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Chapter 1 Solutions

1.1 (B) The utilization of energy is not of concern in our study. If you use energy to power your car, or your car seat is your own decision.

1.2 (C) All properties are assumed to be uniformly distributed throughout the volume. 1.3 (D) 1.4 (B) When the working fluid crosses the boundary, it is a control volume, as during intake and

exhaust. The ice plus the water forms the system of (C). The entire atmosphere forms the system of (D).

1.5 (C) An extensive property doubles if the mass doubles. Temperature is the same for the entire

room or half the room. 1.6 (D) A process may be very fast, humanly speaking, but molecules move very rapidly so an engine

operating at 4000 rpm is not thermodynamically fast. All sudden expansion processes and combustion processes are non-equilibrium processes. Air leaving a balloon is thermodynamically slow.

1.7 (B) If force, length, and time had been selected as the three primary dimensions, the newton

would have been selected and mass expressed in terms of the other three. But, in Thermo-dynamics, the newton is expressed as kg·m/s2.

1.8 (D) 2 2 3W J/s N m/s (kg m/s ) m/s kg m /s= = ⋅ = ⋅ ⋅ = ⋅ 1.9 (A) 34 000 000 000 N = 34 × 109 N = 34 GN (or 34 000 MN.)

1.10 (A) 36 3

10 kg 1250 kg/m8000 10 m

ρ mV −

= = =×

33

3Hg

3water

1 1 0 0008 m /kg1250 kg/m

1250 kg/m 1 251000 kg/m

ρρ

ρ

.

.

Vvm

SG

= = = =

= = =

1.11 (D) We must know if the surface is horizontal, vertical, on an angle? The surface cannot be

assumed to be horizontal just because it is drawn that way on the paper. (Sometimes problems aren’t fair. This is an example of such a problem.)

1.12 (C) 22 4 2 2

36cos30 kN 1559 kN/m or 1560 kPa200 cm 10 m /cm−

°= = =

×nFPA

1.13 (A) Use Eq. 1.13 to convert to pascals:

3 2

2 3 2

13 6 1000 kg/m 9 81 m/s 0 42 m

56 030 kg/m s or 56.03 10 N/m or 56.03 kPa

ρ ( . ) . .p gH= = × × ×

= ⋅ ×

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1.14 (C) 0 = + =∑ F PA Kx mg

2 20.05 400 0.2 40 9.81. 39 780 N/m or 39.8 kPa gageπ× × + × = × ∴ =P P

The atmospheric pressure acts down on the top and up on the bottom of the cylinder and hence cancels out.

1.15 (B) We do not sense the actual temperature but the temperature gradient between our skin and

the water. As our skin heats up, the water feels cooler so we increase the water temperature until it feels warm again. This is done until out skin temperature ceases to change. An object feels cool if its temperature is less than out skin temperature. If that’s the case, a temperature difference occurs between our skin and the object over a very small distance, creating a temperature gradient skin object( ) /T T x− Δ .

1.16 (B) The energy equation states that at the position of maximum compression, the kinetic energy

of the vehicle will be zero and the potential energy of the spring will be maximum, that is,

12m 2V = 21

2Kx . (The velocity must be expressed in m/s.)

2

2 61 80 1000 12000 0.1 . 98.8 10 N/m or 100 MN/m2 3600 2

×⎛ ⎞× × = × × ∴ = ×⎜ ⎟⎝ ⎠

K K

If the mass is in kg, the velocity in m/s, and x in meters, K will be in N/m. But, check the units to make sure. Get used to always using N, kg, m, and s and the units will work out. You don’t have to always check all those units. It takes time and on a multiple-choice test, there are usually problems left over when time runs out.

1.17, 1.18, and 1.19. The Internet has the answers! 1.20 True. Thermodynamics presents how energy is transferred, stored, and transformed from one

form to another. If you use it to dry your hair, power your car, or store it in a battery, we don’t really care. Just use it any way that allows you to enjoy life!

1.21 Energy derived from coal is not sustainable since coal will eventually not be available, even

though that may take 500 years. If an energy source is not available indefinitely, it is not sustainable.

1.22 Consult the Internet. 1.23 A large number of engineers were required when the industrial revolution occurred. 1.24 Trains were traveling the rails in the mid-1800s so mechanical engineers were needed, not to

drive the trains, but to design them! Coal was mined with a pick and shovel until the late 1800’s. Power plants and automobiles also came near the end of the 1800’s.

1.25 It’s CO2 and it keeps things very cold. Check it out on the Internet.

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1.26 i) A system, ii) a control volume, iii) a system, iv) a system. No fluid crosses the boundary of a system. Fluid crosses the boundary of a control volume.

1.28 The number of molecules in a cubic meter of air at sea level is 16 9 25(3 10 ) 10 3 10× × = × .

123

25 34 10 molecules . 0.00002 m or 0.02 mm3 3 10 molecules/m

V r rπ= = ∴ =×

1.29 Catsup is not a fluid. It is a pseudo plastic or a shear-thinning liquid, whatever that is! A fluid

always moves if acted upon by a shear. A plastic can resist a shear but then moves when the shear is sufficiently large. Catsup is like that: first it won’t move, then it suddenly comes.

1.30 From Wikipedia, 1 stone = 6.35 kg (= 14 lbm). 6.3 stones 6.3 6.35 40 kg∴ = × =

1.31 The units using Newton’s 2nd law are simpler:

2ftlbf slugs

= × is simpler than 2

232.2 ft/slbf lbm

32.2 ft-lbm/lbf-s= ×

The conversion between mass and weight does not require the use of a gravitational constant when using the slug as the mass unit in the English system.

1.32 Volume is extensive since it increases when the mass is increased, other properties remaining

constant.

1.33 0.000998 0.001008% change 100 0.992% or 1 %

0.001008−

= × = − −

1.34 3 3ice water

1 1 57.2 lbm/ft . 62.4 lbm/ft0.01747

ρ ρ= = = =v

. So, ice is lighter than water at

32ºF, so ice floats. If ice was heavier than water, it would freeze from the bottom up. That would be rather disastrous. Fish as well as skaters would have a problem. You can speculate as to the consequences.

The system andc.v. are identical

The system is  the airinside plus that whichhas exited. The c.v.extends to the exit ofthe balloon nozzle.

Control  surface

1.27 Before After

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1.35 Hg13 600 13.61000

SG = =

33 2

kg m13 600 9.81 2 m 266 800 Nm s

W Vγ ⎛ ⎞= = × × =⎜ ⎟⎝ ⎠

lbf266 800 N 0.2248 59,980 lbfN

W = × =

1.36 31 1) 0.2 kg/m , 0.2 2 0.4 kg, 0.4 9.8 3.92 N5

a m V W mgv

ρ ρ= = = = = × = = = × =

31 1) 0.5 m /kg, 2 2 4 kg, 4 9.8 39.2 N2

b v m V W mgρρ

= = = = = × = = = × =

3 32 1) 0.002 m /kg, 500 kg/m , 1000 9.8 9800 N1000 0.002

Vc v W mgm

ρ= = = = = = = × =

3 31000 102 1 1) 102 kg, 51 kg/m , 0.0196 m /kg9.8 2 51

W md m vg V

ρρ

= = = = = = = = =

1.37 31 1) 0.02 lbm/ft , 0.02 20 0.4 lbm, 50

a m Vv

ρ ρ= = = = = × =

2

232.2 ft/s 0.4 lbm 0.4 lbf

32.2 ft-lbm/lbf-sc

gW mg

= = × =

3

2

2

1 1) 50 ft /lbm, 0.02 20 0.4 lbm, 0.02

32.2 ft/s 0.4 lbm 0.4 lbf32.2 ft-lbm/lbf-sc

b v m V

gW mg

ρρ

= = = = = × =

= = × =

2

2

3 3

32.2 ft/s) 1000 lbm 1000 lbf32.2 ft-lbm/lbf-s

20 1 1 0.02 ft /lbm, 50 lbm/ft1000 0.02

c

gc W mgVvm v

ρ

= = × =

= = = = = =

2

2

3 3

32.2 ft-lbm/lbf-s) 500 lbf 500 lbm32.2 ft/s

20 1 1 0.04 ft /lbm, 25 lbm/ft500 0.04

cgd m Wg

Vvm v

ρ

= = × =

= = = = = =

This problem should demonstrate the difficulty using English units with lbm and lbf! Note that

lbm and lbf are numerically equal at sea level where g = 32.2 ft/s2, which will be true for problems of interest in our study. In space travel, g is not 32.2 ft/s2.

1.38 31 1 0.25 kg/m4v

ρ = = = , water

0.25 0.000251000

xSG ρρ

= = = , 3

38 m 2 kg

4 m /kgVmv

= = =

2

m2 kg 9.81 19.62 Ns

W mg= = × = . (We used N = kg·m/s2.)

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1.39 31 1 5 ft /lbm0.2

vρ

= = = , 3

3water

0.2 lbm/ft 0.0032162.4 lbm/ft

xSG ρρ

= = =

2

232.2 ft/s0.2 20 4 lbm, 4 lbm 4 lbf

32.2 ft-lbm/lbf-sc

gm V W mg

ρ= = × = = = × =

1.40 Only (ii) can be considered a quasi-equilibrium process. Process (i) uses a temperature

distribution in the room to move the heated air to other locations in the room, i.e., the temperature is not uniform. When the membrane in process (iii) is removed, a sudden expansion occurs, which cannot be considered a quasi-equilibrium process.

1.41 From Table B-1 in the Appendix, we observe that. So,

3

3

1.225 kg/mi) 0.0012251000 kg/m

= =SG

3

3

0.6012 1.225 kg/mii) 0.000 7361000 kg/m

×= =SG

3

3

0.3376 1.225 kg/miii) 0.000 4141000 kg/m

×= =SG

1.42 From Table B-1 in the Appendix, we observe find the local atmospheric pressure. First,

2

2 22 2 2

kg cm m 2.1 100 9.81 206 000 N/m or 206 kPa gagecm m s

= × × =gP

(We used N = kg·m/s2.)

i) 206 kPa 101 kPa 307 kPa= + =P ii) 206 kPa 101 0.887 kPa 296 kPa= + × =P iii) 206 kPa 101 0.5334 kPa 260 kPa= + × =P

(We could have used Patm = 101.3 kPa or even 100 kPa since extreme accuracy is not of interest) 1.43 Refer to Fig. 1.6 and Eq. 1.14. The pressure in the tire would be P2 and P1 would be open to the

atmosphere:

2 2

2 2gage 2 2 2 2

kg cm m kg m/s3 4 100 9 81 334 000 334 000 N/mcm m m

Ps

⋅= × × = =. .

2 1P P−2 3 2

N kg m 334 000 1000 13 6 9 81 2 50 m or 2500 mmm m s

g h h hρ= Δ = × × × Δ ∴Δ =. ( . ) . . .

1.44 2 3 2

N kg m 100 000 786 9 81 13 0 mm m s

P gh h hρ= = × × ∴ =. . . . (We used N = kg·m/s2.)

1.45 kPa10 atm 100 1000 kPa. 1000 kPa 100 kPa 900 kPa

atm= ⋅ = ∴ = − =gP P

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1.46 water 3 2

kg m1000 9.81 0.25 m 2453 Pa 2.45 kPa gagem s

ρΔ = Δ = × × = =P g h

Hg Hg Hg Hg. 2453 (1000 13.6) 9.81 . 0.0184m or 0.724 in.ρΔ = Δ = × × × Δ ∴Δ =P g h h h 1.47 A measured pressure is a gage pressure.

a) 13 600 9 81 0 10 13 340 Pa or 13.34 kPa gageP ghρΔ = = × × =. .

b) 13 600 9 81 0 28 37 360 Pa or 37.36 kPa gageP ghρΔ = = × × =. .

1.48 a) abs gage atm 5 0.371 14.7 10.45 psia or 1505 psfa= + = + × =P P P

b) abs gage atm 20 0.371 14.7 25.45 psia or 3665 psfa= + = + × =P P P

1.49 Consult the Internet 1.50 Consult the Internet 1.51 R F 460 120 460 580 RT T° = ° + = + = °( ) ( ) 1.52 C K) 273 3 273 270 CT T° = − = − = − °( ) ( 1.53 R) F) 460 400 460 860 RT T° = ° + = + = °( ( 1.54 (K) 37 273 310 K= + =T 1.55 Use Eq. 1.20:

a) 0 0( )/ 4220(25 60)/298 3330 3000 677 β − − ×= = = ΩT T T TR R e e

b) 0 0( )/ 4220(25 120)/298 3930 3000 97.8 β − − ×= = = ΩT T T TR R e e

c) 0 0( )/ 4220(25 180)/298 4530 3000 23.6 β − − ×= = = ΩT T T TR R e e

1.56 Use V V TβΔ = Δ .

a) 2

30 00018 40 00018 0 003 20 0 016 m or 16 mm4 3

H Hπ π= × × × ∴ =. . . . .

b) 2

30 00018 40 00018 0 003 40 0 032 m or 32 mm4 3

H Hπ π= × × × ∴ =. . . . .

c) 2

30 00018 40 00018 0 003 60 0 048 m or 48 mm4 3

H Hπ π= × × × ∴ =. . . . .

1.57 =V mi 5280 ft/mi60 88 ft/shr 3600 s/hr

× = , 2

=c

mKEg

2V

22

2500 lbm 88 300 600 ft-lbf2 32 2 ft-lbm/lbf-s

,.

= × =×

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300 600 ft-lbfor 386 Btu778 ft-lbf/Btu

=,

The English unit on energy is most often the Btu (some authors use BTU).

1.58 12

+ =KE PE m 2V + mgh = 2 61 5000 80 5000 9.81 1000 65 10 N m 65 MJ2

× × + × × = × ⋅ =

1.59 At 10 000 m, 6 29.81 3.32 10 10 000 9.777 m/s−= − × × =g

6

surface6

10 km

140 000 9.81 1.373 10 N

140 000 9.777 1.369 10 N

= = × = ×

= = × = ×

W mg

W mg

10 000 10 0006

0 0

140 000(9.81 3.32 10 )−= = − ×∫ ∫PE mgdh h dh

26 10 10

10

10 000140 000 9.81 10 000 3.32 10 1.373 10 0.0023 102

1.371 10 N m or 13.71 GJ

−⎛ ⎞= × − × × = × − ×⎜ ⎟⎜ ⎟

⎝ ⎠

= × ⋅

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