INTRODUCTION OF CHEMICAL ENGINEERING THERMODYNAMICS

SIXTH EDITION IN SI UNITS

SOLUTION MANUAL

J.M. SMITH H.C. VAN NESS M.M. ABBOTT

Chapter 1

1.1. Problem 1What is the value of gc and what are its units in a system in which the second, the foot, and the pound mass are defined as in Sec. 1.2, and the unit of force is the poundal, defined as the force required to give l(lbm) an acceleration of l(ft)(s)-2?

Solution

Newton’s second law expresses force as the product of mass and acceleration. However, a dimensional constant is required to adjust the unit of force, so Newton’s law is written as:

F= 1gc

ma

Replace the given values of force, acceleration and mass in the equation.

1 (poundal )= 1gc

(1Lbm)(1ft

s2)

gc=1Lbm ∙ ft

s2 ∙ poundal

This system of units is the English-system equivalent to the SI system.

1.2. Problem 2Electric current is the fundamental electrical dimension in SI; its unit is the ampere (A). Determine units for the following quantities, as combinations of fundamental SI units.

(a) Electric power; (b) Electric charge; (c) Electric potential difference;

(d) Electric resistance; (e) Electric capacitance.

Solution

The power is the rate at which work is done from a system to its surroundings or vice versa (in an electric system is defined as the work done by an electric circuit). The SI unit of power is the watts (W).

W= Js= N ∙m

s=

( kg ∙ms2

) ∙m

s= kg ∙m2

s3

The electric charge is defined as the charge transported by a steady current and its unit is the coulomb (C).

C=A ∙ s

The electric potential difference or voltage is the electric potential energy per unit charge. Keep in mind that the potential is the capacity for doing work depending on the position. The unit of voltage is volts (V)

V= JC

=N ∙mA ∙ s

=( kg ∙m

s2) ∙m

A ∙ s= kg ∙m2

s3 ∙ A

The electric resistance is defined as the opposition to the passage of an electric current through a conductor. The unit of resistance is the Ohm ( ).Ω

Ω=VA

= kg ∙m2

s3 ∙ A2

The electric capacitance is the ability of a body to store charge and its unit is the farad (F).

F=CV

= A ∙ s

( kg ∙m2

s3 ∙ A)= A2 ∙ s4

kg ∙m2

1.3. Problem 3Liquid/vapor saturation pressure Psat is often represented as a function of temperature by an equation of the form:

log10 Psat (torr )=a− bT (°C )+c

Here, parameters a, b, and c are substance-specific constants. Suppose it is required to represent Psat

by the equivalent equation:

ln P sat (kpa )=A− BT (K )+C

Show how the parameters in the two equations are related.

Solution

The natural logarithm and the common logarithm (base 10) are related as:

log10 x=ln x

ln 10(Eq .1)

The relationship between the pressure units Kpa and Torr is:

100 Kpa=750.061 torr (Eq .2)

The relationship between Celsius degrees and Kelvin is:

K=° C+273.15(Eq .3)

By Eq. 1

log10 Psat (torr )=ln Psat (torr )

ln10

By Eq. 2

log10 Psat (torr ) ∙ ln 10=ln(P¿¿ sat( torr) ∙100Kpa

750.061torr)¿

log10(P sat (torr ))∙ ln 10=ln(P¿¿ sat (Kpa) ∙ 750.061 torr100 Kpa

)¿

log10(P sat ( torr ))∙ ln 10=ln(P¿¿ sat (Kpa))+ ln ( 750.061100

¿)¿¿

log10 (Psat (torr ) ) ∙ ln10−ln (750.061100

)=ln(P ¿¿ sat (Kpa )¿)(Eq .4 )¿¿

The given equation for the saturation pressure log10 Psat (torr ) is:

log10 Psat (torr )=a− bT (°C )+c

Replacing this equation into Eq.4 gives

(a− bT (° C )+c ) ln 10−ln (

750.061100

)=ln (P¿¿ sat (Kpa )¿)¿¿

a ln10−ln( 750.061100 )+ b

T (° C )+c∙ ln10=ln(P ¿¿ sat (Kpa )¿)¿¿

By Eq. 3

a ln10−ln( 750.061100 )+ b

T (K )−273.15+c∙ ln 10=ln (P¿¿ sat (Kpa )¿)¿¿

a ln10−ln( 750.061100 )+ b ln10

T (K )−273.15+c=ln (P ¿¿ sat (Kpa )¿)¿¿

Comparing the last expression with the given equation for ln Psat (kpa), shows that:

A=a ln 10−ln( 750.061100 ) ,B=b ln 10 ,C=c−273.15

1.4. Problem 4At what absolute temperature do the Celsius and Fahrenheit temperature scales give the same numerical value? What is the value?

Solution

The relationship between the Celsius and Fahrenheit scale is

T (° F )=95T (° C )+32

The same temperature value is obtained at

T=95T +32

−45

T=32

T=−40

1.5. Problem 5Pressures up to 3000 bar are measured with a dead-weight gauge. The piston diameter is 4 mm. What is the approximate mass in kg of the weights required?

Solution

The pressure in a dead weight gauge is given by the formula:

P= FA

In an ideal dead-weight gauge, the force (F) is exerted only by the weights (the product of mass and the local gravity) and A is the transversal area, replacing both F and A in the last equation gives.

P= mg

π d2

4

Pπ d2

4=mg

Pπ d2

4 g=m (Eq .1 )

Beforehand, a conversion of units is necessary

3000 ¿ 105 Pa

1 ¿=3 x108Pa=3 x 108( Nm2 )¿

4 mm=1 x10−3m1mm

=4 x10−3m

Replacing the given values in Eq. 1

3 x108 ∙π (4 x10−3)2

4 ∙9.807=384.41 Kg

1.6. Problem 6Pressures up to 3000 atm are measured with a dead-weight gauge. The piston diameter is 0.17 (in). What is the approximate mass in (lbm) of the weights required?

Solution

This problem is similar to exercise 1.6; the same equation can be applied.

Pπ d2

4 g=m

Beforehand, a conversion of units is necessary

3000atm=101325 Pa1atm

=303975000 Pa

0.17∈¿ 2.54 cm

1∈¿ x1m

100cm=0.004318m ¿

Replacing the given values

303975000 ∙π (0.004318)2

4 ∙9.807=453.89 Kg

In Lbm

453.89 Kg ∙1Lbm

0,45359 Kg=1000,6757 Lbm

1.7. Problem 7The reading on a mercury manometer at 298.15 K (25°C) (open to the atmosphere at one end) is 56.38 cm. The local acceleration of gravity is 9.832 m/s2. Atmospheric pressure is 101.78 kPa. What is the absolute pressure in kPa being measured? The density of mercury at 298.15 K (25°C) is 13.534 g / cm3

Solution

The gauge pressure measured by the mercury manometer is given by the equation:

Pg=ρgh=(13.534g

cm3 ∙1Kg

1000 g∙1003 cm3

1m3 ∙9.832ms2 ∙56.38cm ∙

1m100cm )=75022,77 Pa

The absolute pressure being measured is the addition of gauge and atmospheric pressure

P|¿|=Pg+Patm=75,02Kpa+101,78Kpa=176,8Kpa¿

1.8. Problem 8Liquids that boil at relatively low temperatures are often stored as liquids under their vapor pressures, which at ambient temperature can be quite large. Thus, n-butane stored as a liquid/vapor system is at a pressure of 2.581 bar for a temperature of 300 K. Large scale storage (>50 m3) of this kind is sometimes done in spherical tanks. Suggest two reasons why.

Solution

1. The pressure inside the spherical tank is uniform within the tank wall.2. The tensile stress within the tank wall is uniform everywhere, with no sites of stress

concentration. Moreover, the maximum stress within the tank is kept at minimum.

The sphere is the solid which given a volume has the smallest surface area. Therefore:

3. The quantity of metal needed to construct the tank is minimized.4. The surface area that must be isolated against heat transfer by solar radiation is

minimized.

1.9. Problem 9The first accurate measurements of the properties of high-pressure gases were made by E. H. Amagat in France between 1869 and 1893. Before developing the dead-weight gauge, he worked in a mine shaft, and used a mercury manometer for measurements of pressure to more than 400 bar. Estimate the height of manometer required.

Solution

The temperature in a mine shaft depends mainly on how deep is the mine due to the heat released from the earth crust and the ambient temperature on the surface. However, it is reasonable to assume a standard temperature (25°C) at which the mining process won’t be life-threatening to mine workers.

The density of mercury at 25°C is 13.534 g/cm3

Pressure measured by the manometer is obtained by the formula: P= ρgh,

Solving for h and replacing values:

h= Pρg

=400 ∙

1x 105Pa1 ¿

13543Kgm3 ∙9.802

ms2

=301,52m ¿

1.10. Problem 10An instrument to measure the acceleration of gravity on Mars is constructed of a spring from which is suspended a mass of 0.40 kg. At a place on earth where the local acceleration of gravity is 9.81 m/s2, the spring extends 1.08 cm. When the instrument package is landed on Mars, it radios the information that the spring is extended 0.40 cm. What is the Martian acceleration of gravity?

Solution

According to Hooke’s law: F=kx , where K is the spring constant. First calculate the spring constant based on the earth measurements.

On earth: mg=kx→0.4 kg ∙9.81m

s2=k ∙0.0108m→k=363.3 3

Nm

The spring constant remains the same regardless of the planet; calculate Mars gravity with the spring measurements.

On mars: mg=kx→0.4 kg ∙g=363.33Nm

∙0.004 m→g=3,63m

s2

1.11. Problem 11The variation of fluid pressure with height is described by the differential equation:

dPdz

=−ρg

Here, p is specific density and g is the local acceleration of gravity. For an ideal gas, p = MP/RT, where M is molar mass and R is the universal gas constant. Modeling the atmosphere as an isothermal column of ideal gas at 283.15 K (10°C), estimate the ambient pressure in Denver, where z = l (mi1e) relative to sea level. For air, take M = 29 g/mol; values of R are given in App. A.

Solution

Replace the ideal gas equation into the given differential equation.

dPdz

=−MPRT

g

Now, solve the differential equation (remember that temperature and molar mass are also constant)

∫ dPP

=−MgRT ∫ dz

ln|P|=−MgRT

z+C

At sea level (z = 0 miles) the pressure is 1 atm = 101325 Pa, so

ln|101325|=C

At 1 mile above sea level, the pressure is

ln P=−MgRT

z+ln 101325

lnP

101325=−Mg

RTz

P=101325 ∙ e(−Mg

RTz )

MgRT

z=29

gmol

∙1kg

1000gr∙9.8

m

s2

8.314m3Pamol K

∙283.15 K∙1600m=0.193

P=101325 Pa ∙e−0.193=83527.26 Pa=0.8243atm

1.12. Problem 12A 70 W outdoor security light burns, on average, 10 hours a day. A new bulb costs $5.00, and the lifetime is about 1000 hours. If electricity costs $0.10 per kWh, what is the yearly price of "security," per light?

Solution

Assuming there are 365 days per year and that the security light will work every day, the total working hours are 3650. The yearly cost per light is the sum of the cost of consumed electricity (variable) and the cost of buying new bulbs as necessary (fixed).

The fixed cost is:

36501000

≅ 4newbulbs→4 ∙ $5.00=$20.00

*The number of total bulbs must be rounded up as you can’t buy a fraction of a bulb.

The variable cost is:

0.1$

kW ∙h∙3650h∙0.07kW=$25,55

The total cost is:

CTotal=$20.00+$25.55=$ 45.55year

1.13. Problem 13A gas is confined in a 0.47-m-diameter cylinder by a piston, on which rests a weight. The mass of the piston and weight together is 150 kg. The local acceleration of gravity is 9.813 m/s2, and atmospheric pressure is 101.57 kPa.

(a) What is the force in newtons exerted on the gas by the atmosphere, the piston, and the weight, assuming no friction between the piston and cylinder?

(b) What is the pressure of the gas in kPa?(c) If the gas in the cylinder is heated, it expands, pushing the piston and weight upward.

If the piston and weight are raised 0.83 m, what is the work done by the gas in kJ? What is the change in potential energy of the piston and weight?

Solution

Figure 1. Basic piston

A. The total force exerted on the gas is the sum of the atmospheric pressure and the weight of the piston and mass together. The exerted force by the atmosphere is:

Fatm=P ∙ A=P ∙π4∙ d2=101.57 kpa ∙

1000Pa1kPa

∙π4∙(0,47m)2=17621.83 N

The force done by the piston and mass together is:

Fweigh t=M ∙ g=150 Kg ∙9.813m

s2=1471,95 N

The total force exerted on the gas is:

FTotal=Fatm+Fweigh t=19093.78 N

B. The pressure is the total force divided the effective area (transversal).

P= FA

= Fπ4∙d2

= 19093.78 Nπ4∙(0,47m)2

=110054.13Pa=1,086 atm

C. The work done by the gas is the change of volume of the gas at constant pressure.

W=−∫ PdV

In a cylinder, the following formula applies. V=A trans ∙ Z→dV=A transdZ

W=−∫0

0.83

P A transdZ

W=−P A trans ( 0,83−0 )=110054.13Pa∙π4∙(0,47m)2 ∙0,83m=−15847,84 J=−15,848kJ

The change on potential energy is: ∆ E p=mg∆ l=150 Kg∗9,813m

s2∗0,83m=1221,72J

1.14. Problem 14Verify that the SI unit of kinetic and potential energy is the joule.

Solution

Joule is work done when a force of 1 newton acts through a distance of 1 meter. 1 J = 1Nm

The fundamental units of the kinetic energy are:

E k=12m(kg)(u (ms ))

2

=kg ∙m2

s2 =N ∙m=J

The fundamental units of the potential energy are:

E k=m(kg) ∙ g (ms2 )∙ l(m)=kg ∙m

s2 ∙m=N ∙m=J

1.15. Problem 15 An automobile having a mass of 1250 kg is traveling at 40 m/s. What is its kinetic energy in kJ? How much work must be done to bring it to a stop?

Solution

E k=12mu2=1

2(1250 kg )(40

ms)

2

=1000000 J=1000kJ

The necessary work to stop the car has to be equal or greater than its kinetic energy.

1.16. Problem 16 The turbines in a hydroelectric plant are fed by water falling from a 50-m height. Assuming 91% efficiency for conversion of potential to electrical energy, and 8% loss of the resulting power in transmission, what is the mass flowrate of water required to power a 200 W light bulb?

Solution

First calculate the electric power produced by the turbines.

Pele=200W0,92

=217,39W

Then, calculate the total potential energy

EP=217,39W

0,91=238,89W

Finally, calculate the total mass flowrate.

Ep (W )=m( kgs ) g(ms2 )h (m ) , solving for m→m=Ep

gh

m=238.89(W )

9.8(ms2 ) ∙50(m)=0,4875

kgs

1.17. Problem 17Below is a list of approximate conversion factors, useful for "back-of-the-envelope" estimates. None of them is exact, but most are accurate to within about 10%. Use Table A. 1 (App. A) to ∓establish the exact conversions.

1atm≈1 ¿1Btu≈1kj1hp≈0.75kW1 inch≈2.5cm1 lbm≈0,5kg1mile ≈1.6 km1quart ≈1 liter

1 yard≈1m

Add your own items to the list. The idea is to keep the conversion factors simple and easy to remember.

Solution

The exact conversions are

1atm=1.01325 ¿1Btu≈1,05504 kj1hp≈0.745701kW1 inch≈2.54cm1 lbm≈0,4535kg1mile ≈1.60934 km1quart ≈0,94635 liter

1 yard≈0,9144m

1.18. Problem 18Consider the following proposal for a decimal calendar. The fundamental unit is the decimal year (Yr), equal to the number of conventional (SI) seconds required for the earth to complete a circuit of the sun. Other units are defined in the table below. Develop, where possible, factors for converting decimal calendar units to conventional calendar units. Discuss pros and cons of the proposal.

Decimal Calendar Unit Symbol Definition

Second Sc 10−6Yr

Minute Mn 10−5Yr

Hour Hr 10−4Yr

Day Dy 10−3Yr

Week Wk 10−2Yr

Month Mo 10−1Yr

Year Yr

Solution

The year may not be the best fundamental unit as the number of days may change according to a specific year. However, a reasonable and approximate assumption is that

1 year=364 days

Yr=1 year=364 Days1 year

∙24 hoursDays

∙60 Minutes

1hour∙60 seconds1 Minutes

=31449600

This assumption makes a year exactly to 52/7 days weeks and therefore, a month would

have 36412

=3013Days and a week would have

30.337

=413Days.

Decimal Calendar Unit Conversion factor

Sc 10−6Yr=31.449600 Seconds

Mn 10−5Yr=314.49600 Seconds

Hr 10−4Yr=3144.9600Seconds

Dy 10−3Yr=31449.600 Seconds

Wk 10−2Yr=314496.00 Seconds

Mo 10−1Yr=3144960.0 Seconds

Decimal Calendar Unit Conversion factor

Conventional Calendar Unit

Sc1 Sc

31.449600SecondSecond

Mn1Mn

314.496∙60 Seconds

1MinuteMinute

Hr1Hr

3144.9600Seconds∙3600 Seconds

1 HourHour

Dy1 Dy

31449.600Seconds∙3600 Seconds

1 Hour∙24 Hours

1 DayDay

Wk 1Wk314496.00Seconds

∙86400 Seconds

1Day∙4

13Days

1WeekWeek

Mo 1 Mo3144960.0Seconds

∙86400 Seconds

1Day∙30

13Days

1 MonthMonth

Chapter 2

2.1. Problem 1

A nonconducting container filled with 25 kg of water at 293.15 K (20°C) is fitted with a stirrer, which is made to turn by gravity acting on a weight of mass 35 kg. The weight falls slowly through a distance of 5 m in driving the stirrer. Assuming that all work done on the weight is transferred to the water and that the local acceleration of gravity is 9.8 m/s2, determine:

(a) The amount of work done on the water.

(b) The internal-energy change of the water.

(c) The final temperature of the water, for which Cp = 4.18 kJ/ kg-C.

(d) The amount of heat that must be removed from the water to return it to its initial temperature.

(e) The total energy change of the universe because of: (1) the process of lowering the weight,

(f) The process of cooling the water back to its initial temperature, and (3) both processes together.

Solution

The system is defined as the water contained in the container.

(a) The work done on the system is the work done by the stirrer on the water. Assuming there are no dissipative forces, the work done will be equal to the change of potential energy of the weight.

W stirrer=mg∆ z=(35kg )(9.8m

s2 )(5m )=1715 J

(b) Considering the container is nonconducting, the heat losses are negligible; the first law of thermodynamics applied to the closed system gives:

∆U=W+Q(Q=0)∆U=W →∆U=1715 J

(c) The volume of water does not change significantly during the process, therefore is reasonable to assume a constant volume process and the following definition of heat capacity can be applied.

C v=ndUdT

n∫T1

T2

C v dT=∫U 1

U 2

dU

nC v (T 2−T 1 )=∆U

T 2=∆UnC v

+T1→T 2=1.715kJ

25kg ∙4,18kJ

kgC °

+20 °C→20,016 ° C

(d) Because the system returns to its initial state, the change of internal energy is 0; the first law neglecting the change of potential and kinetic energy reduces to:

∆U=Q+W →∆U=00+W=0→Q=−W=−1715J

(e) The energy of the universe remains unchanged regardless of the process on the system. For all three cases, the change of internal energy is 0.

2.2. Problem 2

Rework Prob. 2.1 for an insulated container that changes in temperature along with the water

and has a heat capacity equivalent to 5 kg of water. Work the problem with:

(a) The water and container as the system; (b) The water alone as the system.

Solution (a) The work done on both systems remains the same: W = 1715J, (b) If the water and container is taken as the system, then ∆U=W →∆U=1715 J

However if only the water is taken as the system, then there will be lost heat due to the temperature change of the container, ∆U Water=W−Q

It is reasonable to assume that the volume of the container will not change significantly and the heat can be modeled as:

Q=∆UContainer=n∫T 1

T 2

C vdT=nC vContainer (T 2−T1 )

The first law of thermodynamics becomes:

∆U water=W−nC vContainer (T 2−T1 )

(c) Water + container:

∆U total=∆U container+∆U water=ncontainer∫T 1

T 2

C vContainerdT +nWater∫

T1

T2

C vWaterdT

∆U Total=ncontainerC vContainer (T2−T 1 )+nWaterC v

Water (T 2−T 1 )

Solving for T2

T 2=∆UTotal+ (ncontainerC v

Container+nWaterC vWater ) ∙T 1

ncontainerCvContainer+nWaterC v

Water

T 2=1.715 kJ+(5kg ∙4.18

kJkg ∙° C

+25kg ∙ 4.18kJ

kg ∙ °C ) ∙20° C

5kg ∙4.18kJ

kg ∙ °C+25kg ∙4.18

kJkg ∙° C

=20,0136° C

Water:

∆U Water=W−Q→∆U Water=W−ncontainerC vContainer (T2−T 1 )

Assuming the volume of water remains constant

∆U Water=n∫T1

T2

C vWaterdT=nwaterC v

Water(T2−T1)

Replacing on the first law

nwaterCvWater (T 2−T1 )=W−ncontainerC v

Container (T2−T 1 )

Solving for T 2:

T 2=W +(ncontainerC v

Container+nWaterC vWater ) ∙ T1

ncontainerC vContainer+nWaterC v

Water

T 2=1.715 kJ+(5kg ∙4.18

kJkg ∙° C

+25kg ∙ 4.18kJ

kg ∙ °C ) ∙20° C

5kg ∙4.18kJ

kg ∙ °C+25kg ∙4.18

kJkg ∙° C

=20,0136° C

The same final temperature is obtained regardless of what is taken as the system. With the final temperature, the change of internal energy of water may be calculated.

∆U Water=W−ncontainerC vContainer (T2−T1 )

∆U Water=1.715kJ−5kg ∙4,18kJ

kg°C(20,0136° C−20 °C )

∆U Water=1,429kJ

(d) For both systems, the same amount of heat must be removed to reach its initial temperature.

∆U=Q+W →∆U=00+W=0→Q=−W=−1715J

(e) The same as problem 2.1, the total internal energy change of the universe is zero for all the processes.

2.3. Problem 3

An egg, initially at rest, is dropped onto a concrete surface and breaks. With the egg treated as the system,

(a) What is the sign of W?

(b) What is the sign of EΔ p?

(c) What is ΔEk?

(d) What is U'Δ ?

(e) What is the sign of Q?

In modeling this process, assume the passage of sufficient time for the broken egg to return to its initial temperature. What is the origin of the heat transfer of part (e)?

Solution

The first law of thermodynamics may be written as:

∆U+∆ Ep+∆Ek=W+Q

(a) There is no work done on the system, neither the system does work to its surroundings, therefore W=0.

(b) The egg falls down from a higher surface, the elevation decreases and the sign of potential energy is negative (-)

(c) Since the egg is at rest in both its initial and final state, ∆ Ek=0 (d) Assuming the egg does not get scrambled, the internal energy does not change, U = 0Δ(e) ∆U+∆ Ep+∆Ek=W+Q

∆ E p=Q

Since the change of potential energy is negative, the heat must also be negative. A closer look to the process indicates that the potential energy turns into kinetic while the egg is falling, then just before it strikes the concrete the kinetic energy appears instantly as internal energy, thus raising its temperature. Heat transfer to the surroundings lowers the temperature until the egg reaches its initial state.

2.4. Problem 4

An electric motor under steady load draws 9.7 amperes at 110 volts, delivering 0.93 kW of mechanical energy. What is the rate of heat transfer from the motor, in kW?

Solution

The total electric power supplied by the motor is calculated as:

Pmotor=i ∙ E=9,7 A ∙110V=1067W

Assuming no dissipative forces from friction or other sources, the power delivered by the motor must be equal to the work done by the motor and the heat released by the motor.

Pmotor=Wmotor+Qmotor

Solving for Q

Qmotor=Pmotor−Wmotor

Qmotor=1067W−930W

Qmotor=137W

2.5. Problem 5

One mole of gas in a closed system undergoes a four-step thermodynamic cycle. Use the data given in the following table to determine numerical values for the missing quantities, i.e., "fill in the blanks."

Step U/JΔ Q/J W/J

12 -200 ? -6000

23 ? -3800 ?

34 ? -800 300

41 4700 ? ?

12341

? ? -1400

Solution

Equation 2.3 may be applied assuming there are no significant changes on the kinetic and potential energy.

∆U=Q+W

In the step 1-2, heat can be calculated directly.

∆U 12=Q 12+W 12

Q12=−200−(−6000)→Q 12=5800 J

In step 3-4, internal energy can be calculated directly.

∆U 34=Q34+W 34

∆U 34=−800+300→∆U 34=−500J

As in a thermodynamic cycle, the system returns to its initial state, the total change of internal energy is zero and the total heat can be calculated.

Q12341=∆U 12341−W 12341

Q12341=0−(−1400)→Q12341=1400J

Likewise, the sum of internal energy of all the steps in the process must also be zero.

∆U 12341=∆U 12+∆U 23+∆U 34+∆U 41

∆U 23=−(∆U 12+∆U 34+∆U 41 )+∆U 12341

∆U 23=−(−200±500+4700 )+0→∆U 23=−4000 J

Now, work done on step 2-3 can be calculated

∆U 23=Q23+W 23

W 23=−4000−(−3800 )→W 23=−200J

The total work done is equal to the work done on each step

W 12341=W 12+W 23+W 34+W 41

W 41=−(W 12+W 23+W 34 )+W 12341

W 41=−(−6000±200+300 )+−1400→W 41=4500 J

Finally, calculate the heat on step 4-1

∆U 41=Q41+W 41

Q41=4700−(4500)→Q 41=200J

Step U/JΔ Q/J W/J

12 -200 5800 -6000

23 -4000 -3800 -200

34 -500 -800 300

41 4700 200 4500

12341

0 1400 -1400

2.6. Problem 6

Comment on the feasibility of cooling your kitchen in the summer by opening the door to the electrically powered refrigerator.

Consider that the cooling process of a refrigerator is not a ideal process, not all the electrical energy entering the refrigerator can be turn into work done by the motor. Some heat must be released, this happens on the motor and the condenser of the refrigerator. The amount of heat released must be greater the absorbed heat (see Chapter 6), thus the temperature will rise instead of fall. Additionally, the refrigerator will burn out within days.

2.7. Problem 7

A renowned laboratory reports quadruple-point coordinates of 10.2 Mbar and 297.25 K (24.1°C) for four-phase equilibrium of allotropic solid forms of the exotic chemical

-miasmone. Evaluate the claim.Β

Solution

The phase rule states that F=2−π+n, according to the laboratory report a complex yet pure substance (N=1) is in a four-phase equilibrium (π=4¿. If this is true, then the degrees of freedom will be F=2−4+1=−1. This result is impossible; the claim is invalid.

2.8. Problem 8

A closed, nonreactive system contains species 1 and 2 in vapor/liquid equilibrium. Species 2 is a very light gas, essentially insoluble in the liquid phase. The vapor phase contains both species 1 and 2. Some additional moles of species 2 are added to the system, which is then restored to its initial T and P. As a result of the process, does the total number of moles of liquid increase, decrease, or remain unchanged?

Solution

The phase rule states F=2−π+N=2−2+2=2. Specification of T and P fixes the intensive state of the system, so the phase compositions are also fixed. Since the liquid phase is pure species 1, an addition of species 2 must increase its amount on the vapor phase. However, as phase compositions remain fixed some liquid moles evaporate to restore the initial vapor phase compositions, thus decreasing the total number of moles of liquid.

2.9. Problem 9

A system comprised of chloroform, 1,4-dioxane, and ethanol exists as a two-phase vapor/liquid system at 323.15 K (50°C) and 55 kPa. It is found, after the addition of some pure ethanol, that the system can be returned to two-phase equilibrium at the initial T and P. In what respect has the system changed, and in what respect has it not changed?

Solution

The phase rule states F=2−π+N=2−2+3=3. Specification of T and P leaves one degree of freedom, thus the phase compositions may be changed for a given T and P. With the addition of pure ethanol, the intensive state of the system changes completely as the phase composition and molar amount of the phases are affected. Only T and P remain the same within the system.

2.10. Problem 10

For the system described in Pb. 2.9:

(a) How many phase-rule variables in addition to T and P must be chosen so as to fix the compositions of both phases?

(b) If the temperature and pressure are to remain the same, can the overall composition of the system be changed (by adding or removing material) without affecting the compositions of the liquid and vapor phases?

Solution

(a) F=3, with fixed T and P, one extra intensive variable must be specified to fix the intensive state of the system.

(b) If vapor with the same vapor phase composition enters or leaves the system, the two phase compositions remain unaffected. Likewise if a liquid with the same liquid phase composition. However, the overall composition will be affected at all cases as it depends on the relative amount of each phase. Only if an azeotropic composition is achieved, when the two phase composition are the same, is possible to add or remove material without altering the overall composition of the system.

2.11. Problem 11

A tank containing 20 kg of water at 293.15 K (20°C) is fitted with a stirrer that delivers work to the water at the rate of 0.25 kW. How long does it take for the temperature of the water to rise to 303.15 K (30°C) if no heat is lost from the water? For water, Cp = 4.18 kJ/ kg-°C.

Solution

∆U∆ t

=W →∆ t=∆UW

∆U=m∙C p∙∆T← AssumingCpconstant

∆U=20kg ∙4.18kJ

kg ∙ °C∙ (30° C−20 °C )→836 kJ

∆ t= 836kJ

0,25kJs

=3345 s→55,7min→0,92 8hr

2.12. Problem 12

Heat in the amount of 7.5 kJ is added to a closed system while its internal energy decreases by 12 kJ. How much energy is transferred as work? For a process causing the same change of state but for which the work is zero, how much heat is transferred?

Solution

For a closed system with negligible kinetic and potential energy changes, the equation 2.3. becomes valid.

∆U=Q+W (a)

−12kJ=7.5kJ+WW=−19.5kJ

(b) ∆U=Q→Q=−12kJ

2.13. Problem 13

A steel casting weighing 2 kg has an initial temperature of 773.15 K (500°C); 40 kg of water initially at 298.15 K (25°C) is contained in a perfectly insulated steel tank weighing 5 kg. The casting is immersed in the water and the system is allowed to come to equilibrium. What is its final temperature? Ignore any effect of expansion or contraction, and assume constant specific heats of 4.18 kJ/kg-K for water and 0.50 kJ/kg-K for steel.

Solution

No heat and no work enters nor leaves the system thus, there is no change of internal energy

∆U total=0

The total change of internal energy is the sum of all internal energy changes within the system. C = casting, T = steel tank, W = water.

mt ∙Cpsteel ∙∆T t+mw ∙C p

water ∙∆T w+mc ∙C psteel ∙∆Tc=0

mt ∙Cpsteel ∙(T 2−T1

Tank )+mw ∙Cpwater ∙(T 2−T 1

Water )+mc ∙C psteel ∙ (T2−T 1

Casting )=0

Solving for T2

T 2=mt ∙C p

steel ∙ T 1Tank+mw ∙Cp

water ∙ T 1Water+mc ∙C p

steel ∙T 1Casting

mt ∙Cpsteel+mw ∙Cp

water+mc ∙Cpsteel

T 2=5 kg ∙0.50

kJkg ∙ K

∙298.15 K+40kg ∙4.18kJ

kg ∙ K∙298.15 K+2kg ∙0,50

kJkg ∙ K

∙773.15 K

5kg ∙0.50kJ

kg ∙ K+40kg ∙ 4.18

kJkg ∙ K

+2kg ∙0,50kJ

kg ∙ K

T 2=300,93 K→27.78 °C

2.14. Problem 14

An incompressible fluid ( ρ = constant) is contained in an insulated cylinder fitted with a frictionless piston. Can energy as work be transferred to the fluid? What is the change in internal energy of the fluid when the pressure is increased from P1 to P2?

Solution

If the fluid density remains constant through the compression process, then the process is considered a V-constant process, for which the work is zero. Furthermore if the cylinder is perfectly insulated, no heat is transferred and U = 0.Δ

2.15. Problem 15

One kg of liquid water at 298.15 K (25°C):

(a) Experiences a temperature increase of 1 K. What is UΔ t, in kJ?

(b) Experiences a change in elevation z. Δ The change in potential energy EΔ p is the

same as U' Δ for part (a). What is z, Δ in meters?

(c) Is accelerated from rest to final velocity u. The change in kinetic energy EΔ k is the

same as UΔ t for part (a). What is u in m/s?

Compare and discuss the results of the three preceding parts.

Solution

(a) Water at 25°C has a heat capacity of 4.18 kJ

kg ∙ K; an increase of 1 K does not change its

heat capacity significantly, thus the internal energy can be calculated as:

∆U=m∫T 1

T 2

C v dT=mC v∆T=(1kg ) ∙(4.18kJ

kg ∙K )∙ (1K )=4.18 kJ

(b)

∆ E p=mg∆ z→∆z=∆E p

mg= 4180J

(1kg ) ∙(9.8ms2 )

=426.53m

(c)

∆ Ek=12mu2→u=√ 2∆ Ek

m=√ 2 ∙4180J

1kg=91.43

ms

2.16. Problem 16

An electric motor runs "hot” under load, owing to internal irreversibilities. It has been suggested that the associated energy loss be minimized by thermally insulating the motor casing. Comment critically on this suggestion.

Solution

The “hotness” of an electric motor is due to mechanical and electrical irreversibilities that increase its internal energy, which elevates the temperature of the motor. The temperature continues to rise until the system reaches a thermal equilibrium with the surroundings. Insulating the motor will not decrease its internal irreversibilities and merely causes an increase of the motor temperature, which may cause internal damage.

2.17. Problem 17

A hydroturbine operates with a head of 50 m of water. Inlet and outlet conduits are 2 m in diameter. Estimate the mechanical power developed by the turbine for an outlet velocity of 5 m/s.

Solution

The general mass balance is:

d mcv

dt+∆ (m)fs=0

Assuming a steady state, the mass balance reduces to:

∆ (m)fs=0

Since the turbine has only one input and one output stream:

∆ (m)fs=0→m2−m1=0→m1=m2 (Constant mass flowrate)

Calculate the mass flowrate (assume a water density of 1000kg

m3 )

m2=uAρ→A=π D2

4

m2=uρπ D2

4=(5 m

s )∙ (1000kgm3 ) ∙( π ∙22m2

4 )=15708kgs

The general heat balance for an open system (such as a hydroturbine) is:

d (mU )cvdt

+∆(H+ 12u2+gz)m=Q+W

Consider the following reasonable assumptions of the process

No significant increase in temperature (∆H ≅ 0) No significant increase in velocity (∆u≅ 0) Negligible heat loss to the surroundings. (Q≅ 0)

Steady state (d mcv

dt=

0∧d mU cv

dt=0)

The heat balance simplifies to:

W=m g∆ z

W=15708kgs∙9.8

m

s2∙50m=7696kW

2.18. Problem 18

Liquid water at 453.15 K (180°C) and 1002.7 kPa has an internal energy (on an arbitrary scale) of 762.0 kJ/kg and a specific volume of 1.128 cm3/g.

(a) What is its enthalpy?

(b) The water is brought to the vapor state at 573.15 K (300°C) and 1500 kPa, where its internal energy is 2784.4 kJ/kg and its specific volume is 169.7 cm3 g-'. Calculate U Δ and H Δfor the process.

Solution

(a)

H 1=U 1+P1V 1→762kJkg

+ (1002.7kPa ) ∙(0.001128m3

kg )=763.13kJkg

(b)

∆U=2784.4−762=2022.4kJkg

H 2=U 2+P2V 2→2784.4kJkg

+ (1500kPa ) ∙(0.1697m3

kg )=3038.95kJkg

∆ H=3038.95−763.13=2275.82kJkg

2.19. Problem 19

A solid body at initial temperature T0 is immersed in a bath of water at initial temperature Tw0.

Heat is transferred from the solid to the water at a rate Q = K (Tw - T), where K is a constant and Tw and T are instantaneous values of the temperatures of the water and solid. Develop an expression for T as a function of time t. Check your result for the limiting cases, t = 0 and t = ∞. Ignore effects of expansion or contraction, and assume constant specific heats for both water and solid.

Solution

Write the general energy balance for the solid immersed in the water bath.

d (mU )soliddt

+[∆ (H+12u2+gz )m]

fs

=Q+W

Assume the following

The process occurs within a closed system. Negligible potential and kinetic energy. No work transferred into the system. Symbols with subscripts s refer to the solid as well as symbols with subscripts w refer

to the water bath.

The general energy balance reduces to:

d (mU )sdt

=Q

Since the mass of the solid remains constant

ms

dU s

dt=Q→Q=k (T w−T s )→ms

dU s

dt=k (T w−T s )

Applying the chain rule

ms

dU s

dT s

∙dT s

dt=k (T w−T s )→C s=

dU s

d T s

→msC s ∙d T s

dt=k (T w−T s )

Now an expression of T w as a function of T s is required. To obtain this expression consider that the decrease of internal energy within the solid equals the increase of internal energy in the water. Mathematically, it is expressed as

∆U s=−∆Uw

Assuming constant specific heats for both water and solidmsC s(T s−T s0)=−mwCw(T w−T w 0)

Solving for Tw

T w=T w 0−msC s

mwCw

(T s−T s0)

Replace this expression into the reduced energy balance

msC s ∙dT s

dt=k (T w−T s )→msC s ∙

d T s

dt=k (T w 0−

msC s

mwCw(T s−T s0 )−T s)

msC s ∙dT s

dt=k (T w 0−T s−

msC s

mwCw(T s−T s0 ))

dT s

dt=k (

T w 0−T s

msC s

−T s−T s0

mwCw

)→dT s

dt=k (

T w 0

msC s

−T s

msC s

−T s

mwCw

+T s0

mwCw

)

dT s

dt=−kT s( 1

msC s

+ 1mwCw

)+k ( T s0

mwCw

+T w 0

msC s)

Define: β=k ( 1msC s

+ 1mwCw

) , α=k ( T s0

mwCw

+T w 0

msC s), where both and are constants.α β

dT s

dt=α−β T s

Integration from 0 to t and from Ts0 to Ts yields

∫T s0

T s dT s

α−βT s

=∫0

t

dt

−1β

ln (α−βT s)|Ts0

Ts

= t|0t

−1β

ln( α−β T s

α−β T s0)=t →

α−βT s

α−βT s0

=exp (−βt)

Solving for Ts

T s=(T¿¿ s0−αβ)exp (−βt )+ α

β¿

For t = 0

T s=(T ¿¿ s0−αβ)exp (0 )+ α

β→T s=(T ¿¿ s0−α

β)+ α

β→T s=T s0¿¿

For t = ∞

T s=limt →∞

(T ¿¿ s0−αβ)exp (−βt )+ α

β→T s=lim

t→∞

(T¿¿ s0−αβ)

exp ( βt)+αβ

¿¿

T s=0+ αβ→T s=

αβ

Another way to find the result at t = ∞ is to think that when enough time passes, a thermal equilibrium is achieved and the temperature of the solid equals the temperature of the water bath.

T w=T w 0−msC s

mwCw(T s−T s0 )→T s=T w 0−

msC s

mwCw

(T s−T s0)

Solving for Ts

T s=T w 0mwCw+T s0msC s

mwCw+msC s

However

αβ=

k ( T s0

mwCw

+T w0

msC s)

k ( 1m sC s

+ 1mwCw )

→αβ=

(T s0msC s+Tw 0mwCw

msC smwCw)

(mwCw+msC s

msC smwCw)

→αβ=

T s0msC s+T w 0mwCw

mwCw+msC s

Therefore,

T s=αβ

2.20. Problem 20!!!

A list of common unit operations follows:

(a) Single-pipe heat exchanger; (b) Double-pipe heat exchanger; (c) Pump;

(d) Gas compressor: (e) Gas turbine; (f) Throttle valve: (g) Nozzle.

Develop a simplified form of the general steady-state energy balance appropriate for each operation. State carefully, and justify, any assumptions you make.

2.21. Problem 21

The Reynolds number Re is a dimensionless group which characterizes the intensity of a flow.

For large Re, a flow is turbulent; for small Re, it is laminar. For pipe flow, ¿uρDμ

, where D is

pipe diameter and is dynamic viscosity.μ

(a) If D and ρ are fixed, what is the effect of increasing mass flowrate m on Re?

(b) If m and ρ are fixed, what is the effect of increasing D on Re?

Solution

For a cylindrical pipe, the cross-sectional area is A=π D2

4

m=uAρ→A=π D2

4→m=uπ D2 ρ

4

Solving for u

u= 4 m

π D2ρ

Replace u in the definition of Reynolds number.

ℜ=uρDμ

→ℜ= 4 m

π D2 ρ

ρDμ

→ℜ= 4 mπDμ

(a) Clearly, an increase of mass flowrate results in an increase of Re.(b) Clearly, an increase of the pipe diameter results in a decrease of Re

2.22. Problem 22

An incompressible (ρ = constant) liquid flows steadily through a conduit of circular cross-section and increasing diameter. At location 1, the diameter is 2.5 cm and the velocity is 2 m/s ; at location 2, the diameter is 5 cm.

(a) What is the velocity at location 2?

(b) What is the kinetic-energy change (J/kg) of the fluid between locations 1 and 2?

Solution

(a) A circular cross-section area is calculated as: A=π D2

4

The mass balance of the incompressible flow gives:

∆ m=0→m1=m2→u1 A1 ρ=u2 A2 ρ→ ρ=ct→u1 A1=u2 A2

u2=u1

A1

A2

→u2=u1

π D12

4π D2

2

4

→u2=u1

D12

D22 →u2=(2

ms ) (

2.5100 )

2

m2

( 5100 )

2

m2

=0.5ms

(b)

∆ Ek

m=1

2u2

2−12u1

2→∆Ek=12

(u22−u1

2 )→∆Ek=12

((0.5)2−22 )¿ m2

s2 →−1.875Jkg

2.23. Problem 23

A stream of warm water is produced in a steady-flow mixing process by combining 1.0 kg/s of cool water at 298.15 K (25°C) with 0.8 kg/s of hot water at 348.15 K (75°C). During mixing, heat is lost to the surroundings at the rate of 30 kW. What is the temperature of the warm-water stream? Assume the specific heat of water constant at 4.18 kJ/ kg-K.

Solution

Write the general energy balance and mass balance for the system

Energy balance :d (mU )cv

dt+[∆(H+ 1

2u2+gz )m ]

fs

=Q+W

Mass balance :d(m)cvdt

+¿¿

Assume the following

The process is at steady state Negligible potential and kinetic energy. No work transferred into the system. Constant heat capacity for both streams

The energy balance and mass balance reduces to

Reduced energybalance : [∆ (H )m ]fs=Q→m3 H 3−(m2H 2+m1H 1)=Q

Reduced massbalance : ¿¿Combine the energy and mass balances

¿¿

m2 ( H 3−H 2 )+m1 (H 3−H 1 )=Q→∆H=C p∆T→m2C p (T 3−T 2 )+m1C p (T3−T1 )=Q

Solve for T3

T 3=Q+m2C pT 2+m1C pT 1

m2C p+m1C p

T 3=−30kW+0.8

kgs∙4.18

kJkg ∙K

∙348.15+1kgs

∙4.18kJ

kg ∙ K∙298.15 K

0.8kgs∙4.18

kJkg ∙K

+1kgs∙4.18

kJkg ∙ K

=316,38 K

2.24. Problem 24

Gas is bled from a tank. Neglecting heat transfer between the gas and the tank, show that mass and energy balances produce the differential equation:

dU

H ´−U=dm

m

Here, U and m refer to the gas remaining in the tank; H' is the specific enthalpy of the gas leaving the tank. Under what conditions can one assume H' = H?

Solution

Write the general energy balance and mass balance for the system (the tank)

Energy balance :d (mU )cv

dt+[∆(H+ 1

2u2+gz )m ]

fs

=Q+W

Mass balance :d(m)cvdt

+¿¿

Assume/consider the following

Negligible potential and kinetic energy. No work transferred into the system. No heat transferred into the system. Only one stream leaving the tank

The reduced energy and mass balances are

Reduced energybalance :d (mU )

dt+H ' m1=0

ReducedMass balance :dmdt

+m1=0

Eliminate m1from the energy balance by the mass balance

d (mU )dt

−H ' dmdt

=0→mdUdt

+U dmdt

−H ' dmdt

=0→mdUdt

+(U−H ' ) dmdt

=0

mdUdt

=(H ´−U ) dmdt

→1

(H ´−U )dUdt

= 1m

dmdt

→multiply by dt→dU

H '−U=dm

m

Presumption of H’ = H is only valid assuming uniform conditions throughout the tank. This requires the absence of any gradients in the gas in the tank.

2.25. Problem 25

Water at 301.15 K (28°C) flows in a straight horizontal pipe in which there is no exchange of either heat or work with the surroundings. Its velocity is 14 m/s in a pipe with an internal diameter of 2.5 cm until it flows into a section where the pipe diameter abruptly increases. What is the temperature change of the water if the downstream diameter is 3.8 cm? If it is 7.5 cm]? What is the maximum temperature change for an enlargement in the pipe?

Solution

Assume a steady-state system

The material balance for a horizontal pipe is straightforward

m1=m2→ρ1u1 A1=ρ1u2 A2

Assume an incompressible fluid ( = constant) and calculate the final velocity:ρ

ρ u1 A1=ρ u2 A2→u2=u1

A1

A2

→A=π D2

4→u2=u1

D 12

D 22

The reduced energy balance with negligible work, heat and potential energy is

∆ (H+ 12u2)m=0

As mass flowrate is constant and can’t be zero, the equation reduces to

∆ (H+ 12u2)=0

H 2−H 1+12

(u22−u1

2)=0→u2=u1

D 12

D 22 →H 2−H 1+

12 ((u1

D12

D22 )

2

−u12)=0

H 2−H 1=−12 ((u1

D12

D22 )

2

−u12)→∆H=

−12

u12( D1

4

D24 −1)→∆H=

12u1

2(1−D1

4

D24 )

Assume a constant heat capacity for the water (Cp = 4.18 kJ/kg-K)

∆ H=Cp∆T ↔∆H=12u1

2(1− D14

D24 )→Cp∆T=

12u1

2(1− D14

D24 )

∆T=12u1

2

C p (1−( D1

D2)

4

)

Case D2 = 3.8cm

∆T=12u1

2

C p (1−( D1

D2)

4

)→ 12¿¿¿

Case D2 = 7.5cm

∆T=12u1

2

C p (1−( D1

D2)

4

)→ 12¿¿¿

Case D2 = ∞

∆T= limD2→∞

12u1

2

C p (1−( D1

D2)

4

)→ 12u1

2

Cp

→12¿¿¿

The temperature change is negligible, thus the assumptions of constant density and constant heat capacity are completely reasonable.

2.26. Problem 26

Fifty (50) kmol per hour of air is compressed from P1 = 1.2 bar to P2 = 6.0 bar in a steady-flow compressor. Delivered mechanical power is 98.8 kW. Temperatures and velocities are:

T 1=300 KT 2=520 K

u1=10m / su2=3.5m / s

Estimate the rate of heat transfer from the compressor. Assume for air that Cp = 72

R and that

enthalpy is independent of pressure.

Solution

The material (molar) balance for the compressor is

∆ n=0↔n1=n2↔ρ1u1 A1=ρ1u2 A2 constant

Additionalinformation→n=50kmolhr

=13. 8mols

/molecularweight=29g

mol

The energy balance with negligible potential energy

∆ (H+ 12u2)n=Q+W

Solving for Q

Q= n(∆ H+ 12∆u2)−W →∆H=C p∆T→Q=n(C p∆T+ 1

2∆u2)−W

Q= n(C p∆T+ 12∆u2)−W

Q=13.8mols

∙( 72∙8.314

Jmol ∙ K

∙ (520−300 )+ 12

(3.52−102 ) m2

s2 ∙29

1000kgmol

❑)−98800Js

Q=−9904.06J /s=−9.9kW

2.27. Problem 27

Nitrogen flows at steady state through a horizontal, insulated pipe with inside diameter of 38.1 mm. A pressure drop results from flow through a partially opened valve. Just upstream from the valve the pressure is 690kPa, the temperature is 322.15 K (49"C), and the average velocity is 6.09 m/s. If the pressure just downstream from the valve is 138 kPa, what is the temperature? Assume for nitrogen that PV/ T is constant, Cv = (5/2) R, and Cp = (7/2) R.

Solution

The mass balance is straightforward

m1=m2→u1 A1

V 1

=u2 A2

V 2

→theinitial∧final areasare the same→u1

V 1

=u2

V 2

u2=V 2

V 1

u1

The following equations comes as a consequence of the assumption of a constant PV/T.

P1V 1

T 1

=P2V 2

T 2

→solving forV 2→V 2=V 1 ∙T 2

T 1

∙P1

P2

Replacing this expression on the later equation

u2=V 2

V 1

u1→V 2=V 1 ∙T 2

T 1

∙P1

P2

→u2=u1

T2

T1

P1

P2

The energy balance for the horizontal insulated pipe is:

∆ (H+ 12u2)m=0→as mcannot be zero→∆(H+ 1

2u2)=0

∆ (H+ 12u2)=0↔∆H=−1

2¿

Assuming that enthalpy’s dependency of pressure is negligible and constant heat capacity

dH=CpdT →∆H=C p∆T

Replacing this expression on the energy balance.

C p∆T=−12

¿

Replacing the final velocity expression on the energy balance

u2=u1

T2

T1

P1

P2

→Cp∆T=−12

¿

C p(T2−T1)=12u1

2(1−(T 2

T 1

P1

P2)

2

)

Use Newton’s method to find T 2

Objective function: C p (T 2−T 1 )−12u1

2(1−(T2

T1

P1

P2)

2

)=0

Initial value: T 2=T1=322.15

Iterative solution: T 2=¿321.72 K

2.28. Problem 28

Water flows through a horizontal coil heated from the outside by high-temperature flue gases. As it passes through the coil the water changes state from liquid at 200 kPa and 353.15 K (80°C) to vapor at 100 kPa and 398.15 K (125°C). Its entering velocity is 3 m/s and its exit velocity is 200 m/s. Determine the heat transferred through the coil per unit mass of water. Enthalpies of the inlet and outlet streams are: Inlet: 334.9 kJ/kg; Outlet: 2726.5 kJ/kg.

Solution

The intensive energy balance for the coil is:

∆ (H+ 12u2)=Q

The heat can be calculated directly

Q=∆(H+ 12u2)→Q=H 2−H 1+

(u22−u1

2)2

Q=H 2−H 1+(u2

2−u12 )

2=(2726.5−334.9 ) kJ

kg+

(2002−32 )2

∙1

1000kJkg

=2411,6kJkg

2.29. Problem 29

Steam flows at steady state through a converging, insulated nozzle, 25 cm long and with an inlet diameter of 5 cm. At the nozzle entrance (state I), the temperature and pressure are 598.15 K (325°C) and 700 kPa, and the velocity is 30 m/s. At the nozzle exit (state 2), the steam temperature and pressure are 513.15 K (240°C) and 350 kPa. Property values are:

H 1=3112.5 kJ /kgV 1=388,61kJ /kg

H 2=2945.7 kJ /kgV 2=667,75kJ /kg

What is the velocity of the steam at the nozzle exit, and what is the exit diameter?

Solution

The material balance for the nozzle is:

∆ m=0→m1=m2→m=constant

The energy balance for the insulated nozzle gives:

∆ (H+ 12u2)m=0→constant m→∆(H+ 1

2u2)=0→∆H=

−u22−u1

2

2

Solving for u2

u2=√−2∆ H +u12→u2=√−2 ∙ (2945.7−3112,5 ) ∙1000

m2

s2+302 m

2

s2→578,36

ms

Considering that the mass flowrate is constant, the final diameter can be calculated

m1=m2→u1 A1

V 1

=u2 A2

V 2

→solving for A2→A2=A1 ∙u1

u2

∙V 2

V 1

For a cylindrical nozzle, the cross-section area is: A=π D2/ 4 .

π D22

4=

π D12

4∙u1

u2

∙V 2

V 1

→D22=D1

2 ∙u1

u2

∙V 2

V 1

→D2=√D12∙u1

u2

∙V 2

V 1

→D2=D1 √ u1

u2

∙V 2

V 1

D2=D1 √ u1

u2

∙V 2

V 1

→D2=5cm√ 30578,36

∙667,75388,61

→D2=1,493cm

2.30. Problem 30

In the following take Cv = 20.8 and Cp = 29.1 J/mol-C° for nitrogen gas:

(a) Three moles of nitrogen at 303.15 K (30°C), contained in a rigid vessel, is heated to 523.15 K (250°C). How much heat is required if the vessel has a negligible heat capacity? If the vessel weighs 100 kg and has a heat capacity of 0.5 kJ/kg-C°, how much heat is required?

(b) Four moles of nitrogen at 473.15 K (200°C) is contained in a piston/cylinder arrangement.

How much heat must be extracted from this system, which is kept at constant pressure, to cool it to 313.15 K (40°C) if the heat capacity of the piston and cylinder is neglected?

Solution

(a) A rigid vessel resembles a closed container with constant volume, therefore the following equation may be applied.

n(∆U )=Q

A) Neglecting the heat capacity of the vessel

n (∆U )=Q→n∆U=nCV ∆T→Q=nCV ∆T

Q=3mol ∙20.8J

mol ∙C °∙ (250−30 )=13728J=13,73kJ

B) Considering the heat capacity of the vessel ( ve – Vessel, N – Nitrogen)

n (∆U )=Q→n∆U=nveCVve∆T +nNCV N∆T→Q=nveCV ve∆T +nNCVN ∆T

Q=3mol ∙20.8J

mol ∙C °∙ (250−30 )+100 kg ∙0.5

kJkg ∙C °

∙ (250−30 )=11013,7 kJ

(b) For a closed system at a constant pressure, the following equation applies

n(∆ H )=Q

If the heat capacity of the piston and cylinder is neglected

n (∆ H )=Q→n∆H=nCP∆T→Q=nCP∆T

Q=4mol ∙29.1J

mol ∙C °∙ ( 40−200 )=−18624 J=18,62kJ

2.31. Problem 31In the following take Cv = 21 and Cp = 29.3 kJ/kmol-K for nitrogen gas:(a) 1.5 kmol of nitrogen at 294.15 K (21°C) contained in a rigid vessel, is heated to 450.15 K (177"C). How much heat is required if the vessel has a negligible heat capacity? If it weighs 90.7 kg and has a heat capacity of 0.5 kJ/kg-K, how much heat is required?(b) 2 kmol of nitrogen at 447.15 K (174"C) is contained in a pistol/cylinder arrangement.How much heat must be extracted from this system, which is kept at constant pressure, to cool it to 338.15 K (65"C) if the heat capacity of the piston and cylinder is neglected?

SolutionThis problem is very similar to problem 2.30 so the same equations may be used.

(a) Constant volumeA) Neglecting the heat capacity of the vessel

n (∆U )=Q→n∆U=nCV ∆T→Q=nCV ∆T

Q=1.5 kmol ∙21kJ

kmol ∙ K∙ (450.15−294.15 )K=4914 J=4,91kJ

B) Considering the heat capacity of the vessel ( ve – Vessel, N – Nitrogen)

n (∆U )=Q→n∆U=nveCVve∆T +nNCV N∆T→Q=nveCV ve∆T +nNCVN ∆T

Q=1.5 kmol ∙21kJ

kmol ∙ K∙ (450.15−294.15 )+90.7kg ∙0.5

kJkg ∙C °

∙ (450.15−294.15 )

Q=11988,6 J=11,98 kJ

(b) Constant pressure

A) Neglecting the heat capacity of the vessel

n (∆ H )=Q→n∆H=nCP∆T→Q=nCP∆T

Q=2kmol ∙29.3kJ

kmol ∙C °∙ (338.15−447.15 ) K=−6387,4 J=6,38kJ

2.32. Problem 32Find the equation for the work of a reversible, isothermal compression of 1 mol of gas in a piston/cylinder assembly if the molar volume of the gas is given by:

V= RTP

+b

Where b and R are positive constants.

Solution

For a mechanical reversible process, the work is defined as:

W=−∫V 1

V 2

PdV

To integrate this equation, first solve for P on the molar volume expression

V= RTP

+b→P= RTV−b

Then replace this on the work equation.

W=−∫V 1

V 2

PdV→W=−∫V 1

V 2

RTV −b

dV →W=−RT∫V 1

V 2

1V −b

dV →W=RT ln|V 2−b||V 1−b|

2.33. Problem 33Steam at 14 bar and 588.15 K (315°C) (state 1) enters a turbine through a 75 mm-diameter pipe with a velocity of 3 m/s . The exhaust from the turbine is carried through a 250mm diameter pipe and is at 0.35 bar and 366.15 K (93"C) [state 21. What is the power output of the turbine?

H 1=3074,5kJ /kgV 1=0,1909m3/kgH 2=2871,6 kJ /kgV 2=4,878m3/kg

Solution

Assume the following

Steady-state. Negligible change of potential energy. No heat transfer.

The mass balance of the turbine gives

∆ m=0→m1=m2→mconstant→m1=u1 A1

V 1

→m1=¿¿

m1=0,0694kgs

As a consequence of a constant mass flowrate, the final velocity can be calculated directly.

m2=m1=0,0694kgs→m2=

u2 A2

V 2

→solving for u2→u2=m2 A2

V 2

u2=(0,0694

kgs )∙4,878m3/kg

( π (0,25m )2

4 )=6,899

ms

The energy balance of the process is

∆˙

(H+ 12u2)m=W s

W s=m ¿

W s=0,0694kgs∙((2871,6−3074,5 ) kJ

kg+ 6,892−32

2∙

11000

kJkg

)

W s=−12,74 kW

2.34. Problem 34Carbon dioxide gas enters a water-cooled compressor at the initial conditions P1 = 1.04bar and T1 = 284.15 K (10°C) and is discharged at the final conditions P2 = 35.8 bar and T2 = 366.15 K (93'C). The entering CO2 flows through a 100 mm-diameter pipe with a velocity of 6 m/s, and is discharged through a 25 mm-diameter pipe. The shaft work supplied to the compressor is 12 500 kJ/kmol. What is the heat-transfer rate from the compressor in kW?

H 1=714 kJ /kgV 1=0,5774 m3/kgH 2=768kJ /kgV 2=0,0175m3/kg

Solution

Assume the following

Steady-state. Negligible change of potential energy.

The mass balance of the compressor gives

∆ m=0→m1=m2→mconstant→m1=u1 A1

V 1

→m1=¿¿

m1=0,0816kgs

As a consequence of a constant mass flowrate, the final velocity can be calculated directly.

m2=m1=0,0694kgs→m2=

u2 A2

V 2

→solving for u2→u2=m2 A2

V 2

u2=(0,0816

kgs ) ∙0,0175m3/kg

( π (0,025m )2

4 )=2,909

ms

The intensive energy balance of the process is

∆ (H+ 12u2)=W s+Q

Solving for Q

Q=¿

Q=(768−714) kJkg

+ 2,9092−62

2∙

11000

kJkg

¿−12500kJ

kmol∙

1kmol44kg

Q=−230,07kJkg

2.35. Problem 35Show that W and Q for an arbitrary mechanically reversible non-flow process are given by:

W=∫V dP−∆ (PV )Q=∆ H−∫V dP

Solution

For one hand, in a mechanically reversible process, work is defined as:

dW=−PdV

For the other hand, recall thatd (PV )=PdV+VdP→−PdV =VdP−d (PV )

Whence,dW=VdP−d (PV )→integrating→W=∫VdP−∆(PV )

The first law of thermodynamics for a closed system is:

dU=dQ+dW (Eq .2.4)

dU=dQ+dW→dU=dQ+VdP−d (PV )

The enthalpy is defined as:

H=U+PV (Eq .2.11)

Differentiating

dH=dU+d (PV )→solving for dU→dU=dH−d (PV )

Combining the two expressions for internal energy

dU=dH−d (PV )↔dU=dQ+VdP−d (PV )

dH−d (PV )=dQ+VdP−d (PV )→solving for dQ→dQ=dH−VdP→Q=∆H−∫VdP

2.36. Problem 36One kilogram of air is heated reversibly at constant pressure from an initial state of 300 K and 1 bar until its volume triples. Calculate W, Q, ΔU, and HΔ for the process. Assume for air that PV/T = 83.14 bar-cm3/mol-K and Cp = 29 J/mol-K.

Solution

The initial molar volume is P1V 1

T 1

=83.14 cm3 ∙ ¿mol ∙ K

→V 1=300 K ∙83.14

1 ¿=24942cm3

mol¿¿

Average molecular weight of air: 28 ∙0.79+32∙0.21=28,84 gr /mol

Total moles of air: 1kg ∙1000 gr

1kg∙

1mol28,84 gr

=34,67 mol

Calculate final temperature

P1V 1

T 1

=P2V 2

T 2

→solving for T 2→T 2=T1

P2V 2

P1V 1

→constant P→T 2=T 1

V 2

V 1

→V 2=3V 1

T 2=T1

3V 1

V 1

→T 2=3T1→900K

Recall that for a constant pressure process, equation 2.22 can be applied

∆ H=∫T1

T2

CpdT →∆H=C p∆T →∆H=29J

mol ∙K∙ (900 K−300 K )=17400

Jmol

Heat for a constant pressure process is expressed as:Q=n∆ H→Q=34,67 mol ∙17400 J /mol→Q=603328,71J

Work for a constant pressure process is expressed as:

W=−nP∆V →W=−nP (3V 1−V 1)→W=−n2PV 1

W=−34,67mol ∙2∙100000 Pa∙24924cm3

mol∙

1m3

1003 cm3=−172843,27J

By eq. 2.3.

n∆U=Q+W →n∆U=603328,71−172843,27→∆U=12415,2J

mol

2.37. Problem 37The conditions of a gas change in a steady-flow process from 293.15 K (20°C) and 1000 kPa to 333.15 K (60°C) and 100 kPa. Devise a reversible non-flow process (any number of steps) for accomplishing this change of state, and calculate U Δ and H Δ for the process on the basis of 1 mol of gas. Assume for the gas that PV/T is constant, Cv = (5/2)R, and Cp = (7/2)R.

Solution

This problem can be solved in different ways, the choice of steps is totally arbitrary. However the final change of internal energy and enthalpy must be the same. The chosen steps are:

(a) Heat at constant pressure – V2

(b) Cool at constant volume – P2

RecallPVT

=R→8,314m3 ∙ Pamol ∙ K

Calculate initial volume

P iV i

T i

=R→V f=R ∙T i

Pi

→V f=8,314 ∙293.15K

1000000Pa→V f=0,00243m3/mol

Calculate final volume

P f V f

T f

=R→V f=R ∙T f

P f

→V f=8,314 ∙333.15K

100000Pa→V f=0,02769m3/mol

(a) Step 1-2 / constant pressureKnown data

P1=1000kPa P2=1000kPaT 1=293.15 K

V 1=0,00243m3/molV 2=0,02769m3/mol

Calculate T 2

P2V 2

T 2

=R→T 2=P2V 2

R→T 2=

1000000 Pa ∙0,02769m3/mol8,314 m3 ∙Pa/mol ∙ K

→T 2=3331,5 K

Calculate HΔ

∆ H=Cp∆T →∆H=(7¿¿2)R∆T→∆H=(7¿¿2)(8,314J

mol ∙K )∙ (3331.5−293,15 )¿¿

∆ H=88412,94 J /mol

Calculate UΔ

∆U=∆ H−P∆V→∆U=88412,94J

mol−1000 000 Pa ∙ ( 0,02769−0,00243 ) m3

mol

∆U=63152,10J /mol

(a) Step 2-3 / constant volumeKnown data

P2=1000kPa P3=100kPaT 2=3 33 1.5 KT 3=333. 1 5K

V 2=0,02769m3/molV 3=0,02769m3/mol

Calculate UΔ

∆U=C v∆T→∆U=52∙8,314

Jmol ∙ K

∙ (333.15−3331.5 )K→∆U=−62355J

mol

Calculate HΔ

∆ H=∆U+V ∆ P→∆H=−62355J

mol+0,02769

m3

mol∙(100000Pa−1000000Pa)

∆ H=−87283,28J /mol

Calculate total U and HΔ Δ

∆ H=∆ H 12+∆ H 23=88412,94−87283,28=1129,66 J /mol∆U=∆U 12+∆U 23=63152,1−62355=797,10J /mol