Thermo HW Solutions

7/26/2019 Thermo HW Solutions

1/35

1

1.4 Perform the following unit conversions:

(a) 333

in.61ft1

in.12

L1

ft0353.0L1

(b) Btu616.0kJ1.0551

Btu1

J10

kJ1J506

3

(c)s

lbfft596.99

Btu1

lbfft78.177

s3600

h1

kW1

Btu/h4133kW.1350

(d)min

lb50

min1

s60

kg0.4536

lb1

g10

kg1

s

g783

3

(e)2

32

in.

lbf09.44

kPa1

Pa10

Pa894.86

lbf/in.1kPa043

(f)s

ft54.0

s3600

h1

m1

ft.28083

h

m55

333

(g)s

ft57.45

s3600

h1

m1

ft.28083

km1

m01

h

km05

3

(h) ton1lbf2000

ton1

N.44824

lbf1N8968


2/35


3/35


4/35

1

1.28 A closed system consisting of 4 lb of a gas undergoes a process during which the relation

between pressure and volume is pVn= constant. The process begins withp1= 15 lbf/in.

2, v1=

1.25 ft3/lb and ends with p2= 53 lbf/in.

2, v2= 0.5 ft

3/lb. Determine (a) the volume, in ft

3,

occupied by the gas at states 1 and 2 and (b) the value of n. (c) Sketch Process 1-2 on pressure-

volume coordinates.

KNOWN: Gas undergoes a process from a known initial pressure and specific volume to a

known final pressure and specific volume.

FIND: Determine (a) the volume, in ft3, occupied by the gas at states 1 and 2 and (b) the value

of n. (c) Sketch Process 1-2 on pressure-volume coordinates.

SCHEMATIC AND GIVEN DATA:

ENGINEERING MODEL:

1. The gas is a closed system.

2. The relation between pressure and volume is pVn= constantduring process 1-2.

ANALYSIS:

(a)The specific volume is volume per unit mass. Thus, the volume occupied by the gas can be

determined by multiplying its mass by its specific volume.

V= mv

For state 1

lb

ft25.1)lb4(

3

11 vmV = 5 ft3

For state 2

lb

ft5.0)lb4(

3

22 vmV = 2 ft3

State 1 State 2

m = 4 lb

p1 = 15 lbf/in.2

v

1= 1.25 ft3/lb

p2 = 53 lbf/in.2

v2 = 0.5 ft3/lb

Gas

Gas


5/35

2

(b)The value of ncan be determined by substituting values into the relationship:

p1(V

1)n= constant= p

2(V

2)n

Solving for n

n

V

V

p

p

1

2

2

1

1

2

2

1 lnlnV

Vn

p

p

3

3

2

2

1

2

2

1

ft5ft2ln

lbf/in.35

lbf/in.15ln

ln

ln

VV

p

p

n = 1.38

(c)Process 1-2 is shown on pressure-volume coordinates below:

0

10

20

30

40

50

60

0 1 2 3 4 5 6

Pressure(lbf/in.2)

Volume (ft3)

Process 1-2

State 1

State 2


6/35

1

1.46 As shown in Figure P1.46, an inclined manometer is used to measure the pressure of the

gas within the reservoir. (a) Using data on the figure, determine the gas pressure, in lbf/in.2(b)

Express the pressure as a gage or a vacuum pressure, as appropriate, in lbf/in.2(c) What

advantage does an inclined manometer have over the U-tube manometer shown in Figure 1.7?

KNOWN: A gas contained in a reservoir with inclined manometer attached.

FIND: (a) Pressure of gas within the reservoir, in lbf/in.2 (b) Pressure expressed as gage or

vacuum pressure, as appropriate, in lbf/in.2 (c) Advantage of inclined manometer over the U-

tube manometer.


ENGINEERING MODEL:

1. The gas is a closed system.2. Atmospheric pressure is exerted at the open end of the manometer.

3. The manometer fluid is oil with a density of 54.2 lb/ft3.

ANALYSIS:(a)Applying Eq. 1.11

pgas= patm+ gL

wherepatmis the local atmospheric pressure, is the density of the manometer fluid (oil), gis the

acceleration due to gravity, and Lis the vertical difference in liquid levels. Since level a is the

same as level b, applying trigonometry to determine the vertical difference in liquid levels

between level b and the liquid level at the free surface with the atmosphere yields

pgas= patm+ gL(sin 40o)

Substituting values

Gas

Oil (= 54.2 lb/ft3)

patm = 14.7 lbf/in.2

g= 32.2 ft/s2

40o

15 in.a b


7/35

2

3

3

2

22in.1728

ft1

s

ftlbm32.2

lbf1)40in.)(sin15(

s

ft2.32

ft

lb2.54

in.

lbf7.14

gasp = 15.0 lbf/in.

2

(b)Since the pressure of the gas is greater than atmospheric pressure, gage pressure is given by

Eq. 1.14

p(gage) = p(absolute)patm(absolute) = 15.0 psia14.7 psia = 0.3 psig

(c) The advantage of the inclined manometer is its easier readability since the surface of the

liquid is wider than with a same diameter U-tube manometer. The scale on the inclinedmanometer is much more precise since more graduations are possible compared with the U-tube

manometer.


8/35

3

Substituting values for pressures and specific volume yields

v2=5.0

13

kPa100

kPa250

kg

m5.0

= 3.125 m

3/kg

The volume of the system increased while pressure decreased during the process.

A plot of the process on a pressure versus specific volume graph is as follows:

Pressure versus Specific Volume

100

120

140

160

180

200

220

240

260

0.50 1.00 1.50 2.00 2.50 3.00 3.50

Specific Volume (m^3/kg)

Pressure(k

Pa)


9/35

Problem 2.27

Carbon dioxide (CO2) gas within a piston-cylinder assembly undergoes a process from a state

wherep1= 5 lbf/in.2, V1= 2.5 ft

3to a state wherep2= 20 lbf/in.

2, V2= 0.5 ft

3. The relationship

between pressure and volume during the process is given byp= 23.757.5V, where Vis in ft3

andpis in lbf/in.

2

Determine the work for the process, in Btu.

KNOWN: CO2gas within a piston-cylinder assembly undergoes a process where thep-V

relation is given. The initial and final states are specified.

FIND: Determine the work for the process.


ANALYSIS: The givenp-Vrelation can be used with Eq. 2.17 as follows:

= -3600 ftlbf

= (negative sign denotes energy transfer in.)Alternative SolutionSince thep-Vrelation is linear, Wcan also be evaluated geometrically as the area under the

process line:

= -4.63 Btu

V (ft^3)32.521.510.50

25

20

15

10

5

0

.

.1W

2

CO2

p1= 5 lbf/in.V1= 2.5 ft

3

p2= 20 lbf/in.2

V2= 0.5 ft3

p= 23/757.5 V

ENGINEERING MODEL: (1) The CO2is theclosed system. (2) Thep-Vrelation during the

process is linear. (3) Volume change is the only

work mode.


10/35

Problem 2.34

Carbon monoxide gas (CO) contained within a piston-cylinder assembly undergoes three

processes in series:

Process 1-2: Constant pressure expansion at 5 bar from V1= 0.2 m

3

to V2= 1 m

3

.

Process 2-3: Constant volume cooling from state 2 to state 3 wherep3= 1 bar.

Process 3-1:Compression from state 3 to the initial state during which the pressure-volumerelationship ispV= constant.

Sketch the processes in series onp-Vcoordinates and evaluate the work for each process, in kJ.

KNOWN: Carbon monoxide gas within a piston-cylinder assembly undergoes three processes

in series.

FIND: Sketch the processes in series on ap-Vdiagram and evaluate the work for each process.


ENGINEERING MODEL: (1) The gas is the closed system. (2) Volume change is the onlywork mode. (3) Each of the three processes is specified.

ANALYSIS: Since volume change is the only work mode, Eq. 2.17 applies.

Process 1-2: Constant pressure processes:

(out)

Process 2-3: Constant volume (piston does not move). Thus W23= 0

CO

Process 1-2: Constant pressure expansion at 5

bar from V1= 0.2 m3to V2= 1 m

3.

Process 2-3: Constant volume cooling from

state 2 to state 3 wherep3= 1 bar.Process 3-1:Compression from state 3 to the

initial state during which the pressure-volume

relationshi is V= constant.

125

1

0.2

3

1

p

(bar)

V(m )

pV= constant


11/35

Problem 2.33 (Continued)

Process 3-1: For process 3-1,pV= constant = p1V1. Noting that V3= V2, we get

Inserting values and converting units

= -160.9 kJ (in)

1. The network for the three process is

Wnet= W12+ W23+ W31= (+400) + 0 + (-160.9) = 239.1 kJ (network is positive - out)

1


12/35

PROBLEM 2.37


13/35

PROBLEM 2.46


14/35

Problem 2.56

Each line of the following table gives data for a process of a closed system. Each entry has the

same energy units. Determine the missing entries.

Process Q W E1 E2Ea +50 -20 +70

b +20 +50 +30

c -60 +40 +60d -40 +50 0

e +50 +150 -80

Process a:W= Q- E= +50(+ 70) = -20E=E2E1E2= E+E1= +70 + (-20) = +50

Process b:

Q= E+ W= +30 + (+20) = +50E=E2E1

E1=E2E= +50(+30) = +20

Process c:E=E2E1= +60(+40) = +20Q= E+ W= +20 + (-60) = -40

Process d:W= QE= (-90)0 = -90

E=E2E1

E2= E+E1= 0 +50 = +50

Process e:

E= QW= +50(+150) = -100E1=E2E= (-80)(-100) = +20

Process Q W E1 E2 E

a +50 -20 -20 +50 +70

b +50 +20 +20 +50 +30c -40 -60 +40 +60 +20d -90 -90 +50 +50 0

e +50 +150 +20 -80 -100

E= Q- W


15/35

Problem 2.62

An electric motor draws a current of 10 amp with a voltage of 110 V, as shown in Fig. P2.62.

The output shaft develops a torque of 9.7 Nm and a rotational speed of 1000 RPM. Foroperation at steady state, determine for the motor

(a) the electric power required, in kW.(b) the power developed by the output shaft, in kW.(c) the average surface temperature, Ts, in

oC, if heat transfer occurs by convection to the

surroundings at Tf= 21oC.

KNOWN: Operating data are provided for an electric motor at steady state.

FIND: Determine (a) the electric power required, (b) the power developed by the output shaft,

and (c) average the surface temperature.

ENGINEERING MODEL: (1) The motor is the closed system. (2) The system is at steady

state.

ANALYSIS: (a) Using Eq. 2.21

- (voltage) (current) = - (110 V)(10 amp) = -1.1 kW (in)

(b) Using Eq. 2.20

= (torque) (angular velocity)

= 1.016 kW (out)

(c) To determine the surface temperature, first find the rate of energy transfer by heat using theenergy balance

= (-1.1 kW) + (1.016 kW) = -0.084 kW

The surface temperature of the motor is

10 amp

110 V

T = 9.7 Nm

1000 RPM

Tf= 21oC

Ts

hA = 3.9 W/K

0


16/35


= (-0.084 kW)/(3.9 W/K)

+ 294 K

= 315.5 K = 42.5oC


17/35

Fig. P2.628e

Pick-up motor graphic from Fig. E2.67e


18/35

PROBLEM 2.76

.


19/35

Problem 2.79

A gas undergoes a cycle in a piston-cylinder assembly consisting of the following three

processes:

Process 1-2: Constant pressure,p= 1.4 bar, V1= 0.028 m

3

, W12= 10.5 kJ

Process 2-3: Compression withpV= constant, U3= U2

Process 3-1: Constant volume, U1U3= -26.4 kJ

There are no significant changes in kinetic or potential energy.

(a) Sketch the cycle on ap-Vdiagram.(b) Calculate the net work for the cycle, in kJ.

(c) Calculate the heat transfer for process 1-2, in kJ

KNOWN: A gas undergoes a cycle consisting of three processes.

FIND: Sketch the cycle on ap-Vdiagram and determine the net work for the cycle and the heat

transfer for process 1-2.

SCHEMATIC AND GIVEN DATA:Process 1-2: Constant pressure,p= 1.4 bar, V1= 0.028 m

3,

W12= 10.5 kJ

Process 2-3: Compression withpV= constant, U3= U2

Process 3-1: Constant volume, U1U3= -26.4 kJ

ENGINEERING MODEL: (1) The gas is a closed system. (2) Kinetic and potential energy

effects are negligible. (3) The compression from state 2 to 3 is a polytropic process.

ANALYSIS: (a) Since W12> 0, the process is an expansion. Thus

Gas

1 2

p

V

3

.

.

.


20/35


(b) The net work for the cycle is Wcycle= W12+W23+ W31. W12 = 10.5 kJ, so we need W23.

W23

(*)

where V3= V1has been incorporated. But, we still need to evaluate V2. For Process 1-2 at

constant pressure

W12= or

V2=

+ 0.028 m3= 0.103 m3

Thus, with Eq. (*)

W23=

=-18.78 kJThus

Wcycle = 10.5 kJ + (-18.78 kJ) + 0 = -8.28 kJ

(c) To get Q12, we apply the energy balance to process 1-2: KE+ PE+ (U2U1) = Q12W12

With U2= U3,

Q12= (U3U1) + W12= (+26.4 kJ) + (10.5 kJ) = 36.9 kJ

0

00


21/35

PROBLEM 2.86


22/35

Problem 2.92

A window-mounted room air conditioner removes energy by heat transfer from a room andrejects energy by heat transfer to the outside air. For steady operation, the air conditioner cyclerequires a power input of 0.434 kW and has a coefficient of performance of 6.22. Determine the

rate that energy is removed from the room air, in kW. If electricity is valued at $0.1/kWh,determine the cost of operation for 24 hours of operation.

KNOWN: Steady-state operating data are provided for an air conditioner.

FIND: Determine the rate energy is removed from the room and air the cost of 24 hours ofoperation.


ANALYSIS: Using Eq. 2.45 on a time rate basis

=

(6.22)(0.434 kW) = 2.70 kW

The total amount of electric energy input by work for 24 h of operation is

Wcycle= = (0.434 kW)(24 h) = 10.42 kWh

Thus, the total cost is

Total cost = (10.42 kWh)($0.1/kWh) = $1.04 (for 24 hours)

AirConditioner

Room air

Outside air

= 0.434 kW

Electric cost:

$0.1/kWh

ENGINEERING MODEL:(1) The system shown in the schematic

undergoes a refrigeration cycle.

(2) Energy transfers are positive in thedirections of the arrows.(3) The cycle operates steadily for 24

hours.

(4) Electricity is valued at $0.1/kWh.

RefrigerationCycle, =

6.22


23/35

PROBLEM 3.6

Determine the phase or phases in a system consisting of H2O at the following conditions and

sketch thep-vand T-vdiagrams showing the location of each state.

(a) p= 10 bar, T= 179.9

o

C(b) p= 10 bar, T= 150oC

(c) T= 100oC,p= 0.5 bar

(d) T= 20oC,p= 50 bar

(e) p= 1 bar, T= - 6oC

(a) p= 10 bar, T= 179.9oC

(b) p= 10 bar, T= 150oC

pT

vv

10 bar

179.9oC

(Table A-3)

10 bar

179.9oC

Two-phaseliquid-vapor

mixture

10 bar

179.9oC

150oC

p

vv

T

.

179.9oC

10 bar

150oC.

T


24/35


(c) T= 100

o

C,p= 0.5 bar

(d) T= 20oC,p= 50 bar

(e) p= 1 bar, T= - 6oC

0.003689 bar

-6oC

Tp

v v

-6oC

0.003689 bar (Table A-5)

1 bar

.

p>psat@Tsolid

(Tis below the triple

point temperature)

. 1 bar

.

1.014 bar

0.5 bar100

oC

100oC.

1.014 bar (Table A-2)

0.5 bar

pT

v v

ppsat@T

sub-cooled liquid


25/35

PROBLEM 3.7


26/35

PROBLEM 3.9

COMMENT: As the pressure increases, the difference in specific volume between saturatedvapor and saturated liquid decreases. At the critical pressure, the two states

coincide and the difference is zero.


27/35

PROBLEM 3.16

A 1-m3tank holds a two-phase liquid-vapor mixture of carbon dioxide at 17

oC. The quality of

the mixture is 70%. For saturated carbon dioxide at 17oC, vf= 0.9827 x 10

-3m

3/kg and

vg=

1.756 x 10-2

m3/kg. Determine the masses of saturated liquid and saturated vapor, each in kg.

What is the percent of the total volume occupied by saturated liquid?

First, find the total mass as follows:

vx = vf+x(vg- vf)= 0.9827 x 10-3

+ (0.7)(1.756 x 10-2

- 0.9827 x 10-3

) = 0.01258 m3/kg

Thus

m= V/vx= (1 m3)/(0.01258 m

3/kg) = 79.46 kg

Now, using the definition of quality

mg=x m= (0.7) ( 79.46 kg) = 55.62 kg

mf= (1 -x) m= (1 0.7) (79.46 kg) = 23.84 kg

The volume of saturated liquid is

Vf= vfmf= (0.9827 x 10-3

m3/kg) (23.84 kg) = 0.0234 m

3

The total volume is 1 m3, so the percent of the total volume occupied saturated liquid is 2.34%.

Note: Although the liquid is 30% of the total mass, its specific volume is much less than that of

the vapor. Consequently, the liquid occupies a very small fraction of the total volume.

V= 1 m

saturated vapor

saturated liquid

vf= 0.9827 x 10-

m /kg

vg= 1.756 x 10-2

m3/kg

carbon

dioxide

T= -17oC

x = 0.7


28/35

PROBLEM 3.33

Two kg of Refrigerant 134A undergoes a polytropic process in a piston-cylinder assembly from

an initial state of saturated vapor at 2 bar to a final state of 12 bar, 80oC. Determine the work for

the process, in kJ.

KNOWN: Refrigerant 134A undergoes a polytropic process in a piston-cylinder assembly.

FIND: Determine the work.


ANALYSIS: The work for the polytropic process is determined using Eq. 2.17, withpvn=

constant. Following the procedure of part (a) of Ex. 2.1

W=

(*)

In order to evaluate this expression, we need to determine the specific volumes and the

polytropic exponent, n.

State 1: From Table A-11; v1= vg1= 0.0993 m3/kg

State 2: From Table A-12, at 12 bar, 80oC; v2= 0.02051 m

3/kg

The polytropic exponent is found frompvn= constantas follows.

n= ln(p1/p2) / ln(v2/v1)

n= ln(2/12) / ln(0.02051/0.0993) = 1.136

Inserting values in Eq. (*) and converting units, we get

R-134A

m= 2 kg12 bar

2 bar

80oC pv

n= consta

.

.2

1

v

p

ENGINEERING MODEL: 1. The refrigerant is a

closed system. 2. The process is polytropic:pvn=

constant.


29/35

PROBLEM 3.33 (CONTINUED)

W=

= -69.88 kJ (in)


30/35

PROBLEM 3.47

4. The volume is constant.


31/35

PROBLEM 3.49


32/35

PROBLEM 3.71


33/35

PROBLEM 3.92

Determine the volume, in m3, occupied by 2 kg of H2O at 100 bar, 400

oC, using (a) data from the

compressibility chart, (b) data from the steam tables.

Compare the results of parts (a) and (b) and discuss.

(a) Using the compressibility chart, first we need to determine the reduced pressure andtemperature. From Table A-1:

pc= 220.9 bar and Tc= 647.3 K

pR=p/pc= (100)/(220.9) = 0.45 (Figure A-1): z 0.86

TR= T/Tc= (400 + 273.15)/(647.3) = 1.04

Now, we can calculate the specific volume as follows.

v=

= (0.86)

= 0.0267 m3

/kg

So, the volume is: V= mv= (2)(0.0267) = 0.0534 m3

(b) From Table A-4 at 100 bar, 400oC; v= 0.02641 m

3/kg

Thus, V= mv= (2)(0.02641) = 0.05282 m3

Comments: The compressibility chart gives a fairly accurate value considering the relative

imprecision of reading values from the chart. The percent difference is approximately 1.1%.

Note also that the value ofzis 0.86. Hence, the ideal gas model is not particularly applicable at

this state. The ideal gas model would predict a volume of 0.03105 m3, which is about 15% low.

H2O

m= 2 kgp= 100 barT= 400

oC

V= ??


34/35

PROBLEM 3.105


35/35

PROBLEM 3.141

Documents

Thermo HW Solutions