ch02

Chapter 2 Solutions

Engineering and Chemical Thermodynamics

Wyatt Tenhaeff

Milo Koretsky

Department of Chemical Engineering

Oregon State University

[email protected]

2

2.1

There are many possible solutions to this problem. Assumptions must be made to solve the

problem. One solution is as follows. First, assume that half of a kilogram is absorbed by the

towel when you dry yourself. In other words, let

kg 5.02

OHm

Assume that the pressure is constant at 1.01 bar during the drying process. Performing an energy

balance and neglecting potential and kinetic energy effects reveals

hq ˆˆ

Refer to the development of Equation 2.57 in the text to see how this result is achieved. To find

the minimum energy required for drying the towel, assume that the temperature of the towel

remains constant at K 298.15Cº 25 T . In the drying process, the absorbed water is

vaporized into steam. Therefore, the expression for heat is

lOH

vOH

hhq22

ˆˆˆ

where is vOH

h2

ˆ is the specific enthalpy of water vapor at bar 01.1P and K 15.298T and

lOH

h2

ˆ is the specific enthalpy of liquid water at bar 01.1P and K 15.298T . A hypothetical

path must be used to calculate the change in enthalpy. Refer to the diagram below

P = 1 [atm]

liquid vapor

liquid vapor

ˆ h 1

P

ˆ h 2

ˆ h 3

ˆ h

Psat

1 atm

3.17 kPa

By adding up each step of the hypothetical path, the expression for heat is

Cº 25ˆbar 01.1 C,º 25ˆ

Cº 25ˆCº 25ˆbar 1.01 C,º 25 ˆCº 25ˆ

ˆ

,

,,,

321

22

2222

satv

OH

v

OH

satl

OH

satv

OH

l

OH

satl

OH

hh

hhhh

hhhq

3

However, the calculation of heat can be simplified by treating the water vapor as an ideal gas,

which is a reasonable assumption at low pressure. The enthalpies of ideal gases depend on

temperature only. Therefore, the enthalpy of the vapor change due to the pressure change is

zero. Furthermore, enthalpy is weakly dependent on pressure in liquids. The leg of the

hypothetical path containing the pressure change of the liquid can be neglected. This leaves

kPa 3.17 C,º 25ˆ kPa .173 C,º 25ˆˆ22

lOH

vOH

hhq

From the steam tables:

kg

kJ 2.2547ˆ ,

2

satvOH

h (sat. H2O vapor at 25 ºC)

kg

kJ 87.104ˆ ,

2

satlOH

h (sat. H2O liquid at 25 ºC)

which upon substitution gives

kg

kJ 3.2442q

Therefore,

kJ 2.1221kg

kJ 442.32kg 5.0

Q

To find the efficiency of the drying process, assume the dryer draws 30 A at 208 V and takes 20

minutes (1200 s) to dry the towel. From the definition of electrical work,

kJ 7488s 1200V 208A 30 IVtW

Therefore, the efficiency is

%3.16%100kJ 7488

kJ 221.21%100

W

Q

There are a number of ways to improve the drying process. A few are listed below.

Dry the towel outside in the sun.

Use a smaller volume dryer so that less air needs to be heated.

Dry more than one towel at a time since one towel can’t absorb all of the available

heat. With more towels, more of the heat will be utilized.

4

2.3 In answering this question, we must distinguish between potential energy and internal energy.

The potential energy of a system is the energy the macroscopic system, as a whole, contains

relative to position. The internal energy represents the energy of the individual atoms and

molecules in the system, which can have contributions from both molecular kinetic energy and

molecular potential energy. Consider the compression of a spring from an initial uncompressed

state as shown below.

Since it requires energy to compress the spring, we know that some kind of energy must be

stored within the spring. Since this change in energy can be attributed to a change of the

macroscopic position of the system and is not related to changes on the molecular scale, we

determine the form of energy to be potential energy. In this case, the spring’s tendency to restore

its original shape is the driving force that is analogous to the gravity for gravitational potential

energy.

This argument can be enhanced by the form of the expression that the increased energy takes. If

we consider the spring as the system, the energy it acquires in a reversible, compression from its

initial uncompressed state may be obtained from an energy balance. Assuming the process is

adiabatic, we obtain:

WWQE

We have left the energy in terms of the total energy, E. The work can be obtained by integrating

the force over the distance of the compression:

2

2

1kxkxdxdxFW

Hence:

2

2

1kxE

We see that the increase in energy depends on macroscopic position through the term x.

It should be noted that there is a school of thought that assigns this increased energy to internal

energy. This approach is all right as long as it is consistently done throughout the energy

balances on systems containing springs.

5

2.4

For the first situation, let the rubber band represent the system. In the second situation, the gas is

the system. If heat transfer, potential and kinetic energy effects are assumed negligible, the

energy balance becomes

WU

Since work must be done on the rubber band to stretch it, the value of the work is positive. From

the energy balance, the change in internal energy is positive, which means that the temperature

of the system rises.

When a gas expands in a piston-cylinder assembly, the system must do work to expand against

the piston and atmosphere. Therefore, the value of work is negative, so the change in internal

energy is negative. Hence, the temperature decreases.

In analogy to the spring in Problem 2.3, it can be argued that some of the work imparted into the

rubber band goes to increase its potential energy; however, a part of it goes into stretching the

polymer molecules which make up the rubber band, and the qualitative argument given above

still is valid.

6

2.5

To explain this phenomenon, you must realize that the water droplet is heated from the bottom.

At sufficiently high temperatures, a portion of the water droplet is instantly vaporized. The

water vapor forms an insulation layer between the skillet and the water droplet. At low

temperatures, the insulating layer of water vapor does not form. The transfer of heat is slower

through a gas than a liquid, so it takes longer for the water to evaporate at higher temperatures.

7

2.6

Apartment

System

Fri

dge

+-W

Q

HOT

Surr.

If the entire apartment is treated as the system, then only the energy flowing across the apartment

boundaries (apartment walls) is of concern. In other words, the energy flowing into or out of the

refrigerator is not explicitly accounted for in the energy balance because it is within the system.

By neglecting kinetic and potential energy effects, the energy balance becomes

WQU

The Q term represents the heat from outside passing through the apartment’s walls. The W term

represents the electrical energy that must be supplied to operate the refrigerator.

To determine whether opening the refrigerator door is a good idea, the energy balance with the

door open should be compared to the energy balance with the door closed. In both situations, Q

is approximately the same. However, the values of W will be different. With the door open,

more electrical energy must be supplied to the refrigerator to compensate for heat loss to the

apartment interior. Therefore,

shutajar WW

where the subscript “ajar” refers the situation where the door is open and the subscript “shut”

refers to the situation where the door is closed. Since,

shutajar QQ

shutshutshutajarajarajar WQUWQU

shutajar TT

The refrigerator door should remain closed.

8

2.7 The two cases are depicted below.

Let’s consider the property changes in your house between the following states. State 1, when

you leave in the morning, and state, the state of your home after you have returned home and

heated it to the same temperature as when you left. Since P and T are identical for states 1 and 2,

the state of the system is the same and U must be zero, so

0 WQU

or

WQ

where -Q is the total heat that escaped between state 1 and state 2 and W is the total work that

must be delivered to the heater. The case where more heat escapes will require more work and

result in higher energy bills. When the heater is on during the day, the temperature in the system

is greater than when it is left off. Since heat transfer is driven by difference in temperature, the

heat transfer rate is greater, and W will be greater. Hence, it is cheaper to leave the heater off

when you are gone.

9

2.8

The amount of work done at constant pressure can be calculated by applying Equation 2.57

QH

Hence,

hmQH ˆ

where the specific internal energy is used in anticipation of obtaining data from the steam tables.

The mass can be found from the known volume, as follows:

kg 0.1

kg

m0.0010

L

m001.0L1

ˆ 3

3

v

Vm

As in Example 2.2, we use values from the saturated steam tables at the same temperature for

subcooled water at 1 atm. The specific enthalpy is found from values in Appendix B.1:

kg

kJ 15.314

kg

kJ 87.104

kg

kJ 02.419C25at ˆC100at ˆˆ o

1,o

2, ll uuu

Solve for heat:

kJ 15.314kg

kJ 05.314kg 0.1ˆ

umQ

and heat rate:

kW 52.0

min

s 60min. 01

kJ 15.314

t

QQ

This value is the equivalent of five strong light bulbs.

10

2.9

(a) From Steam Tables:

kg

kJ 8.2967ˆ1u (100 kPa, 400 ºC)

kg

kJ 8.2659ˆ2u (50 kPa, 200 ºC)

kg

kJ 0.308ˆˆˆ 12 uuu

(b) From Equations 2.53 and 2.63

2

1

2

1

12

T

T

P

T

T

v dTRcdTcuuu

From Appendix A.2

)( 322 ETDTCTBTARcP

2

1

1322T

T

dTETDTCTBTARu

Integrating

)(

4)

11()(

3)(

2))(1(

41

42

12

31

32

21

2212 TT

E

TTDTT

CTT

BTTARu

The following values were found in Table A.2.1

0

1021.1

0

1045.1

470.3

4

3

E

D

C

B

A

Substituting these values and using

11

K 473.15K) 15.273200(

K 673.15K) 15.273400(

Kmol

J 314.8

2

1

T

T

R

provides

kg

kJ 1.308

J 1000

kJ 1

kg 1

g 1000

O]H g[ 0148.18

O]H [mol 1

mol

J 5551ˆ

mol

J 5551

2

2u

u

The values in parts (a) and (b) agree very well. The answer from part (a) will serve as the basis

for calculating the percent difference since steam table data should be more accurate.

%03.0%1000.308

1.308308%

Difference

12

2.10

(a) Referring to the energy balance for closed systems where kinetic and potential energy are

neglected, Equation 2.30 states

WQU

(b) Since internal energy is a function of temperature only for an ideal gas (Equation 2.4) and the

process is isothermal

0U

According to Equation 2.77

1

211

1

2 lnlnP

PRTn

P

PnRTW

From the ideal gas law:

1111 VPRTn

1

211 ln

P

PVPW

Substitution of the values from the problem statement yields

J 940

bar 8

bar 5lnm 105.2Pa 108 335

W

W

The energy balance is

J 940

J 0

Q

WQ

(c) Since the process is adiabatic

0Q

The energy balance reduces to

WU

13

The system must do work on the surroundings to expand. Therefore, the work will be negative

and

12

0

0

2

1

TT

TcnU

U

v

T

T

T2 will be less than 30 ºC

14

2.11

(a) (i).

2

P [b

ar]

v [m3/mol]

1

2

3

0.01 0.030.02

1

Path

A

Path B

(ii).

Since internal energy is a function of temperature only for an ideal gas (Equation 2.4) and the

process is isothermal

0u

Equation 2.48 states that enthalpy is a function of temperature only for an ideal gas. Therefore,

0h

Performing an energy balance and neglecting potential and kinetic energy produces

0 wqu

For an isothermal, adiabatic process, Equation 2.77 states

1

2lnP

PnRTW

or

1

2lnP

PRT

n

Ww

Substituting the values from the problem statement gives

15

bar 3

bar 1lnK )15.27388(

Kmol

J 8.314w

mol

J 3299w

Using the energy balance above

mol

J 3299wq

(b) (i). See path on diagram in part (a)

(ii).

Since the overall process is isothermal and u and h are state functions

0u

0h

The definition of work is

dvPw E

During the constant volume part of the process, no work is done. The work must be solved for

the constant pressure step. Since it is constant pressure, the above equation simplifies to

)( 12 vvPdvPw EE

The ideal gas law can be used to solve for 2v and 1v

mol

m 010.0

Pa 103

K )15.27388(K

molJ 314.8

mol

m 030.0

Pa 101

K )15.27388(K

molJ 314.8

3

5

1

11

3

5

2

22

P

RTv

P

RTv

Substituting in these values and realizing that 1PPE since the process is isobaric produces

16

mol

m 010.0

mol

m 030.0Pa) 103(

335w

mol

J 6000w

Performing an energy balance and neglecting potential and kinetic energy results in

0 wqu

mol

J 6000wq

17

2.12

First, perform an energy balance. No work is done, and the kinetic and potential energies can be

neglected. The energy balance reduces to

QU

We can use Equation 2.53 to get

2

1

T

T

vdTcnQ

which can be rewritten as

2

1

T

T

PdTcnQ

since the aluminum is a solid. Using the atomic mass of aluminum we find

mol 3.185

mol

kg 0.02698

kg 5

n

Upon substitution of known values and heat capacity data from Table A.2.3, we get

K 15.323

K 15.294

3 1049.1486.2Kmol

J 314.8mol 3.185 dTTQ

kJ 61.131Q

18

2.13

First, start with the energy balance. Potential and kinetic energy effects can be neglected.

Therefore, the energy balance becomes

WQU

The value of the work will be used to obtain the final temperature. The definition of work

(Equation 2.7) is

2

1

V

V

EdVPW

Since the piston expands at constant pressure, the above relationship becomes

12 VVPW E

From the steam tables

kg

m 02641.0ˆ

3

1v (10 MPa, 400 ºC)

33

111 m 07923.0kg

m 02641.0kg) 3(ˆ

vmV

Now 2V and 2v are found as follows

3

6

312 m 4536.0

Pa 100.2

J 748740m 07923.0

EP

WVV

kg

m 1512.0

kg 3

m 4536.0ˆ

33

2

22

m

Vv

Since 2v and 2P are known, state 2 is constrained. From the steam tables:

Cº 4002 T

kg

m 0.1512 bar, 20

3

Now U will be evaluated, which is necessary for calculating Q . From the steam tables:

19

kg

kJ 2.2945ˆ2u

kg

m 0.1512 bar, 20

3

kg

kJ 4.2832ˆ1u Cº 400 bar, 001

kJ 4.338kg

kJ 4.2832

kg

kJ 2.2945kg 3ˆˆ

121

uumU

Substituting the values of U and W into the energy equation allows calculation of Q

WUQ

J 1009.1J 748740 [J] 338400 6Q

20

2.14

In a reversible process, the system is never out of equilibrium by more than an infinitesimal

amount. In this process the gas is initially at 2 bar, and it expands against a constant pressure of

1 bar. Therefore, a finite mechanical driving force exists, and the process is irreversible.

To solve for the final temperature of the system, the energy balance will be written. The piston-

cylinder assembly is well-insulated, so the process can be assumed adiabatic. Furthermore,

potential and kinetic energy effects can be neglected. The energy balance simplifies to

WU

Conservation of mass requires

21 nn

Let 21 nnn

The above energy balance can be rewritten as

2

1

2

1

V

V

E

T

T

v dVPdTcn

Since vc and EP are constant:

1212 VVPTTnc Ev

2V and 1T can be rewritten using the ideal gas law

2

22

P

nRTV

nR

VPT 11

1

Substituting these expressions into the energy balance, realizing that 2PPE , and simplifying

the equation gives

nR

VPP

T

2

7

2

5112

2

Using the following values

21

Kmol

barL 08314.0

mol 0.1

L 10

bar 1

bar 2

1

2

1

R

n

V

P

P

results in

K 2062 T

To find the value for work, the energy balance can be used

12 TTncUW v

Before the work can be calculated, 1T must be calculated

K 241

Kmol

barL 08314.0mol 1

L 10bar 2111

nR

VPT

Using the values shown above

J 727W

22

2.15

The maximum work can be obtained through a reversible expansion of the gas in the piston.

Refer to Section 2.3 for a discussion of reversible processes. The problem states that the piston

assembly is well-insulated, so the heat transfer contribution to the energy balance can be

neglected, in addition to potential and kinetic energy effects. The energy balance reduces to

WU

In this problem, the process is a reversible, adiabatic expansion. For this type of process,

Equation 2.90 states

11221

1VPVP

kW

From the problem statement (refer to problem 2.13),

bar 1

L 10

bar 2

2

1

1

P

V

P

To calculate W, 2V must be found. For adiabatic, reversible processes, the following

relationship (Equation 2.89) holds:

constPV k

where k is defined in the text. Therefore,

kkV

P

PV

1

12

12

Noting that 5

7

v

P

c

ck and substituting the proper values provides

L 4.162 V

Now all of the needed values are available for calculating the work.

J 900barL 9 W

From the above energy balance,

J 900U

23

The change in internal energy can also be written according to Equation 2.53:

2

1

T

T

vdTcnU

Since vc is constant, the integrated form of the above expression is

122

5TTRnU

Using the ideal gas law and knowledge of 1P and 1V ,

K 6.2401 T

and

K 3.1972 T

The temperature is lower because more work is performed during the reversible expansion.

Review the energy balance. As more work is performed, the cooler the gas will become.

24

2.16

Since the vessel is insulated, the rate of heat transfer can be assumed to be negligible.

Furthermore, no work is done on the system and potential and kinetic energy effects can be

neglected. Therefore, the energy balance becomes

0ˆ u

or

12 ˆˆ uu


kg

kJ 2.2619ˆ1u (200 bar, 400 ºC)

kg

kJ 2.2619ˆ2u

The values of 2u and 2P constrain the system. The temperature can be found from the steam

tables using linear interpolation:

Cº 5.3272 T

kg

kJ 2.2619ˆ ,bar 100 2u

Also at this state,

kg

m 02012.0ˆ

3

2v

Therefore,

33

2 m 020.0kg

m 02012.0kg 0.1

mvVvessel

25

2.17

Let the entire tank represent the system. Since no heat or work crosses the system boundaries,

and potential and kinetic energies effects are neglected, the energy balance is

0u

Since the tank contains an ideal gas

K 300

0

12

12

TT

TT

The final pressure can be found using a combination of the ideal gas law and conservation of

mass.

22

2

11

1

VP

T

VP

T

We also know

12 2VV

Therefore,

bar 52

12

PP

26

2.18

(a) First, as always, simplify the energy balance. Potential and kinetic energy effects can be

neglected. Therefore, the energy balance is

WQU

Since, this system contains water, we can the use the steam tables. Enough thermodynamic

properties are known to constrain the initial state, but only one thermodynamic property is

known for the final state: the pressure. Therefore, the pressure-volume relationship will be used

to find the specific volume of the final state. Since the specific volume is equal to the molar

volume multiplied by the molecular weight and the molecular weight is constant, the given

expression can be written

constvP 5.1ˆ

This equation can be used to solve for 2v .

5.1

1

5.11

2

12 ˆˆ

v

P

Pv

Using

bar 100

kg

m 1.0

kg 10

m 1.0 ˆ

bar 20

2

33

1

1

P

v

P

gives

kg

m 0.0342 ˆ

3

2v

Now that the final state is constrained, the steam tables can be used to find the specific internal

energy and temperature.

kg

kJ 6.3094ˆ

K 7.524

2

2

u

T

To solve for the work, refer to the definition (Equation 2.7).

27

2

1

V

V

EdVPW

or

2

1

ˆ

ˆ

ˆˆ

v

v

E vdPw

Since the process is reversible, the external pressure must never differ from the internal pressure

by more than an infinitesimal amount. Therefore, an expression for the pressure must be

developed. From the relationship in the problem statement,

constvPvP 5.1

115.1 ˆˆ

Therefore, the expression for work becomes

2

1

2

1

ˆ

ˆ5.1

5.111

ˆ

ˆ5.1

5.111 ˆ

ˆ

1ˆˆ

ˆ

ˆˆ

v

v

v

v

vdv

vPvdv

vPw

Integration and substitution of proper values provides

kg

kJ 284

kg

mbar 840.2ˆ

3

w

kJ 2840kg

kJ 284kg 10

W

A graphical solution is given below:

28

To solve for Q , U must first be found, then the energy balance can be used.

kJ 4918 kg

kJ 8.2602

kg

kJ 6.3094kg 10ˆˆ 12

uumU

Now Q can be found,

kJ 2078kJ 2840kJ 4918 WUQ

(b)

Since the final state is the same as in Part (a), U remains the same because it is a state function.

The energy balance is also the same, but the calculation of work changes. The pressure from the

weight of the large block and the piston must equal the final pressure of the system since

mechanical equilibrium is reached. The calculation of work becomes:

2

1

ˆ

ˆ

ˆ

v

v

E vdmPW

All of the values are known since they are the same as in Part (a), but the following relationship

should be noted

2PPE

Substituting the appropriate values results in

kJ 6580W

Again we can represent this process graphically:

29

Now Q can be solved.

kJ 1662kJ 6580kJ 4918 WUQ

(c) This part asks us to design a process based on what we learned in Parts (a) and (b). Indeed, as is

characteristic of design problems there are many possible alternative solutions. We first refer to

the energy balance. The value of heat transfer will be zero when

WU

For the same initial and final states as in Parts (a) and (b),

kJ 4918 UW

There are many processed we can construct that give this value of work. We show two

alternatives which we could use:

Design 1:

If the answers to Part (a) and Part (b) are referred to, one can see that two steps can be used: a

reversible compression followed by an irreversible compression. Let the subscript “i" represent

the intermediate state where the process switches from a reversible process to an irreversible

process. The equation for the work then becomes

J 4918000ˆˆ

1ˆ)ˆˆ(

ˆ

ˆ5.1

5.11122

1

i

v

v

i vdv

vPvvPmW

Substituting in known values (be sure to use consistent units) allows calculation of iv :

kg

m 0781.0ˆ

3

iv

The pressure can be calculated for this state using the expression from part (a) and substituting

the necessary values.

bar 0.29

5.11

1

i

ii

P

v

vPP

30

Now that both iP and iv are known, the process can be plotted on a P-v graph, as follows:

Design 2:

In an alternative design, we can use two irreversible processes. First we drop an intermediate

weight on the piston to compress it to an intermediate state. This step is followed by a step

similar to Part (b) where we drop the remaining mass to lead to 100 bar external pressure. In this

case, we again must find the intermediate state. Writing the equation for work:

J 4918000)ˆˆ()ˆˆ( 221 iii vvPvvPmW

However, we again have the relationship:

5.1

11

i

iv

vPP

Substitution gives one equation with one unknown vi:

J 4918000)ˆˆ()ˆˆ( 221

5.1

11

ii

i

vvPvvv

vPmW

There are two possible values vi to the above equation.

Solution A:

kg

m 043.0ˆ

3

iv

31

which gives

bar 8.70iP

This solution is graphically shown below:

Solution B

kg

m 0762.0ˆ

3

iv

which gives

bar 0.30iP

This solution is graphically shown below:

32

2.19

(a)

Force balance to find k:

Piston

Fspring=kx Fatm=PatmA

Fgas=PgasAFmass=mg

Pgas Patm mgA kxA

since V=Ax

Pgas Patm mgA kV

A2

since V is negative. Now solve:

22

3

2

255

m 1.0

m 02.0

m 1.0

m/s 81.9kg 2040Pa 101Pa 102

k

m

N1001.5 4k

Work can be found graphically (see P-V plot) or analytically as follows: Substituting the

expression in the force balance above:

WA Patm mgA kV

A2

dV

f

i

f

i

V

V

V

V

atmA VdA

VkdV

A

mgPW

2

2

0m02.0

m 1.0

N/m 1001.5m 05.003.0Pa1000.3

223

22

425

AW

33

J 105 3AW

V [m 3]

1

2

3

0.01 0.03 0.05

Patm

mgA

kV

A2

-W 10squares0.5kJ

square = 5kJ

Work

P [b

ar]

2

1

=

(b)

You need to find how far the spring extends in the intermediate (int) position. Assume

PVn=const (other assumptions are o.k, such as an isothermal process, and will change the answer

slightly). Since you know P and V for each state in Part (a), you can calculate n.

PiP f

ViV f

n

or 105 Pa

2105

Pa 0.03 m3

0.05 m3

n

n 1.35

Now using the force balance

Pgas,int Patm mgA kV

A2

with the above equation yields:

Pint PiViVint

n Patm

mgA

k VintVi A2

This last equality represents 1 equation. and 1 unknown (we know k), which gives

3int m 0385.0V

Work can be found graphically (see P-V plot) or analytically using:

34

WB PgasVi

V int

dV PgasVint

V f

dV

expanding as in Part (a)

WB 2105 N

m2

Vi

V int

dV 5106 N

m5 V

V i

Vint

d V 3105 N

m2

Vint

V f

dV 5106 N

m5 V

Vint

V i f

d V

Therefore,

kJ85.3BW

just like we got graphically.

V [m 3]

1

2

3

0.01 0.03 0.05

Patm

mgA

P [b

ar]

2

1

Work

mgA

(c)

The least amount of work is required by adding differential amounts of mass to the piston. This

is a reversible compression. For our assumption that PVn = const, we have the following

expression:

f

i

f

i

V

V

V

V

gasrevC dVV

constdVPW

35.1,

Calculate the constant from the initial state

35

335.13 5 1075.1m05.0Pa101. const

Therefore,

WC,rev 1.75103

.05

.03

dV

V1.35 1.75103

0.35V0.35

.05

.03 J2800

36

2.20

Before this problem is solved, a few words must be said about the notation used. The system

was initially broken up into two parts: the constant volume container and the constant pressure

piston-cylinder assembly. The subscript “1” refers to the constant volume container, “2” refers

the piston-cylinder assembly. “i" denotes the initial state before the valve is opened, and “f”

denotes the final state.

To begin the solution, the mass of water present in each part of the system will be calculated.

The mass will be conserved during the expansion process. Since the water in the rigid tank is

saturated and is in equilibrium with the constant temperature surroundings (200 ºC), the water is

constrained to a specific state. From the steam tables,

kPa 8.1553

kg

kJ 3.2595ˆ

kg

kJ 64.850ˆ

kg

kJ 12736.0ˆ

kg

kJ 001156.0ˆ

,1

,1

,1

,1

sat

vi

li

vi

li

P

u

u

v

v

(Sat. water at 200 ºC)

Knowledge of the quality of the water and the overall volume of the rigid container can be used

to calculate the mass present in the container.

vi

li vmvmV ,11,111 ˆ95.0ˆ05.0

Using the values from the steam table and 31 m 5.0V provides

kg 13.41 m

Using the water quality specification,

kg 207.005.0

kg 92.395.0

11

11

mm

mm

l

v

For the piston-cylinder assembly, both P and T are known. From the steam tables

37

kg

kJ 9.2638ˆ

kg

m 35202.0ˆ

,2

3

,2

i

i

u

v

(600 kPa, 200 ºC)

Enough information is available to calculate the mass of water in the piston assembly.

kg 284.0ˆ2

22

v

Vm

Now that the initial state has been characterized, the final state of the system must be determined.

It helps to consider what physically happens when the valve is opened. The initial pressure of

the rigid tank is 1553.8 kPa. When the valve is opened, the water will rush out of the rigid tank

and into the cylinder until equilibrium is reached. Since the pressure of the surroundings is

constant at 600 kPa and the surroundings represent a large temperature bath at 200 ºC, the final

temperature and pressure of the entire system will match the surroundings’. In other words,

kg

kJ 9.2638ˆˆ ,2 if uu (600 kPa, 200 ºC)

Thus, the change in internal energy is given by

li

li

vi

viif

li

vi

umumumummmU,1,1,1,1,22,1,12 ˆˆˆˆ)(

Substituting the appropriate values reveals

kJ 0.541U

To calculate the work, we realize the gas is expanding against a constant pressure of 600 kPa

(weight of the piston was assumed negligible). From Equation 2.7,

)( ifE

V

V

E VVPdVPW

f

i

where

333

3,2,1,12

m 6.0m 5.0m 0.1

m 55.1ˆ)(

Pa 600000

i

il

iv

if

E

V

vmmmV

P

38

Note: iv ,2ˆ was used to calculate fV because the temperature and pressure are the same

for the final state of the entire system and the initial state of the piston-cylinder assembly.

The value of W can now be evaluated.

kJ 570W

The energy balance is used to obtain Q.

kJ 1111kJ 570kJ 0.541 WUQ

39

2.21

A sketch of the process follows:

The initial states are constrained. Using the steam tables, we get the following:

State 1,A State 1,B

p 10 [bar] 20 [bar]

T 700 [oC] 250 [

oC]

v 0.44779 [m3/kg] 0.11144 [m

3/kg]

u 3475.35 [kJ/kg] 2679.58 [kJ/kg]

V 0.01 m3 0.05 m

3

m V

v 0.11 [kg] 0.090 [kg]

All the properties in the final state are equal. We need two properties to constrain the system:

We can find the specific volume since we know the total volume and the mass:

kg

m 30.0

kg 0.20

m 06.0 33

,1,1

,1,12

BA

BA

mm

VVv

We can also find the internal energy of state 2. Since the tank is well insulated, Q=0. Since it is

rigid, W=0. An energy balance gives:

U QW 0

Thus,

U2 U1 m1,Au1, A m1,Bu1,B

or

kg

kJ 3121

,1,1

,1,1,1,1

2

22

BA

BBAA

mm

umum

m

Uu

We have constrained the system with u2 and v2, and can find the other properties from the steam

Tables. Very close to

T2 = 500 [oC] and P2 = 1200 [kPa]

Thus,

40

m 267.0kg 0.09kg

m 30.0

3

,2,2,2

AAA mvV and

m 167.01.0267.0 x

41

2.22

We start by defining the system as a bubble of vapor rising through the can. We assume the

initial temperature of the soda is 5 oC. Soda is usually consumed cold; did you use a reasonable

estimate for T1? A schematic of the process gives:

where the initial state is labeled state 1, and the final state is labeled state 2. To find the final

temperature, we perform an energy balance on the system, where the mass of the system (CO2 in

the bubble) remains constant. Assuming the process is adiabatic and potential and kinetic energy

effects are negligible, the energy balance is

wu

Expressions for work and internal energy can be substituted to provide

1212 vvPdvPTTc EEv

where cv = cP – R. Since CO2 is assumed an ideal gas, the expression can be rewritten as

1

122

1

1

2

212

P

TPTR

P

T

P

TRPTTc Ev

where the equation was simplified since the final pressure, P2, is equal to the external pressure,

PE. Simplifying, we get:

1

212 11

Pc

RPT

c

RT

vv

or

K 23711

212

P

v

v c

c

Pc

RPTT

42

2.23

The required amount of work is calculated as follows:

VPW

The initial volume is zero, and the final volume is calculated as follows:

3333 m0436.0ft 54.1ft 5.03

4

3

4 ππrV

Assuming that the pressure is 1 atm, we calculate that

J4417m 0m 0436.0Pa1001325.1 33 5 W

This doesn’t account for all of the work because work is required to stretch the rubber that the

balloon is made of.

43

2.24

(a)

Since the water is at its critical point, the system is constrained to a specific temperature,

pressure, and molar volume. From Appendix B.1

kg

m 003155.0ˆ

3

cv

Therefore,

kg 17.3

kg

m 003155.0

m 01.0

ˆ 3

3

cv

Vm

(b)

The quality of the water is defined as the percentage of the water that is vapor. The total volume

of the vessel can be found using specific volumes as follows

vlvvll vxmvmxvmvmV ˆˆ1ˆˆ

where x is the quality of the water. To solve for the quality, realize that starting with saturated

water at a pressure of 1 bar constrains the water. From the steam tables,

kg

m 001043.0ˆ

kg

m 6940.1ˆ

3

3

l

v

v

v

(sat. H2O at P = 1 bar)

Now the quality can be found

00125.0x

Thus, the quality of the water is 0.125%.

(c)

To determine the required heat input, perform an energy balance. Potential and kinetic energy

effects can be neglected, and no work is done. Therefore,

QU

44

where

vl uxmumxumU112 ˆˆ1ˆ


kg

kJ 58.2029ˆ2u (H2O at its critical point)

kg

m 1.2506ˆ

kg

m 33.417ˆ

3

1

3

1

v

l

u

u

(sat. H2O at P=1 bar)

Evaluation of the expression reveals

J 1010.5kJ 6.5102 6U

45

2.25

(a)

Consider the air in ChE Hall to be the system. The system is constant volume, and potential and

kinetic energy effects can be neglected. Furthermore, disregard the work. The energy balance is

qdt

du

since the temperature of the system changes over time. Using the given expression for heat

transfer and the definition of dU , the expression becomes

surrv TThdt

dTc

We used a negative sign since heat transfer occurs from the system to the surroundings. If vc is

assumed constant, integration provides

ChtTTc surrv ln

where C is the integration constant. Therefore,

t

c

h

surrveCTT 1

where C1 is a constant. Examining this equation reveals that the temperature is an exponential

function of time. Since the temperature is decreasing, we know that the plot of temperature vs.

time shows exponential decay.

time

Tem

pera

ture

T0

Tsurr

46

(b)

Let time equal zero at 6 PM, when the steam is shut off. At 6 PM, the temperature of the hall is

22 ºC. Therefore,

t

c

h

surrveCTT 1

)0(1Cº 2 Cº 22 eC

Cº 201 C

After 10 PM, ( hr 4t ), the temperature is 12 ºC.

hr) 4(

Cº 20Cº 2 Cº 21 vc

h

e

1-hr 173.0vc

h

At 6 AM, hr 12t . Substitution of this value into the expression for temperature results in

Cº 5.4T

47

2.26

The gas leaving the tank does flow work as it exits the valve. This work decreases the internal

energy of the gas – lowering the temperature. During this process, water from the atmosphere

will become supersaturated and condense. When the temperature drops below the freezing point

of water, the water forms a solid.

Attractive interactions between the compressed gas molecules can also contribute to this

phenomena, i.e., it takes energy to pull the molecules apart as they escape; we will learn more of

these interactions in Chapter 4.

48

2.27

Mass balance

inoutin mmmdt

dm

Separating variables and integrating:

t

in

m

m

dtmdm0

2

1

or

t

indtmmm0

12

Energy balance

Since the potential and kinetic energy effects can be neglected, the open system, unsteady state

energy balance is

out

s

in

ininoutoutsys

WQhmhmdt

dU

The process is adiabatic and no shaft work is done. Furthermore, there is only one inlet stream

and not outlet stream. Therefore, the energy balance simplifies to

ininsys

hmdt

dU

The following math is performed

in

t

inin

t

inin

U

U

hmmumumUU

dtmhdthmdU

ˆˆˆ12112212

00

2

1

where the results of the mass balance were used. Both 2m and 1m can be calculated by dividing

the tank volume by the specific volume

49

11

22

ˆ

ˆ

v

Vm

v

Vm

Substitution of these relationships and simplification results in

0ˆ

ˆ

ˆ

ˆ

1

1

2

2

v

hu

v

hu inin


kg

m 19444.0ˆ

kg

kJ 6.2583ˆ

3

1

1

v

u

(sat. H2O vapor at 1 MPa)

kg

kJ 2.3177ˆ

inh (6 MPa, 400 ºC)

There are still two unknowns for this one equation, but the specific volume and internal energy

are coupled to each other. To solve this problem, guess a temperature and then find the

corresponding volume and internal energy values in the steam tables at 6 MPa. The correct

temperature is the one where the above relationship holds.

Cº 600T : Expression = 4427.6

Cº 500T : Expression = 1375.9

Cº 450T : Expression = -558.6

Interpolation between 500 ºC and 450 ºC reveals that the final temperature is

Cº 4.4642 T

50

2.28

We can pick room temperature to be 295 K

Tin T1 295 K

Mass balance

inoutin nnndt

dn


t

in

n

n

dtndn0

2

1

or

t

indtnnn0

12

Energy balance

Neglecting ke and pe, he unsteady energy balance, written in molar units is written as:

WQhnhndt

dUoutoutinin

sys

The terms associated with flow out, heat and work are zero.

ininsys

hndt

dU

Integrating both sides with respect to time from the initial state where the pressure is 10 bar to

the final state when the tank is at a pressure of 50 bar gives:

dtnhdthndU

t

inin

t

inin

U

U

00

2

1

since the enthalpy of the inlet stream remains constant throughout the process. Integrating and

using the mass balance above:

inhnnunun 121122

51

Now we do some math:

inhnnunun 121122

inin hunhun 1122

By the definition of h

11 RTuRTuvPuh inininininin

so

1111112122 TRnuunRTnuun

1112122 RTnRTnTTcn v

Since RRcc Pv2

3

n23

2T2 T1 n2T1 n1T1

or

111222 253 TnTnTn

dividing by n1:

111

22

1

2 253 TTn

nT

n

n

Using the ideal gas law:

21

12

1

2

TP

TP

n

n

so

1121

122

21

12 253 TTTP

TPT

TP

TP

or

52

[K] 434

23

5

1

2

1

12

2

P

P

P

TP

T

(b) Closed system

u qw q

kg

kJ 9.28

2

5ˆ 1212 TT

MW

RTT

MW

cu v

kg

kJ 9.28q

(c)

P2T2 P3T3

bar 343

223

T

TPP

53

2.29

Mass balance

inoutin nnndt

dn


t

in

n

n

dtndn00

2

1

or

t

indtnn0

2

Energy balance

Neglecting ke and pe, the unsteady energy balance, in molar units, is written as:

WQhnhndt

dUoutoutinin

sys

The terms associated with flow out and heat are zero.

Whndt

dUinin

sys

Integrating both sides with respect to time from the empty initial state to the final state gives:

WnhWdtnhdtWdthndU in

t

inin

tt

inin

U

U

2

000

2

1

since the enthalpy of the inlet stream remains constant throughout the process. The work is

given by:

22122 )( vPnvvPnW extext

54

2222 vPhnun extin

Rearranging,

222 vPvPuvPhu extinininextin

22 vPvPuu extininin

2

22

P

TPRRTTTc ext

ininv

so

K 333

2

2

in

extv

v T

RP

Pc

RcT

55

2.30

valve maintains pressure in system constant

T1 = 200 oC

x1 = 0.4

V = 0.01 m3

v

l

Mass balance

outoutin mmmdt

dm


t

out

m

m

dtmdm0

2

1

or

t

outdtmmm0

12

Energy balance

Qhmdt

dUoutout

sys

ˆ

Integrating

t tt

outoutoutout

um

um

dtQdtmhdtQhmdU0 00

ˆ

ˆ

ˆˆ22

11

Substituting in the mass balance and solving for Q

outhmmumumQ ˆˆˆ121122

56

We can look up property data for state 1 and state 2 from the steam tables:

kg

m 0.0511274.04.0001.6.0ˆˆ)1(ˆ

3

1 gf vxvxv

kg

m 1274.0ˆ

3

2v

So the mass in each state is:

kg 196.0

kg

m 0.051

m 01.0

ˆ 3

3

1

11

v

Vm

kg 0785.0

kg

m 0.1274

m 01.0

ˆ 3

3

2

22

v

Vm

kg 1175.012 mm

And for energy and enthalpy

kg

kJ 54915.25974.064.8506.0ˆˆ)1(1 gf uxuxu

kg

kJ 3.2595ˆ

2u

kg

kJ 2.2793ˆ

outh

Solving for heat, we get

kJ 228ˆˆˆ121122 outhmmumumQ

57

2.31

Consider the tank as the system.

Mass balance

inoutin mmmdt

dm


t

in

m

m

dtmdm0

2

1

or

t

indtmmm0

12

Energy balance

Since the potential and kinetic energy effects can be neglected, the open system, unsteady state

energy balance is

out

s

in

ininoutoutsys

WQhmhmdt

dU

The process is adiabatic and no shaft work is done. Furthermore, there is only one inlet stream

and not outlet stream. Therefore, the energy balance simplifies to

ininsys

hmdt

dU

The following math is performed

in

t

inin

t

inin

U

U

hmumU

dtmhdthmdU

ˆˆ2222

000

2

1

where the results of the mass balance were used. Thus,

inhu ˆˆ2


58

kg

kJ 5.3632ˆ2u (9 MPa, 800 ºC)

so

kg

kJ 5.3632ˆ

inh

We can use the value of inh and the fact that the steam in the pipe is at 9 MPa to find the

temperature.

Cº 600inT

59

2.32

(a)

First, the energy balance must be developed. Since the problem asks how much energy is stored

in the battery after 10 hours of operation, the process is not steady-state. Let the battery be the

system. Potential and kinetic energy effects can be neglected. Furthermore, heating of the

battery as it is charged can be ignored. The energy balance is

ssys

WQdt

dU

No shaft work is performed, but electrical is supplied to the battery, which must be accounted for

in sW . The value of Q is given explicitly in the problem statement. Both of these values remain

constant over time, so integration provides

tWQU s

From the problem statement

s 36000

kW 1

kW 5

t

Q

Ws

Substituting these values allows the calculation of the amount of energy stored:

MJ 144kJ 000,144 U

(b)

To calculate the velocity of the falling water, an energy balance must be developed with the

water passing through the electricity generator (probably a turbine) as the system, where the

water enters with a velocity 1V

and leaves with a negligible velocity, which will be approximated

as 0. Assume that potential energy changes can be neglected. Furthermore, assume that the

temperature of the water does not change in the process, so the change in internal energy is zero.

Also, view the process as adiabatic. The energy balance reduces to

riverK WE

where riverW is the power of the flowing water. The actual power being provided by the stream

can be calculated using the efficiency information. Let represent the efficiency.

60

kW 105.0

kW 5

sriver

river

s

WW

W

W

The value of riverW should be negative since the water is supplying work that is stored electrical

energy. Therefore, the energy balance becomes

W 10000 KE

This expression can be rewritten as

W 100002

1 21

22 VVm

From the problem statement and the assumptions made,

s

m 0

s

kg 200

2V

m

Therefore,

s

m 101V

There are a number of reasons for the low conversion efficiency. A possible potential energy

loss inherent in the design of the energy conversion apparatus decreases the efficiency. Heat is

lost to the surroundings during conversion. Some of the energy is also lost due to friction (drag)

effects.

61

2.33

Considering the turbine to be the system, rearrangement of the steady-state, open system energy

balance provides

out

s

in

inpkinoutpkout WQeehneehn )()(

Performing a mass balance reveals

21 nnnn outin

Assuming the rate of heat transfer and potential energy effects are negligible and realizing that

there is one inlet and one outlet allows the simplification of the above equation to

1,2,122 KKs eehhnW

12 hh can be rewritten using Equations 2.58 and Appendix A.2

2

1

32212

T

T

p dTETDTCTBTARdTchh

Since the quantity 1,2, kk ee is multiplied by n , it is rewritten as follows for dimensional

homogeneity

2

1

2

21,2, )(2

1VVMWee airkk

To solve for n , the ideal gas law is used

2

222

21222

RT

VPn

RTnVP

To solve for the volumetric flow rate, the fluid velocity must be multiplied by the cross-sectional

area

4

2

2

22

VDV

The energy balance is now

62

21

22

32222

2

2

2 )(2

1

4

2

1

VVMWdTETDTCTBTARVD

RT

PW air

T

T

s

Substituting values from Table A.2.1 and the problem statement results in

[MW] -4.84[W] 1084.4 6 W

63

2.34

First, a sketch of the process is useful:

q

30 bar

100 oC

20 bar

150 oC

To find the heat in we will apply the 1st law. Assuming steady state, the open system energy

balance with one stream in and one stream out can be written:

Qhhn 210

which upon rearranging is:

12 hhn

Q

Thus this problem reduces to finding the change in the thermodynamic property, enthalpy from

the inlet to the outlet. We know 2 intensive properties at both the inlet and outlet so the values

for the other properties (like enthalpy!) are already constrained. From Table A.2.1, we have an

expression for the ideal gas heat capacity:

263 10392.410394.14424.1 TTR

cp

with T in (K). Since this expression is limited to ideal gases any change in temperature must be

under ideal conditions. From the definition of heat capacity:

2

1

26312 10392.410394.14424.1

T

T

dTTThhn

Q

By integrating and substituting the temperatures, we obtain:

mol

J5590

n

Q

64

2.35

A schematic of the process follows:

To solve for nWs / we need a first law balance. With negligible eK and eP, the 1st law for a

steady state process becomes:

sWQhhn 210

If heat transfer is negligible,

hn

Ws

We can calculate the change in enthalpy from ideal gas heat capacity data provided in the

Appendix.

2

1

2

1

263 10824.810785.28213.1

T

T

T

T

ps dTTTRdTch

n

W

Integrate and evaluate:

mol

J 5358

n

Ws

65

2.36

(a)

First start with the energy balance. Nothing is mentioned about shaft work, so the term can be

eliminated from the energy balance. The potential and kinetic energy effects can also be

neglected. Since there is one inlet and one outlet, the energy balance reduces to

1122 hnhnQ

A mass balance shows

12 nn

so the energy balance reduces to

121 hhnQ

Using the expressions from Appendix A.1, the energy balance becomes

2

1

3221

T

T

dTETDTCTBTARnQ

Using

0

10031.0

0

10557.0

376.3

5

3

E

D

C

B

A

Kmol

J 314.8R

s

mol 201n

K 15.773

K 15.373

2

1

T

T

gives

kW 1.245W 245063 Q

66

(b) To answer this question, think about the structure of n-hexane and carbon monoxide. N-hexane

is composed of 20 atoms, but carbon monoxide has two. One would expect the heat capacity to

be greater for n-hexane since there are more modes for molecular kinetic energy (translational,

kinetic, and vibrational). Because the heat capacity is greater and the rate of heat transfer is the

same, the final temperature will be less.

67

2.37 First start with the energy balance around the nozzle. Assume that heat transfer and potential

energy effects are negligible. The shaft work term is also zero. Therefore, the energy balance

reduces to

0)()( 1122 KK ehnehn


21 nn

On a mass basis, the energy balance is

22

212,1,12

2

1ˆˆˆˆ VVeehh KK

Since the steam outlet velocity is much greater than the velocity of the inlet, the above

expression is approximately equal to

2212

2

1ˆˆ Vhh

The change in enthalpy can be calculated using the steam tables.

kg

J 109.2827 3

1h (10 bar, 200 ºC)

kg

J 105.2675 3

2h (sat. H2O(v) at 100 kPa)

Therefore,

s

m 5522V

To solve for the area, the following relationship is used

2

2

v

VAm


68

kg

m 6940.1ˆ

3

2v

Now all but one variable is known.

23 m 1007.3 A

69

2.38

First start with the energy balance around the nozzle. Assume that heat transfer and potential

energy effects are negligible. The shaft work term is also zero. Therefore, the energy balance

reduces to

0)()( 1122 kk ehnehn

The molar flow rates can be eliminated from the expression since they are equal. Realizing that

1,2, KK ee since the velocity of the exit stream is much larger than the velocity of the inlet

stream simplifies the energy balance to

2,12 kehh

Using Appendix A.2 and the definition of kinetic energy

22

32212

83

2

1

)(2

1VMWdTETDTCTBTARhh HC

T

T

From Table A.2.1

0

0

10824.8

10785.28

213.1

6

3

E

D

C

B

A

It is also important that the units for the molecular weight and universal gas constant are

consistent. The following values were used

Kmol

J 314.8R

mol

kg 0441.0)(

83HCMW

Integration of the above expression and then solving for 2T provides

K 2.4192 T

70

2.39

First start an energy balance around the diffuser. Assume that heat transfer and potential energy

effects are negligible. The shaft work term is also zero. The energy balance reduces to

0)()( 1122 kk ehnehn

A mass balance reveals

21 nn

The molar flow rates can be eliminated from the expression. Using the definitions of enthalpy

and the kinetic energy, the equation can be rewritten as

21

22)(

2

12

1

VVMWdTc air

T

T

P

The temperature and velocity of the outlet stream are unknown, so another equation is needed to

solve this problem. From the conservation of mass,

2

222

1

111

1

11

T

VAP

T

VAP

T

VP

where A2, the cross-sectional area of the diffuser outlet, is twice the area of the inlet. Therefore,

11

2

2

12

2

1V

T

T

P

PV

Using Appendix A.2 and the above expression, the energy balance becomes

21

2

11

2

2

1322

2

1)(

2

12

1

VVT

T

P

PMWdTETDTCTBTAR air

T

T

Substituting values from the problem statement provides an equation with one unknown:

K 3812 T

Therefore,

s

m 111m/s300

K15.343

K 381

bar 5.1

bar 1

2

12V

71

2.40 To find the minimum power required for the compressor, one must look at a situation where all

of the power is used to raise the internal energy of the air. None of the power is lost to the

surroundings and the potential and kinetic energy effects must be neglected. Therefore, the

energy balance becomes

sWhnhn 22110


21 nn

The energy balance reduces to

121 hhnWs

Using Equation 2.58 and Appendix A.2, the equation becomes

2

1

3221

T

T

s dTETDTCTBTARnW

Table A.2.1 and the problem statement provide the following values

0

1600

0

10575.0

355.3

3

E

D

C

B

A

K 300

s

mol 50

1

1

T

n

To find the work, we still need T2. We need to pick a reasonable process to estimate T2. Since

the heat flow is zero for this open system problem, we choose an adiabatic, reversible piston

situation. For this situation,

.constPV k

Since we are assuming the air behaves ideally, we can rewrite the equation as

72

kk

P

RTnP

P

RTnP

2

222

1

111

kkkk TPTP2

121

11

Substituting values from the problem statement, we obtain

K 579

bar 10

bar1K300

7/5

5/71

5/715/7

2

T

Substitute this value into the expression for the work and evaluate:

kW 4.417sW

73

2.41

(a)

Perform a mass balance:

outnnn 21

Apply the ideal gas law:

outnRT

VP

RT

VP

2

22

1

11

Substitute values from the problem statement:

outn

15.293314.8

105.2102

15.373314.8

105101 3535

mol/s 366.0outn

(b)

No work is done on the system, and we can neglect potential and kinetic energy effects. We will

assume the process is also adiabatic. The energy balance reduces to

out

out

outin

in

in hnhn 0

022112211 hhnhhnhnhnhn outoutoutout

We can calculate the enthalpy difference from the given ideal heat capacity:

010324.5267.310324.5267.3

15.293

32

15.373

31

outout

TT

dTTRndTTRn

Again, we must calculate the molar flow rates from the ideal gas law. Upon substitution and

evaluation, we obtain

K 329outT

74

2.42

(a)

Since the temperature, pressure, and volumetric flow rate are given, the molar flow rate is

constrained by the ideal gas law.

Note: min/cm 1/sm 1067.1 338

mol/s 1045.7K 15.273KJ/mol 314.8

/sm 1067.1Pa100135.1 7385

RT

VPn

To recap, we have shown

mol/s 1045.7SCCM 1 7

(b)

Assumptions: N2 is an ideal gas

All power supplied by the power supply is transferred to the N2

Uniform temperature radially throughout sensor tube

Kinetic and potential energy effects negligible in energy balance

Let x represent the fraction of N2 diverted to the sensor tube, and sn represent the molar flow

rate through the sensor tube. Therefore, the total molar flow rate, totaln , is

x

nn s

total

We can use temperature and heat load information from the sensor tube to find the molar flow

rate through the sensor tube. First, perform an energy balance for the sensor tube:

QhhnH inouts

The enthalpy can be calculated with heat capacity data. Therefore,

2

1

T

T

P

s

dTc

Qn

Now, we can calculate the total molar flow rate.

75

2

1

T

T

P

total

dTcx

Qn

To find the flow rate in standard cubic centimeters per minute, apply the conversion factor found

in Part (a)

mol/s 1045.7

SCCM 1)SCCM(

72

1

T

T

P

total

dTcx

Qv

(c)

To find the correction factor for SiH4, re-derive the expression for flow rate for SiH4 and then

divide it by the expression for N2 for the same power input, temperatures, and fraction of gas

diverted to the sensor tube.

2

1

4

2

1

2

2

1

2

2

1

4

2

4

,

,

7

,

7

,

,

,

mol/s 1045.7

SCCM 1

mol/s 1045.7

SCCM 1

T

T

SiHP

T

T

NP

T

T

NP

T

T

SiHP

Ntotal

SiHtotal

dTc

dTc

dTcx

Q

dTcx

Q

v

vFactor

If we assume that heat capacities are constant, the conversion factor simplifies:

4

2

,

,

SiHP

NP

c

cFactor

Using the values in Appendix A.2.2 at 298 K, we get

67.0Factor

76

2.43

(a)

It takes more energy to raise the temperature of a gas in a constant pressure cylinder. In both

cases the internal energy of the gas must be increased. In the constant pressure cylinder work,

Pv work must also be supplied to expand the volume against the surrounding’s pressure. This is

not required with a constant volume.

(b)

As you perspire, sweat evaporates from your body. This process requires latent heat which cools

you. When the water content of the environment is greater, there is less evaporation; therefore,

this effect is diminished and you do not feel as comfortable.

77

2.44

From the steam tables at 10 kPa:

Kmol

J 001434.0516.3

Kkg

kJ 0006624.06241.1 RTT

dT

dhc

PP

Now compare the above values to those in Appendix A.2.

T(K) h

323.15 2592.6

373.15 2687.5

423.15 2783

473.15 2879.5

523.15 2977.3

573.15 3076.5

673.15 3279.5

773.15 3489

873.15 3705.4

973.15 3928.7

1073.15 4159.1

1173.15 4396.4

1273.15 4640.6

1373.15 4891.2

1473.15 5147.8

1573.15 5409.7

h vs. T

y = 0.0003x2 + 1.6241x + 2035.7

R2

= 1

0

1000

2000

3000

4000

5000

6000

0 500 1000 1500 2000

T

en

thalp

y,

h

Series1

Poly. (Series 1)

A B

Steam Tables 3.516 0.001434

Appendix A 3.470 0.001450

% difference 1.3 1.4

78

2.45

For throttling devices, potential and kinetic energy effects can be neglected. Furthermore, the

process is adiabatic and no shaft work is performed. Therefore, the energy balance for one inlet

and one outlet is simplified to

2211 hnhn

which is equivalent to

2211ˆˆ hmhm

Since mass is conserved

221ˆˆ hmh


kg

kJ 3.3398ˆ

1h (8 MPa, 500 ºC)

kg

kJ 3.3398ˆ

2h

Now that we know 2u and 2P , 2T is constrained. Linear interpolation of steam table data gives

Cº 4572 T

79

2.46

(a)

An expression for work in a reversible, isothermal process was developed in Section 2.7.

Equation 2.77 is

1

2lnP

PnRTW

Therefore,

1

2lnP

PRTw

Evaluating the expression with

kPa 500

kPa 100

K 300

Kmol

J 314.8

1

2

P

P

T

R

gives

kg

J 4014w

(b)

Equation 2.90 states

121

TTk

Rw

Since the gas is monatomic

Rc

Rc

v

P

2

3

2

5

and

80

3

5k

2T can be calculated by applying the polytropic relation derived for adiabatic expansions. From

Equation 2.89

constPv

constPV

k

k

By application of the ideal gas law

P

RTv

Since R is a constant, substitution of the expression for P into the polytropic relation results in

kkkk

kk

TPTP

constTP

21

21

1

1

1

This relation can be used to solve for 2T .

K 6.1572 T

Now that 2T is known, value of work can be solved.

kg

J 9.1775w

81

2.47

(a)

The change in internal energy and enthalpy can be calculated using

Cº 0 ,atm 1ˆCº 100 ,atm 1ˆˆ


ll

ll

uuu

hhh

We would like to calculate these values using the steam tables; however, the appendices don’t

contain steam table data for liquid water at 0.0 ºC and 1 atm. However, information is provided

for water at 0.01 ºC and 0.6113 kPa. Since the enthalpy and internal energy of liquid water is

essentially independent of pressure in this pressure and temperature range, we use the steam

table in the following way

kg

kJ 02.419Cº 100 ,atm 1ˆlh

kg

kJ 0Cº 01.0 ,kPa .61130ˆCº 0 ,atm 1ˆ ll hh

kg

kJ 91.418Cº 100 ,atm 1ˆlu

kg

kJ 0Cº 01.0 ,kPa .61130ˆCº 0 ,atm 1ˆ ll uu

Therefore,

kg

kJ 02.419h

kg

kJ 91.418u

(b)

The change in internal energy and enthalpy can be calculated using



lv

lv

uuu

hhh


82

kg

kJ 5.2506ˆ

kg

kJ 02.419ˆ

kg

kJ 0.2676ˆ

v

l

v

u

h

h

kg

kJ 91.418ˆlu

Therefore,

kg

kJ 59.2087ˆ

kg

kJ 99.2256ˆ

u

h

The change in internal energy for the process in Part (b) is 5.11 times greater than the change in

internal energy calculated in Part (a). The change in enthalpy in Part (b) is 5.39 times greater

than the change in enthalpy calculated in Part (a).

83

2.48

To calculate the heat capacity of Ar, O2, and NH3 the following expression, with tabulated values

in Table A.2.1, will be used,

322 ETDTCTBTAR

cP

where T is in Kelvin. From the problem statement

K 300T

and from Table 1.1

Kmol

J 314.8R

To find the A-E values, Table A.2.1 must be referred to.

Formula A 310B 610C 510D 910E

Ar - - - - -

O2 3.639 0.506 0 -0.227 0

NH3 3.5778 3.02 0 -0.186 0

The values are not listed for Ar since argon can be treated as a monatomic ideal gas with a heat

capacity independent of temperature. The expression for the heat capacity is

Rc ArP

2

5,

Now that expressions exist for each heat capacity, evaluate the expressions for K 300T .

Kmol

J 560.35

Kmol

J 420.29

Kmol

J 785.20

3

2

,

,

,

NHP

OP

ArP

c

c

c

By examining the heat capacity for each molecule, it should be clear that the magnitude of the

heat capacity is directly related to the structure of the molecule.

84

Ar

Since argon is monatomic, translation is the only mode through which the atoms can

exhibit kinetic energy.

O2

Translation, rotation, and vibration modes are present. Since oxygen molecules are

linear, the rotational mode of kinetic energy contributes RT per mol to the heat capacity.

NH3

Translation, rotation, and vibration modes are present. Ammonia molecules are non-

linear, so the rotation mode contributes 3RT/2 per mole to the heat capacity.

The vibration contributions can also be analyzed for oxygen and ammonia, which reveals that the

vibration contribution is greatest for ammonia. This is due to ammonia’s non-linearity.

85

2.49

For a constant pressure process where potential and kinetic energy effects are neglected, the

energy balance is given by Equation 2.57:

HQ

The change in enthalpy can be written as follows

12 hhmH


kg

kJ 6.2584kPa 10at vapor sat.ˆ

2 hh

kg

kJ 81.191kPa 10at liquid sat.ˆ

1 hh

Therefore,

kg

kJ 81.191

kg

kJ 6.2584kg 2Q

kJ 6.4785Q

We can find the work from its definition:

f

i

V

V

EdVPW

The pressure is constant, and the above equation can be rewritten as follows

12 ˆˆ vvmPW E


kg

m 674.14kPa 10at vapor sat.ˆˆ

3

2 vv

kg

m 00101.0kPa 10at liquid sat.ˆˆ

3

1 vv

Therefore,

86

kJ 5.293kg

m 00101.0

kg 674.14kg2Pa10000

3

W

87

2.50

First, perform an energy balance on the system. Potential and kinetic energy effects can be

neglected. Since nothing is mentioned about work in the problem statement, W can be set to

zero. Therefore, the energy balance is

UQ


12 mm

where

vl mmm111

Now the energy balance can be written as

vvll umumumuumQ111121121 ˆˆˆ)ˆˆ(

Since two phases coexist initially (water is saturated) and 1P is known, state 1 is constrained.

From the saturated steam tables

kg

kJ 9.2437ˆ

kg

kJ 79.191ˆ

1

1

v

l

u

u

(sat. H2O at 10 kPa)

As heat is added to the system, the pressure does not remain constant, but saturation still exists.

One thermodynamic property is required to constrain the system. Enough information is known

about the initial state to find the volume of the container, which remains constant during heating,

and this can be used to calculate the specific volume of state 2.

vl

vvll

mm

vmvm

m

V

m

Vv

11

1111

1

1

2

22

ˆˆ

From the saturated steam tables

88

kg

m 674.14ˆ

kg

m 001010.0ˆ

3

1

3

1

l

l

v

v

(sat. H2O at 10 kPa)

Therefore,

kg

m 335.1ˆ

3

2v

The water vapor is now constrained. Interpolation of steam table data reveals

kg

kJ 6.2514ˆ2u

Now that all of the required variables are known, evaluation of the expression for Q is possible.

kJ 11652Q

89

2.51

Let the mixture of ice and water immediately after the ice has been added represent the system.

Since the glass is adiabatic, no work is performed, and the potential and kinetic energies are

neglected, the energy balance reduces

0H

We can split the system into two subsystems: the ice (subscript i) and the water (subscript w).

Therefore,

0 wwii hmhmH

and

wwii hmhm

We can get the moles of water and ice.

kg 399.0

kg

m 001003.0

m .00040

ˆ 3

3

w

ww

v

Vm

mol 15.22

mol

kg 0180148.0

kg 399.0

2

OH

ww

MW

mn

mol 55.5

2

OH

ii

MW

mn

Now, let’s assume that all of the ice melts in the process. (If the final answer is greater than 0

ºC, the assumption is correct.) The following expression mathematically represents the change

in enthalpy.

Cº 25º 0Cº 100 ,,, fwPwfwPfusiPi TcnCTchcn

Note: Assumed the heat capacities are independent of temperature to obtain this

expression.

From Appendix A.2.3

Rc iP 196.4,

Rc wP 069.9,

and

mol

kJ0.6fush

90

Substitution of values into the above energy balance allows calculation of Tf.

Cº 12.3fT

(Our assumption that all the ice melts is correct.)

(b)

To obtain the percentage of cooling achieved by latent heat, perform the following calculation

iwfwPw

fusilatent

TTcn

hnFraction

,,

911.0

Cº 25Cº 12.3Kmol

J 314.8069.9mol 15.22

mol

J 6000mol 55.5

latentFraction

%1.91latentPercent

91

2.52


vl nnn112

To develop the energy balance, neglect kinetic and potential energy. Also, no shaft work is

performed, so the energy balance becomes

QU

The energy balance can be expanded to

vvllvl unununnQ1111211

If the reference state is set to be liquid propane at 0 ºC and 4.68 bar, the internal energies become

dTRcuu

uu

u

Pvap

vapv

l

2T

K 273

2

1

1

Cº 0

Cº 0

0

Once the change in internal energy for vaporization and temperature of state 2 is determined, Q

can be solved. As the liquid evaporates, the pressure increases. At state 2, where saturated

propane vapor is present, the ideal gas law states

2

22

v

RTP

To find 2v , assume that lv vv11

. The volume of the rigid container is

mol

m 00485.0

3

1

1111

P

RTnvnV vvv

Therefore,

mol

m 00243.0

3

11

2 vl nn

Vv

92

Also, since the propane is saturated, 2P and 2T are not independent of each other. They are

related through the Antoine Equation,

CT

BAP

sat

sat

ln

where

2PPsat and 2TT sat

Substitution provides,

CT

BA

v

RT

22

2ln

Using values from Table A.1.1 and

Kmol

mbar 10314.8

35R

K 7.3012 T

To find vapu , refer to the definition of enthalpy.

lvlvvap PvuPvuhhh )(

Since lv vv , the above the change in internal energy of vaporization can be written as

RThPvhu vapvvapvap

Therefore,

mol

kJ 39.14Cº 0vapu

Evaluation of the following equation after the proper values have been substituted from Table

A.2.1

Cº 00Cº 011

K 301.7

K 273

11 vapvl

Pvapvl unndTRcunnQ

93

gives

kJ 1.18Q

94

2.53

The equation used for calculating the heat of reaction is given in Equation 2.72. It states

ifirxn hvh

This equation will be used for parts (a)-(e). Since the heat of reaction at 298 K is desired, values

from Appendix A.3 can be used.

(a)

First the stoichiometric coefficient must be determined for each species in the reaction.

1

1

1

1

)(

)(

)(

)(

2

2

2

4

gOH

gCO

gO

gCH

v

v

v

v

From Tables A.3.1 and A.3.2

mol

kJ 82.241

mol

kJ 51.393

mol

kJ 0

mol

kJ 81.74

)(298,

)(298,

)(298,

)(298,

2

2

2

4

gOHf

gCOf

gOf

gCHf

h

h

h

h

From Equation 2.72, the equation for the heat of reaction is

)(298,)()(298,)()(298,)()(298,)(298,

22

22

22

44 gOHfgOHgCOfgCOgOfgOgCHfgCHrxn hvhvhvhvh

mol

kJ 82.241

mol

kJ 51.393

mol

kJ 0

mol

kJ 81.74298,rxnh

mol

kJ 52.560298,rxnh

Now that a sample calculation has been performed, only the answers will be given for the remaining

parts since the calculation process is the same.

95

(b)

mol

kJ 53.604298,rxnh

(c)

mol

kJ 12.206298,rxnh

(d)

mol

kJ 15.41298,rxnh

(e)

mol

kJ 38.905298,rxnh

96

2.54

The acetylene reacts according to the following equation

C2H2(g) + (5/2)O2(g) 2CO2(g) + H2O(g)

(a)

First, choose a basis for the calculations.

mol 122HCn

Calculate the heat of reaction at 298 K using Equation 2.72 and Appendix A.3

ifirxn hvh

OHfCOfOfHCfrxn hhhhh

22222

25.2

J 10255.1

mol

J 10255.1

6298,

6

22

rxnHCrxn

rxn

hnH

h

The required amount of oxygen is calculated as follows

mol 5.2)(5.2 11 222 HCO nn

The compositions for both streams are

Streams 22

HCn 2

On 2

Nn 2

COn OHn2

1 (Inlet) 1 2.5 0 0 0

2 (Outlet) 0 0 0 2 1

From Table A.2.2

Species A 310B 610C 510D 910E

C2H2 6.132 1.952 0 -1.299 0

O2 3.639 0.506 0 -0.227 0

CO2 5.457 1.045 0 -1.157 0

H2O 3.470 1.45 0 0.121 0

Integration of the following equation provides an algebraic expression where only 2T is

unknown.

97

02

298

2298, T

iiPirxn dTcnH

Substituting the proper values into the expression gives

K 61692 T

(b)

The calculations follow the procedure used in Part (a), but now nitrogen is present. The basis is

mol 122HCn

The heat of reaction is the same as in Part (a), but the gas composition is different. Since

stoichiometric amount of air is used,

mol 5.212On

mol 40.9

2

2

22 11

airO

NON

y

ynn

The composition of the streams are summarized below

Streams 22

HCn 2

On 2

Nn 2

COn OHn2

1 1 2.5 9.40 0 0

2 0 0 9.40 2 1

From Appendix A.2


C2H2 6.132 1.952 0 -1.299 0

O2 3.639 0.506 0 -0.227 0

CO2 5.457 1.045 0 -1.157 0

H2O 3.470 1.45 0 0.121 0

N2 3.280 0.593 0 0.04 0

Therefore,

K 27922 T

98

(c)

Now excess air is present, so not all of the oxygen reacts. The heat of reaction remains the same

because only 1 mole of acetylene reacts. Since the amount of air is twice the stoichiometric

amount

mol 512On

mol 80.18

2

2

22 11

airO

NON

y

ynn

The compositions of the streams are summarized below

Streams 22

HCn 2

On 2

Nn 2

COn OHn2

1 1 5 18.8 0 0

2 0 2.5 18.8 2 1

The table of heat capacity data in Part (b) will be used for this calculation. Using the expression

shown in Part (a)

K 17872 T

99

2.55

(a)

The combustion reaction for propane is

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

For all subsequent calculations, the basis is one mole of propane. The heat of reaction is

calculated as follows

OHfCOfOfHCfrxn hhhhh

22283

435

J 10044.2

mol

J 10044.2

6298,

6

83

rxnHCrxn

rxn

hnH

h

The required amount of oxygen for complete combustion of propane is

mol 5)(5 11 832 HCO nn

mol 81.18

2

2

22 11

airO

NON

y

ynn

The stream compositions are listed below

Streams 83

HCn 2

On 2

Nn 2

COn OHn2

1 (Inlet) 1 5 18.8 0 0

2 (Outlet) 0 0 18.8 3 4

From Table (a)2.2


N2 3.280 0.593 0 0.04 0

CO2 5.457 1.045 0 -1.157 0

H2O 3.470 1.45 0 0.121 0

Now all of the necessary variables for the following equation are known, except 2T .

02

298

2298, T

iiPirxn dTcnH

Solving the resulting expression provides

K 23742 T

100

(b)

The combustion reaction for butane is

C4H10(g) + (13/2)O2(g) 4CO2(g) + 5H2O(g)

For all subsequent calculations, the basis is one mole of butane. The heat of reaction is

calculated as shown in Part (a)

J 10657.2 6298,

rxn

H

The moles of nitrogen and oxygen in the feed stream are calculated according to the method in

Part (a). The compositions are

Streams 104

HCn 2

On 2

Nn 2

COn OHn2

1 (Inlet) 1 6.5 24.5 0 0

2 (Outlet) 0 0 24.5 4 5

The Pc data listed in Part (a) can also be used for this reaction since there is no remaining

butane.

K 23762 T

(c)

The combustion reaction for pentane is

C5H12(g) + 8O2(g) 5CO2(g) + 6H2O(g)

The basis is one mole of pentane. The heat of reaction is calculated as shown in Part (a).

J 10272.3 6298, rxnH

The moles of nitrogen and oxygen in the feed stream are calculated according to the method in

Part (a). The compositions are listed below

Streams 125

HCn 2

On 2

Nn 2

COn OHn2

1 1 8 30.1 0 0

2 0 0 30.1 5 6

Substitution of the values into the expression used to find 2T and subsequent evaluation results

in

K 23822 T

The adiabatic flame temperatures are nearly identical in all three cases.

101

2.56

The equation for the combustion of methane is

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

Using Equation 2.72

mol

J 1002.8 5

rxnh

The basis for this problem is

mol 114CHn

Also, let represent the fractional conversion of methane. Therefore, the composition of the

product gas leaving the reactor is

mol 2

mol

mol 52.7

mol 12

mol 11

2

2

2

2

2

2

2

2

2

4

OH

CO

N

O

CH

n

n

n

n

n

Furthermore, the heat of reaction is calculated as follows

J 1002.8 5298, rxnH

From Table A.2.2


CH4 1.702 9.081 -2.164 0 0

O2 3.639 0.506 0 -0.227 0

CO2 5.457 1.045 0 -1.157 0

H2O 3.470 1.45 0 0.121 0

N2 3.280 0.593 0 0.04 0

After substitution of the outlet composition values, heat capacity data, and the heat of reaction

into the following equation

102

0

1273

298

2298, i

iPirxn dTcnH

integration provides an equation with one unknown: . Solving the equation gives

=0.42

Since the fractional conversion is 0.42, 58% of the methane passed through the reactor unburned.

103

2.57

For the entire cycle,

kJ 50

0

12

31231211

U

UUUU

From state 1 to state 2

kJ 40012

121212

Q

WQU


kJ 023

232323

Q

WQU


kJ 25031

313131

W

WQU

Hence, the completed table is

Process kJ U kJ W kJ Q

State 1 to 2 -50 -400 350

State 2 to 3 800 800 0

State 3 to 1 -750 -250 -500

To determine if this is a power cycle or refrigeration cycle, look at the overall heat and work,

11W and 11Q .

kJ 150

kJ 150

31231211

31231211

QQQQ

WWWW

Since work is done on the system to obtain a negative value of heat, which means that heat is

leaving the system, this is a refrigeration cycle.

104

2.58

Refer to the graph of the Carnot cycle in Figure E2.20. From this graph and the description of

Carnot cycles in Section 2.9, it should be clear that state 3 has the lowest pressure of all 4 states,

and state 1 has the highest pressure. States 1 and 2 are at the higher temperature. States 3 and 4

have the lower temperature. Since both the temperature and pressure are known for states 1 and

3, the molar volume can be calculated using

P

RTv

The table below summarizes the known thermodynamic properties.

State K T bar P /molm 3v

1 1073 60 0.00149

2 1073

3 298 0.2 0.124

4 298

For each step of the process, potential and kinetic energy effects can be neglected. The step from

state 1 to state 2 is a reversible, isothermal expansion. Since it is isothermal, the change in

internal energy is 0, and the energy balance becomes

1212 WQ

From Equation 2.77,

1

2112 ln

P

PnRTW

where 12W is the work done from state 1 to state 2. The value of 2P is not known, but

recognizing that the process from state 2 to 3 is an adiabatic expansion provides an additional

equation. The polytropic relationship can be employed to find 2P . A slight modification of

Equation 2.89 provides

constPvk

From the ideal gas law

v

RTP

Combining this result with the polytropic expression and noting that R is constant, allows the

expression to be written as

105

constTvk 1

Therefore,

1

1

13

2

32

kkvT

Tv

Substituting the appropriate values (k=1.4) gives

mol

m 00504.0

3

2v

Applying the ideal gas law

bar 7.172

22

v

RTP

Now, 12W can be calculated.

kJ 89.1012 W

Calculation of 34W follows a completely analogous routine as calculation for 12W . The

following equations were used to find the necessary properties

mol

m 0367.0

31

1

11

4

14

kkvT

Tv

bar 675.04

44

v

RTP

Now the following equation can be used

3

4334 ln

P

PnRTW

which gives

kJ 01.334 W

106

For an adiabatic, reversible process, Equation 2.90 states

121

TTk

nRW

This equation will be used to calculate the work for the remaining processes.

kJ 11.161

2323

TTk

nRW

kJ 11.161

4141

TTk

nRW

To find the work produced for the overall process, the following equation is used

41342312 WWWWWnet

Evaluating this expression with the values found above reveals

kJ 88.7netW

Therefore, 7.88 kilojoules of work is obtained from the cycle.

The efficiency of the process can be calculated using Equation 2.98:

72.089.10

88.7

H

net

Q

W

since kJ 89.1012 WQH . Alternatively, if we use Equation E2.20D.

H

C

T

T1

where 43 TTTC and 21 TTTH . Upon substitution of the appropriate values

72.0

107

2.59

Since this is a refrigeration cycle, the direction of the cycle described in Figure 2.17 reverses.

Such a process is illustrated below:

States 1 and 2 are at the higher temperature. States 3 and 4 have the lower temperature. Since

the both the temperature and pressure are known for states 2 and 4, the molar volume can be

calculated using

P

RTv

The following table can be made

State K T bar P /molm 3v

1 1073

2 1073 60 0.00149

3 298

4 298 0.2 0.124

For each step of the process, potential and kinetic energy effects can be neglected. The process

from state 1 to state 2 is a reversible, isothermal expansion. Since it is isothermal, the change in

internal energy is 0, and the energy balance becomes

1212 WQ

108

From Equation 2.77,

1

212 ln

P

PnRTW H

where 12W is the work done from state 1 to state 2. The value of 1P is not known, but

recognizing that the process from state 4 to 1 is an adiabatic compression provides an additional

relation. The polytropic relationship can be employed to find 1P . A slight modification of

Equation 2.89 provides

constPvk

From the ideal gas law

v

RTP

Combining this result with the polytropic expression and noting that R is constant allows the

expression to be written as

constTvk 1

Therefore,

1

1

1

4

1

41

kkv

T

Tv

Substituting the appropriate values (k=1.4) gives

mol

m 00504.0

31

1

1

4

1

41

kkv

T

Tv

bar 7.171

11

v

RTP

Now, 12W can be calculated.

kJ 9.1012 W

Calculation of 34W follows a completely analogous routine as the calculation for 12W . The

following equations were used to find the necessary properties:

109

mol

m 0367.0

3

3v

Applying the ideal gas law

bar 675.03

33

v

RTP

Now the following equation can be used

3

434 ln

P

PnRTW C

which gives

kJ 0.341 W

Equation 2.90 can be used to determine the work for adiabatic, reversible processes. This

equation will be used to calculate the work for the remaining processes.

kJ 11.161

23

HC TTk

nRW

kJ 11.161

41

CH TTk

nRW

To find the work produced for the overall process, the following equation is used

41342312 WWWWWnet

Evaluating this expression with the values found above reveals

kJ 88.7netW

Therefore, 7.88 kJ of work is obtained from the cycle. The coefficient of performance is defined

in Equation 2.99 as follows

net

C

W

QCOP

where CQ is the equal to 34Q . From the energy balance developed for the process from state 3

to state 4

110

kJ 01.33434 WQ

Therefore,

382.0kJ 88.7

kJ 01.3COP

111

2.60

(a)

The Pv path is plotted on log scale so that the wide range of values fits (see Problem 1.13)

logP

v

1100

0.0752

3

4

log v

(b)

The work required to compress the liquid is the area under the Pv curve from state 3 to state 4.

Its sign is positive. The power obtained from the turbine is the area under the curve from state 1

to 2. Its sign is negative. The area under the latter curve is much larger (remember the log

scale); thus the net power is negative.

(c)

First, perform a mass balance for the entire system:

mmmmm 4321

Since no work is done by or on the boiler, the energy balance for the boiler is

HQhmhm 41ˆˆ

Similarly, the energy balance for the condenser is

CQhmhm 23ˆˆ

To find the necessary enthalpies for the above energy balances, we can use the steam tables:

kg

kJ 5.3424ˆ

1h (520 ºC, 100 bar)

112

kg

kJ 2.2334

kg

kJ 8.257490.0

kg

kJ 77.16810.0ˆ

2h

(sat. liq at 7.5 kPa) (sat. vap. at 7.5 kPa)

kg

kJ 77.168ˆ

3h (sat. liquid at 0.075 bar)

kg

kJ 81.342ˆ

4h (subcooled liquid at 80 ºC, 100 bar)

Now, we can calculate the heat loads:

kW 308169kg

kJ 81.342

kg

kJ 5.3424kg/s 100

HQ

kW 216543kg

kJ 2.2334

kg

kJ 68.771kg/s 100

CQ

(d)

Use Equation 2.96:

0 netnet QW

From Part (c), we know

kW 91626kW 216543kW 308169 netQ

Therefore,

kW 91626netW

(e)

Using the results from Parts (c) and (d):

297.0kW 081693

kW 91626η

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