Introduction to Thermodynamics
Introduction to Thermodynamics

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Solutions Manual for

Thermodynamics: An Engineering Approach Seventh Edition in SI Units

Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011

Chapter 3 PROPERTIES OF PURE SUBSTANCES

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3-2

Pure Substances, Phase Change Processes, Property Diagrams

3-1C A vapor that is about to condense is saturated vapor; otherwise it is superheated vapor.

3-2C No.

3-3C The temperature will also increase since the boiling or saturation temperature of a pure substance depends on pressure.

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 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.

3-6C Yes.

3-7C 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.

3-8C At supercritical pressures, there is no distinct phase change process. The liquid uniformly and gradually expands into a vapor. At subcritical pressures, there is always a distinct surface between the phases.

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3-3

Property Tables

3-9C A perfectly fitting pot and its lid often stick after cooking as a result of the vacuum created inside as the temperature and thus the corresponding saturation pressure inside the pan drops. An easy way of removing the lid is to reheat the food. When the temperature rises to boiling level, the pressure rises to atmospheric value and thus the lid will come right off.

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

3-11C 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 No. Quality is a mass ratio, and it is not identical to the volume ratio.

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3-4

3-19C The compressed liquid can be approximated as a saturated liquid at the given temperature. Thus TfPT @, vv ≅ .

3-20C 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.

3-21 Complete the following table for H2 O:

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

110 350 0.001051 Compressed liquid

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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 0.0512

250.00

500.00 – –

110.00

350.00 100 0.08666

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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 10.00

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

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3-7

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

101

102

103

104

105

v [m /kg]3

P [k

Pa]

250 C

170 C

110 C

75 C

Steam

0 500 1000 1500 2000 2500 3000100

101

102

103

104

105

h [kJ/kg]

P [k

Pa]

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 10.00

500

1000

1500

2000

2500

3000

3500

4000

s [kJ/kg-K]

h [k

J/kg

]

8600 kPa

2600 kPa

500 kPa

45 kPa

Steam

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3-8

3-23 Complete the following table for H2 O:

T, °C P, kPa h, kJ / kg x Phase description

120.21 200 2045.8 0.7 Saturated mixture

140 361.53 1800 0.565 Saturated mixture

177.66 950 752.74 0.0 Saturated liquid

80 500 335.37 – – – Compressed liquid

350.0 800 3162.2 – – – Superheated vapor

3-24 Complete the following table for Refrigerant-134a:

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

-12 320 0.000750 Compressed liquid

30 770.64 0.0065 Saturated mixture

18.73 550 0.03741 Saturated vapor

60 600 0.04139 Superheated vapor

3-25 Complete the following table for water:

P, kPa T, oC v, m3/kg h, kJ/kg Condition description and quality, if

applicable 200 120.2

0.8858 2706.3 x = 1, Saturated vapor

270.3 130 1959.3 x = 0.650, Two-phase mixture

201.8

400 1.5358 3277.0 Superheated vapor

800

30 0.001004* 125.74* Compressed liquid

450

147.90 – – Insufficient information

* Approximated as saturated liquid at the given temperature of 30oC

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3-93-26 Complete the following table for Refrigerant-134a:

T, °C P, kPa h, kJ /kg x Phase description 21.55 600 180 0.545 Saturated mixture -10 200.74 162.13 0.6 Saturated mixture -14 500 33.40 – – – Compressed liquid 70 1200 300.61 – – – Superheated vapor 44 1131 272.95 1.0 Saturated vapor

3-27 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 1000kN 1

/4m) (0.25)m/s kg)(9.81 (12

kPa 88/4 ππD

gmPPP p

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(33

22

3311

V

vV

vV

m

m

kJ/kg 17.4

(b) The specific volume and enthalpy of R-134a at the initial state of 90.4 kPa and -10°C and at the final state of 90.4 kPa and 15°C are (from EES)

v1 = 0.2302 m3/kg h1 = 247.76 kJ/kg

R-134a 0.85 kg -10°C

Q

v 2 = 0.2544 m3/kg h2 = 268.16 kJ/kg

The initial and the final volumes and the volume change are

(c) The total enthalpy change is determined from

=−=−=∆ kJ/kg 247.76)6kg)(268.1 85.0()( 12 hhmH

13)-A (Table /kgm 0.030

kPa 8003 C60°=

⎭⎬⎫

==

TPv

3-28 The temperature of R-134a at a specified state is to be determined.

Analysis Since the specified specific volume is higher than vg for 800 kPa, this is a superheated vapor state. From R-134a tables,

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3-10

3-29 A rigid container that is filled with water is cooled. The initial temperature and the final pressure are to be determined.

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

H2O 2 MPa 1 kg

150 L

/kgm 150.0kg 1

m 150.0 33

21 ====mV

vv

6)-A (Table /kgm 150.0

MPa 213

1

1 C395°=⎭⎬⎫

==

TPv

4)-A (Table C40

C40 @sat 232

2 kPa 7.385==⎬⎫°=

°PPTv

Q

The initial state is superheated vapor. The temperature is determined to be

P

1

2

This is a constant volume cooling process (v = V /m = constant). The final state is saturated mixture and thus the pressure is the saturation pressure at the final temperature:

v /kgm 150.01 ⎭==v

3-30 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

R-134a -40°C 10 kg

1.348 m3

/kgm 1348.0m 348.1 33====

Vvv

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

13)-A (Table /kgm 1348.0

kPa 20023

2

2 C66.3°=⎭⎬⎫

==

TPv

kg 1021 m

The initial state is determined to be a mixture, and thus the pressure is the saturation pressure at the given temperature

P

1

2The final state is superheated vapor and the temperature is determined by interpolation to be

v

3-31 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

===mV

v

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 180.02=+=+=

=−

−=

−=

)73.190)(6008.0(43.65

6008.0/kgm )0007930.0049403.0(

/kgm )0007930.003.0(3

3

fgf

fg

f

xhhh

xv

vv

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3-11

3-32 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

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13)-A ⎭

/kgm 21686 3

331 m 0.021686/kg)m 6kg)(0.2168 1.0( ==

3312 m 0.03036)m 021686.0(4.14.1 ==V

(Table C25

kPa 200 /kgm 0.11646 3=⎬

⎫°=

=v

TP

3-33 A spring-loaded piston-cylinder device is filled with R-134a. The water now undergoes a process until its volume increases by 40%. The final temperature and the enthalpy are to be determined.

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

.0)0007142.027090.0)(80.0(0007142.011 =−+=+= fgf x vvv

and the initial volume will be P

1

2

1= vV m

With a 40% increase in the volume, the final volume will be

=V

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

m 1227.0m ) 3=

p m) 30.0(021686.003036.0(4

4/ 22−

=∆

=∆

=∆ππDA

x VV

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

kPa 11.46kN/m 46.11m) 30.0(

m) 12274/

22 ==

ππDpp

kPa 02.8146.1156.6912 =+=∆+= PPP

.0(kN/m) 6.6(42 =∆

=∆

=∆

=∆xk

Axk

AFP

The initial pressure is

11)-A (Table kPa 56.69C34- @sat 1 == °PP

The final pressure is then

and the final specific volume is

/kgm 003036.0kg 1.0

m 0.03036 33

22 ===

mV

v

EES) (from kJ/kg C30.9°

At this final state, the temperature and enthalpy are

/kgm 003036.0

kPa 02.81

1

13

2

2

13.73=−=

⎭⎬⎫

==

hTP

v

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3-12

3-34 A piston-cylinder device that is filled with water is cooled. The final pressure and volume of the water are to be determined.

H2O 350°C 1 kg

0.1546 m3

Analysis The initial specific volume is

/kgm 1546.0kg 1

m 1546.0 33

11 ===

mV

v

This is a constant-pressure process. The initial state is determined to be superheated vapor and thus the pressure is determined to be

6)-A (Table /kgm 1546.0

C035213

1

1 MPa 1.8==⎭⎬⎫

=°=

PPTv

22

P

1 2 The saturation temperature at 1.8 MPa is 207.11°C. Since the final temperature is less than this temperature, the final state is compressed liquid. Using the incompressible liquid approximation,

4)-A (Table /kgm 001043.0 3C100 @ 2 == °fvv

v The final volume is then

3m 0.001043=== /kg)m 001043.0)(kg 1( 3vV m

3-35 A piston-cylinder device that is filled with R-134a is heated. The final volume is to be determined.

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

R-134a -26.4°C 10 kg

1.595 m3

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/kgm 1595.0kg 101 ===

mv

m 595.1 33V

13)-A C1002 ⎭°=T

3m

The initial state is determined to be a mixture, and thus the pressure is the saturation pressure at the given temperature

12)-A (Table kPa 100C26.4- @sat 1 == °PP

The final state is superheated vapor and the specific volume is P

2 1 (Table/kgm 30138.0 kPa 100 3

22 =⎬

⎫=v

P

The final volume is then

v 3.0138=== /kg)m 30138.0)(kg 10( 322 vV m

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3-13

3-36 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

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6)-A kPa 100 ⎫=P

3m 7.22=== /kg)m 4062.2)(kg 3( 3vV m

(Table/kgm 4062.2 C250

3 =⎭⎬°=

vT Water

3 kg 100 kPa 250°C

The volume of the container is then

3-37 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 R-134a 300 kPa 10 kg 14 L

/kgm 0014.0m 014.0 33

21 ====V

vv

C0.61°== kPa 300 @sat 1 TT

kg 10m

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

P

2

1

and

kJ/kg 52.54)13.198

009321.0/kgm )

/kgm )0007736.00014.0(3

31

1

=

=−

=−

= fxv

vv

kJ 545.2=== )kJ/kg 52.54)(kg 10(11 mhH

C21.55°== kPa 600 @sat 2 TT

)(009321.0(67.52

0007736.0067978.0(

11 +=+=

fgf

fg

hxhh v

The total enthalpy is then

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

kJ/kg 64.84)90.180)(01733.0(51.81

01733.3

32

2

=+=+=

=−

−=

−=

fg

f

hxhh

xv

vv

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

0/kgm )0008199.0034295.0(

/kgm )0008199.00014.0(

22 fgf

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3-14

3-38 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.

R-134a 200 kPa 100 kg

12.322 m3

Analysis The initial specific volume is

/kgm 12322.0kg 100m 322.12 3

3

1 ===mV

v

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

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

1

1 =⎭⎬=

uv

kPa 200 ⎫=P

P

1 2

The final specific volume is

/kgm 06161.02

/m 12322.02

33

12 ===

kgvv

12)

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

v

-A (Table kPa 200 @sat 2 C10.09°−== TT

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

kJ/kg 61.152)21.186)(6140.0(28.38

6140.0/kgm )0007533.0

/kgm )0007533.006161.0(

22

3

32

=+=+=

=−

−−

fgf

f

uxuu

vv

099867.0(2 ==

fgx

v

Hence, the change in the internal energy is

kJ/kg 110.47−=−=−=∆ 08.26361.15212 uuu

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3-15

3-39 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

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6)-A 1 ⎭

2 ⎭

5)-A /kgm 6058.03 ⎭=v

Water 300 kPa 250°C

(TablekJ/kg 9.2728 C250

kPa 3001

1 =⎬⎫

°==

uTP

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

5)-A (Table/kgm 6058.0 vapor)(sat. 1

kPa 300 32

2 =⎬⎫

==

vxP

100 kPa

300 kPa

2

1

3

v

TProcess 2-3 is a constant-volume process. Thus, 250°C

(TablekJ/kg 3.1163 kPa 100

333 =⎬

⎫=u

P

The overall change in internal energy is

kJ/kg 1566=−=−=∆ 3.11639.272831 uuu

3-40 The local atmospheric pressure, and thus the boiling temperature, changes with the weather conditions. The change in the boiling temperature corresponding to a change of 5 mm of mercury in atmospheric pressure is to be determined.

Properties The saturation pressures of water at 95 and 100°C are 84.609 and 101.42 kPa, respectively (Table A-4). One mm of mercury is equivalent to 1 mmHg = 0.1333 kPa (inner cover page).

Analysis A change of 5 mm of mercury in atmospheric pressure corresponds to

kPa 0.6665mmHg 1

kPa 0.1333mmHg) 5( =⎟⎟⎠

⎞⎜⎜⎝

⎛=∆P

P ± 5 mmHg

At about boiling temperature, the change in boiling temperature per 1 kPa change in pressure is determined using data at 95 and 100°C to be

C/kPa 2974.0kPa )609.8442.101(

C)95100(°=

−°−

=∆∆

PT

C0.20°°=∆°=∆ =kPa) 665C/kPa)(0.6 2974.0(C/kPa) 2974.0(boiling PT

Then the change in saturation (boiling) temperature corresponding to a change of 0.147 psia becomes

which is very small. Therefore, the effect of variation of atmospheric pressure on the boiling temperature is negligible.

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3-16

3-41 A person cooks a meal in a pot that is covered with a well-fitting lid, and leaves the food to cool to the room temperature. It is to be determined if the lid will open or the pan will move up together with the lid when the person attempts to open the pan by lifting the lid up.

Assumptions 1 The local atmospheric pressure is 1 atm = 101.325 kPa. 2 The weight of the lid is small and thus its effect on the boiling pressure and temperature is negligible. 3 No air has leaked into the pan during cooling.

Properties The saturation pressure of water at 20°C is 2.3392 kPa (Table A-4).

Analysis Noting that the weight of the lid is negligible, the reaction force F on the lid after cooling at the pan-lid interface can be determined from a force balance on the lid in the vertical direction to be

PA +F = PatmA

or,

P

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)N/m 1 = Pa 1 (since =Pam 6997=

Pa )2.2339325,101(4

m) 3.0(

22

2

N 6997

−=π

))(4/()( 2 −=−= π PPDPPAF atmatm

N 78.5=)m/s kg)(9.81 8(== mgW

2.3392 kPa

Patm = 1 atm

The weight of the pan and its contents is

2

which is much less than the reaction force of 6997 N at the pan-lid interface. Therefore, the pan will move up together with the lid when the person attempts to open the pan by lifting the lid up. In fact, it looks like the lid will not open even if the mass of the pan and its contents is several hundred kg.

3-42 Water is boiled at 1 atm pressure in a pan placed on an electric burner. The water level drops by 10 cm in 45 min during boiling. The rate of heat transfer to the water is to be determined.

Properties The properties of water at 1 atm and thus at a saturation temperature of Tsat = 100°C are hfg = 2256.5 kJ/kg and vf = 0.001043 m3/kg (Table A-4).

Analysis The rate of evaporation of water is

H2O 1 atm

kg/s 001742.0kg 704.4

0.001043

evap

evapevap

===

===

mm

mff

&

vv

s 6045

kg 704.4m) 0.10](4/m) 0.25([)4/(

evap

22

×∆

=

t

LD ππV

kW 3.93=== kJ/kg) .5kg/s)(2256 001742.0(evap fghmQ &&

Then the rate of heat transfer to water becomes

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3-17

3-43 Water is boiled at a location where the atmospheric pressure is 79.5 kPa in a pan placed on an electric burner. The water level drops by 10 cm in 45 min during boiling. The rate of heat transfer to the water is to be determined.

Properties The properties of water at 79.5 kPa are Tsat = 93.3°C, hfg = 2273.9 kJ/kg and vf = 0.001038 m3/kg (Table A-5).

Analysis The rate of evaporation of water is

H2O 79.5 kPa

kg/s 001751.0s 6045

kg 727.4

kg 727.40.001038

m) 0.10](4/m) 0.25([)4/(

evapevap

22evap

evap

=∆

=

====

tm

m

LDm

ff

&

ππvv

V

kW 3.98

kW 86.9=kJ/h 780,312kJ/kg) .0kg/h)(2406 130(

evap

==

= fghmQ &&

Then the rate of heat transfer to water becomes

=== kJ/kg) .9kg/s)(2273 001751.0(evap fghmQ &&

3-44 Saturated steam at Tsat = 40°C 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 30°C.

Properties The properties of water at the saturation temperature of 40°C 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 40°C condenses, the rate of heat transfer from the steam to the cooling water in the tube is determined directly from

40°C

L = 35 m D = 3 cm

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3-18

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 ρ = 1000 kg/m3.

Analysis The pressure at the bottom of the 5-cm pan is the saturation pressure corresponding to the boiling temperature of 98°C: 40 cm

5 cm (Table A-4) kPa [email protected] == oPP

The pressure difference between the bottoms of two pans is

kPa 3.43kPa 1

=⎟⎟⎠

⎞⎛=P∆

C99.0°== kPa [email protected] TT

skg/m 1000m) )(0.35m/s )(9.807kg/m (1000

223

⎜⎜⎝ ⋅

=hgρ

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

(Table A-5)

3-46 A vertical piston-cylinder device is filled with water and covered with a 20-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

P

Patm

W = mg

or,

kPa 119.61skg/m 1000

kPa 1m 0.01

)m/s kg)(9.81 (20kPa) (100 22

2atm

=⎟⎟⎠

⎞⎜⎜⎝

⋅+=

+=A

mgPP

C104.7°== kPa [email protected]

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

(Table A-5)

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3-19

3-47 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 H2O

90°C/kgm 0.12

kg 15m 1.8 3

3===

mV

v

3

C202.9°== = /kgm [email protected] 3gv

TT

When the liquid is completely vaporized the tank will contain saturated vapor only. Thus, T

1

2 /kgm 0.122 == gvv

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

(Table A-4) v

3-48 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 [email protected]

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

kg 7.395=+=+= 852.2543.4gft mmm

/kgm 0.3521C020kPa 006 3

22

2 =⎭⎬⎫

==

voTP

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

===

===

kg .8522/kgm 0.3156

m 0.9

kg 543.4/kgm 0.001101

m 0.005

3

3

3

3

g

gg

f

ff

m

m

v

V

v

V

P

1 2

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

(Table A-6)

Then,

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3-20

3-49 Problem 3-48 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.

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

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]

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

101

102

103

104

105

106

v [m3/kg]

P [k

Pa] 200°C

SteamIAPWS

1 2

P=600 kPa

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 [kg

]

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3-21

3-50 Superheated water vapor cools at constant volume until the temperature drops to 120°C. 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

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

/kgm 16356.0C025

MPa 1.4 31

1

1 =⎭⎬⎫

==

voTP

kPa 198.67== [email protected] oPP

(Table A-6)

At 120°C, vf = 0.001060 m3/kg and vg = 0.89133 m3/kg. 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,

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

0.1825=−−

=−

=001060.089133.0001060.016356.02

2fg

fxv

vv

kJ/kg 905.7=×+=+= 1.22021825.081.503fgf xhhh

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

2

1

H2O 1.4 MPa 250°C

T

v

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3-22

3-51 Problem 3-50 is reconsidered. The effect of initial pressure on the quality of water at the final state as the pressure varies from 700 kPa to 2000 kPa is to be investigated. The quality is to be plotted against initial pressure, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. T[1]=250 [C] P[1]=1400 [kPa] T[2]=120 [C] 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 [kPa]

x2

800 1000 1200 1400 1600 1800 20000.1

0.15

0.2

0.25

0.3

0.35

0.4

P[1] [kPa]

x[2]

700 800 900

1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000

0.3766 0.3282 0.2904 0.2602 0.2355 0.2149 0.1975 0.1825 0.1696 0.1582 0.1482 0.1392 0.1312 0.124

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

100

200

300

400

500

600

700

v [m3/kg]

T [°

C]

1400 kPa

198.7 kPa

0.05 0.1 0.2 0.5

SteamIAPWS

1

2

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3-23

3-52 The properties of compressed liquid water at a specified state are to be determined using the compressed liquid tables, and also by using the saturated liquid approximation, and the results are to be compared.

Analysis Compressed liquid can be approximated as saturated liquid at the given temperature. Then from Table A-4,

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error) (4.53% kJ/kg .02335error) (1.35%kJ/kg 97.334

error) (0.90% /kgm 0.001029

C80@

C80@

3C80@

=≅=≅=≅

°

°

°

f

f

f

hhuuvv

kJ/kg 0.9035kJ/kg .50330

/kgm 0.00102

C08MPa 20

3

===

⎭⎬⎫

°==

huT

Pv

T = 80°C ⇒

From compressed liquid table (Table A-7),

The percent errors involved in the saturated liquid approximation are listed above in parentheses.

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3-24

3-53 Problem 3-52 is reconsidered. Using EES, the indicated properties of compressed liquid are to be determined, and they are to be compared to those obtained using the saturated liquid approximation.

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

“Given” T=80 [C] P=20000 [kPa] “Analysis” Fluid\$=’steam_iapws’ “Saturated liquid assumption” v_f=volume(Fluid\$, T=T, x=0) u_f=intenergy(Fluid\$, T=T, x=0) h_f=enthalpy(Fluid\$, T=T, x=0) “Compressed liquid treatment” v=volume(Fluid\$, T=T, P=P) u=intenergy(Fluid\$, T=T, P=P) h=enthalpy(Fluid\$, T=T, P=P) “Percentage Errors” error_v=100*(v_f-v)/v error_u=100*(u_f-u)/u error_h=100*(h-h_f)/h SOLUTION error_h=4.527 error_u=1.351 error_v=0.8987 Fluid\$=’steam_iapws’ h=350.90 [kJ/kg] h_f=335.02 [kJ/kg] P=20000 [kPa] T=80 [C] u=330.50 [kJ/kg] u_f=334.97 [kJ/kg] v=0.00102 [m^3/kg] v_f=0.001029 [m^3/kg]

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3-25

3-54 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,

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

C179.88°== MPa [email protected]

/kgm 0.25799C300

MPa 1.0 31

1

1 =⎭⎬⎫

==

voTP

/kgm .097750)001127.019436.0(5.0001127.0

5.0MPa 1.0

3

222

2

=−×+=

+=⎭⎬⎫

==

fgf xxP

vvv

3m 20.128−=−=−= /kgm0.25799)5kg)(0.0977 (0.8)(∆ 312 vvV m

/kgm 79270.0 [email protected] === °gvvv

MPa 0.30=⎭⎬⎫

=°=

131

1

/kgm 0.79270C025

PTv

H2O 300°C 1 MPa

(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

(Table A-6) T

2

1

Thus, v

3-55 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

T °C

2

1 25 (Table A-4)

H2O T1= 250°C

P1 = ?

since the vapor starts condensing at 150°C. Then from Table A-6, 15

v

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3-26

3-56 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 200°C 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

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33 m 001619.0/kg)m =

3m 0.006476== )001619.0(4V

11 001157.0kg)( 4.1(== vV m

Water 1.4 kg, 200°C

sat. liq.

The volume at the final state is

Q (b) The final state properties are

kg/m 004626.0kg 1.4

m 0.006476 3V 32 ===

mv

kJ/kg 5.22011kg/m 004626.0

2

2

2

2

32

==

°=

⎪⎭

⎪⎬⎫

==

uPT

xkPa 21,367

C371.3v

kJ 1892

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

(c) The total internal energy change is determined from

==−=∆ kJ/kg 850.46)-kg)(2201.5 4.1()( 12 uumU

7)-A (Table kJ/kg 07.602

/kgm C140

3

=⎭°= hT

4)-A (TablekJ/kg 16.

/kgm 001080 3

3-57 The errors involved in using the specific volume and enthalpy of water by the incompressible liquid approximation are to be determined.

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

0010679.0 MPa 20 =

⎬⎫=P v

Based upon the incompressible liquid approximation,

589 .0

C140

MPa 20

C140 @

C140 @

=≅=≅

⎭⎬⎫

°==

°

°

f

f

hhTP vv

The errors involved are

2.14%

1.13%

=×−

=

−=×−

=

10007.602

16.58907.602(enthalpy)Error Percent

1000010679.0

001080.00010679.0 volume)(specificError Percent

which are quite acceptable in most engineering calculations.

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3-27

3-58 A piston-cylinder device that is filled with R-134a is heated. The volume change is to be determined.

Analysis The initial specific volume is

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13)-A (Table/kgm 33608.0 C20

kPa 60 31

1 =⎭⎬⎫

°−==

vTP

33 m 0.033608/kg)m 8kg)(0.3360 100.0( === vV m

13)-AkPa 60 ⎫=P

33 m 0.050410===

3m 0.0168=−=−=∆ 0336080050410012 ..VVV

/kgm 0.037625kJ/kg .87327

C012kPa 008

3==

⎭⎬⎫

==

vu

TP

o

R-134a 60 kPa -20°C 100 g

1

and the initial volume is

11

At the final state, we have

P

1

v

(Table/kgm 0.50410 C100

32

2

2 =⎭⎬°=

vT

2 22 /kg)m 0kg)(0.5041 100.0(vV m

The volume change is then

3-59 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).

R-134a 2 kg

800 kPa 120°C

Analysis The total volume and internal energy are determined from

kJ 655.7

m 0.0753 3

======

kJ/kg) kg)(327.87 (2/kg)m 25kg)(0.0376 (2 3

muUmvV

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3-28

3-60 A rigid vessel contains R-134a at specified temperature. The pressure, total internal energy, and the volume of the liquid phase are to be determined.

Analysis (a) The specific volume of the refrigerant is

R-134a 10 kg -20°C

/kgm 0.05kg 10m 0.5 3

3===

mV

v

At -20°C, vf = 0.0007362 m3/kg and vg = 0.14729 m3/kg (Table A-11). Thus the tank contains saturated liquid-vapor mixture since vf < v < vg , and the pressure must be the saturation pressure at the specified temperature,

kPa 132.82== − [email protected] oPP

(b) The quality of the refrigerant-134a and its total internal energy are determined from

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kJ 904.2=== kJ/kg) kg)(90.42 (10muUfgf =×+=+=

=−

−=

−=

kJ/kg .429045.1930.336125.39

0.33610.00073620.14729

0.00073620.05

xuuu

xfg

f

v

vv

3m 0.00489===

=×−=−=

/kg)m 362kg)(0.0007 (6.639

kg 6.639100.3361)(1)1(3

fff

tf

m

mxm

vV

(c) The mass of the liquid phase and its volume are determined from

3-61 The Pessure-Enthalpy diagram of R-134a showing some constant-temperature and constant-entropy lines are obtained using Property Plot feature of EES.

-100 0 100 200 300 400 500101

102

103

104

105

h [kJ/kg]

P [k

Pa] 70°C

40°C

10°C

-10°C

-30°C

0.2

0

.3

0.5

0.8

1

1.2

kJ/

kg-K

R134a

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3-29

3-62 A rigid vessel that contains a saturated liquid-vapor mixture is heated until it reaches the critical state. The mass of the liquid water and the volume occupied by the liquid at the initial state are to be determined.

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

(last row of Table A-4) /kgm 0.003106 321 === crvvv

The total mass is

vcr

cp T kg .6096

m 0.33

3===

Vm /kgm 0.003106v

H2O 150°C At 150°C, vf = 0.001091 m3/kg and vg = 0.39248 m3/kg (Table A-

4). Then the quality of water at the initial state is

0.0051490.0010910.392480.0010910.0031061

1 =−−

=−

=fg

fxv

vv v

Then the mass of the liquid phase and its volume at the initial state are determined from

3m 0.105

kg 96.10

===

=−=−=

/kg)m 91kg)(0.0010 (96.10

96.60)0.005149)((1)1(3

1 tf

m

mxm

vV

fff

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3-30

Ideal Gas

3-63C Mass m is simply the amount of matter; molar mass M is the mass of one mole in grams or the mass of one kmol in kilograms. These two are related to each other by m = NM, where N is the number of moles.

3-64C 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.

3-65C 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.

3-66C 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.

3-67 The specific volume of nitrogen at a specified state is to be determined.

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

Properties The gas constant of nitrogen is R = 0.2968 kJ/kg⋅K (Table A-1).

Analysis According to the ideal gas equation of state,

/kgm 0.495 3=+⋅⋅

==kPa 300

K) 273K)(227/kgmkPa (0.2968 3

PRT

v

3-68 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 = 0.2598 kPa⋅m3/kg⋅K (Table A-1).

Analysis The definition of the specific volume gives

/kgm 1.0kg 0.9m 0.09 3

3===

mV

v

Using the ideal gas equation of state, the temperature is

K 231=⋅⋅

==K/kgmkPa 0.2598

/kg)m kPa)(0.1 (6003

3

RPT v

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3-31

3-69 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/kg⋅K (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

PmRT

V

3-70 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

He D = 9 m

27°C 200 kPa

333 m 7.381m) (4.534

34

=== ππ 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

TRPN

u

V

kg123

Then the mass of He can be determined from

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

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3-32

3-71 Problem 3-70 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.

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

“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 7 9 11 13 150

100

200

300

400

500

600

D [m]

m [

kg]

P=200 kPa

P=100 kPa

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

PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.

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3-33

3-72 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.

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

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 5.846K) K)(298/kgmkPa (0.287

)m kPa)(1.0 (500

kPa 200

K) K)(308/kgmkPa kg)(0.287 (5

3

3

1

1

3

1

11

=⋅⋅

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

=⋅⋅

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

AA

BB

RTPm

PRTm

V

V 3m2.21

kg 10.8465.05.846m 3.212.211.0 3

=+=+==+=+=

BA

BA

mmmVVV

A B

× Air

m = 5 kg T = 35°C

P = 200 kPa

Air V = 1 m3

T = 25°C P = 500 kPa

Thus,

Then the final equilibrium pressure becomes

kPa284.1 m 3.21

K) K)(293/kgmkPa kg)(0.287 (10.8463

32

2 =⋅⋅

==V

mRTP

3-73 An elastic tank contains air at a specified state. The volume is doubled at the same pressure. The initial volume and the final temperature are to be determined.

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

Analysis According to the ideal gas equation of state,

C309K 582

m 0.0930 3

°==⎯→⎯+

=⎯→⎯=

=

+⋅⋅=

=

22

1

2

1

2

3

K )273(182

K )273K)(18/kmolmkPa 4kmol)(8.31 1(kPa) (225

TT

TT

TnRP u

V

V

V

V

V

PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.

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3-34

3-74 An ideal gas in a rigid tank is cooled to a final gage 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 at constant volume,

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

21

2211

11

TP

TP

mmVV

=

=

21 VV =

Ideal gas 1227°C 200 kPa (gage)

Patm = 100 kPa Q Since

Then,

[ ] C477°==+

=

K 750kPa kPa 100)(50

2

2

1

1

PTP

TP

1112

12

32 VVVV =+== PP

++==

100)(200K )273(1227

2

212 P

TT

3-75 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,

Evacuated 2V1

Ideal gas

927°C V1

Q Applying these,

= 11 mm

[ ] C3327°==+====

=

=

K 3600K )273927333

11

11

1

212

2

2

1

1

2

22

1

11

TTTT

TT

TP

TP

V

V

V

V

VV

VV

preparation. If you are a student using this Manual, you are using it without permission.

3-35

3-76 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.

Argon 0.6 kg

0.05 m3

550 kPa

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

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

22112 VV

VPP

RT=⎯→⎯211V P

RTP

m ==

which when solved for final pressure becomes

kPa 275===== )kPa 550(5.05.011 PPPPVV

kPa310100210atm1 =+=+= PPP g

2 1

11

212 VV

3-77 An automobile tire is inflated with air. The pressure rise of air in the tire when the tire is heated and the amount of air that must be bled off to reduce the temperature to the original value are 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 = 0.287 kPa.m3/kg.K (Table A-1).

Analysis Initially, the absolute pressure in the tire is Tire

25°C

Treating air as an ideal gas and assuming the volume of the tire to remain constant, the final pressure in the tire can be determined from

kPa 336kPa) (310K 298K 323

11

22

21=== P

TT

TT2211 ⎯→⎯= PPP VV

Thus the pressure rise is

∆P P P= − = − =2 1 336 310 26 kPa

The amount of air that needs to be bled off to restore pressure to its original value is

kg 0.0070=−=−=∆

=⋅⋅

==

=⋅⋅

==

0.08360.0906

kg 0.0836K) K)(323/kgmkPa (0.287

)m kPa)(0.025 (310

kg 0.0906K) K)(298/kgmkPa (0.287

)m kPa)(0.025 (310

21

3

3

2

12

3

3

1

11

mmmRTPm

RTPm

V

V

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3-36

Compressibility Factor

3-78C It represent the deviation from ideal gas behavior. The further away it is from 1, the more the gas deviates from ideal gas behavior.

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.

3-81 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

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

error) (67.0%/kgm 0.01917 3===kPa 15,000

K) P

v⋅⋅ K)(623.15/kgmkPa (0.4615 3RT

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

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

65.0K 673

0.453MPa 22.06

MPa 10

=⎪⎪⎬

⎫===

ZTPPPcr

R

error) (8.5%

H2O 15 MPa 350°C

1.04

K 647.1 ⎪⎪⎭

===T

Tcr

R

Thus,

/kgm 0.01246 3=== /kg)m 1917(0.65)(0.0 3idealvv Z

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

} /kgm 0.01148 3=°== v C035

MPa 15TP

preparation. If you are a student using this Manual, you are using it without permission.

3-37

3-82 Problem 3-81 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.

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

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]

Per

cent

Err

or [

%]

Ideal GasIdeal Gas

Compressibility FactorCompressibility Factor

Steam at 15 MPa

Spec

ific

Volu

me

PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.

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3-38

3-83 The specific volume of R-134a is to be determined using the ideal gas relation, the compressibility chart, and the R-134a 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 refrigerant-134a are, from Table A-1,

R = 0.08149 kPa·m3/kg·K, Tcr = 374.2 K, Pcr = 4.059 MPa

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

)(kPa 900

K) K)(343/kgmkPa (0.08149 3error 13.3%/kgm 0.03105 3=

⋅⋅==

PRT

v

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

894.0.9170

K 374.2K 343

0.222MPa 4.059

MPa 0.9

=

⎪⎪⎭

⎪⎪⎬

===

===

Z

TTT

PPP

crR

crR

error)(1.3%/kgm 0.02776 3=== /kg)m 03105(0.894)(0. 3idealvv Z

R-134a 0.9 MPa

70°C

Thus,

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

} /kgm 0.027413 3=°== vC07

MPa .90TP

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.

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

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

error) (3.7%/kgm 0.09533 3=⋅⋅

==kPa 3500

K) K)(723/kgmkPa (0.4615 3

PRT

v

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

H2O 3.5 MPa 450°C

961.0.121

K 647.1K 723

0.159MPa 22.06

MPa 3.5

=

⎪⎪⎭

⎪⎪⎬

===

===

Z

TTT

PPP

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

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3-39

3-85 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-1,

R = 0.2765 kPa·m3/kg·K, Tcr = 305.5 K, Pcr = 4.48 MPa

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

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

960.0 Z 123.0kPa 550

1.025K 305.5

11

1

cr

11

=⎪

⎪⎪⎬

===

===

PP

TT

R

R

kPa 4480

K 313

cr ⎪⎭

P

TEthane 550 kPa

40°C

Q

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

/kgm 1511.0K) K)(313/kg 3=⋅

kPa 550mkPa 2765(0.960)(0. 3

1

1121

⋅===

PRTZ

vv

At the final state,

0.1 01.8

kPa) K)/(4480 K)(305.5/kgm 0.1511

1.810K 305.5

K 553

23actual2,

cr

22

=

⎪⎪⎭

⎪⎪⎬

=⋅

==

===

ZTT

TR

vv

/kgmkPa (0.2765 3crcr

2⋅/PRTR

Thus,

kPa 1012=⋅⋅

==/kgm 0.1511

K) K)(553/kgmkPa 65(1.0)(0.273

3

2

222 v

RTZP

preparation. If you are a student using this Manual, you are using it without permission.

3-40

3-86 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,

R = 0.2964 kPa·m3/kg·K, Tcr = 282.4 K, Pcr = 5.12 MPa

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

56.0 Z 977.0

MPa 5.12MPa 5

1.038K 282.4

K 293

11

1

cr

11

=

⎪⎪

⎪⎪⎬

===

===

PP

P

TT

T

R

R

Ethylene 5 MPa 20°C

cr ⎭

Q 961.0 Z

977.0

1.675KR 282.4K 473

1

12

cr

22 =

⎪⎭

⎪⎬

==

===

RR

R

PPTT

T

The specific volume change is

[ ]−⋅⋅

=

−=∆

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

K/kgmkPa 0.2964

)(

3

1122 TZTZPR

v

/kgm 0.0172 3=

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3-41

3-87 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.

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

Analysis (a) From the ideal gas equation,

K 1246=+== )2)(K 273350(1

212 v

vTT

Water 350°C

sat. vapor

(b) The pressure of the steam is

Q kPa 529,16=== PPP [email protected] °

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

75.0 ,593.0 ZMPa 16.529

963.0KR 647.1K 623

11cr

11

==⎪⎪⎬

⎫===

R

R

PTT

Tv

88.0 50.1)75.0(22

749.02

12

12 =⎭⎬⎫

=====

ZPP

RR

RR

vv

749.0MPa 22.06cr

11 ⎪

⎪⎭

===R PP

At the final state,

Thus,

K 826====kPa 22,060

K) .1(1.50)(6470.88

kPa 16,529

cr

cr2

2

2

2

222 P

TZP

RZP

T Rvv

/kgm 008806.0 1

C350 31

1

1 =⎭⎬⎫

=°=

vxT

K 750=°=⎭⎬⎫

===

C477 /kgm 01761.02

kPa 529,1623

12

2 TP

vv

(c) From the superheated steam table,

(Table A-4)

(from Table A-6 or EES)

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3-42

3-88 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,

R = 0.5182 kPa·m3/kg·K, Tcr = 191.1 K, Pcr = 4.64 MPa

Analysis From the ideal gas equation,

K 450=== )5.1)(K 300(1

212 v

vTT

Methane 8 MPa 300 K

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

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

80.0 ,88.0 Z570.1

K 191.1K 300

11cr

11

==⎪⎪⎬

⎫===

R

R TT

Tv

975.0 2.1)80.0(5.15.1

724.12

12

12 =⎭⎬⎫

=====

ZPP

RR

RR

vv

Q

724.1

MPa 4.64MPa 8

cr

11 ⎪

⎪⎭

===R PP

P

At the final state,

Thus,

K 406=kPa 4640

K) 10.975cr22 PZRZ

===(1.2)(191.kPa 8000cr2222

2TPP

T 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.

3-89 The percent 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,

MPa7.39andK304.2 crcr == PT

CO2

5 MPa 25°C

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

69.00.980

K 304.2K 298

0.677MPa 7.39

MPa 5

cr

cr =

⎪⎪⎭

⎪⎪⎬

===

===

Z

TTT

PPP

R

R

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

44.9%or 0.44.90.69

1111Error ideal −=−=−=−

=Zv

vv

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3-43

3-90 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)

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

Analysis

3 MPa 500 K 2 kg/s

CO2 450 K

(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 3kPa) (3000P=

⋅⋅==

K) K)(500/kgmkPa (0.1889 31

11 RT

ρ

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)

9791.01.64

K 304.2K 500

0.407MPa 7.39

11

1,

1

=

⎪⎪⎭

⎪⎪⎬

===

===

Z

TTT

PP

crR

crR

MPa 3 ⎫P

9656.01.48

K 304.2K 450

0.407MPa 7.39

MPa 3

22

2,

2

=

⎪⎪⎭

⎪⎪⎬

===

===

Z

TTT

PPP

crR

crR

/kgm 0.06165 3=⋅⋅

==kPa) (3000

K) K)(500/kgmkPa 89 3

11 P

Vkg/s)(0.18 (0.9791)(211 RTmZ && Thus,

3mkg/ 32.44=⋅ K) K)(500⋅

==/kgmkPa .1889(0.9791)(0

kPa) (30003

11

11 RTZ

/kgm 0.05472 3=⋅⋅

==kPa) (3000

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

2

222 P

RTmZ &&V

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3-44

3-91 The specific volume of nitrogen gas is to be determined using the ideal gas relation and the compressibility chart. The errors involved in these two approaches are also to be determined.

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

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

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

)error %4.86( kPa 10,000

K) K)(150/kgmkPa (0.2968 3/kgm0.004452 3=

⋅⋅==

PRT

v

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

N2

10 MPa 150 K

54.01.19

K 126.2K 150

=

⎪⎪⎭

⎬===

Z

TTTcr

R

2.95MPa 3.39

MPa 10

⎪⎪⎫

===PPPcr

R

Thus,

error) (0.7% /kg)m 04452(0.54)(0.0 3ideal /kgm0.002404 3=== vv Z

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3-45

Other Equations of State

3-92C 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.

3-93 Carbon monoxide is heated in a piston-cylinder device. The final volume of the carbon monoxide is to be determined using the ideal gas equation and the Benedict-Webb-Rubin equation of state.

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

CO 1000 kPa

200°C

R = 0.2968 kPa·m3/kg·K, M = 28.011 kg/kmol

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

3m 0.0140=⋅⋅

==kPa1000

K) K)(473/kgmkPa kg)(0.2968 (0.10022

3

PmRT

V

(b) Using the coefficients of Table 3-4 for carbon dioxide and the given data, the Benedict-Webb-Rubin equation of state for state 2 is

)/0060.0exp(0060.01)473(

10054.1000135.071.3

71.3473314.8002632.014473

10673.89.135473314.805454.0)473)(314.8(

)/exp(11

2223

5

6

322

5

2

222

2363

222

2

0020

2

22

vvvv

vvv

vvvvvvv

−⎟⎠

⎞⎜⎝

⎛ +×

+

−××+⎟

⎟⎠

⎞⎜⎜⎝

⎛ ×−−××+=

−⎟⎠

⎞⎜⎝

⎛ +++−

+⎟⎟⎠

⎞⎜⎜⎝

⎛−−+= γγα

TcaaTbR

TC

ATRBTR

P uu

u

1000

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

/kmolm 952.3 32 =v

Then,

3m 0.0141===

===

/kg)m 1411.0kg)( 100.0(

/kgm 1411.0kg/kmol 011.28

/kmolm 952.3

322

33

22

vV

vv

m

M

PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.

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3-46

3-94 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.

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

Methane 80 kPa 20°C

kPa 156.5===K 293K 573kPa) 80(

1

212 T

TPP

Q

(b) The specific molar volume of the methane is

/kmolm 41.29kPa 80

K) K)(293/kmolmkPa (8.314 33

1

121 =

⋅⋅===

PTRuvv

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

−⎟⎠

⎞⎜⎝

⎛+

×+

××+

−××+

−⎟⎠

⎞⎜⎝

⎛ +++−

+⎟⎟⎠

⎞⎜⎜⎝

⎛−−+=

)29.41/0060.0exp(29.41

0060.01)573(29.41

10578.229.41

10244.100.5

29.4100.5573314.8003380.0

)/exp(11

22236

32

222

2363

222

2

0020

2

22 v

vvvvvvγγα

TcaaTbR

TC

ATRBTR

P uu

u

⎟⎟⎠

⎞⎜⎜⎝

⎛ ×−−××+=

29.411

57310286.291.187573314.804260.0

29.41)573)(314.8(

54

2

6

kPa 162.0=

PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.

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3-47

3-95 Carbon monoxide is heated in a rigid container. The final pressure of the CO is to be determined using the ideal gas equation and the Benedict-Webb-Rubin equation of state.

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

R = 0.2968 kPa·m3/kg·K, M = 28.011 kg/kmol CO

101 kPa 21°C

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

kPa 240.5===K 700kPa) 101(2T

PPK 2941

12 T

The specific molar volume of the CO in SI units is

/kmolm 20.24kPa 101

K) K)(294/kmolmkPa (8.314 33

1

121 =

⋅⋅===

PTRuvv

(b) The specific molar volume of the CO in SI units is

/kmolm 20.24kPa 101

K) K)(294/kmolmkPa (8.314 33

1

121 =

⋅⋅===

PTRuvv

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

kPa 240.8=

−⎟⎠⎞

⎜⎝⎛ +

×+

××+

−××+⎟

⎟⎠

⎞⎜⎜⎝

⎛ ×−−××+=

−⎟⎠⎞

⎜⎝⎛ +++

−+⎟

⎟⎠

⎞⎜⎜⎝

⎛−−+=

)24.20/0060.0exp(24.20

0060.01)700(24.20

10054.124.20

10350.171.3

24.2071.3700314.8002632.0

24.201

70010673.887.135700314.805454.0

24.20)700)(314.8(

)/exp(11

2223

5

6

4

322

5

222

2363

222

2

0020

2

22 v

vvvvvvγγα

TcaaTbR

TCATRBTRP u

uu

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3-48

3-96 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-1)

R = 0.08149 kPa·m3/kg·K, Tcr = 374.2 K, Pcr = 4059 kPa

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

K 287.2=⋅⋅

==K/kgmkPa 0.08149

/kg)m 5kPa)(0.019 (12003

3

RPT v

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

/kgm 0.000939K) K)(374.2/kgmkPa (0.08149

kPa/kgm 09663.0kPa) (64)(4059

K) (374.2K)/kgmkPa 49(27)(0.08164

27

33

2622322

=⋅⋅

==

⋅=⋅⋅

==

cr

cr

cr

RTb

PTR

a

kPa 405988 ×crP

Then,

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

( ) ( ) K 331.2=−⎟⎞

⎜⎛

+=−⎟⎞

⎜⎛ += 0.0009390.01950.09663120011 baPT v

K) (343/kgm 0.0195kPa 0012

3 C70°=⎭⎬⎫

== TP

v

⎟⎠

⎜⎝⎠⎝ (0.0195)0.08149 22R v

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

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3-49

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)

N2

0.041884 m3/kg 150 K

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

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

)error .3%6( /kgm 0.041884

kPaK) K)(150/kgmkPa (0.29683

31063=

⋅⋅==

vRTP

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

/kmolKm104.2 334 ⋅×=c

0.050761.17330.0069110.050461

133.1931.17330.026171136.23151

=⎟⎠⎞

⎜⎝⎛ −−=⎟

⎠⎞

⎜⎝⎛ −=

=⎟⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛ −=

bBB

aAA

o

o

v

v

since

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

Substituting,

( ) ( )( )

error) e(negligibl1.1733133.1930.050761.1733

1501.1733104.21

(1.1733)150

23

4

2

kPa 1000.4=

−+⎟⎟⎠

⎞⎜⎜⎝

××

−×

⎠⎝ vvv T8.3141 232 =−+⎟

⎞⎜⎛ −= v

ABcTRP u

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3-50

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.

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

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

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3-51

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

CO2 1 MPa 200°C

/kgm 0.08935kPa 1000

K) K)(473/kgmkPa (0.1889 33

1

11 =

⋅⋅==

PRT

v

According to process specification,

/kgm 03577.0kPa 3000kPa 1000/kg)m 08935.0( 3

2.1/13

/1

2

112 =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎞⎜⎜⎝

⎛=

n

PP

vv

The final temperature is then

K 568=⋅⋅

==K/kgmkPa 0.1889/kg)m 77kPa)(0.035 (3000

3

322

2 RP

Tv

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

/kgm 0.0009720kPa 73908

K) K)(304.2/kgmkPa (0.18898

kPa/kgm 0.1885kPa) (64)(7390

K) (304.2K)/kgmkPa 9(27)(0.18864

27

33

cr

cr

26223

cr

2cr

2

⋅⋅==

⋅=⋅⋅

==

PRT

b

PTR

a

Applying the van der Waals equation to the initial state,

)473)(1889.0()0009720.0(1885.01000

)(

2

2

=−⎟⎠

⎞⎜⎝

⎛ +

=−⎟⎠

⎞⎜⎝

⎛ +

vv

vv

RTbaP

/kgm 0.08821 31 =v

Solving this equation by trial-error or by EES gives

According to process specification,

/kgm 03531.0kPa 3000kPa 1000/kg)m 08821.0( 3

2.1/13

/1

2

112 =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎞⎜⎜⎝

⎛=

n

PP

vv

Applying the van der Waals equation to the final state,

T..

RTbaP

)1889.0()0009720.0035310(035310

1885.03000

)(

2

2

=−⎟⎠

⎞⎜⎝

⎛+

=−⎟⎠

⎞⎜⎝

⎛ + vv

K 573=2T

Solving for the final temperature gives

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3-52

Special Topic: Vapor Pressure and Phase Equilibrium

3-100 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 20°C, and 1.706 kPa at 15°C (Table A-4).

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

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

kPa 0.936

H2O 15°C

kPa 1.706=== °[email protected]@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) 339.2)(4.0(PPP °[email protected]@satair , airTv φφ

3-101 The vapor pressure in the air at the beach when the air temperature is 30°C is claimed to be 5.2 kPa. The validity of this claim is to be evaluated.

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

WATER

Analysis The maximum vapor pressure in the air is the saturation pressure of water at the given temperature, which is

kPa 4.247=== °[email protected]@satmax , airPPP Tv

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

3-102 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.

Properties The saturation pressure of water at 20°C is 2.339 kPa (Table A-4). Patm, 20°C Analysis The vapor pressure of air over the swimming pool is

POOL kPa0.9357kPa) 339.2)(4.0( [email protected]@satair , air

==== PPP °Tv φφ

Phase equilibrium will be established when the vapor pressure at the water surface equals the vapor pressure of air far from the surface. Therefore,

kPa0.9357air ,surface water , == vv PP

C6.0°=== kPa [email protected]@satwater vTTT Pand

Discussion Note that the water temperature drops to 6.0°C in an environment at 20°C when phase equilibrium is established.

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3-53

3-103 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 10°C is to be evaluated.

Properties The saturation pressure of water at 35°C is 5.629 kPa (Table A-4).

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

kPa940.3kPa) 629.5)(7.0([email protected]@satair , air ==== °PPP Tv φφ

C28.7°=== TTT

kPa 0.8187kPa) 3392.2)(35.0([email protected]@satair , air ==== °PPP Tv φφ

air ,surface water , vv

C4.1°=== kPa [email protected]@satwater vTTT P

kPa 1.27==° kPa) 17.3)(4.0([email protected]φ

kPa 1.29

Analysis The vapor pressure of air is 35°C

70%

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

kPa [email protected]@satsat vP

That is, the vapor in the air will condense at temperatures below 28.7°C. Noting that no condensation is observed on the can, the claim that the drink is at 10°C is false.

3-104 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 20°C is 2.3392 kPa (Table A-4). Analysis The vapor pressure of air in the room is

20°C

35%

Thermos bottle

Phase equilibrium will be established when the vapor pressure at the water surface equals the vapor pressure of air far from the surface. Therefore,

kPa0.8187 == PP

and

Discussion Note that the water temperature drops to 4.1°C in an environment at 20°C when phase equilibrium is established.

3-105 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 20°C, and 3.17 kPa at 25°C (Table A-4).

Analysis The vapor pressures in the two rooms are

Room 1: == @sat11 1PP Tv φ

Room 2: ==== [email protected] 2PPP Tv ° kPa) 339.2)(55.0([email protected]φφ

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

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3-54

Review Problems

3-106 Nitrogen gas in a rigid tank is heated to a final gage pressure. The final temperature is to be determined.

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

Analysis According to the ideal gas equation of state at constant volume,

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

[ ] C602°==++

+==

=

=

=

K 875kPa 100)(100kPa 100)(250K )273(227

Since

2

212

2

2

1

1

21

2211

11

PP

TT

TP

TP

PT

Pmm

VV

VV=

21 T

Nitrogen gas227°C

100 kPa (gage)

Patm = 100 kPa Q

3-107 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

3 MPa 500 K

0.4 kmol/s

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

/sm 0.5543 3===kPa 3000

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

1 PTRN u

&&V

kg/s 17.60===K) /kg.K)(500kPa.m (0.1889)/m 3kPa)(0.554 (3000

3

3

1

111

sRTPm V&

&

The density is

3kg/m 31.76===/s)m 5543.0(

kg/s) (17.603

1

11

V&&m

ρ

(b) The volume flow rate at the exit is

/sm 0.4988=kPa 3000

3==K) 50/kmol.K)(4kPa.m 314kmol/s)(8. 4.0( 3

22 P

TRN u&

&V

CO2 450 K

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3-55

3-108 The cylinder conditions before the heat addition process is specified. The pressure after the heat addition process is to be determined.

Combustion chamber 1.4 MPa 450°C

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 3627=⎟⎠

⎜⎝ +

== kPa) 1400(K 2734501

1

22 P

TP ⎞⎛ + K 2731600T

3-109 The cylinder conditions before the heat addition process is specified. The temperature after the heat addition process is to be determined.

Assumptions 1 The contents of cylinder is approximated by the air properties. 2 Air is an ideal gas.

Combustion chamber 950 K 75 cm3

Analysis The ratio of the initial to the final mass is

2322

12222

1AFAF

2

1 =+

=+

=mm

The final temperature may be determined from ideal gas relation

K 1817=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛== K) 950(

cm 75cm 150

2322

3

3

11

2

2

12 T

mm

TV

V

11)-A (Table0

C40

1

1 ⎫=

°−=Tv

2

2

==P

v

3-110 A rigid container that is filled with R-13a is heated. The initial pressure and the final temperature are to be determined.

R-134a -40°C 1 kg

0.090 m3

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,

/kgm 090. C40- @sat 13 kPa 51.25==⎭⎬ °PP

P

2

1

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

1 C50°=

⎭⎬⎫

=T

v

v

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3-56

3-111 The volume of chamber 1 of the two-piston cylinder shown in the figure 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/kg⋅K (Table A-1).

P1A1

P2A2

Analysis Since the water vapor in chamber 2 is condensing, the pressure in this chamber is the saturation pressure,

(Table A-4) kPa 1555C200 @sat 2 == °PP

Summing the forces acting on the piston in the vertical direction gives

kPa 8.248104kPa) 1555(

22

1

22

1

221 =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎞⎜⎜⎝

⎛==

DD

PAA

PP

According to the ideal gas equation of state,

3m 3.95=+⋅⋅

=kPa 248.8

K) 273K)(200/kgmkPa kg)(2.0769 (1 3

11 P

C48 @sat 2 °

1121322

132

)( APAAPAPFFF

=−+=+

=mRT

V

3-112 The volume of chamber 1 of the two-piston cylinder shown in the figure 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 Since R-134a in chamber 2 is condensing, the pressure in this chamber is the saturation pressure,

(Table A-11) kPa 6.1253== PP

Summing the forces acting on the piston in the vertical direction gives

F1

F3

F2

which when solved for P1 gives

⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

1

23

1

221 1

AA

PAA

PP

since the areas of the piston faces are given by the above equation becomes

4/2DA π=

kPa 6.611

12

1

23

2

1

221

=

⎥⎥⎤

⎢⎢⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−+⎟⎟

⎞⎜⎜⎝

⎛=

DD

PDD

PP

851kPa) 200(

85kPa) 6.1253(

22

⎥⎥⎦

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−+⎟

⎠⎞

⎜⎝⎛=

⎦⎣

According to the ideal gas equation of state,

3m 0.0301=+⋅⋅

=kPa 611.6

K) 273K)(48/kgmkPa kg)(0.287 (0.2 3

11 P=

mRTV

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3-57

3-113 The difference in the volume of chamber 1 for two cases of pressure in chamber 3 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 Since R-134a in chamber 2 is condensing, the pressure in this chamber is the saturation pressure,

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

kPa 6.1253== PP

FFF =+

(Table A-11) C48 @sat 2 °

Summing the forces acting on the piston in the vertical direction gives

F1

F3

F2

1121322

132

)( APAAPAP =−+

which when solved for P1 gives

⎟⎟⎠

⎜⎜⎝−+=

1

23

1

221 1

AP

APP

⎞⎛ AA

4/2DA π=

since the areas of the piston faces are given by the above equation becomes

kPa 4.733

851kPa) 400(

85kPa) 6.1253(

1

22

2

1

23

2

1

221

=

⎥⎥⎦

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−+⎟

⎠⎞

⎜⎝⎛=

⎥⎥

⎢⎢

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+⎟⎟

⎞⎜⎜⎝

⎛=

DD

PDD

PP

According to the ideal gas equation of state,

3m 0.0251=+⋅⋅

kPa 733.4K) 273K)(48/kgmkPa kg)(0.287 (0.2 3

11 P

3m 0.005=−=−=∆ 0251.00301.012 VVV

==mRT

V

For a chamber 3 pressure of 200 kPa, the volume of chamber 1 was determined to be 0.0301 m3. Then the change in the volume of chamber 1 is

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3-58

3-114 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,

R = 0.2765 kPa·m3/kg·K, Tcr = 305.5 K, Pcr = 4.48 MPa

Analysis From the ideal gas equation,

K 596.8=== )6.1)(K 373(1

212 v

TTv

Ethane 10 MPa 100°C

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

Q

35.0 ,61.0 Z 232.

221.1K 305.5

11

cr

11

cr1

==

⎪⎪⎭

⎪⎪⎬

===

===

R

R

R

PP

P

TT

v

83 232.212 ⎫== PP RR

2MPa 4.48

MPa 10

K 3731T

At the final state,

.0 56.0)35.0(6.16.1 2

12=

⎭⎬===

ZRR vv

Thus,

K 460====kPa 4480

K) .5(0.56)(3050.83

kPa 10,000

cr

cr2

2

2

2

222 P

TZP

RZP

T 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.

3-115 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 0.92K) K)(293/kgmkPa (0.2968

)m kPa)(20 (400

kg 136.6K) K)(296/kgmkPa(0.2968

)m kPa)(20 (600

3

3

2

22

3

3

1

11

=⋅⋅

==

=⋅⋅

==

RTP

m

RTP

m

V

V

kg 44.6=−=−= 0.92136.621 mmm

N2600 kPa

23°C 20 m3

Thus the amount of N2 that escaped is

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3-59

3-116 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,

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

12 vv = H2O

V = 4 L m = 2 kg T = 50°C

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/kgm 0.002m 0.0044 v

VvV <===⎯→⎯= L

3 Thus, liquid.

kg 2m

./kgm 0.2kg 2m 0.4400 cr

33

vV

vV >===⎯→⎯=m

L Thus, vapor.

3-117 Two rigid tanks that contain hydrogen at two different states are connected to each other. Now a valve is opened, and the two gases are allowed to mix while achieving thermal equilibrium with the surroundings. The final pressure in the tanks is to be determined.

Properties The gas constant for hydrogen is 4.124 kPa·m3/kg·K (Table A-1).

Analysis Let’s call the first and the second tanks A and B. Treating H2 as an ideal gas, the total volume and the total mass of H2 are

A B

kg 0.22180.05630.1655

kg 0.0563K) K)(323/kgmkPa (4.124

)m kPa)(0.5 (150

kg 0.1655K) K)(293/kgmkPa (4.124

)m kPa)(0.5 (400

m 1.00.50.5

3

3

1

1

3

3

1

1

3

=+=+=

=⋅⋅

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

=⋅⋅

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

=+=+=

BA

BB

AA

BA

mmm

RTP

m

RTP

m

V

V

VVV

× H2

V = 0.5 m3

T=50°C P=150 kPa

H2V = 0.5 m3

T=20°C P=400 kPa

Then the final pressure can be determined from

kPa 264=3 ⋅⋅

== 32

m 1.0K) K)(288/kgmkPa kg)(4.124 (0.2218

V

mRTP

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3-60

3-118 Problem 3-117 is reconsidered. The effect of the surroundings temperature on the final equilibrium pressure in the tanks is to be investigated. The final pressure in the tanks is to be plotted versus the surroundings temperature, and the results are to be discussed.

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

“Given Data” V_A=0.5 [m^3] T_A=20 [C] P_A=400 [kPa] V_B=0.5 [m^3] T_B=50 [C] P_B=150 [kPa] {T_2=15 [C]} “Solution” R=R_u/MOLARMASS(H2) R_u=8.314 [kJ/kmol-K] V_total=V_A+V_B m_total=m_A+m_B P_A*V_A=m_A*R*(T_A+273) P_B*V_B=m_B*R*(T_B+273) P_2*V_total=m_total*R*(T_2+273)

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

P2 [kPa]

T2 [C]

-10 -5 0 5 10 15 20 25 30240

250

260

270

280

T2 [C]

P2

[kPa

]

240.6 245.2 249.7 254.3 258.9 263.5 268

272.6 277.2

-10 -5 0 5

10 15 20 25 30

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3-61

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. TIRE

200 kPa 0.035 m3

Properties The local atmospheric pressure is 90 kPa.

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

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

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=kPa290kPa310=

1

2

1

2

2

22

1

11

P

P

T

T

T

P

T

P���

VV

MPa22.06K,647.1,K/kgmkPa0.4615 crcr3 ����� PTR

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

3-120 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.

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

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

H2O0.02 m3/kg

400�C

kPa15,529���

��/kgm0.02

K)K)(673/kgmkPa(0.46153

3

vRT

P

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

57.01.48

K)K)(647.1/kgmkPa(0.4615kPa)0/kg)(22,06m(0.02

/

040.1K647.1

K673

3

3

crcr

actual

cr�

���

���

���

��

���

R

R

R

P

PRT

T

TT

vv

kPa12,574���� 060,2257.0crPPP R

kPa12,576����

��� PT

/kgm0.02C400

3v

Thus,

(c) From the superheated steam table,

(from EES)

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12,515 kPa

3-62

3-121 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 .5833/kgm 0.0008934

m 0.033

3

1

1 ===vV

m

3m 1.91==== /kg)m 0kg)(0.0568 (33.58 322tank vVV m

Evacuated R-134a

P=1.2 MPaV =0.03 m3

since

/kgm 0.0008934 3MPa [email protected] == fvv

At the final state (Table A-13),

/kgm 0.05680 C30kPa 004 3

22

2 =⎭⎬⎫

°==

vTP

Thus,

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3-63

3-122 Problem 3-121 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.

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

“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 [m3]

m [kg]

500 600 700 800 900

1000 1100 1200 1300 1400 1500

2.114 2.078 2.045 2.015 1.986 1.958 1.932 1.907 1.883 1.859 1.836

37.23 36.59 36.01 35.47 34.96 34.48 34.02 33.58 33.15 32.73 32.32

500 700 900 1100 1300 15001.8

1.85

1.9

1.95

2

2.05

2.1

2.15

P1 [kPa]

Vol

2 [m

3 ]

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3-64

3-123 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.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

ρ = 581 kg / m3

T T= = − °sat @ atm1 42.1 C

kg 2.905)m )(0.005kg/m (581 33 === Vρm

Q mhfgabsorbed (2.905 kg)(427.8 kJ / kg)= =

Properties The properties of propane at 1 atm are Tsat = -42.1°C, , 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,

Leak

Propane 5 L

20°C

The initial mass of liquid propane is

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

= 1243 kJ

ρ = 593 8. kg / m3

T T= = − °sat @ atm1 11.7 C

kg2.969)m )(0.005kg/m (593.8 33 === Vρm

Q mhfgabsorbed (2.969 kg)(367.1 kJ / kg)= =

3-124 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 = -11.7°C, , 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,

Leak

Isobutane 5 L

20°C

The initial mass of liquid isobutane is

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

= 1090 kJ

3-125 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

Helium 77ºC

110 kPa gage

Q

kPa .8343K)27377(K)273300(kPa) 100110(

1

212 =

++

+==TT

PP

kPa 244=−

Then the gage pressure becomes

=−= 1008.343atm2gage,2 PPP

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3-65

3-126 The first eight virial coefficients of a Benedict-Webb-Rubin gas are to be obtained.

Analysis The Benedict-Webb-Rubin equation of state is given by

)/1 236322

000 v

vvvvvγα

++−

+⎟⎠⎞

⎜⎝⎛ −−+=

aaTbRTCATRBTRP u

uu exp(1 22 v

γ−⎟

⎠⎞

⎜⎝⎛ +

Tc

Expanding the last term in a series gives

….!3

16

3

4

2+−

vv

γγ!2

11)/exp( 22 +−=−

vv

γγ

Substituting this into the Benedict-Webb-Rubin equation of state and rearranging the first terms gives

29

2

2762532000 )1(

!21)1()1(

Tc

Tca

TcP uuu

vvvvvvvγγγγαγ +

++

−++

+++=2/ aTbRTCATBRTR −−−

The virial equation of state is

…)()()()()()()()(98765432 vvvvvvvvvThTgTfTeTdTcTbTaTR

P u ++++++++=

Comparing the Benedict-Webb-Rubin equation of state to the virial equation of state, the virial coefficients are

2

2

2

2

2000

)1(1)(

/)1()(

)(/)1()(

0)()(

/)(

TcTh

TcTf

aTeTcTd

TcaTbRTb

TCATBRTa

u

u

γγ

γγ

αγ

+=

+=

=+=

=−=

−−=

0)(Tg =

!2

3-127 The table is completed as follows:

P, kPa T, oC v, m3/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 180oC

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3-66

3-128 The table is completed as follows:

P, kPa T, oC v, m3/kg u, kJ/kg Condition description and quality, if

applicable 200 120.2

0.8858 2529.1 Saturated vapor

232.23 125 0.5010 1831.0 x = 0.650, Two-phase mixture

7829

400 0.0352 2967.2 Superheated vapor

1000

30 0.001004* 125.73* Compressed liquid

120.90

105 – –

Insufficient information

* Approximated as saturated liquid at the given temperature of 30oC

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3-67

3-129 Water at a specified state is contained in a tank. It is now cooled. The process will be indicated on the P-v and T- v diagrams.

Analysis The properties at the initial and final states are

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

6)-A (Table/kgm 7964.0 C250

kPa 300 31

1

1 =⎭⎬⎫

°==

vTP

5)-A (TableC35.111 /kgm 7964.0

kPa 15023

12

2 °=⎭⎬⎫

===

TP

vv

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 [k

Pa]

250°C 111,4°C

SteamIAPWS

1

2 150 kPa

300 kPa

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

100

200

300

400

500

600

700

v [m3/kg]

T [°

C]

300 kPa

150 kPa

SteamIAPWS

1

2

250°C

111.35°C

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3-68

3-130 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

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

5)-A (Table

6)-A (TableC8.517 /kgm 6058.0

kPa 60023

3

2 °=⎭⎬⎫

==

TPv

Water 300 kPa

0.5 m3/kg

5)-A (TableC5.133 /kgm 5.0

kPa 30013

1

1 °=⎭⎬⎫

==

TPv

C5.133 /kg,m 6058.0 vap.)(sat. 1

kPa 3002

32

2

2 °==⎭⎬⎫

==

TxP

v Q

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 [k

Pa] 517.8°C

158.8°C

133.5°C

SteamIAPWS

1 2

3

300 kPa

600 kPa

0.5

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

517.8°C

158.8°C

133.5°C

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3-69

3-131 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

K 547=⎟⎠⎞

6.1/6.0

⎜⎝⎛+=⎟⎟

⎞⎜⎜⎝

⎛=

− /1

1

212 kPa 6895

kPa 13,790K) 273(149nn

PP

TT

Argon 6895 kPa

149°C

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

/kmolKm10.995

0.023281.78021301

334 ⋅×=

⎟⎞

⎜⎛ −=⎟

⎞⎜⎛ −=

c

aAA

010.039311 ⎟⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛ −=

⎠⎝⎠⎝bBB o

o

vv

vv

Substituting these coefficients into the Beattie-Bridgeman equation and using data (P = 6895 kPa, T = 422.2 K, Ru = 8.314 kJ/kmol·K)

( )232

1v

vvv

ABTcTR

P u −+⎟⎠

⎞⎜⎝

⎛ −=

and solving using an equation solver such as EES gives

/lbmolft 201.8/kmolm 5120.0 33 ==v

From the polytropic equation

/kmolm 3319.021/kmol)m (0.5120 3

6.1/13

/1

2

112 =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎞⎜⎜⎝

⎛=

n

PP

vv

Substituting this value into the Beattie-Bridgeman equation and using data (P = 13,790 kPa and Ru = 8.314 kJ/kmol·K),

( )232

1v

vvv

ABTcTR

P u −+⎟⎠

⎞⎜⎝

⎛ −=

K 532 2 =T

and solving using an equation solver such as EES gives

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3-70

Fundamentals of Engineering (FE) Exam Problems

3-132 A rigid tank contains 2 kg of an ideal gas at 4 atm and 40°C. 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) 71°C (b) 44°C (c) -100°C (d) 20°C (e) 172°C

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).

“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”

3-133 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 25°C, the air temperature after the trip is

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

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).

“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”

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3-71

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

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

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).

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”

PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.

. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

3-72

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

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

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=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

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).

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”

PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.

. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

3-73

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

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).

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 500°C. The mass of the steam is

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

PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.

. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

3-74

3-140 Consider a sealed can that is filled with refrigerant-134a. The contents of the can are at the room temperature of 25°C. 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) 0°C (b) -29°C (c) -16°C (d) 5°C (e) 25°C