13 (4003

3

R

PT

v

(b) From the compressibility chart (Fig. A-15a),

03.1

1.15R) R)(673.6/lbmftpsia (0.10517

psia) 7/lbm)(588.ft (0.1384

/

0.679psia 588.7

psia 400

3

3actual

R

crcrR

crR

T

PRT

P

PP

vv

Thus,

R 694 6.67303.1crRTTT

(c) From the superheated refrigerant table (Table A-13E),

R) (700 F240/lbmft 0.1384

psia 4003

T

P

v



3-38

3-82 The specific volume of steam is to be determined using the ideal gas relation, the compressibility chart, and the steam

tables. The errors involved in the first two approaches are also to be determined.

Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1,

R = 0.4615 kPa·m3/kg·K, Tcr = 647.1 K, Pcr = 22.06 MPa


error) (67.0%/kgm 0.01917 3

kPa 15,000

K) K)(623.15/kgmkPa (0.4615 3

P

RTv

(b) From the compressibility chart (Fig. A-15),

65.0

1.04K 647.1

K 673

0.453MPa 22.06

MPa 10

Z

T

TT

P

PP

crR

crR

Thus,

error) (8.5%/kgm 0.01246 3 /kg)m 1917(0.65)(0.0 3idealvv Z

(c) From the superheated steam table (Table A-6),

/kgm 0.011483

v

C035MPa 15

TP

H2O

15 MPa

350C



3-39

3-83 Problem 3-82 is reconsidered. The problem is to be solved using the general compressibility factor feature of

EES (or other) software. The specific volume of water for the three cases at 15 MPa over the temperature range of 350°C to

600°C in 25°C intervals is to be compared, and the % error involved in the ideal gas approximation is to be plotted against

temperature.


P=15 [MPa]*Convert(MPa,kPa) {T_Celsius= 350 [C]} T=T_Celsius+273 "[K]" T_critical=T_CRIT(Steam_iapws) P_critical=P_CRIT(Steam_iapws) {v=Vol/m} P_table=P; P_comp=P;P_idealgas=P T_table=T; T_comp=T;T_idealgas=T v_table=volume(Steam_iapws,P=P_table,T=T_table) "EES data for steam as a real gas" {P_table=pressure(Steam_iapws, T=T_table,v=v)} {T_sat=temperature(Steam_iapws,P=P_table,v=v)} MM=MOLARMASS(water) R_u=8.314 [kJ/kmol-K] "Universal gas constant" R=R_u/MM "[kJ/kg-K], Particular gas constant" P_idealgas*v_idealgas=R*T_idealgas "Ideal gas equation" z = COMPRESS(T_comp/T_critical,P_comp/P_critical) P_comp*v_comp=z*R*T_comp "generalized Compressibility factor" Error_idealgas=Abs(v_table-v_idealgas)/v_table*Convert(, %) Error_comp=Abs(v_table-v_comp)/v_table*Convert(, %)

Errorcomp [%] Errorideal gas [%] TCelcius [C]

9.447 2.725

0.4344 0.5995 1.101 1.337 1.428 1.437 1.397 1.329 1.245

67.22 43.53 32.21 25.23 20.44 16.92 14.22 12.1

10.39 8.976 7.802

350 375 400 425 450 475 500 525 550 575 600

300 350 400 450 500 550 6000

10

20

30

40

50

60

70

TCelsius [C]

Pe

rce

nt

Err

or

[%

]

Ideal GasIdeal Gas

Compressibility FactorCompressibility Factor

Steam at 15 MPa

Sp

ecif

ic V

olu

me



3-40

3-84 The specific volume of steam is to be determined using the ideal gas relation, the compressibility chart, and the steam

tables. The errors involved in the first two approaches are also to be determined.




error) (3.7%/kgm 0.09533 3

kPa 3500

K) K)(723/kgmkPa (0.4615 3

P

RTv


961.0

.121K 647.1

K 723

0.159MPa 22.06

MPa 3.5

Z

T

TT

P

PP

crR

crR

Thus,

error) (0.4%/kgm 0.09161 3 /kg)m 09533(0.961)(0. 3idealvv Z

(c) From the superheated steam table (Table A-6),

/kgm 0.09196 3

v C450

MPa .53TP

3-85 Somebody claims that oxygen gas at a specified state can be treated as an ideal gas with an error less than 10%. The

validity of this claim is to be determined.

Properties The critical pressure, and the critical temperature of oxygen are, from Table A-1,

T Pcr cr 154.8 K and 5.08 MPa

Analysis From the compressibility chart (Fig. A-15),

79.0

1.034K 154.8

K 160

0.591MPa 5.08

MPa 3

Z

T

TT

P

PP

crR

crR

Then the error involved can be determined from

%..Z

626790

11

11Error ideal

v

vv

Thus the claim is false.

H2O

3.5 MPa

450C

O2

3 MPa

160 K



3-41

3-86E Ethane in a rigid vessel is heated. The final pressure is to be determined using the compressibility chart.

Properties The gas constant, the critical pressure, and the critical temperature of ethane are, from Table A-1E,

R = 0.3574 psia·ft3/lbm·R, Tcr = 549.8 R, Pcr = 708 psia

Analysis From the compressibility chart at the initial state (from Fig. A-15 or EES. We used EES throughout the solution.),

977.0 Z

0706.0psia 708

psia 50

1.019R 549.8

R 560

1

cr

11

cr

11

P

PP

T

TT

R

R

The specific volume does not change during the process. Then,

/lbmft 91.3psia 50

R) R)(560/lbmftpsia 3574(0.977)(0. 33

1

1121

P

RTZvv

At the final state,

0.1

1.14psia) R)/(708 R)(549.8/lbmftpsia (0.3574

/lbmft3.911

1.819R 549.8

R 1000

2

3

3

crcr

actual2,2

cr

22

Z

/PRT

T

TT

R

R

vv

Thus,

psia 91.4

/lbmft 3.91

R) R)(1000/lbmftpsia 74(1.0)(0.353

3

2

222 v

RTZP

Ethane

50 psia

100F

Q



3-42

3-87 Ethylene is heated at constant pressure. The specific volume change of ethylene is to be determined using the

compressibility chart.

Properties The gas constant, the critical pressure, and the critical temperature of ethane are, from Table A-1,


Analysis From the compressibility chart at the initial and final states (Fig. A-15),

56.0 Z

977.0MPa 5.12

MPa 5

1.038K 282.4

K 293

1

cr

11

cr

11

P

PP

T

TT

R

R

961.0 Z

977.0

1.675KR 282.4

K 473

1

12

cr

22

RR

R

PP

T

TT

The specific volume change is

/kgm 0.0172 3

K) 293)(56.0(K) 473)(961.0(kPa 5000

K/kgmkPa 0.2964

)(

3

1122 TZTZP

Rv

3-88 The % error involved in treating CO2 at a specified state as an ideal gas is to be determined.

Properties The critical pressure, and the critical temperature of CO2 are, from Table A-1,

MPa 7.39andK 304.2 crcr PT

Analysis From the compressibility chart (Fig. A-15),

84.0

1.25K 304.2

K 380

0.947MPa 7.39

MPa 7

cr

cr

Z

T

TT

P

PP

R

R

Then the error involved in treating CO2 as an ideal gas is

19.0% or 0.1900.84

11

11Error ideal

Zv

vv

Ethylene

5 MPa

20C

Q

CO2

7 MPa

380 K



3-43

3-89 Water vapor is heated at constant pressure. The final temperature is to be determined using ideal gas equation, the

compressibility charts, and the steam tables.



Analysis (a) From the ideal gas equation,

K 1246 )2)(K 273350(1

212 v

vTT

(b) The pressure of the steam is

kPa 529,16Csat@35021 PPP

From the compressibility chart at the initial state (Fig. A-15),

75.0 ,593.0 Z

749.0MPa 22.06

MPa 16.529

963.0KR 647.1

K 623

11

cr

11

cr

11

R

R

R

P

PP

T

TT

v

At the final state,

88.0 50.1)75.0(22

749.02

12

12

Z

PP

RR

RR

vv

Thus,

K 826kPa 22,060

K) .1(1.50)(647

0.88

kPa 16,529

cr

cr2

2

2

2

222

P

T

Z

P

RZ

PT Rvv

(c) From the superheated steam table,

/kgm 008806.0 1

C350 31

1

1

v

x

T (Table A-4)

K 750

C477

/kgm 01761.02

kPa 529,1623

12

2T

P

vv (from Table A-6 or EES)

Water

350C

sat. vapor

Q



3-44

3-90 Methane is heated at constant pressure. The final temperature is to be determined using ideal gas equation and the

compressibility charts.

Properties The gas constant, the critical pressure, and the critical temperature of methane are, from Table A-1,


Analysis From the ideal gas equation,

K 540 )8.1)(K 300(1

212 v

vTT


63.0 ,86.0 Z

16.2MPa 4.64

MPa 10

57.1K 191.1

K 300

11

cr

11

cr

11

R

R

R

P

PP

T

TT

v

At the final state,

42.0 134.1)63.0(8.18.1

16.22

12

12

Z

PP

RR

RR

vv (Fig. A-15)

Thus,

K 1112kPa 4640

K) 1.1(1.134)(19

0.42

kPa 10,000

cr

cr2

2

2

2

222

P

T

Z

P

RZ

PT Rvv

Of these two results, the accuracy of the second result is limited by the accuracy with which the charts may be read.

Accepting the error associated with reading charts, the second temperature is the more accurate.

Methane

10 MPa

300 K

Q



3-45

3-91 CO2 gas flows through a pipe. The volume flow rate and the density at the inlet and the volume flow rate at the exit of

the pipe are to be determined.

Properties The gas constant, the critical pressure, and the critical temperature of CO2 are (Table A-1)


Analysis

(a) From the ideal gas equation of state,

error) (2.1%/kgm 0.06297 3

kPa) (3000

K) K)(500/kgmkPa 89kg/s)(0.18 (2 3

1

11

P

RTm&&V

error) (2.1%mkg/ 31.76 3

K) K)(500/kgmkPa (0.1889

kPa) (30003

1

11

RT

P

error) (3.6%/kgm 0.05667 3

kPa) (3000

K) K)(450/kgmkPa 89kg/s)(0.18 (2 3

2

22

P

RTm&&V

(b) From the compressibility chart (EES function for compressibility factor is used)

9791.0

1.64K 304.2

K 500

0.407MPa 7.39

MPa 3

11

1,

1

Z

T

TT

P

PP

crR

crR

9656.0

1.48K 304.2

K 450

0.407MPa 7.39

MPa 3

22

2,

2

Z

T

TT

P

PP

crR

crR

Thus, /kgm 0.06165 3

kPa) (3000

K) K)(500/kgmkPa 89kg/s)(0.18 (0.9791)(2 3

1

111

P

RTmZ &&V

3mkg/ 32.44

K) K)(500/kgmkPa .1889(0.9791)(0

kPa) (30003

11

11

RTZ

P

/kgm 0.05472 3

kPa) (3000

K) K)(450/kgmkPa 89kg/s)(0.18 (0.9656)(2 3

2

222

P

RTmZ &&V

3 MPa

500 K

2 kg/s

CO2 450 K



3-46

3-92 The pressure of R-134a is to be determined using the ideal gas relation, the compressibility chart, and the R-134a

tables.

Properties The gas constant, the critical pressure, and the critical temperature of refrigerant-134a are, from Table A-1,


Analysis The specific volume of the refrigerant is

/kgm 0.016773kg 1

m 0.016773 33

m

Vv


kPa 1861

/kgm 0.016773

K) K)(383/kgmkPa (0.081493

3

vRT

P


39.0

2.24kPa) K)/(4059 K)(374.2/kgmkPa (0.08149

/kgm 0.016773

1.023K 374.2

K 383

3

3

crcr

actual

cr

R

R

R

P

/PRT

T

TT

vv

Thus,

kPa 1583 kPa) 4059(0.39)(crPPP R

(c) From the superheated refrigerant table (Table A-13),

kPa 1600

P

T

/kgm 0.016773

C1103v

R-134a

0.016773 m3/kg

110C



3-47

Other Equations of State

3-93C The constant a represents the increase in pressure as a result of intermolecular forces; the constant b represents the

volume occupied by the molecules. They are determined from the requirement that the critical isotherm has an inflection

point at the critical point.

3-94 The pressure of nitrogen in a tank at a specified state is to be determined using the ideal gas, van der Waals, and

Beattie-Bridgeman equations. The error involved in each case is to be determined.

Properties The gas constant, molar mass, critical pressure, and critical temperature of nitrogen are (Table A-1)

R = 0.2968 kPa·m3/kg·K, M = 28.013 kg/kmol, Tcr = 126.2 K, Pcr = 3.39 MPa

Analysis The specific volume of nitrogen is

/kgm 0.0327kg 100

m 3.27 33

m

Vv


error) (5.5%kPa 1588

/kgm 0.0327

K) K)(175/kgmkPa (0.2968

3

3

vRT

P

(b) The van der Waals constants for nitrogen are determined from

aR T

P

bRT

P

cr

cr

cr

cr

27

64

8

2 2(27)(0.2968 kPa m / kg K) (126.2 K)

(64)(3390 kPa)0.175 m kPa / kg

(0.2968 kPa m / kg K)(126.2 K)

8 3390 kPa0.00138 m / kg

3 2 26 2

33

Then,

error) (0.7%kPa 1495

22 (0.0327)

0.175

0.001380.0327

1750.2968

vva

b

RTP

(c) The constants in the Beattie-Bridgeman equation are

/kmolKm104.2

0.050840.9160

0.0069110.050461

132.3390.9160

0.026171136.23151

334

c

bBB

aAA

o

o

v

v

since /kmolm 0.9160/kg)m .0327kg/kmol)(0 (28.013 33 vv M . Substituting,

error) (0.07%kPa 1504

23

4

2

232

0.9160

132.3390.050840.9160

1750.9160

104.21

(0.9160)

5178.314

1v

vvv

AB

T

cTRP u

N2

0.0327 m3/kg

175 K



3-48

3-95 Methane is heated in a rigid container. The final pressure of the methane is to be determined using the ideal gas

equation and the Benedict-Webb-Rubin equation of state.


kPa 156.5K 293

K 573kPa) 80(

1

212

T

TPP

(b) The specific molar volume of the methane is

/kmolm 45.30kPa 80

K) K)(293/kmolmkPa (8.314 33

1

121

P

TRuvv

Using the coefficients of Table 3-4 for methane and the given data, the Benedict-Webb-Rubin equation of state for state 2

gives

kPa 156.5

)30.45/0060.0exp(30.45

0060.01

)573(30.45

10578.2

30.45

10244.100.5

30.45

00.5573314.8003380.0

30.45

1

573

10286.291.187573314.804260.0

30.45

)573)(314.8(

)/exp(11

2

223

5

6

4

322

6

2

222

3632

222

0020

2

22 v

vvvvvv

T

caaTbR

T

CATRB

TRP u

uu

Methane

80 kPa

20C

Q



3-49

3-96E The temperature of R-134a in a tank at a specified state is to be determined using the ideal gas relation, the van der

Waals equation, and the refrigerant tables.

Properties The gas constant, critical pressure, and critical temperature of R-134a are (Table A-1E)

R = 0.1052 psia·ft3/lbm·R, Tcr = 673.6 R, Pcr = 588.7 psia


R 435

R/lbmftpsia 0.1052

/lbm)ft 44psia)(0.11 (4003

3

R

PT

v

(b) The van der Waals constants for the refrigerant are determined from

/lbmft 0.01504psia 588.78

R) R)(673.6/lbmftpsia (0.1052

8

psia/lbmft 3.596psia) (64)(588.7

R) (673.6R)/lbmftpsia 2(27)(0.105

64

27

33

2622322

cr

cr

cr

cr

P

RTb

P

TRa

Then,

R 638

0.015040.1144

(0.3479)

3.591400

0.1052

1122

ba

PR

T vv

(c) From the superheated refrigerant table (Table A-13E),

)( F100/lbmft 0.1144

psia 0403 R 660

T

P

v



3-50

3-97 The pressure of nitrogen in a tank at a specified state is to be determined using the ideal gas relation and the

Beattie-Bridgeman equation. The error involved in each case is to be determined.

Properties The gas constant and molar mass of nitrogen are (Table A-1)

R = 0.2968 kPa·m3/kg·K and M = 28.013 kg/kmol


)error .3%6( /kgm 0.041884

K) K)(150/kgmkPa (0.29683

3

kPa1063

vRT

P

(b) The constants in the Beattie-Bridgeman equation are

/kmolKm104.2

0.050761.1733

0.0069110.050461

133.1931.1733

0.026171136.23151

334

c

bBB

aAA

o

o

v

v

since

/kmolm 1.1733/kg)m .041884kg/kmol)(0 (28.013 33 vv M .

Substituting,

error) e(negligibl

1.1733

133.1930.050761.1733

1501.1733

104.21

(1.1733)

1508.3141

23

4

2232

kPa 1000.4

vv

vv

AB

T

cTRP u

N2

0.041884 m3/kg

150 K



3-51

3-98 Problem 3-97 is reconsidered. Using EES (or other) software, the pressure results of the ideal gas and Beattie-

Bridgeman equations with nitrogen data supplied by EES are to be compared. The temperature is to be plotted versus

specific volume for a pressure of 1000 kPa with respect to the saturated liquid and saturated vapor lines of nitrogen over the

range of 110 K < T < 150 K.


Function BeattBridg(T,v,M,R_u) v_bar=v*M "Conversion from m^3/kg to m^3/kmol" "The constants for the Beattie-Bridgeman equation of state are found in text" Ao=136.2315; aa=0.02617; Bo=0.05046; bb=-0.00691; cc=4.20*1E4 B=Bo*(1-bb/v_bar) A=Ao*(1-aa/v_bar) "The Beattie-Bridgeman equation of state is" BeattBridg:=R_u*T/(v_bar**2)*(1-cc/(v_bar*T**3))*(v_bar+B)-A/v_bar**2 End T=150 [K] v=0.041884 [m^3/kg] P_exper=1000 [kPa] T_table=T; T_BB=T;T_idealgas=T P_table=PRESSURE(Nitrogen,T=T_table,v=v) "EES data for nitrogen as a real gas" {T_table=temperature(Nitrogen, P=P_table,v=v)} M=MOLARMASS(Nitrogen) R_u=8.314 [kJ/kmol-K] "Universal gas constant" R=R_u/M "Particular gas constant" P_idealgas=R*T_idealgas/v "Ideal gas equation" P_BB=BeattBridg(T_BB,v,M,R_u) "Beattie-Bridgeman equation of state Function"

PBB [kPa] Ptable [kPa] Pidealgas [kPa] v [m3/kg] TBB [K] Tideal gas [K] Ttable [K]

1000 1000 1000 1000 1000 1000 1000

1000 1000 1000 1000 1000 1000 1000

1000 1000 1000 1000 1000 1000 1000

0.01 0.02 0.025 0.03 0.035 0.04 0.05

91.23 95.52 105 116.8 130.1 144.4 174.6

33.69 67.39 84.23 101.1 117.9 134.8 168.5

103.8 103.8 106.1 117.2 130.1 144.3 174.5

10-3 10-2 10-170

80

90

100

110

120

130

140

150

160

v [m3/kg]

T [

K]

1000 kPa

Nitrogen, T vs v for P=1000 kPa

EES Table ValueEES Table Value

Beattie-BridgemanBeattie-Bridgeman

Ideal GasIdeal Gas



3-52

3-99 Carbon dioxide is compressed in a piston-cylinder device in a polytropic process. The final temperature is to be

determined using the ideal gas and van der Waals equations.

Properties The gas constant, molar mass, critical pressure, and critical temperature of carbon dioxide are (Table A-1)

R = 0.1889 kPa·m3/kg·K, M = 44.01 kg/kmol, Tcr = 304.2 K, Pcr = 7.39 MPa

Analysis (a) The specific volume at the initial state is

/kgm 0.08935kPa 1000

K) K)(473/kgmkPa (0.1889 33

1

11

P

RTv

According to process specification,

/kgm 03577.0kPa 3000

kPa 1000/kg)m 08935.0( 3

2.1/13

/1

2

112

n

P

Pvv

The final temperature is then

K 568

K/kgmkPa 0.1889

/kg)m 77kPa)(0.035 (30003

322

2R

PT

v

(b) The van der Waals constants for carbon dioxide are determined from

/kgm 0.0009720kPa 73908

K) K)(304.2/kgmkPa (0.1889

8

kPa/kgm 0.1885kPa) (64)(7390

K) (304.2K)/kgmkPa 9(27)(0.188

64

27

33

cr

cr

26223

cr

2cr

2

P

RTb

P

TRa

Applying the van der Waals equation to the initial state,

)473)(1889.0()0009720.0(1885.0

1000

)(

2

2

vv

vv

RTba

P

Solving this equation by trial-error or by EES gives

/kgm 0.08821 31 v

According to process specification,

/kgm 03531.0kPa 3000

kPa 1000/kg)m 08821.0( 3

2.1/13

/1

2

112

n

P

Pvv

Applying the van der Waals equation to the final state,

T..

RTba

P

)1889.0()0009720.0035310(035310

1885.03000

)(

2

2

v

v

Solving for the final temperature gives

K 5732T

CO2

1 MPa

200C



3-53

3-100 The temperature of steam in a tank at a specified state is to be determined using the ideal gas relation, van der Waals

equation, and the steam tables.

Properties The gas constant, critical pressure, and critical temperature of steam are (Table A-1)


Analysis The specific volume of steam is

/kgm 0.3520kg 2.841

m 1 33

m

Vv


K457.6 K/kgmkPa 0.4615

/kg)m kPa)(0.352 (6003

3

R

PT

v

(b) The van der Waals constants for steam are determined from

/kgm 0.00169kPa 22,0608

K) K)(647.1/kgmkPa (0.4615

8

kPa/kgm 1.705kPa) 0(64)(22,06

K) (647.1K)/kgmkPa 5(27)(0.461

64

27

33

2622322

cr

cr

cr

cr

P

RTb

P

TRa

Then,

K 465.9

0.001690.352

(0.3520)

1.705600

0.4615

1122

ba

PR

T vv

(c) From the superheated steam table (Tables A-6),

K) 473( /kgm 00.352

MPa 0.63

C200T

P

v

H2O

1 m3

2.841 kg

0.6 MPa



3-54

3-101 Problem 3-100 is reconsidered. The problem is to be solved using EES (or other) software. The temperature of

water is to be compared for the three cases at constant specific volume over the pressure range of 0.1 MPa to 1 MPa in 0.1

MPa increments. The %error involved in the ideal gas approximation is to be plotted against pressure.


Function vanderWaals(T,v,M,R_u,T_cr,P_cr) v_bar=v*M "Conversion from m^3/kg to m^3/kmol" "The constants for the van der Waals equation of state are given by equation 3-24" a=27*R_u^2*T_cr^2/(64*P_cr) b=R_u*T_cr/(8*P_cr) "The van der Waals equation of state gives the pressure as" vanderWaals:=R_u*T/(v_bar-b)-a/v_bar**2 End m=2.841[kg] Vol=1 [m^3] {P=6*convert(MPa,kPa)} T_cr=T_CRIT(Steam_iapws) P_cr=P_CRIT(Steam_iapws) v=Vol/m P_table=P; P_vdW=P;P_idealgas=P T_table=temperature(Steam_iapws,P=P_table,v=v) "EES data for steam as a real gas" {P_table=pressure(Steam_iapws, T=T_table,v=v)} {T_sat=temperature(Steam_iapws,P=P_table,v=v)} MM=MOLARMASS(water) R_u=8.314 [kJ/kmol-K] "Universal gas constant" R=R_u/MM "Particular gas constant" P_idealgas=R*T_idealgas/v "Ideal gas equation" "The value of P_vdW is found from van der Waals equation of state Function" P_vdW=vanderWaals(T_vdW,v,MM,R_u,T_cr,P_cr) Error_idealgas=Abs(T_table-T_idealgas)/T_table*Convert(, %) Error_vdW=Abs(T_table-T_vdW)/T_table*Convert(, %)

P [kPa] Tideal gas [K] Ttable [K] TvdW [K] Errorideal gas [K]

100 76.27 372.8 86.35 79.54

200 152.5 393.4 162.3 61.22

300 228.8 406.7 238.2 43.74

400 305.1 416.8 314.1 26.8

500 381.4 425 390 10.27

600 457.6 473 465.9 3.249

700 533.9 545.3 541.8 2.087

800 610.2 619.1 617.7 1.442

900 686.4 693.7 693.6 1.041

1000 762.7 768.6 769.5 0.7725



3-55

100 200 300 400 500 600 700 800 900 1000

0

10

20

30

40

50

60

70

80

90

100

P [kPa]

Err

or v

dW

[%

]van der Waals

Ideal gas

10-3 10-2 10-1 100 101 102 103200

300

400

500

600

700

800

900

1000

v [m3/kg]

T [

K]

600 kPa

0.05 0.1 0.2 0.5

T vs. v for Steam at 600 kPa

Steam Table

Ideal Gas

van der Waals



3-56

10-3 10-2 10-1 100 101 102 103200

300

400

500

600

700

800

900

1000

v [m3/kg]

T [

K]

6000 kPa

T vs v for Steam at 6000 kPa

Steam Table

Ideal Gas

van der Waals

100 200 300 400 500 600 700 800 900 1000

50

100

150

200

250

300

350

400

450

500

550

600

650

700

750

800

P [kPa]

Tta

ble

[K

]

Steam tablevan der WaalsIdeal gas



3-57

Special Topic: Vapor Pressure and Phase Equilibrium

3-102 The vapor pressure in the air at the beach when the air temperature is 30C is claimed to be 5.2 kPa. The validity of

this claim is to be evaluated.

Properties The saturation pressure of water at 30C is 4.247 kPa (Table A-4).

Analysis The maximum vapor pressure in the air is the saturation pressure of water

at the given temperature, which is

kPa 4.247 Csat@30@satmax , airPPP Tv

which is less than the claimed value of 5.2 kPa. Therefore, the claim is false.

3-103 A glass of water is left in a room. The vapor pressures at the free surface of the water and in the room far from the

glass are to be determined.

Assumptions The water in the glass is at a uniform temperature.

Properties The saturation pressure of water is 2.339 kPa at 20C, and 1.706 kPa at 15C (Table A-4).

Analysis The vapor pressure at the water surface is the saturation pressure of water at the

water temperature,

kPa 1.706 Csat@15@satsurface water , waterPPP Tv

Noting that the air in the room is not saturated, the vapor pressure in the room far from the

glass is

kPa 0.936 kPa) 339.2)(4.0(Csat@20@satair , airPPP Tv

3-104 The temperature and relative humidity of air over a swimming pool are given. The water temperature of the swimming

pool when phase equilibrium conditions are established is to be determined.

Assumptions The temperature and relative humidity of air over the pool remain constant.


Analysis The vapor pressure of air over the swimming pool is

kPa902.1kPa) 170.3)(6.0(Csat@25@satair , air PPP Tv

Phase equilibrium will be established when the vapor pressure at the water surface

equals the vapor pressure of air far from the surface. Therefore,

kPa902.1air ,surface water , vv PP

and C16.7 kPa 902.1@sat@satwater vTTT P

Discussion Note that the water temperature drops to 16.7C in an environment at 25C when phase equilibrium is

established.

H2O

15C

Patm, 25C

POOL

30C

WATER



3-58

3-105 A person buys a supposedly cold drink in a hot and humid summer day, yet no condensation occurs on the drink. The

claim that the temperature of the drink is below 10C is to be evaluated.


Analysis The vapor pressure of air is

kPa940.3kPa) 629.5)(7.0(Csat@35@satair , air PPP Tv

The saturation temperature corresponding to this pressure (called the dew-point

temperature) is

C28.7 kPa [email protected]@satsat vTTT P

That is, the vapor in the air will condense at temperatures below 28.7C. Noting that no condensation is observed on the can,

the claim that the drink is at 10C is false.

3-106 Two rooms are identical except that they are maintained at different temperatures and relative humidities. The room

that contains more moisture is to be determined.

Properties The saturation pressure of water is 2.339 kPa at 20C, and 3.17 kPa at 25C (Table A-4).

Analysis The vapor pressures in the two rooms are

Room 1: kPa 1.27 kPa) 17.3)(4.0(Csat@251@sat11 1PPP Tv

Room 2: kPa 1.29 kPa) 339.2)(55.0(Csat@202@sat22 2PPP Tv

Therefore, room 1 at 30C and 40% relative humidity contains more moisture.

3-107E A thermos bottle half-filled with water is left open to air in a room at a specified temperature and pressure. The

temperature of water when phase equilibrium is established is to be determined.

Assumptions The temperature and relative humidity of air over the bottle remain constant.

Properties The saturation pressure of water at 60F is 0.2564 psia (Table A-4E).

Analysis The vapor pressure of air in the room is

psia 0.08973psia) 2564.0)(35.0(Fsat@70@satair , air PPP Tv

Phase equilibrium will be established when the vapor pressure at the water surface

equals the vapor pressure of air far from the surface. Therefore,

psia0.08973air ,surface water , vv PP

and

F32.3 psia 08973.0@sat@satwater vTTT P

Discussion Note that the water temperature drops to 32.3F in an environment at 60F when phase equilibrium is

established.

Thermos

bottle

60F

35%

35C

70%



3-59

Review Problems

3-108E Water in a pressure cooker boils at 260F. The absolute pressure in the pressure cooker is to be determined.

Analysis The absolute pressure in the pressure cooker is the saturation

pressure that corresponds to the boiling temperature,

psia 35.45F260@sat PP

3-109 Carbon dioxide flows through a pipe at a given state. The volume and mass flow rates and the density of CO2 at the

given state and the volume flow rate at the exit of the pipe are to be determined.

Analysis

(a) The volume and mass flow rates may be determined from ideal gas relation as

/sm 0.5543 3kPa 3000

K) 00/kmol.K)(5kPa.m 314kmol/s)(8. 4.0( 31

1P

TRN u&

&V

kg/s 17.60K) /kg.K)(500kPa.m (0.1889

)/m 3kPa)(0.554 (30003

3

1

111

s

RT

Pm

V&&

The density is

3kg/m 31.76/s)m 5543.0(

kg/s) (17.603

1

11

V&&m

(b) The volume flow rate at the exit is

/sm 0.4988 3kPa 3000

K) 50/kmol.K)(4kPa.m 314kmol/s)(8. 4.0( 32

2P

TRN u&

&V

3 MPa

500 K

0.4 kmol/s

CO2 450 K

H2O

260F



3-60

3-110 A tank contains argon at a specified state. Heat is transferred from argon until it reaches a specified temperature. The

final gage pressure of the argon is to be determined.

Assumptions 1 Argon is an ideal gas.

Properties The local atmospheric pressure is given to be 100 kPa.

Analysis Noting that the specific volume of argon in the tank remains constant,

from ideal gas relation, we have

kPa 196.9K)273600(

K)273300(kPa) 100200(

1

212

T

TPP

Then the gage pressure becomes

kPa 96.9 1009.196atm2gage,2 PPP

3-111 The cylinder conditions before the heat addition process is specified. The pressure after the heat addition process is

to be determined.

Assumptions 1 The contents of cylinder are approximated by the air properties.

2 Air is an ideal gas.

Analysis The final pressure may be determined from the ideal gas relation

kPa 3358

kPa) 1200(

K 273450

K 27317501

1

22 P

T

TP

3-112 A rigid container that is filled with R-13a is heated. The initial pressure and the final temperature are to be

determined.

Analysis The initial specific volume is 0.090 m3/kg. Using this with

the initial temperature reveals that the initial state is a mixture. The

initial pressure is then the saturation pressure,

11)-A (Table /kgm 090.0

C40C40- @sat 13

1

1 kPa 51.25

PP

T

v

This is a constant volume cooling process (v = V /m = constant). The final

state is superheated vapor and the final temperature is then

13)-A (Table /kgm 090.0

kPa 28023

12

2 C50

T

P

vv

R-134a

-40°C

1 kg

0.090 m3

P

v

1

2

Combustion

chamber

1.2 MPa

450C

Argon

600ºC

200 kPa gage

Q



3-61

3-113 The refrigerant in a rigid tank is allowed to cool. The pressure at which the refrigerant starts condensing is to be

determined, and the process is to be shown on a P-v diagram.

Analysis This is a constant volume process (v = V /m = constant), and the specific

volume is determined to be

/kgm 0.117kg 1

m 0.117 33

m

Vv

When the refrigerant starts condensing, the tank will contain saturated

vapor only. Thus,

/kgm 0.117 32 gvv

The pressure at this point is the pressure that corresponds to this vg value,

kPa 170 /kgm 0.117@sat2 3

gvPP

3-114E A piston-cylinder device that is filled with water is cooled. The final pressure and volume of the water are to be

determined.


/lbmft 649.2lbm 1

ft 649.2 33

11

m

Vv

This is a constant-pressure process. The initial state is determined to be

superheated vapor and thus the pressure is determined to be


F400213

1

1 psia 180

PP

T

v

The saturation temperature at 180 psia is 373.1°F. Since the final

temperature is less than this temperature, the final state is compressed

liquid. Using the incompressible liquid approximation,

4E)-A (Table /lbmft 01613.0 3F100 @ 2 fvv

The final volume is then

3ft 0.01613 /lbm)ft 01613.0)(lbm 1( 322 vV m

P

v

2 1

H2O

400°F

1 lbm

2.649 ft3

P

v

2

1

R-134a

240 kPa



3-62

3-115 Ethane is heated at constant pressure. The final temperature is to be determined using ideal gas equation and the

compressibility charts.

Properties The gas constant, the critical pressure, and the critical temperature of ethane are, from Table A-1,


Analysis From the ideal gas equation,

K 596.8 )6.1)(K 373(1

212 v

vTT


35.0 ,61.0 Z

232.2MPa 4.48

MPa 10

221.1K 305.5

K 373

11

cr

11

cr

11

R

R

R

P

PP

T

TT

v

At the final state,

83.0 56.0)35.0(6.16.1

232.22

12

12

Z

PP

RR

RR

vv

Thus,

K 460kPa 4480

K) .5(0.56)(305

0.83

kPa 10,000

cr

cr2

2

2

2

222

P

T

Z

P

RZ

PT Rvv

Of these two results, the accuracy of the second result is limited by the accuracy with which the charts may be read.

Accepting the error associated with reading charts, the second temperature is the more accurate.

3-116 A large tank contains nitrogen at a specified temperature and pressure. Now some nitrogen is allowed to escape, and

the temperature and pressure of nitrogen drop to new values. The amount of nitrogen that has escaped is to be determined.

Properties The gas constant for nitrogen is 0.2968 kPa·m3/kg·K (Table A-1).

Analysis Treating N2 as an ideal gas, the initial and the final masses in the tank are determined to be

kg 84.60K) K)(278/kgmkPa (0.2968

)m kPa)(13 (400

kg 62.90K) K)(290/kgmkPa(0.2968

)m kPa)(13 (600

3

3

2

22

3

3

1

11

RT

Pm

RT

Pm

V

V

Thus the amount of N2 that escaped is

kg 29.8 84.6062.9021 mmm

Ethane

10 MPa

100C

Q

N2

600 kPa

17C

13 m3



3-63

3-117 Superheated refrigerant-134a is cooled at constant pressure until it exists as a compressed liquid. The changes in total

volume and internal energy are to be determined, and the process is to be shown on a T-v diagram.

Analysis The refrigerant is a superheated vapor at the initial state and a compressed liquid at the final state. From Tables A-

13 and A-11,

/kgm 0.0008160

kJ/kg 78.85

C02

MPa .21

/kgm 0.019502

kJ/kg 277.23

C07

MPa .21

3

C20@2

C20@2

2

2

31

1

1

1

f

fuu

T

P

u

T

P

vv

v

Thus,

(b) 3m 0.187 /kgm 0.019502)160kg)(0.0008 (10)( 312 vvV m

(c) kJ 1984 kJ/kg 277.23)kg)(78.85 (10)( 12 uumU

3-118 The rigid tank contains saturated liquid-vapor mixture of water. The mixture is heated until it exists in a single phase.

For a given tank volume, it is to be determined if the final phase is a liquid or a vapor.

Analysis This is a constant volume process (v = V /m = constant), and thus the final specific volume will be equal to the

initial specific volume,

12 vv

The critical specific volume of water is 0.003106 m3/kg. Thus if the final specific

volume is smaller than this value, the water will exist as a liquid, otherwise as a vapor.

cr3

3

/kgm 0.002kg 2

m 0.0044 v

VvV

mL Thus, liquid.

./kgm 0.2kg 2

m 0.4400 cr

33

vV

vV m

L Thus, vapor.

H2O

V = 4 L

m = 2 kg

T = 50C

R-134a

70C

1.2 MPa

T

v

2

1



3-64

3-119 The pressure in an automobile tire increases during a trip while its volume remains constant. The percent increase in

the absolute temperature of the air in the tire is to be determined.

Assumptions 1 The volume of the tire remains constant. 2 Air is an ideal gas.

Properties The local atmospheric pressure is 90 kPa.

Analysis The absolute pressures in the tire before and after the trip are

P P P

P P P

1

2

200 90 290

220 90 310

gage,1 atm

gage,2 atm

kPa

kPa

Noting that air is an ideal gas and the volume is constant, the ratio of absolute

temperatures after and before the trip are

1.069=kPa 290

kPa 310=

1

2

1

2

2

22

1

11

P

P

T

T

T

P

T

P

VV

Therefore, the absolute temperature of air in the tire will increase by 6.9% during this trip.

3-120 A piston-cylinder device contains steam at a specified state. Steam is cooled at constant pressure. The volume change

is to be determined using compressibility factor.

Properties The gas constant, the critical pressure, and the critical temperature of steam are


Analysis The exact solution is given by the following:

/kgm 95986.0C150

kPa 200

/kgm 31623.1C300

kPa 200

32

2

31

1

v

v

T

P

T

P

(Table A-6)

3m 0.07128 /kg0.95986)m3kg)(1.3162 (0.2)( 321exact vvmV

Using compressibility chart (EES function for compressibility factor is used)

9956.0

.8860K 647.1

K 273300

0.0091MPa 22.06

MPa 0.2

11

1,

1

Z

T

TT

P

PP

crR

crR

9897.0

65.0K 647.1

K 273150

0.0091MPa 22.06

MPa 0.2

22

2,

2

Z

T

TT

P

PP

crR

crR

33

2

222

33

1

111

m 1932.0kPa) (200

K) 273K)(150/kgmkPa kg)(0.4615 .2(0.9897)(0

m 2633.0kPa) (200

K) 273K)(300/kgmkPa kg)(0.4615 .2(0.9956)(0

P

mRTZ

P

mRTZ

V

V

1.7%m 0.07006 3 :Error ,1932.02633.021chart VVV

Tire

200 kPa

0.035 m3

Steam

0.2 kg

200 kPa

300C

Q



3-65

3-121 The temperature of steam in a tank at a specified state is to be determined using the ideal gas relation, the

generalized chart, and the steam tables.


MPa 22.06 K, 647.1 ,K/kgmkPa 0.4615 crcr3 PTR


kPa 15,529

/kgm 0.02

K) K)(673/kgmkPa (0.46153

3

vRT

P

(b) From the compressibility chart (Fig. A-15a),

57.0

1.48K) K)(647.1/kgmkPa (0.4615

kPa) 0/kg)(22,06m (0.02

/

040.1K 647.1

K 673

3

3

crcr

actual

cr

R

R

R

P

PRT

T

TT

vv

Thus,

kPa 12,574 060,2257.0crPPP R

(c) From the superheated steam table,

kPa 12,515

PT

/kgm 0.02C400

3v (from EES)

3-122 One section of a tank is filled with saturated liquid R-134a while the other side is evacuated. The partition is

removed, and the temperature and pressure in the tank are measured. The volume of the tank is to be determined.

Analysis The mass of the refrigerant contained in the tank is

kg 96.34/kgm 0.0008580

m 0.033

3

1

1 v

Vm

since

/kgm 0.0008580 3MPa 9.0@1 fvv

At the final state (Table A-13),

/kgm 0.07997 C20

kPa 028 32

2

2

v

T

P

Thus,

3m 2.80 /kg)m 7kg)(0.0799 (34.96 322tank vVV m

R-134a

P =0.9 MPa

V =0.03 m3

Evacuated

H2O

0.02 m3/kg

400C



3-66

3-123 Problem 3-122 is reconsidered. The effect of the initial pressure of refrigerant-134 on the volume of the tank is

to be investigated as the initial pressure varies from 0.5 MPa to 1.5 MPa. The volume of the tank is to be plotted versus the

initial pressure, and the results are to be discussed.


"Given Data" x_1=0.0 Vol_1=0.03 [m^3] P_1=1200 [kPa] T_2=30 [C] P_2=400 [kPa] "Solution" v_1=volume(R134a,P=P_1,x=x_1) Vol_1=m*v_1 v_2=volume(R134a,P=P_2,T=T_2) Vol_2=m*v_2

P1 [kPa]

Vol2 [m

3]

500 600 700 800 900

1000 1100 1200 1300 1400 1500

2.977 2.926 2.88 2.837 2.796 2.757 2.721 2.685 2.651 2.617 2.584

500 700 900 1100 1300 15002.5

2.6

2.7

2.8

2.9

3

P1 [kPa]

Vo

l 2

[m3]



3-67

3-124 A propane tank contains 5 L of liquid propane at the ambient temperature. Now a leak develops at the top of the tank

and propane starts to leak out. The temperature of propane when the pressure drops to 1 atm and the amount of heat

transferred to the tank by the time the entire propane in the tank is vaporized are to be determined.

Properties The properties of propane at 1 atm are Tsat = -42.1C, 581 kg /m3 , and hfg = 427.8 kJ/kg (Table A-3).

Analysis The temperature of propane when the pressure drops

to 1 atm is simply the saturation pressure at that temperature,

T T sat @ atm1 42.1 C

The initial mass of liquid propane is

kg 2.905)m )(0.005kg/m (581 33 Vm

The amount of heat absorbed is simply the total heat of vaporization,

Q mh fgabsorbed (2.905 kg)(427.8 kJ /kg) 1243 kJ

3-125 An isobutane tank contains 5 L of liquid isobutane at the ambient temperature. Now a leak develops at the top of the

tank and isobutane starts to leak out. The temperature of isobutane when the pressure drops to 1 atm and the amount of heat

transferred to the tank by the time the entire isobutane in the tank is vaporized are to be determined.

Properties The properties of isobutane at 1 atm are Tsat = -11.7C, 593 8. kg /m3 , and hfg = 367.1 kJ/kg (Table A-3).

Analysis The temperature of isobutane when the pressure drops to 1 atm is simply the saturation pressure at that

temperature,

T T sat @ atm1 11.7 C

The initial mass of liquid isobutane is

kg2.969)m )(0.005kg/m (593.8 33 Vm

The amount of heat absorbed is simply the total heat of vaporization,

Q mh fgabsorbed (2.969 kg)(367.1 kJ /kg) 1090 kJ

3-126 A tank contains helium at a specified state. Heat is transferred to helium until it reaches a specified temperature. The

final gage pressure of the helium is to be determined.

Assumptions 1 Helium is an ideal gas.

Properties The local atmospheric pressure is given to be 100 kPa.

Analysis Noting that the specific volume of helium in the tank remains

constant, from ideal gas relation, we have

kPa 2.366K)27337(

K)273300(kPa) 100140(

1

212

T

TPP

Then the gage pressure becomes

kPa 266 1002.366atm2gage,2 PPP

Isobutane

5 L

20C

Leak

Propane

5 L

20C

Leak

Helium

37ºC

140 kPa gage

Q



3-68

3-127 The table is completed as follows:

P, kPa T, oC v, m

3/kg u, kJ/kg Condition description and quality, if

applicable

300 250

0.7921 2728.9 Superheated vapor

300

133.52 0.3058 1560.0 x = 0.504, Two-phase mixture

101.42

100 - - Insufficient information

3000

180 0.001127* 761.92

*

Compressed liquid

* Approximated as saturated liquid at the given temperature of 180

oC



3-69

3-128 Water at a specified state is contained in a piston-cylinder device fitted with stops. Water is now heated until a final

pressure. The process will be indicated on the P-v and T- v diagrams.

Analysis The properties at the three states are

5)-A (TableC5.133 /kgm 5.0

kPa 30013

1

1

T

P

v

5)-A (TableC5.133 /kg,m 6058.0 vap.)(sat. 1

kPa 3002

32

2

2

T

x

Pv

6)-A (TableC8.517 /kgm 6058.0

kPa 60023

3

2

T

P

v

Using Property Plot feature of EES, and by adding state points we obtain following diagrams.

10-4 10-3 10-2 10-1 100 101 102100

101

102

103

104

105

106

v [m3/kg]

P [

kP

a] 517.8°C

158.8°C

133.5°C

SteamIAPWS

1 2

3

0.5

300 kPa

600 kPa

10-3 10-2 10-1 100 101 1020

100

200

300

400

500

600

700

v [m3/kg]

T [

°C]

600 kPa

300 kPa

SteamIAPWS

1 2

3

0.5

517.8C

158.8C

133.5C

Water

300 kPa

0.5 m3/kg

Q



3-70

3-129E Argon contained in a piston-cylinder device at a given state undergoes a polytropic process. The final temperature is

to be determined using the ideal gas relation and the Beattie-Bridgeman equation.

Analysis (a) The polytropic relations for an ideal gas give

R 986

6.1/6.0/1

1

212

psia 1000

psia 2000R) 460(300

nn

P

PTT

(b) The constants in the Beattie-Bridgeman equation are expressed as

/kmolKm10.995

010.039311

0.023281.78021301

334

c

bBB

aAA

o

o

vv

vv

Substituting these coefficients into the Beattie-Bridgeman equation and using data in SI units (P = 1000 psia = 6895 kPa,

T=760 R = 422.2 K, Ru = 8.314 kJ/kmol∙K)

232

1v

vvv

AB

T

cTRP u

and solving using an equation solver such as EES gives

/lbmolft 201.8/kmolm 5120.0 33 v

From the polytropic equation

/kmolm 3319.02

1/kmol)m (0.5120 3

6.1/13

/1

2

112

n

P

Pvv

Substituting this value into the Beattie-Bridgeman equation and using data in SI units (P = 2000 psia = 13790 kPa and Ru =

8.314 kJ/kmol∙K),

232

1v

vvv

AB

T

cTRP u

and solving using an equation solver such as EES gives

R 958 K 2.5322 T

Argon

1000 psia

300F



3-71

3-130E The specific volume of nitrogen at a given state is to be determined using the ideal gas relation, the Benedict-Webb-

Rubin equation, and the compressibility factor.

Properties The properties of nitrogen are (Table A-1E)

R = 0.3830 psia·ft3/lbm·R, M = 28.013 lbm/lbmol, Tcr = 227.1 R, Pcr = 492 psia


/lbmft 0.3447 3

psia 400

R) R)(360/lbmftpsia (0.3830 3

P

RTv

(b) Using the coefficients of Table 3-4 for nitrogen and the given data in SI units, the

Benedict-Webb-Rubin equation of state is

)/0053.0exp(0053.0

1)200(

10379.710272.154.2

54.2200314.8002328.01

200

10164.873.106200314.804074.0

)200)(314.8(2758

)/exp(11

2

223

4

6

4

322

5

2

2

2236320

00

vvvv

vvv

vvvvvvv

T

caaTbR

T

CATRB

TRP u

uu

The solution of this equation by an equation solver such as EES gives

/kmolm 5666.0 3v

Then,

/lbmft 0.3240 3

/kgm 1

/lbmft 16.02

kg/kmol 013.28

/kmolm 5666.03

33

M

vv

(c) From the compressibility chart (Fig. A-15),

94.0 Z

813.0psia 492

psia 400

585.1R 227.1

R 360

cr

cr

P

PP

T

TT

R

R

Thus,

/lbmft 0.3240 3 )/lbmft 0.3447)(94.0( 3idealvv Z

Nitrogen

400 psia, -100F



3-72

3-131 A hot air balloon with 3 people in its cage is hanging still in the air. The average temperature of the air in the balloon

for two environment temperatures is to be determined.

Assumptions Air is an ideal gas.


Analysis The buoyancy force acting on the balloon is

N 44,700m/skg1

N 1)m )(4189m/s )(9.8kg/m (1.089

kg/m 1.089K) K)(288/kgmkPa (0.287

kPa 90

m4189/3m)(1043/4

2

323

balloonair cool

3

3air cool

333balloon

V

V

gF

RT

P

r

B

The vertical force balance on the balloon gives

gmmm

WWWFB

)( peoplecageairhot

peoplecageairhot

Substituting,

2

2airhot

m/skg 1

N 1)m/s kg)(9.8 195kg 80(N 44,700 m

which gives

kg 4287airhot m

Therefore, the average temperature of the air in the balloon is

K 306.5

K)/kgmkPa kg)(0.287 (4287

)m kPa)(4189 (903

3

mR

PT

V

Repeating the solution above for an atmospheric air temperature of 30C gives 323.6 K for the average air temperature in

the balloon.

Hot air balloon

D = 20 m

mcage = 80 kg

Patm = 90 kPa

T = 15C



3-73

3-132 Problem 3-131 is to be reconsidered. The effect of the environment temperature on the average air temperature

in the balloon when the balloon is suspended in the air is to be investigated as the environment temperature varies from -

10°C to 30°C. The average air temperature in the balloon is to be plotted versus the environment temperature.


"Given Data:" "atm---atmosphere about balloon" "gas---heated air inside balloon" g=9.807 [m/s^2] d_balloon=20 [m] m_cage = 80 [kg] m_1person=65 [kg] NoPeople = 6 {T_atm_Celsius = 15 [C]} T_atm =T_atm_Celsius+273 "[K]" P_atm = 90 [kPa] R=0.287 [kJ/kg-K] P_gas = P_atm T_gas_Celsius=T_gas - 273 "[C]" "Calculated values:" P_atm= rho_atm*R*T_atm "rho_atm = density of air outside balloon" P_gas= rho_gas*R*T_gas "rho_gas = density of gas inside balloon" r_balloon=d_balloon/2 V_balloon=4*pi*r_balloon^3/3 m_people=NoPeople*m_1person m_gas=rho_gas*V_balloon m_total=m_gas+m_people+m_cage "The total weight of balloon, people, and cage is:" W_total=m_total*g "The buoyancy force acting on the balloon, F_b, is equal to the weight of the air displaced by the balloon." F_b=rho_atm*V_balloon*g "From the free body diagram of the balloon, the balancing vertical forces must equal the product of the total mass and the vertical acceleration:" F_b- W_total=m_total*a_up a_up = 0 "The balloon is hanging still in the air"

Tatm,Celcius [C] Tgas,Celcius [C]

-10 17.32

-5 23.42

0 29.55

5 35.71

10 41.89

15 48.09

20 54.31

25 60.57

30 66.84

-10 -5 0 5 10 15 20 25 300

10

20

30

40

50

60

70

80

90

100

Tatm,Celsius [C]

Tg

as

,Ce

lsiu

s

[C]

3 people

6 people9 people



3-74

3-133 A hot air balloon with 2 people in its cage is about to take off. The average temperature of the air in the balloon for

two environment temperatures is to be determined.

Assumptions Air is an ideal gas.

Properties The gas constant of air is R = 0.287 kPa.m3/kg.K.

Analysis The buoyancy force acting on the balloon is

N 34,029m/skg 1

N 1)m )(3054m/s )(9.8kg/m (1.137

kg/m 1.137K) K)(285/kgmkPa (0.287

kPa 93

m 3054/3m) (943/4

2

323

balloonaircool

3

3aircool

333balloon

V

V

gF

RT

P

r

B

The vertical force balance on the balloon gives

gmmm

WWWFB

)( peoplecageairhot

peoplecageairhot

Substituting,

2

2airhot

m/skg 1

N 1)m/s kg)(9.81 140kg 120(N 34,029 m

which gives

kg 3212airhot m

Therefore, the average temperature of the air in the balloon is

K 308

K)/kgmkPa kg)(0.287 (3212

)m kPa)(3054 (933

3

mR

PT

V

Repeating the solution above for an atmospheric air temperature of 25C gives 323 K for the average air temperature in the

balloon.

Hot air balloon

D = 18 m

mcage = 120 kg

Patm = 93 kPa

T = 12C



3-75

Fundamentals of Engineering (FE) Exam Problems

3-134 A 300-m3 rigid tank is filled with saturated liquid-vapor mixture of water at 200 kPa. If 25% of the mass is liquid

and the 75% of the mass is vapor, the total mass in the tank is

(a) 451 kg (b) 556 kg (c) 300 kg (d) 331 kg (e) 195 kg

Answer (a) 451 kg

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES

screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).

V_tank=300 "m3" P1=200 "kPa" x=0.75 v_f=VOLUME(Steam_IAPWS, x=0,P=P1) v_g=VOLUME(Steam_IAPWS, x=1,P=P1) v=v_f+x*(v_g-v_f) m=V_tank/v "kg" "Some Wrong Solutions with Common Mistakes:" R=0.4615 "kJ/kg.K" T=TEMPERATURE(Steam_IAPWS,x=0,P=P1) P1*V_tank=W1_m*R*(T+273) "Treating steam as ideal gas" P1*V_tank=W2_m*R*T "Treating steam as ideal gas and using deg.C" W3_m=V_tank "Taking the density to be 1 kg/m^3"

3-135 Water is boiled at 1 atm pressure in a coffee maker equipped with an immersion-type electric heating element. The

coffee maker initially contains 1 kg of water. Once boiling started, it is observed that half of the water in the coffee maker

evaporated in 10 minutes. If the heat loss from the coffee maker is negligible, the power rating of the heating element is

(a) 3.8 kW (b) 2.2 kW (c) 1.9 kW (d) 1.6 kW (e) 0.8 kW

Answer (c) 1.9 kW



m_1=1 "kg" P=101.325 "kPa" time=10*60 "s" m_evap=0.5*m_1 Power*time=m_evap*h_fg "kJ" h_f=ENTHALPY(Steam_IAPWS, x=0,P=P) h_g=ENTHALPY(Steam_IAPWS, x=1,P=P) h_fg=h_g-h_f "Some Wrong Solutions with Common Mistakes:" W1_Power*time=m_evap*h_g "Using h_g" W2_Power*time/60=m_evap*h_g "Using minutes instead of seconds for time" W3_Power=2*Power "Assuming all the water evaporates"



3-76

3-136 A 1-m3 rigid tank contains 10 kg of water (in any phase or phases) at 160C. The pressure in the tank is

(a) 738 kPa (b) 618 kPa (c) 370 kPa (d) 2000 kPa (e) 1618 kPa

Answer (b) 618 kPa



V_tank=1 "m^3" m=10 "kg" v=V_tank/m T=160 "C" P=PRESSURE(Steam_IAPWS,v=v,T=T) "Some Wrong Solutions with Common Mistakes:" R=0.4615 "kJ/kg.K" W1_P*V_tank=m*R*(T+273) "Treating steam as ideal gas" W2_P*V_tank=m*R*T "Treating steam as ideal gas and using deg.C"

3-137 Water is boiling at 1 atm pressure in a stainless steel pan on an electric range. It is observed that 2 kg of liquid

water evaporates in 30 minutes. The rate of heat transfer to the water is

(a) 2.51 kW (b) 2.32 kW (c) 2.97 kW (d) 0.47 kW (e) 3.12 kW

Answer (a) 2.51 kW



m_evap=2 "kg" P=101.325 "kPa" time=30*60 "s" Q*time=m_evap*h_fg "kJ" h_f=ENTHALPY(Steam_IAPWS, x=0,P=P) h_g=ENTHALPY(Steam_IAPWS, x=1,P=P) h_fg=h_g-h_f "Some Wrong Solutions with Common Mistakes:" W1_Q*time=m_evap*h_g "Using h_g" W2_Q*time/60=m_evap*h_g "Using minutes instead of seconds for time" W3_Q*time=m_evap*h_f "Using h_f"



3-77

3-138 Water is boiled in a pan on a stove at sea level. During 10 min of boiling, its is observed that 200 g of water has

evaporated. Then the rate of heat transfer to the water is

(a) 0.84 kJ/min (b) 45.1 kJ/min (c) 41.8 kJ/min (d) 53.5 kJ/min (e) 225.7 kJ/min

Answer (b) 45.1 kJ/min



m_evap=0.2 "kg" P=101.325 "kPa" time=10 "min" Q*time=m_evap*h_fg "kJ" h_f=ENTHALPY(Steam_IAPWS, x=0,P=P) h_g=ENTHALPY(Steam_IAPWS, x=1,P=P) h_fg=h_g-h_f "Some Wrong Solutions with Common Mistakes:" W1_Q*time=m_evap*h_g "Using h_g" W2_Q*time*60=m_evap*h_g "Using seconds instead of minutes for time" W3_Q*time=m_evap*h_f "Using h_f"

3-139 A rigid 3-m3 rigid vessel contains steam at 4 MPa and 500C. The mass of the steam is

(a) 3 kg (b) 9 kg (c) 26 kg (d) 35 kg (e) 52 kg

Answer (d) 35 kg



V=3 "m^3" m=V/v1 "m^3/kg" P1=4000 "kPa" T1=500 "C" v1=VOLUME(Steam_IAPWS,T=T1,P=P1) "Some Wrong Solutions with Common Mistakes:" R=0.4615 "kJ/kg.K" P1*V=W1_m*R*(T1+273) "Treating steam as ideal gas" P1*V=W2_m*R*T1 "Treating steam as ideal gas and using deg.C"



3-78

3-140 Consider a sealed can that is filled with refrigerant-134a. The contents of the can are at the room temperature of

25C. Now a leak developes, and the pressure in the can drops to the local atmospheric pressure of 90 kPa. The

temperature of the refrigerant in the can is expected to drop to (rounded to the nearest integer)

(a) 0C (b) -29C (c) -16C (d) 5C (e) 25C

Answer (b) -29C



T1=25 "C" P2=90 "kPa" T2=TEMPERATURE(R134a,x=0,P=P2) "Some Wrong Solutions with Common Mistakes:" W1_T2=T1 "Assuming temperature remains constant"

3-141 A rigid tank contains 2 kg of an ideal gas at 4 atm and 40C. Now a valve is opened, and half of mass of the gas is

allowed to escape. If the final pressure in the tank is 2.2 atm, the final temperature in the tank is

(a) 71C (b) 44C (c) -100C (d) 20C (e) 172C

Answer (a) 71C



"When R=constant and V= constant, P1/P2=m1*T1/m2*T2" m1=2 "kg" P1=4 "atm" P2=2.2 "atm" T1=40+273 "K" m2=0.5*m1 "kg" P1/P2=m1*T1/(m2*T2) T2_C=T2-273 "C" "Some Wrong Solutions with Common Mistakes:" P1/P2=m1*(T1-273)/(m2*W1_T2) "Using C instead of K" P1/P2=m1*T1/(m1*(W2_T2+273)) "Disregarding the decrease in mass" P1/P2=m1*T1/(m1*W3_T2) "Disregarding the decrease in mass, and not converting to deg. C" W4_T2=(T1-273)/2 "Taking T2 to be half of T1 since half of the mass is discharged"



3-79

3-142 The pressure of an automobile tire is measured to be 190 kPa (gage) before a trip and 215 kPa (gage) after the trip at

a location where the atmospheric pressure is 95 kPa. If the temperature of air in the tire before the trip is 25C, the air

temperature after the trip is

(a) 51.1C (b) 64.2C (c) 27.2C (d) 28.3C (e) 25.0C

Answer (a) 51.1C



"When R, V, and m are constant, P1/P2=T1/T2" Patm=95 P1=190+Patm "kPa" P2=215+Patm "kPa" T1=25+273 "K" P1/P2=T1/T2 T2_C=T2-273 "C" "Some Wrong Solutions with Common Mistakes:" P1/P2=(T1-273)/W1_T2 "Using C instead of K" (P1-Patm)/(P2-Patm)=T1/(W2_T2+273) "Using gage pressure instead of absolute pressure" (P1-Patm)/(P2-Patm)=(T1-273)/W3_T2 "Making both of the mistakes above" W4_T2=T1-273 "Assuming the temperature to remain constant"

3-143 … 3-145 Design and Essay Problems

3-145 It is helium.

Chapter 3 PROPERTIES OF PURE SUBSTANCESmars.umhb.edu/~wgt/engr2345/old/Cengel_8th/Chapter_3...Pure...

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