chapter 3 solutions · 2018-01-10 · chapter 3 solutions 3.1 (c) ... 0.05 mg p a π × == = × 2...

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Chapter 3 Solutions 3.1 (C) Since the net change in height is zero. 3.2 (D) A spring being extended is such a work mode so (A) is not correct. And, (B) and (C)

are not correct. 3.3 (B) A stirrer cannot return to its original state by itself as can a spring or a compressed

gas. 3.4 (B) This is a unit of power, which is a work rate.

3.5 (C) 3

3 3

0.2 m2 0.2 m2 1

1 0.04 m 0.04 m

1 12 2 2 40 kJ0.2 0.04

W PdV V dV V− − ⎛ ⎞⎡ ⎤= = = − = − + =⎜ ⎟⎣ ⎦ ⎝ ⎠∫ ∫

3.6 (C) Pressure is constant at 2000 kPa. State 1 is a saturated liquid at 2000 kPa. From

Table C-2 we get v1 = vf = 0.001177 m3/kg. At state 2 the water is superheated at 400°C. From Table C-3 we get v2 = 0.1512 m3/kg. So,

31 2 2 kg 2000 kPa (0.1512 0.0012) m 600 kJW mP v− = Δ = × × − =

3.7 (B) 6rev rad40 N m 400 2 10 min 1.005 10 N m 1005 kJmin rev

W T tω π⎛ ⎞= Δ = ⋅ × × × = × ⋅ =⎜ ⎟⎝ ⎠

3.8 (A) Work is power times time: 12 V 10 A 600 s 72 000 J 72 kJW Vi t= Δ = × × = = 3.9 (B) The initial absolute pressure in the cylinder is

280 9.81 99 920 Pa gage or 200 kPa absolute

0.05mg

PA π

×= = =

×

2

2 2kg m/s NUnits: Pa

m m

⎛ ⎞⋅= =⎜ ⎟⎜ ⎟

⎝ ⎠

The work (on the piston) is the sum of the work to raise the piston PΔV plus the work to compress the spring Kx2/2:

2 2 3 2 22

1 kN 1 kN200 ( 0.05 0.2) m 8 0.2 m2 2 mm

0.314 0.160 0.474 kJ

W P V Kx π= Δ + = × × × + × ×

= + =

3.10 (A) Convection causes the heat to rise thereby drawing in cool air from the periphery.

Conduction is not involved. 3.11 (C) The R-factors can be summed for an over-all R factor. It is

2 ototal (1 0.4 1) 2.4 m C / WiR R= = + + = ⋅∑

abc

Typewritten text

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o2

2 o30 C(0.8 2) m 20 J/s

2.4 m C / WTQ AR

Δ= = × × =

⋅

3.12 (A) At state 1 the water is a subcooled liquid at 20°C. From Table C-1

v1 = vf = 0.001002 m3/kg

At state 2 the water is a saturated vapor at 100 kPa. From Table C-2

v2 = vg = 1.694 m3/kg ΔV = 2 kg (1.694 – 0.001002) = 3.386 m3

W = P·ΔV = 100 kPa·3.386 m3 = 338.6 kJ 3.13 (D) Assume that the room temperature is 295 K and that the room air pressure is 100 kPa. The

ideal gas law gives 100 (4 10 20) 945 kg

0.287 295PVmRT

× × ×= = =

×

Heat Generation = 50 × 0.1 kW + 100 × 0.117 kW = 16.7 kW

Qtotal = 16.7 kW × 900 s = 15 030 kJ

Use Table B-2 to find Cv for air:

( ) o15 030 kJ 944.5 kg 0.717 kJ/kg C . 22.2 CvQ mC T T T= = Δ = × ⋅° Δ ∴Δ = 3.14 (A) First, calculate the mass of air:

200 8 16.74 kg 0.287 333

PVmRT

×= = =

×

16.74 0.717 (400 – 60) 4080 kJvQ mC T= Δ = × × = 3.15 (C) For this isothermal (constant temperature) process at 200°C:

1

2

kJ 600ln 4 kg 0.287 473 K ln 1030 kJkg K 4000

PQ W mRTP

⎛ ⎞ ⎛ ⎞= = = × × × = −⎜ ⎟ ⎜ ⎟⋅ ⎝ ⎠⎝ ⎠ 3.16 (B) For this adiabatic process

12 1

1 2

kT vT v

−⎛ ⎞

= ⎜ ⎟⎝ ⎠

so ( )1.4 12 293 10 736 K 463 CoT −= × = =

3.17 o 200 kJ (cos30 ) 2 m. 115.5 kNW Fd F F= = = × ∴ = 3.18 1 00 lbf 3 ft 300 ft-lbfW Fd= = × =


3.19 The pressure is 2 24000 lbf 141.5 psi

3 inWPA π

= = =×

4000 lbf 7ft 28,000 ft-lbfW Fd= = × =

3.20 All pressures are assumed to be absolute unless otherwise stated. So, the gage pressure holding up the piston is 100 kPa (an atmospheric pressure of 100 kPa would act on the top of the piston). A force balance (the weight is mg) on the piston gives

2 22

kN 78.5100 ( 0.5 ) m 78.5 kN. 8002 kg9.81m

mg PA mπ= = × × = ∴ = =

The work to move the piston 5 m is then

(8002 9.81) N 5 m 392 500 JW Fd= = × × = 3.21 The force is 15 lbf/in 180 lbf/ftF x x= = . The work is

5050 50 2

0 0 0

180 180 225,000 ft lbf2xW F dx xdx

⎡ ⎤= ⋅ = = ⋅ = ⋅⎢ ⎥

⎣ ⎦∫ ∫

3.22

8 8 8/5 /50

0 0

10 10 5 197.7 Jx xW Fdx e dx e⎡ ⎤= = = =⎣ ⎦∫ ∫

3.23 ( )1.21.2 31 1 140 kPa 0.8 m 107.1.PV = = The work is found by integration, using 1.2 ,PV C=

0.0040.004 0.004 0.0041 0.2

0.80.80.8 0.8

1 107.1 1056 kJ1 1 1.2

n n nW PV dV CV dV C V Vn

− − + −⎡ ⎤ ⎡ ⎤= = = = = −⎢ ⎥ ⎣ ⎦− −⎣ ⎦∫ ∫

3.24 a) 3 1.4 1.411 1 1

1

10 48.3 510 85.53 ft . 20 144 85.53 1,460,00020 144

mRTV PVP

× ×= = = = × × =

×

1.432 2

22 2

1,460,000 67.7 ft10 48.3 560

P V VP V

⎫= ⎪ ∴ =⎬= × × ⎪⎭

67.767.7 67.71.4 1

85.585.5 85.567.70.485.5

11 1.4

1,460,000 60,300 ft-lbf1 1.4

n nW PV dV CV dV C V

V

− − +

−

⎡ ⎤= = = ⎢ ⎥−⎣ ⎦

⎡ ⎤= = −⎣ ⎦−

∫ ∫

b) 3 1.8 1.811 1 1

1

10 48.3 510 85.53 ft . 20 144 85.53 8,654,00020 144

mRTV PVP

× ×= = = = × × =

×

1.832 2

22 2

8,654,000 76.1 ft10 48.3 560

P V VP V

⎫= ⎪ ∴ =⎬= × × ⎪⎭


76.176.1 76.11.8 1

85.585.5 85.576.10.885.5

11 1.8

8,654,000 30,100 ft-lbf1 1.8

n nW PV dV CV dV C V

V

− − +

−

⎡ ⎤= = = ⎢ ⎥−⎣ ⎦

⎡ ⎤= = −⎣ ⎦−

∫ ∫

c) 3 2 211 1 1

1

10 48.3 510 85.53 ft . 20 144 85.53 21,070,00020 144

mRTV PVP

× ×= = = = × × =

×

232 2

22 2

21,070,000 77.9 ft10 48.3 560

P V VP V

⎫= ⎪ ∴ =⎬= × × ⎪⎭

77.977.9 77.92 1

85.585.5 85.577.9185.5

11

21,070,000 24,100 ft-lbf

nW PV dV CV dV C V

V

− −

−

⎡ ⎤= = = ⎢ ⎥−⎣ ⎦

⎡ ⎤= − = −⎣ ⎦

∫ ∫

3.25 3 31 2

1000 1000.5787 ft , 0.05787 ft ,1728 1728

V V= = = =

1 1

1

(40 144) 0.5787 0.0948 lbm53.3 660

PVmRT

× ×= = =

×

For this process, 1 1 40 144 1000 /1728 3333PV = × × = , giving the work as follows:

[ ]0.05787 0.05787

0.0578710.5787

0.5787 0.5787

0.05787ln 3333ln 7670 ft-lbf0.5787

W PVdV CV dV C V−= = = = = −∫ ∫

3.26 Since Asurface = 4πr2, 24 .dV r drπ∴ = The work is

( )2

1

4 42 2 4 3 2

1.2 1.2

5 5 4 4 3 3

120 2 1.2 4 8 2.4 58.56

1 2.4 58.56 8 (4 1.2 ) (4 1.2 ) (4 1.2 ) 31900 N m or 31.9 kJ5 4 3

V

V

W PdV r r dr r r r drπ π

π

⎡ ⎤ ⎡ ⎤= = − − = − + −⎣ ⎦⎣ ⎦

⎡ ⎤= − − + − − − = − ⋅ −⎢ ⎥⎣ ⎦

∫ ∫ ∫

3.27 2

1

0.20.22 1

11

1 15 5 5 20 kJ0.2 1

V

V

W PdV V dV V− − ⎛ ⎞⎡ ⎤= = = − = − − = −⎜ ⎟⎣ ⎦ ⎝ ⎠∫ ∫

3.28 This problem can be solved graphically using ,i av i iW P V= Δ :

Region I: 10 to 20 m3 the average pressure is 100 kPa: 1-2 100 10 1000 kJavW P V= Δ = × = Region II: constant pressure of 150 kPa from 20 m3 to 30 m3:

2-3 150 10 1500 kJavW P V= Δ = × = Region III: 30 m3 to 35 m3 the average pressure is 125 kPa: 3-4 125 5 625 kJavW P V= Δ = × =


The work for the entire region is the sum of these three processes:

W = 1000 + 1500 + 625 = 3125 kJ 3.29 We assume linear processes between readings and solve graphically with increment ,av i iW P V= Δ :

(112 0.1 130 0.1 138 0.1 140 0.1 138 0.1 132 0.1 123 0.1) 144 13.150 ft-lbfW = × + × + × + × + × + × + × × =

3.30 [ ]2

1

0.10.1 22

0 0

N25 25 25 0.005 m 0.125 J2 m

x

x

xW Kxdx xdx⎡ ⎤

= − = − = − = − ⋅ = −⎢ ⎥⎣ ⎦

∫ ∫

3.31

[ ]2

1

99 22

4 4

lbf25 25 25 32.5 -in 812.5 in-lbf 67.7 ft-lbf2 in

x

x

xW Kxdx xdx⎡ ⎤

= − = − = − × = − = − = −⎢ ⎥⎣ ⎦

∫ ∫

3.32 2

1

0.150.15 32 2 3

20 0

N100 100 100 0.001125 m 0.1125 J3 m

x

x

xW Kx dx x dx⎡ ⎤

⎡ ⎤= − = − = − = − = −⎢ ⎥ ⎣ ⎦⎢ ⎥⎣ ⎦∫ ∫

3.33 2

1

2 2

0 0

N m10 10 10 49.3 J2 rad 2

W K d dπθ π

θ

θ πθ θ θ θ⎡ ⎤ ⎡ ⎤⋅

= − = − = − = − = −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

∫ ∫

Remember that radians are a nondimensional measure of angle. 3.34 Since the cylinder is horizontal, no work is done to raise the piston. The given pressure is

assumed to be absolute so the gage pressure in the cylinder is 200 kPa. It increases to 800 kPa gage. The force on the spring is

2 2 21 1 22

kN200 ( 0.1 ) m 6.283 kN, 800 ( 0.1 ) 25.13 kNm

F PA Fπ π= = × × = = × × =

The displacement of the spring is

11 2

6.283 kN 25.130.0157 m, 0.0628 m400 kN/m 400

Fx xK

= = = = =

All the work goes to stretch the spring so we find.

2 2 2 2 22 1

1 1 kN( ) 400 (0.0628 0.0157 ) m 0.739 kJ or 739 J2 2 m

W K x x= − = × × − =

We could have expressed the pressure force acting on the face of the piston as a function of x and then integrated but that work all goes to increase the energy contained in the spring.


3.35 For a rotating shaft, 200 2 rad 4 N m 83.78 J/s

60 sW T πω ×

= = ⋅ =

The work done in 1200 s is J83.8 1200 s 100 500 J or 100.5 kJs

W = × =

3.36 In 20 seconds the mass drops a distance of h = 0.04 m/s × 20 s = 0.8 m. The change in the

potential energy of the mass is

400 N 0.8 m 320 JW F d= × = × =

The work rate is 320 N·m/20 s = 16 W. If this power is delivered to the shaft, we have

8 60 2 16. 8 rad/s or 76.4 rpm2

Tω ω ωπ

×= = ∴ = =

3.37 Equation (3.13) also shows that V = iR. So, the power provided by an electrical

resistance can also be written, by using i = V/R,

2 2 212 VPower 14.4 W10

VVIR

= = = =Ω

3.38 22

W (80 20) C W45 135 000 135 kW/mm C 0.02 m m

Q TkA x

Δ − °⎛ ⎞= = = =⎜ ⎟Δ ⋅°⎝ ⎠

3.39 [ ]2

2

(1.2 2.4) m 22 ( 20) C60.5 W

2 m C/WA TQR

× − − °Δ= = =

⋅°

6J hr s J60.5 24 3600 5.23 10 5230 kJ/days day hr day

Q⎛ ⎞⎛ ⎞ ⎛ ⎞= = × =⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠

For 20 windows: 6total

J5230 20 105 10 105 000 kJ/dayday

Q = × = × =

3.40 Heat transfer occurs due to convection:

( ) ( )2 22

Btu30 0.333 ft 170 80 F 941 Btu/ss-ft - F

Q hA T π⎛ ⎞= Δ = × × − ° =⎜ ⎟°⎝ ⎠

3.41 ( )2 210 000 W 1 m 60 20 C. 250 W/m CQ h h= = × × − ° ∴ = ⋅°


3.42 If the human body is at 98.6oF (558.6oR) (be careful to use English units),

( ) ( )4 4 8 4 4 4 2body 2 4

Btu0.1714 10 558.6 510 R 50.9 Btu/hr-fthr-ft - R

Q T TA

σ −∞

⎛ ⎞= − = × − ° =⎜ ⎟°⎝ ⎠

If the body temperature rises to 102oF (562oR), the heat rate is

( ) ( )4 4 8 4 4 4 2

body 2 4Btu0.1714 10 562 510 R 55.0 Btu/hr-ft

hr-ft - RQ T TA

σ −∞

⎛ ⎞= − = × − ° =⎜ ⎟°⎝ ⎠

55.0 50.9%increase 100 8.06%50.9

−= × =

3.43 2 2 24 4 0.5 3.14 mA rπ π= = × =

If we assume outer space to be at 0oK, there results

( )( )44 8 2sat 2 4

W5.67 10 3.14 m 223 K 440 Wm K

Q ATσ −⎛ ⎞= = × =⎜ ⎟⋅⎝ ⎠

3.46 Since the container is rigid, the volume is fixed. The first law gives

2 2 2. 200 4( 83.9). 134 kJ/kg and 32Q m u u u T C= Δ = − ∴ = = °

Table C-1 will give us the specific internal energy as uf for the compressed liquid. An interpolation gives the desired temperature, or the IRC Calculator can be used (as it was here). Table C-1 was used to estimate T2 and the Calculator was used by selecting the temperature with x2 = 0 until u2 = 134 kJ/kg.

No work is done since the container is rigid.

3.47 In order for the temperature to remain constant there must be no net heat flow to or from the

engine. Thus if 20 kW is removed from the engine, the engine must be developing 20 kW of heat.

3.48 A system boundary is placed around the system, i.e. the turbine/condenser; the 1st law

demands that net in out 100 MW 75 MW 25 MWW Q Q= − = − =

3.49 The first law supplies the answer:

(8000 4000) 5000 1000 BtuU Q WΔ = − = − − = −∑ ∑

3.50 Information is given for three of the processes so, using Q W U− = Δ we can calculate

Process 1→2: 1-2 1-2Q W U= + Δ 200 kJ= − Process 3→4: 3-4 3-4 600 200 400 kJW Q U= − Δ = − = Process 4→1: 4-1 4-1Q W= 600 kJU+ Δ = −


Cycle: ( ) 0iUΔ =∑ for the cycle: So, 3 2 400 kJU U− = Process 2→3: 2-3 2-3 600 400 1000 kJQ W U= + Δ = + = Observe that Q W=∑ ∑ = 800 kJ for the cycle.

3.51 Information is given for three of the processes so, using Q W U− = Δ we calculate the following:

Process 1→2: 1-2 1-2Q W U= + Δ 200 Btu= − Process 3→4: 3-4 3-4 600 200 400 BtuW Q U= − Δ = − = Process 4→1: 4-1 4-1Q W= 600 BtuU+ Δ = − Cycle: ( ) 0iUΔ =∑ for the cycle: So, 3 2 400 BtuU U− = Process 2→3: 2-3 2-3 600 400 1000 BtuQ W U= + Δ = + = Observe that Q W=∑ ∑ = 800 Btu for the cycle.

3.52 Since the container is insulated, there is no heat transfer with the surroundings, so the energy

heat transfer with the surroundings, so the energy in the container remains constant. The mass of one liter of water is 1 kg (1000 L = 1 m3). Ten ice cubes have a total volume of 100 mL (0.0001 m3); their specific heat is 2.1 kJ/kg·ºC (Table B-3). The mass of the 10 ice cubes is found by using the specific volume at −20oC in Table C-5:

0.0001 0.0920 kg0.001087

ii

i

Vmv

= = =

a) The 1st law states that the heat lost by the water is gained by the ice. Assume that all the ice melts (hi = 320 kJ/kg is the heat fusion):

water cools ice warms up ice melts ice water warms up

, 2 , , 2(20 ) (20) ( 0)w p w i p i i i i p wm C T mC m h mC T− = + + − ( )2 2 21 4.18 (20 ) 0.092 2.1 20 320 4.18 . 11.0 CT T T× × − = × × + + × ∴ = °

b) The 1st law states that the heat lost by the water is gained by the ice. Assume that

some of the ice (25 cubes have a mass of 0.23 kg) melts (hi = 320 kJ/kg is the heat of fusion) so the final temperature is at 0°C:

water cools ice warms up ice melts

, , ,melts(20 0) (20)w p w i p i i im C mC m h− = + ,melts ,melts1 4.18 (20) 0.23 2.1 20 320. 0.23 kgi im m× × = × × + × ∴ =

All of the ice just melted and the temperature is at 0ºC.

c) The 1st law states that the heat lost by the water is gained by the ice. Assume that some of the ice (100 cubes have a mass of 0.92 kg) melts (hi = 320 kJ/kg is the heat of fusion) so the final temperature is at 0ºC:


, , ,melts(20 0) (20)w p w i p i i im C mC m h− = + ,melts ,melts1 4.18 (20) 0.92 2.1 20 320. 0.14 kgi im m× × = × × + × ∴ =

Only 0.14 kg of the 0.92 kg of ice melted, so the final temperature is 0ºC.


3.53 Since the container is insulated, there is no heat transfer with the surroundings, so the energy

in the container remains constant. The mass of one quart of water is 2.09 lbm (water has a mass of 62.4 lbm/ft3). Ten ice cubes have a total volume of 4 in3 = 0.00232 ft3; their specific heat is approximated to be 0.49 Btu/lbm-°F from Table B-4E. Using the specific volume at 20°F in Table C-5E, the mass of 10 ice cubes is

0.00232 0.133 lbm0.01745

ii

i

Vmv

= = =

a) The 1st law states that the heat lost by the water is gained by the ice. Assume that all the ice melts (hi = 140 Btu/lbm is the heat of fusion):


, 2 , , 2(70 ) (32 20) ( 32)w p w i p i i i i p wm C T mC m h mC T− = − + + −

[ ]2 2 22.09 1.0 (70 ) 0.133 0.49 12 140 1.0 ( 32) . 59 FT T T× × − = × × + + × − ∴ = °

b) The 1st law states that the heat lost by the water is gained by the ice. Assume that all the ice (40 cubes have a mass of 0.532 lbm) melts (hi = 140 Btu/lbm is the heat of fusion):


, 2 , , 2(70 ) (32 20) ( 32)w p w i p i i i i p wm C T mC m h mC T− = − + + −

[ ]2 2 22.09 1.0 (70 ) 0.532 0.49 12 140 1.0 ( 32) . 36 FT T T× × − = × × + + × − ∴ = °

c) The 1st law states that the heat lost by the water is gained by the ice. Assume that some of the ice (200 cubes have a mass of 2.66 lbm) melts (hi = 140 Btu/lbm is the heat of fusion) so the final temperature is at 32°F:


, , ,melts(70 32) (32 20)w p w i p i i im C mC m h− = − +

,melts ,melts2.09 1.0 (38) 2.66 0.49 12 140. 0.46 lbmi im m× × = × × + × ∴ =

Only 0.46 lbm of the 2.66 lbm of ice melted, so the final temperature is 32°F.

3.54 We define the system as both blocks. There is no work done and no heat transfer with the surroundings. The total energy contained in the blocks remains constant but heat is transferred between the blocks. The specific heats are found in Table B-4E and estimated to be Cal = 0.215 Btu/lbm-ºF and Ccu = 0.094 Btu/lbm-ºF. The heat gained by the aluminum is lost by the copper. This is expressed as

al 2 1 cu cu 1 2kJ( ) ( ) Units: kg F kJ

kg Falm C T T m C T T− = − × ×° =⋅°

a) al al 2 cu cu 2 2 2 2( 100) (200 ). 20 0.215( 100) 40 0.094(200 ). 147 Fm C T m C T T T T− = − × − = × − ∴ = °

b) al al 2 cu cu 2 2 2 2( 100) (200 ). 40 0.215( 100) 40 0.094(200 ). 130 Fm C T m C T T T T− = − × − = × − ∴ = °

c) al al 2 cu cu 2 2 2 2( 100) (200 ). 60 0.215( 100) 40 0.094(200 ). 123 Fm C T m C T T T T− = − × − = × − ∴ = °


3.55 The first law can be expressed as .pQ mC T= Δ Use Table B-4 to find Cp = 4.18 kJ/kg·°C (remember, for a liquid or a solid, )p vC C C≅ = . With W = 0, the 1st law gives:

a) 2 2. 3000 100 4.18 ( 15). 22.2 CvQ mC T T T= Δ = × × − ∴ = ° b) 2 2. 5000 100 4.18 ( 15). 27.0 CvQ mC T T T= Δ = × × − ∴ = ° c) 2 2. 8000 100 4.18 ( 15). 34.1 CvQ mC T T T= Δ = × × − ∴ = °

3.56 The tank is rigid so no work is done. The initial state is a superheated vapor. Use Table D-3 to

get the specific internal energy and the specific volume:

31 1 0.11394 m /kg and 244.99 kJ/kgv u= = .

Use the IRC Calculator (or extrapolate from Table D-3) and at

P = 300 kPa and 31/ 1 / 0.11394 8.78 kg/mvρ = = = we find T2 = 153°C. 3.57 Use the ideal gas law to get the initial pressure:

11

50 0.297 293 435 kPa10

mRTPV

× ×= = =

The 1st law: vQ W U mC T− = Δ = Δ

( ) 2 21500 2500 50 0.745 ( 293). = 320 K or 47 CT T− − − = × × − ∴ °

Then 12

50 0.297 320 475 kPa10

mRTPV

× ×= = =

3.58 If we consider both parts as the system, there is no change of mass, no work, and no heat

transfer to the surroundings. The mass and energy contained in the system remains constant. Use the ideal gas law to find the volume for section 1 and section 2:

3 31 1 2 21 2

1 2

10 53.3 660 15 53.3 61030.5 ft , 30.8 ft80 144 110 144

m RT m RTV VP P

× × × ×= = = = = =

× ×

The total volume is 61.3 ft3. The internal energy remains unchanged. This is expressed as

1 vfinal

m CT = 1 2 vT m C+ 2

1 2( ) v

Tm m C+

10 660 15 610 630 R or 170 F10 15

× + ×= = ° °

+

25 53.3 630Then 13,700 psfa 95.1 psia61.3

finalfinal

mRTP

V× ×

= = = =

3.59 If we consider both containers as part of one system there is no change of mass, no work and no

heat transfer with the surroundings. The mass and energy contained in the system remains constant. We use the ideal gas law to find the initial specific volume for each section:


3 31 1 2 21 2

1 2

10 0.287 373 15 0.287 3532.14 m , 1.90 m500 800

m RT m RTV VP P

× × × ×= = = = = =

The total volume is 4.02 m3. The internal energy remains unchanged. This is expressed as

1 vfinal

m CT = 1 2 vT m C+ 2

1 2( ) v

Tm m C+

10 373 15 353 361 K or 88 C10 15

25 0.287 361Then 641 kPa2.14 1.90

finalfinal

total

mRTP

V

× + ×= = °

+

× ×= = =

+

3.60 a) Using the steam tables: The process is a constant-pressure process until the stops are

hit. At the initial state (the weight of the piston is mg; don’t forget the atmospheric pressure):

2 31 1 1 atm 2

64 9.810.1, 0.04 ( 0.1 ) 0.00126 m , 100 000 120 000 Pa0.1

mgx V P PA

ππ

×= = × × = = + = + =

×

i) The water is a saturated mixture at 120 kPa (it has to be absolute, not gage, hence the atmospheric pressure which acts on the top of the piston is added).

ii) The specific volume is

( ) ( ) 31 1

1

1

0.00105 0.1 1.428 0.00105 0.1437 m /kg

0.00126so 0.00877 kg0.1437

f g fv v x v v

Vmv

= + − = + − =

= = =

iii) When the piston hits the stops, the mass and the pressure are the same. The question is: Is the steam still in the quality region? The specific volume is found:

232

20.09 0.1 0.322 m /kg

0.00877Vvm

π× ×= = =

The steam is in the quality region since 0.322 < 1.428 m3/kg. Hence, from Table C-2, 104.8°C.

iv) The steam is in the superheat region with T3 = 500°C and 33 0.322 m /kgv = so Table

C-3 is searched to find 1.0 MPa < P3 < 1.2 MPa. An interpolation yields

30.3541 0.322 0.2 1.0 1.10 MPa or 1100 kPa

0.3541 0.2946P −

= × + =−

v) There is no work after the piston hits the stops so the work results from the raising

of the piston: 2 3

2kN120 ( 0.1 0.05) m 0.188 kJ or 188 Jm

W P V π= × Δ = × × × =

Remember: The pressure has to be absolute. It takes work to not only move the weight of the piston but to also move the atmospheric pressure force acting on the top of the piston.


b) Using the IRC Calculator: The process is a constant-pressure process until the piston hits the stops. At the initial state (the weight of the piston is mg):

2 31 1 1 atm 2

64 9.810.1, 0.04 ( 0.1 ) 0.00126 m , 100 000 120 000 Pa0.1

mgx V P PA

ππ

×= = × × = = + = + =

×

i) P1 = 120 kPa (This is absolute pressure; include the atmospheric pressure on the top area.)

ii) With P1 = 120 kPa and x1 = 0.1, the Calculator gives v1 = 0.144 m3/kg, so

0.00126 0.00875 kg0.144

Vmv

= = =

iii) When the piston just hits the stops, the mass and the pressure are the same. The question is: Is the steam still in the quality region? The specific volume is found:

232

20.09 0.1 0.322 m /kg

0.00877Vvm

π× ×= = =

The steam is in the quality region since 0.322 < 1.428 m3/kg. The Calculator, with 120 kPa and ρ = 3.1 kg/m3 gives T2 = 105°C.

iv) The steam is in the superheat region with

T3 = 500°C and 33 0.322 m /kgv = ( ρ = 3.1 kg/m3)

The Calculator gives the pressure as 1100 kPa.

v) There is no work after the piston hits the stops so the work results from the raising of the piston:

2 32

kN120 ( 0.1 0.05) m 0.188 kJ or 188 Jm

W P V π= × Δ = × × × =

Remember: The pressure has to be absolute. It takes work to not only move the weight of the piston but to also move the atmospheric pressure force acting on the top of the piston.

3.61 The IRC Calculator will be used when convenient. The initial quantities are

2 31 1 1 atm 2

64 9.810.4, 0.04 ( 0.1 ) 0.00126 m , 100 000 120 000 Pa0.1

mgx V P PA

ππ

×= = × × = = + = + =

×

At state 1 the quality is 40%. The Calculator gives v1 = 0.151 m3/kg so m = V1/v1 = 0.00834 kg. If the spring is compressed a distance x, the pressure and specific volume are expressed as

12 1 2

and Kx V AxP P v

A m+

= + = (ρ = 1/v when using the Calculator)


a) For T2 = 300ºC, we assume that the steam is superheated. We must find a value of x that will allow us to find a state 2 where the temperature, pressure, and specific volume match. This trial-and-error method provides

x = 0.044 m or 4.4 cm

b) For T2 = 400ºC, we assume that the steam is superheated. We must find a value of x that will allow us to find a state 2 where the temperature, pressure, and specific volume match. This trial-and-error method provides

x = 0.049 m or 4.9 cm

c) For T2 = 500ºC we assume that the steam is superheated. We must find a value of x that will allow us to find a state 2 where the temperature, pressure, and specific volume match. This trial-and-error method provides

x = 0.055 m or 5.5 cm

3.62 The mass of the water is 1000 0.04 40 kg.m Vρ= = × = The phase change process will take place at a constant pressure of 100 kPa. The water is first heated from state 1 to state 2 at 100°C and then vaporized at state 3:

[ ]2 1 3 2 2 1( ) ( ) ( ) ( )

40 4.18 (100 20) (2506 419) 96 900 kJg fQ m u u m u u mC T T m u u= − + − = − + −

= × − + − =

Units: kJ kJkg °C kJ and kg kJkg C kg

× × = × =⋅°

. These should be quite familiar by now,

right? 3.63 The process is a constant-pressure process at 200 kPa.

State 1: P1 = 200 kPa and T1 = 300°C. The steam is superheated. From Table C-3 we find: h1 = 3071.8 kJ/kg , v1 = 1.316 m3/kg

a) State 2: P2 = P1 = 200 kPa and T2 = 180ºC. From the IRC Calculator, h2 = 2830 kJ/kg: ( ) 50 2830 – 3071.8 12100 kJ

200 50(1.03 1.316) 2860 kJ

Q m h

W Pm v

= Δ = = −

= Δ = × − = −

b) State 2: If P2 = P1 = 200 kPa and T2 = 125ºC the water is superheated. The Calculator gives:

h2 = 2720 kJ/kg 50(2720 – 3071.8) 17 600 kJ

200 50(0.898 1.316) 4180 kJQ m hW Pm v

= Δ = = −= Δ = × − = −

c) State 2: If P2 = P1 = 200 kPa and T2 = 80ºC, the water is a subcooled liquid:

50(335 – 3071.8) 137 000 kJ200 50(0.00103 1.316) 13150 kJ

Q m hW Pm v

= Δ = = −= Δ = × − = −


3.64 State 1: P1 = 120 kPa, m = 0.6 kg, T1 = 25°C or 298 K. Using the ideal gas law:

11

1

0.6 0.287 298 0.428 kg120

mRTVP

× ×= = =

a) V2 = 2V1 = 0.856 m3, 2 22

120 0.856 597 K0.6 0.287

P VTmR

×= = =

×

120 (0.856 0.428) 51.4 kJ0.6 1.0 (597 298) 179 kJp

W P VQ mC T

= Δ = × − == Δ = × × − =

b) V2 = 4V1 = 1.712 m3, 2 22

120 1.712 1193 K0.6 0.287

P vTmR

×= = =

×

120 (1.712 0.428) 154 kJ0.6 1.0 (1193 298) 537 kJp

W P VQ mC T

= Δ = × − == Δ = × × − =

3.65 State 1: x1 = 0.5, P1 = 1 MPa, m = 2 kg, T1 = Tsat = 179.9°C

i) Use Table C-2 to get the water properties:

3 31 1 10.001127 0.5(0.1944 0.001127) 0.0977 kg/m . 2 0.1954 m v V v= + − = ∴ = =

1 762.8 0.5(2778.1 762.8) 1770 kJ/kgh = + − =

State 2: P2 = P1 = 1 MPa, T2 = 600ºC. The water is superheated so use Table C-3: h2 = 3697.9 kJ/kg and v2 = 0.4011 m3/kg so V2 = 0.8022 m3

∴2 (3698 1770) 3860 kJ1000 (0.8022 0.1954) 607 kJ

Q m hW P V

= Δ = × − == Δ = × − =

ii) Use the IRC Calculator: With P1 = 1000 kPa and x1 = 0.5,

v1 = 0.0977 m3/kg and h1 = 1770 kJ/kg. 31 12 0.1954 m V v∴ = =

State 2: With P2 = 1000 kPa and T2 = 600ºC, v2 = 0.401 m3/kg and h2 = 3700 kJ/kg. 3

2 22 0.802 m V v∴ = =

∴2 (3700 1770) 3860 kJ1000 (0.802 0.1954) 607 kJ

Q m hW P V

= Δ = × − == Δ = × − =

The P-V diagram would be a rectangle. 3.66 State 1: P1 = 4 MPa and T1 = 600ºC. The water is superheated and the IRC Calculator will

be used h1 = 3670 kJ/kg, v1 = 0.0989 m3/kg so V1 = 3v1 = 0.2967 m3

a) For V2 = 0.08 m3, 32

2

3 37.5 kg/m0.08

mV

ρ = = = , P2 = 4 MPa. The Calculator gives

h2 = 1980 kJ/kg. Then

3 (1980 3670) 5070 kJQ m h= Δ = × − = −

( )3 34000 kPa 0.08 m 0.2967 m 867 kJW P V= Δ = − = −


b) For V2 = 0.04 m3, 32

2

3 75 kg/m ,0.04

mV

ρ = = = P2 = 4 MPa. The Calculator gives

h2 = 1510 kJ/kg. Then 3 (1510 3670) 6480 kJQ m h= Δ = × − = −

( )3 34000 kPa 0.04 m 0.2967 m 1030 kJW P V= Δ = − = −

c) For V2 = 0.02 m3, 32

2

3 150 kg/m0.02

mV

ρ = = = , P2 = 4 MPa. The Calculator gives

h2 = 1280 kJ/kg. Then 3 (1280 3670) 7170 kJQ m h= Δ = × − = −

( )3 34000 kPa 0.02 m 0.2967 m 1110 kJW P V= Δ = − = −

3.67 When the piston just leaves the stops at state 2 (state 1 is the condition shown), the

absolute pressure in the cylinder is

2 atm 280 9.81 100 000 125 000 Pa or 125 kPa

0.1mgP PA π

×= + = + =

×

The temperature will be

22 2

2125 0.1 0.4 274 K

0.02 0.287P VTmR

π× × ×= = =

× Units:

3 2 3kPa m (kN/m ) m Kkg kJ/kg K kJ/K

× ×= =

× ⋅

State 3: P3 = P2 = 125 kPa and 2 33 0.1 0.8 0.0251 m .V π= × × = The temperature is

23 3

3125 0.1 0.8 547 K

0.02 0.287P VTmR

π× × ×= = =

×

The heat transfer and work done are

[ ]1-3 2 1 3 2 2 1 3 2( ) ( ) ( ) ( )

0.02 0.717 (274 233) 1.0 (547 274) 6.05 kJv pQ m u u m h h mC T T mC T T= − + − = − + −

= × × − + × − =

1-3 3 2( ) 125 (0.0251 0.0251/ 2) 1.57 kJW P V V= − = × − = 3.68 Table B-4 gives us the specific heat of copper, Ccu = 0.39 kJ/kg·°C. and water Cw = 4.18

kJ/kg·°C. The container is insulated so no heat is exchanged with the outside. If we assume no net change in volume and thus no work done, the energy in the container is conserved:

Energy lost by copper = Energy gained by water

Δ = Δcu cu cu w w wm C T m C T

2 2 24 0.39 (200 ) 20 4.18 ( 40). 42.9°CT T T× × − = × × − ∴ =

3.69 i) At constant pressure, the heat transfer is

2 1( ) 20 0.520 (100 10) 936 kJpQ mC T T= − = × × − =


ii) At constant volume, the heat transfer is

2 1( ) 20 0.312 (100 10) 562 kJvQ mC T T= − = × × − = 3.70 Table B-2 gives the R for nitrogen as R = 0.2968 kJ/kg·K. For an isothermal process, the 1st

law states that Q = W since 2 1( ) 0 :vU C T TΔ = − =

a) 2

1

500ln 10 0.2968 513 ln 1673 kJ1500

PQ W mRTP

= = = × × × = −

( kJUnits: kg K kJkg K

× × =⋅

)

When T stands alone (not a difference), invariably it must be absolute temperature.

b) 2

1

500ln 10 0.2968 513 ln 2450 kJ2500

PQ W mRTP

= = = × × × = −

c) 2

1

500ln 10 0.2968 513 ln 3170 kJ4000

PQ W mRTP

= = = × × × = −

3.71 State 1: P1 = 400 kPa, T1 = 200ºC, V1 = 0.08 m3. From the IRC Calculator, v1 = 0.534

m3/kg: 0.08 0.150 kg

0.534Vmv

∴ = = =

State 2: T2 = T1 = 200°C and x2 = 0.5. Use the IRC Calculator to obtain the following values:

P(kPa) 400 600 800 1000 1200 1400 1554 (x = 1) 1554 (x = 0.5) v(m3/kg) 0.534 0.352 0.261 0.206 0.169 0.143 0.127 0.0642

Using a sketch (graph paper would be nice), drawn to scale, estimate the area (kJ/kg) and then multiply by the mass m. Perhaps a cubic could be made to fit the points in the superheat region and then an integration could be accomplished.

3.72 State 1: m = 0.02 kg, x1 = 1, 1 2160 9.81 100 000 149 962 Pa or 150 kPa

0.1P

π×

= + =×

.

The Calculator (Table C-2 could be used) gives u1 = 2520 kJ/kg and v1 = 1.16 m3/kg so that

31 1 0.02 1.16 0.0232 mV mv= = × =

State 2: 2 1 260 0.1150 kPa 341 kPa

0.1k xP PA πΔ ×

= + = + − =×

. The specific volume is

23

2 10.1 0.11.16 =1.317 m /kg0.02

Vv vm

πΔ × ×= + = + (ρ2 = 0.759 kg/m3)

Hence 32 0.02 1.317 0.0263 m .V = × = The IRC calculator gives T2 = 701ºC and u2 = 3480

kJ/kg.


The work and heat transfer follow: 2 2

11 1150 (0.0263 0.0232) 600 0.1 3.46 kJ2 2

W P V kx= Δ + = × − + × × =

0.02 (3480 2520) 3.46 22.7 kJQ m u W= Δ + = × − + =

3.73 The power input is ( ) 200 2 rad0.8 N m 16.8 W60 sec

W T πω ×⎛ ⎞= = ⋅ =⎜ ⎟⎝ ⎠

. The work done on the

system in 40 min is 316.8 (40 60) 40.2 10 J 40.2 kJW = × × = × = .

State 1: m = 10 kg, P1 = 200 kPa, T1 = 25ºC or 298 K

Assuming no heat transfer to the surroundings, the 1st law is: .vW U mC T− = Δ = Δ This is written, recognizing that the work is negative (find vC in Table B-2), as

( ) 2 1 2 240.2 kJ ( – ) 10 0.745 ( 298). 303 K or 30 CvmC T T T T− − = = × × − ∴ = °

3.74 In this problem there is no work done so the 1st law states: .vQ U mC T= Δ = Δ There results

2 1 2 24800 ( – ) 10 0.745 ( 25). 669 CvQ mC T T T T= = = × × − ∴ = °

Note: If T1 is in ºC, the answer will be ºC. If it is in kelvins, the result will be in kelvins. 3.75 State 1: m = 2 kg, x1 = 0.5, P1 = 140 kPa. Since the container is rigid, no work is done.

a) From Table C-2 ( )

( )

31

1

0.001051 0.5 1.237 0.001051 0.619 m /kg

458.2 0.5 2517.3 458.2 1488 kJ/kg

v

u

= + − =

= + − =

State 2: v2 = v1 = 0.619 m3/kg and P2 = 1200 kPa. This is superheat and extrapolation gives

20.619 0.6051 (4681 4465) 4681 4760 kJ/kg0.6051 0.5665

u −= − + =

− and T2 = 1336°C

The heat transfer required is found by applying the 1st law:

2 1( ) 2 (4760 1488) 6540 kJQ m u u= − = × − =

b) Use the IRC Calculator. At x1 = 0.5 and P1 = 140 kPa: 31 10.619 m /kg. 1490 kJ/kg.v u= =

At 1 1 32 2 0.619 =1.616 kg/mvρ − −= = and P2 = 1200 kPa we find u2 = 4760 kJ/kg. The 1st

law is 2 1( ) 2 (4760 1490) 6540 kJQ m u u= − = × − =

3.76 a) State 1: P1 = 400 kPa and x1 = 0.2. From Table C-2 we get

( ) 31 0.001084 0.2 0.4625 0.001084 0.0934 m /kgv = + − =

( )1 604.3 0.2 2553.6 604.3 994.2 kJ/kgu = + − =


State 2: T2 = 200ºC and 32 1 0.0934 m /kgv v= = since the container is rigid.

From Table C-1 we get

( ) 32 2 20.001156 0.1274 0.001156 0.0934 m /kg. 0.731v x x= + − = ∴ =

And then ( )2 850.6 0.731 2595.3 850.6 2126 kJ/kgu = + − =

W = 0 since the container is rigid and 2 1 2126 994 1130 kJ/kgq u u= − = − = . The mass is not available so q is found.

b) Use the IRC Calculator: P1 = 400 kPa and x1 = 0.2. So 3

1 10.0933 m /kg and 994 kJ/kg.v u= =

State 2: T2 = 200ºC and 1 1 32 2 0.0933 =10.75 kg/m . vρ − −= = The Calculator gives

2 22120 kJ/kg and 0.729u x= =

w = 0 since the container is rigid and 2 1 2120 994 1130 kJ/kgq u u= − = − = . The mass is not available so q is given.

3.77 State 1: m = 2 kg, x1 = 0.5 and T1 = 140ºC.

a) From Table C-1 we get

( ) 31 0.00108 0.5 0.5089 0.00108 0.255 m /kgv = + − =

( )1 588.7 0.5 2550 588.7 1569 kJ/kgu = + − =

State 2: T2 = 1000ºC and v2 = v1 = 0.255 m3/kg. From Table C-3, using interpolation: P2 = 2.26 MPa and u2 = 4050 kJ/kg. The 1st law gives

( )2 4047 1569 4955 kJQ = − =

b) Use the IRC Calculator: At x1 = 0.5 and T1 = 140ºC, v1 = 0.255 m3/kg and u1 = 1570 kJ/kg.

State 2: T2 = 1000ºC and 1 1 32 2 0.255 =3.92 kg/m .vρ − −= = The Calculator gives

P2 = 2.3 MPa and u2 = 4050 kJ/kg. ( )2 4050 1570 4960 kJQ∴ = × − =

3.78 State 1: m = 4 kg, x1 = 1, and T1 = 180ºC. From Table C-1 we get v1 = vg = 0.1941 m3/kg, u1 = ug = 2583.7 kJ/kg and P1 = 1.002 MPa

a) State 2: P2 = P1 = Psat = 1.002 MPa. The 1st law is Q = W + ΔU which takes the form, with known quantities,

2 2 2 24000 4 1002( 0.1941) 4( 2584) or 1002 3780v u v u= × − + − + =

This requires a trial-and-error procedure. With P2 = 1.0 MPa, guess T2 and see how close you are.


Referring to Table C-3 at 600ºC there results 3699 ?= 3780. At 700ºC there results

3924 ?= 3780. An acceptable result, since it’s closer to 600ºC than 700ºC, is T2 =

630ºC. Even with the IRC Calculator, it would be trial-and-error.

b) State 2: We know P2 = P1 = 1.0 MPa and, from Table C-2, h1 = 2778 kJ/kg. The 1st law can be written, using enthalpies, as

2 1 2 2( – ) or 4000 4 ( – 2778). 3778 kJ/kgQ m h h h h= = × ∴ =

Using the IRC Calculator, we get T2 = 636°C. Enthalpy is very handy for constant- pressure processes, as we’ve experienced with earlier problems.

3.79 State 1: T1 = 25°C = 298 K, P1 = 200 kPa, m = 0.4 kg, Cp = 1.042 kJ/kg·K. No heat transfer

occurs since the volume is insulated but let’s take the energy input by the paddle wheel as a “heat” input. This energy input is:

( ) ( )200 2 rad" " 0.8 N m 40 60 s 40 200 J or 40.2 kJ60 s

Q T t πω ×⎛ ⎞= Δ = ⋅ × =⎜ ⎟⎝ ⎠

This energy (heat) input is then accounted for as follows:

( )2 1 2 2C . 40.2 0.4 1.042 ( 25). 121 CPQ H m T T T T= Δ = − = × × − ∴ = °

3.80 State 1: T1 = 100°C or 373 K, P1 = 600 kPa.

State 2: T2 = T1 = 373 K, P2 = 1200 kPa For the isothermal process, ΔT = 0 so ΔU = 0. The heat transfer (and work) are

1

2

600ln 0.4 0.297 373 ln 30.7 kJ1200

PQ W mRTP

⎛ ⎞ ⎛ ⎞= = = × × = −⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

3.81 State 1: T1 = 220ºF or 680°R, P1 = 400 psia, V1 = 300 in3 or 0.174 ft3. The mass is

1 1

1

400 144 300 /1728 0.276 lbm53.3 680

PVmRT

× ×= = =

×

1

2

400ln 0.276 53.3 680 ln 20,800 ft-lbf or 26.7 Btu50

PQ W mRTP

⎛ ⎞ ⎛ ⎞= = = × × =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

Units: ft-lbflbm R ft-lbflbm- R

× ×° =°

778 ft-lbf = 1 Btu

3.82 This is an adiabatic quasi-equilibrium process, as occurs in the pistons of an engine.

Use Eq. 3.44: 1

0.412 1

2293 8 673 K or 400 C

kvT Tv

−⎛ ⎞

= = × = °⎜ ⎟⎝ ⎠


The work is found by applying the 1st law:

Q 2 1( ) 0.2 0.717(673 293) 54.5 kJ. 54.5 kJvW m u mC T T W− = Δ = − = × − = ∴ = − 3.83 For this adiabatic quasi-equilibrium process, use Eq. 3.46 with kPV C= and perform

the integration using 1.4200 0.004 0.0879C = × = (k is found in Table B-2).

a) ( )2

1

0.0010.001 1 1.40.4 0.4

0.0040.004

0.08790.0879 0.001 0.004 1.48 kJ1 1.4 0.4

Vk

V

VW PdV C V dV−

− − −⎡ ⎤= = = = − = −⎢ ⎥

− −⎢ ⎥⎣ ⎦∫ ∫

b) ( )2

1

0.00060.0006 1 1.40.4 0.4

0.0040.004

0.08790.0879 .0006 .004 2.27 kJ1 1.4 0.4

Vk

V

VW PdV C V dV−

− − −⎡ ⎤= = = = − = −⎢ ⎥

− −⎢ ⎥⎣ ⎦∫ ∫

c) ( )2

1

0.00040.0004 1 1.40.4 0.4

0.0040.004

0.08790.0879 .0004 .004 3.02 kJ1 1.4 0.4

Vk

V

VW PdV C V dV−

− − −⎡ ⎤= = = = − = −⎢ ⎥

− −⎢ ⎥⎣ ⎦∫ ∫

3.84 State 1: P1 = 100 psia and T1 = 50ºF, so 311

1

53.3 510 1.89 ft /lbm100 144

RTvP

×= = =

×

a) State 2: P2 = 14.7 psia and T2 = 50ºF, so 322

2

53.3 510 12.85 ft /lbm14.7 144

RTvP

×= = =

×

32 1 2 112.85 1.89 10.96 ft /lbm and 0 since 0v v u u T∴ − = − = − = Δ =

b) State 2: P2 = 14.7 psia and n = k = 1.4, so1/ 1/1.4

312 1

2

1001.89 7.43 ft /lbm14.7

kPv vP

⎛ ⎞ ⎛ ⎞= = × =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

and 2 22

(14.7 144) 7.43 295 R53.3

P vTR

× ×= = = ° . There results

32 1 2 1 2 17.43 1.89 5.54 ft /lbm and ( ) 0.171(295 510) 36.8 Btu/lbmvv v u u C T T− = − = − = − = − = −

c) State 2: P2 = 14.7 psia and n = 2.5, so1/ 1/2.5

312 1

2

100 1.89 4.07 ft /lbm14.7

nPv vP

⎛ ⎞ ⎛ ⎞= = × =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

and

2 22

(14.7 144) 4.07 162 R53.3

P vTR

× ×= = = ° . There results

32 1 2 1 2 14.07 1.89 2.18 ft /lbm and ( ) 0.171(162 510) 59.6 Btu/lbmvv v u u C T T− = − = − = − = − = −

3.85 State 1: P1 = 100 kPa T1 = 120ºC,

3 31 11 2

1

0.287 393 1.13 m /kg and 0.141 m /kg100 8

RT vv vP

×= = = = =


a) n = 1.3 so 1.312 1

2100 8 1493 kPa

nvP Pv

⎛ ⎞= = × =⎜ ⎟

⎝ ⎠ and 2 2

21493 0.141 733 K

0.287P vTR

×= = =

2 1 0.717(733 393) 244 kJ/kgu u− = − =

2 2 1 1 1493 0.141 100 1.13 325 kJ/kg1 0.3

P v Pvwn

− × − ×= = = −

− −

244 ( 325) 81 kJ/kgq u w= Δ + = + − = −

b) n = 1.4 so 1.412 1

2100 8 1838 kPa

nvP Pv

⎛ ⎞= = × =⎜ ⎟

⎝ ⎠ and 2 2

21838 0.141 903 K

0.287P vTR

×= = =

2 1 0.717(903 393) 366 kJ/kgu u− = − =

2 2 1 1 1838 0.141 100 1.13 365 kJ/kg1 0.4

P v Pvwn

− × − ×= = = −

− −

366 ( 365) 1 kJ/kgq u w= Δ + = + − = (it is essentially zero, as it should be)

c) n = 1.5 so 1.512 1

2100 8 2263 kPa

nvP Pv

⎛ ⎞= = × =⎜ ⎟

⎝ ⎠ and 2 2

22263 0.141 1112 K

0.287P vTR

×= = =

2 1 0.717(1112 393) 516 kJ/kgu u− = − =

2 2 1 1 2263 0.141 100 1.13 412 kJ/kg1 0.5

P v Pvwn

− × − ×= = = −

− −

516 ( 412) 104 kJ/kgq u w= Δ + = + − =

1 0.41

2 12

1 For this process, 1.4 so 973 424 K or 151 C8

kvn T Tv

−⎛ ⎞ ⎛ ⎞= = = × = °⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠3.86

The work done is found from the 1st law:

W m u Q= − Δ + 0.2 0.717 (151 700) 78.7 kJvmC T= − Δ = − × × − =

3.87 a) Given: n = 2.5. Then 2.5

12 1

2

10500 15.6 kPa,40

nVP PV

⎛ ⎞ ⎛ ⎞= = × =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

1500 10 842 K,

20 0.297T ×

= =×

and 215.6 40 105 K20 0.297

T ×= =

×. The work done, using Eq. 3.51, and heat transfer are

2 2 1 1 15.6 40 500 10 2920 kJ1 1 2.5

P V PVWn

− × − ×= = =

− −

20 0.717 (105 842) 2920 7650 kJvQ U W mC T W= Δ + = Δ + = × × − + = −


b) Given: n = 2.0. Then 2.0

12 1

2

10500 31.25 kPa,40

nVP PV

⎛ ⎞ ⎛ ⎞= = × =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

1500 10 842 K,

20 0.297T ×

= =×

and 231.25 40 210 K20 0.297

T ×= =

×. The work done, using Eq. 3.51, and heat transfer are

2 2 1 1 31.25 40 500 10 3750 kJ1 1 2.0

P V PVWn

− × − ×= = =

− −

20 0.717 (210 842) 3750 5310 kJvQ U W mC T W= Δ + = Δ + = × × − + = −

c) Given: n = 1.4. Then 1.4

12 1

2

10500 71.8 kPa,40

nVP PV

⎛ ⎞ ⎛ ⎞= = × =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

1500 10 842 K,

20 0.297T ×

= =×

and 271.8 40 484 K20 0.297

T ×= =

×. The work done is, using Eq. 3.51, and heat transfer are

2 2 1 1 71.8 40 500 10 5320 kJ1 1 1.4

P V PVWn

− × − ×= = =

− −

20 0.717 (484 842) 5320 186 kJvQ U W mC T W= Δ + = Δ + = × × − + =

This heat transfer is quite small and would approach zero as calculations are carried to more significant digits.

3.88 a) 321 2

2

0.287 473 0.679 m /kg, 0 kJ200

RTv v WP

×= = = = =

[ ]1 1200 1 0.717(200 ). 79 C and 1 1.0 (200 ( 79) 279 kJU T T HΔ = = × − ∴ = − ° Δ = × × − − =

11

1

0.287 194200 kJ and 82 kPa0.679

RTQ U Pv

×= Δ = = = =

b)

1 0.717(400 200) 143 kJ. 1 1.0(400 200) 200 kJ v pU mC T Q H mC TΔ = Δ = × − = = Δ = Δ = × − =

1 400 (0.483 0.339) 57.6 kJ or 200 143 57 kJW mP v W Q U= Δ = × × − = = − Δ = − =

c) 312 1 1

1

0.287 33360 C, 0.478 m /kg200

RTT T vP

×= = ° = = =

12

2 2

200ln . 80 1 0.287 333 ln . 86.6 kPaPW mRT PP P

= = × × × ∴ =

322

2

0.287 333 1.104 m /kg86.6

RTvP

×= = = , Q = W = 80 kJ, ΔU = 0 and ΔH = 0


d) 1 0.4

21 2

1

0.02873 348 K or 75 C.0.2

kvT Tv

−⎛ ⎞ ⎛ ⎞= = × = °⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

1 21 2

1 2

0.287 873 0.287 3481250 kPa, 4990 kPa0.2 0.02

RT RTP Pv v

× ×= = = = = =

1 1.0 (873 348) 525 kJ, 1 0.717 (873 348) 376 kJp vH mC T U mC TΔ = Δ = × × − = Δ = Δ = × × − =

376 kJW U= −Δ = −

3.89 State 1: P1 = 60 psia, V1 10 ft3, T1 = 300ºF. State 2: V2 = 50 ft3.

a) For n = 1.2,

1.2 1.21 1 1

2 12 1

2 2 1 1

10 (60 144) 1060 8.70 psia and 2.13 lbm50 53.3 760

(8.7 50 60 10) 144 118,800 ft-lbf or 153 Btu1 0.2

V PVP P mV RT

P V PVWn

⎛ ⎞ × ×⎛ ⎞= = = = = =⎜ ⎟ ⎜ ⎟ ×⎝ ⎠⎝ ⎠− × − × ×

= = =− −

2 22

(8.7 144) 50 552 R2.13 53.3

P VTmR

× ×= = = °

× 2.13 0.171 (552 760) 153 77.2 Btu vQ U W mC T W= Δ + = Δ + = × × − + =

b) For n = 1.4, (this is the adiabatic, quasi-equilibrium process, so Q should be zero)

1.4 1.41 1 1

2 12 1

2 2 1 1

10 (60 144) 1060 6.304 psia and 2.13 lbm50 53.3 760

(6.304 50 60 10) 144 102,500 ft-lbf or 132 Btu1 0.4

V PVP P mV RT

P V PVWn

⎛ ⎞ × ×⎛ ⎞= = = = = =⎜ ⎟ ⎜ ⎟ ×⎝ ⎠⎝ ⎠− × − × ×

= = =− −

2 22

(6.304 144) 50 399.8 R 2.13 53.3

P VTmR

× ×= = = °

× 2.13 0.171 (399.8 760) 132 0.8 Btu vQ U W mC T W= Δ + = Δ + = × × − + = ≅ 0

c) For n = 1.6,

1.61 1 1

2 12 1

2 2 1 1

10 (60 144) 1060 4.57 psia and 2.13 lbm50 53.3 760

(4.57 50 60 10) 144 89,200 ft-lbf or 115 Btu1 0.6

nV PVP P mV RT

P V PVWn

⎛ ⎞ × ×⎛ ⎞= = = = = =⎜ ⎟ ⎜ ⎟ ×⎝ ⎠⎝ ⎠− × − × ×

= = =− −

2 22

(4.57 144) 50 289.8 R 2.13 53.3

P VTmR

× ×= = = °

×


2.13 0.171 (289.8 760) 115 56 Btu vQ U W mC T W= Δ + = Δ + = × × − + = −

3.90 a) 31 11 2 1

2 1

1 0.287 400For const, 2400 400 K. = = 0.1148 m /kg6 1000

v RTP T T vv P

×= = = × = ∴ =

1 0.41

3 13

1400 195 K6

kvT Tv

−⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

1-2 1 2 1 2-3( ) 2 1000(6 0.1148 0.1148) 1148 kJ, 0W mP v v W= − = × × − = =

3-1 net2 .717 (400 195) 294 kJ. 1148 294 854 kJvW U mC T Q W= −Δ = − Δ = − × × − = − ∴ = = − =∑

b) 31 11 2 1

2 1

1 0.287 400For const, 2400 400 K. = = 0.1531 m /kg6 750

v RTP T T vv P

×= = = × = ∴ =

1 0.41

3 13

1400 195 K6

kvT Tv

−⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

1-2 1 2 1 2-3( ) 2 750(6 0.1531 0.1531) 1148 kJ, 0W mP v v W= − = × × − = =

3-1 2 0.717 (400 195) 294 kJvW U mC T= −Δ = − Δ = − × × − = −

Apply Eq. 3.21: net 1148 294 854 kJQ W= = − =∑

c) 31 11 2 1

2 1

1 0.287 400For const, 2400 400 K. = = 0.2296 m /kg6 500

v RTP T T vv P

×= = = × = ∴ =

1 0.41

3 13

1400 195 K6

kvT Tv

−⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

1-2 1 2 1 2-3( ) 2 500(6 0.2296 0.2296) 1148 kJ, 0W mP v v W= − = × × − = =

3-1 2 .717 (400 195) 294 kJvW U mC T= −Δ = − Δ = − × × − = −

Apply Eq. 3.21: net 1148 294 854 kJQ W= = − =∑

3.91 a) 1.4

32 21 2 1 3 2

2 3

0.287 700 0.2010.201 m /kg, 1000 382 kPa1000 0.4

kRT vv v P P PP v

⎛ ⎞× ⎛ ⎞= = = = = = = × =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

3 31 11 3

382 0.201 382 0.4267.5 K, 532.4 K0.287 0.287

P vPvT TR R

× ×= = = = = =

1-2 2-30, W W Q= = 2 0.717(532.4 700) 240 kJvmC T− Δ = − × − =

3-1 1 1 3 net( ) 2 382 (0.201 0.4) 152 kJ. 0 240 152 88 kJW mP v v Q W= − = × × − = − = = + − =∑

b) 1.4

32 21 2 1 3 2

2 3

0.287 700 0.2680.268 m /kg, 750 428 kPa750 0.4

kRT vv v P P PP v

⎛ ⎞× ⎛ ⎞= = = = = = = × =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

3 31 11 3

428 0.268 428 0.4400 K, 597 K0.287 0.287

P vPvT TR R

× ×= = = = = =

1-2 2-30, W W Q= = 2 0.717(597 700) 148 kJvmC T− Δ = − × − =

3-1 1 1 3 net( ) 2 428 (0.268 0.4) 113 kJ. 0 148 113 35 kJW mP v v Q W= − = × × − = − = = + − =∑


c) 1.4

32 21 2 1 3 2

2 3

0.287 700 0.3350.335 m /kg, 600 468 kPa600 0.4

kRT vv v P P PP v

⎛ ⎞× ⎛ ⎞= = = = = = = × =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

3 31 11 3

468 0.335 468 0.4546 K, 652 K0.287 0.287

P vPvT TR R

× ×= = = = = =

1-2 2-30, W W Q= = 2 0.717(652 700) 68.5 kJvmC T− Δ = − × − =

3-1 1 1 3 net( ) 2 468 (.335 .4) 60.8 kJ. 0 68.5 60.8 7.7 kJW mP v v Q W= − = × × − = − = = + − =∑ 3.92 a) For T = C,

1/ 1/1.4331

3 1 2 33 2

.1 2501000 250 kPa. 0.4 0.1486 m /kg

.4 1000

kPvP P v v

v P⎛ ⎞ ⎛ ⎞= = × = ∴ = = =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

1 1 2 23 1 2

1000 0.1 1000 0.1486348 K, 518 K0.287 0.287

Pv P vT T TR R

× ×= = = = = = =

1-2 2 1( ) 2 1000 (0.1486 0.1) 97.2 kJW mP v v= − = × × − =

2-3W Q= 2 0.717 (348 518) 244 kJvmC T− Δ = − × × − =

13-1 1 net

3

0.1ln 2 0.287 348ln 277 kJ. 97.2 244 277 64 kJ0.4

vW mRT Q Wv

= = × × = − = = + − =∑

b) For T = C, 1/ 1/1.4

3313 1 2 3

3 2

.1 187.5750 187.5 kPa. .4 0.1486 m /kg

.4 750

kPvP P v v

v P⎛ ⎞ ⎛ ⎞= = × = ∴ = = =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

1 1 2 23 1 2

750 0.1 750 0.1486261 K, 388 K0.287 0.287

Pv P vT T TR R

× ×= = = = = = =

1-2 2 1( ) 2 750(0.1486 0.1) 73 kJW mP v v= − = × − =

2-3W Q= 2 0.717(388 261) 182 kJvmC T− Δ = − × − =

13-1 1 net

3

0.1ln 2 0.287 261ln 208 kJ. 73 182 208 47 kJ0.4

vW mRT Q Wv

= = × × = − = = + − =∑

c) For T = C, 1/ 1/1.4

3313 1 2 3

3 2

.1 125500 125 kPa. 0.4 0.1486 m /kg

.4 500

kPvP P v v

v P⎛ ⎞ ⎛ ⎞= = × = ∴ = = =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

1 1 2 23 1 2

500 0.1 500 0.1486174 K, 259 K0.287 0.287

Pv P vT T TR R

× ×= = = = = = =

1-2 2 1( ) 2 500(0.1486 0.1) 49 kJW mP v v= − = × − =

2-3W Q= 2 0.717(174 259) 122 kJvmC T− Δ = − × − =

13-1 1 net

3

0.1ln 2 0.287 174ln 138 kJ. 49 122 138 33 kJ0.4

vW mRT Q Wv

= = × × = − = = + − =∑

3.93 At 3 3

2 2 3240 F, Table C-2E provides 0.016922 ft /lbm, =16.327 ft /lbm, T v v= ° = and P2 = 24.97 psia

a) The work is the area under the P-v diagram, i.e., the area of the rectangle:


[ ]

net3

2

net

lbf ft 4 lbm (24.97 2) 144 (16.327 0.016922) 216,000 ft-lbflbmft

216,000 / 778 278 Btu

W m P v

Q

= Δ × Δ

= × − × × − =

= =

b) The work is the area under the P-v diagram, i.e., the area of the rectangle:

[ ]

net3

2

net


178,200 / 778 229 Btu

W m P v

Q

= Δ × Δ

= × − × × − =

= =

c) The work is the area under the P-v diagram, i.e., the area of the rectangle:

[ ]

net3

2

net


140,600 / 778 181 Btu

W m P v

Q

= Δ × Δ

= × − × × − =

= =

chapter 3 solutions · 2018-01-10 · chapter 3 solutions 3.1 (c) ... 0.05 mg p a π × == = × 2...

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