Solution manual for thermodynamics for engineers 1st edition by kroos

1.3 D 1.4 B When the working fluid crosses the boundary, it is a control volume, as during intake and exhaust.. Temperature is the same for the entire room or half the room.. 1.15 B We

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) W = J/s = ⋅ N m/s = (kg m/s ) m/s ⋅ 2 ⋅ = kg m /s ⋅ 2 3

1.9 (A) 34 000 000 000 N = 34 × 109 N = 34 GN (or 34 000 MN.)

6 3

10 kg

8000 10 m

ρ m

×

3 3

3 Hg

3 water

1250 kg/m

1250 kg/m

1 25

1000 kg/m

ρ ρ ρ

V v m 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.)

36 cos 30 kN

1559 kN/m or 1560 kPa

200 cm 10 m /cm−

°

×

n

F P A

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

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

1.14 (C) ∑ F=0 PA Kx+ =mg

0.05 400 0.2 40 9.81 39 780 N/m or 39.8 kPa gage π

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(Tskin−Tobject) /Δx

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,

1

2mV2 = 1 2

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

2

×

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

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(3 10 ) 10× 16 × 9= ×3 1025

12 3

0.00002 m or 0.02 mm

×

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:

2

ft lbf slug

s

= × is simpler than

2 2

32.2 ft/s lbf 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

0.001008

−

57.2 lbm/ft 62.4 lbm/ft 0.01747

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 and c.v. are identical The system is the airinside plus that which

has exited. The c.v.

extends to the exit of the balloon nozzle.

Control   surface

1.35 Hg 13 600 13.6

1000

266 800 N 0.2248 59, 980 lbf lbf

N

5

v

1 1 3

2

ρ

V

ρ

1.37 ) 1 1 0.02 lbm/ft , 3 0.02 20 0.4 lbm,

50

v

2 2

32.2 ft/s

32.2 ft-lbm/lbf-s

c

g

g

3

2 2

0.02

32.2 ft/s

32.2 ft-lbm/lbf-s

c

g

g

ρ ρ

2 2

32.2 ft/s

32.2 ft-lbm/lbf-s

c

g

g V v

2 2

32.2 ft-lbm/lbf-s

32.2 ft/s

0.04 ft /lbm, 25 lbm/ft

c

g

d m W

g V v

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

0.25 kg/m 4

v

water

0.25 0.00025 1000

x

ρ

3 3

8 m

2 kg

4 m /kg

V m v

2

m

2 kg 9.81 19.62 N

s

W=mg= × = (We used N = kg·m/s2.)

1.39 1 1 5 ft /lbm3

0.2

v

ρ

3 3 water

0.2 lbm/ft

0.00321 62.4 lbm/ft

x

ρ

2 2

32.2 ft/s

32.2 ft-lbm/lbf-s

c

g

g

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/m i) 0.001225

1000 kg/m

SG

3

3

0.6012 1.225 kg/m ii) 0.000 736

1000 kg/m

×

SG

3

3

0.3376 1.225 kg/m iii) 0.000 414

1000 kg/m

×

SG

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

2

2.1 100 9.81 206 000 N/m or 206 kPa gage

g P

(We used N = kg·m/s2.)

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

(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:

3 4 100 9 81 334 000 334 000 N/m

P

s

⋅

P2−P1

334 000 1000 13 6 9 81 2 50 m or 2500 mm

ρ

1.44

100 000 786 9 81 13 0 m

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

1.45 P=10 atm 100 ⋅ kPa =1000 kPa ∴P =1000 kPa 100 kPa− =900 kPa

1.46 water kg3 m2

1000 9.81 0.25 m 2453 Pa 2.45 kPa gage

ρ

Hg Hg 2453 (1000 13.6) 9.81 Hg Hg 0.0184 m or 0.724 in ρ

1.47 A measured pressure is a gage pressure

a) Δ =P ρgh=13 600 9 81 0 10 13 340 Pa or 13.34 kPa gage× × =

b) Δ =P ρgh=13 600 9 81 0 28× × =37 360 Pa or 37.36 kPa gage

1.48 a) Pabs=Pgage+Patm = +5 0.371 14.7× =10.45 psia or 1505 psfa

b) Pabs=Pgage+Patm =20 0.371 14.7+ × =25.45 psia or 3665 psfa

1.49 Consult the Internet 1.50 Consult the Internet 1.51 T(°R)= ° +T( )F 460 120 460= + =580 R°

1.52 T(° =C) T(K) 273 3 273− = − = −270 C°

1.53 T(° = ° +R) T( F) 460=400 460 860 R+ = °

1.54 T (K) = 37 273 310 K + =

1.55 Use Eq 1.20:

a) ( 0 )/ 0 4220(25 60)/ 298 333

b) ( 0 )/ 0 4220(25 120)/ 298 393

c) ( 0 )/ 0 4220(25 180)/ 298 453

1.56 Use Δ =V βV TΔ

a)

2

3

0 00018 0 003 20 0 016 m or 16 mm

b)

2

3

c)

2

3

0 00018 0 003 60 0 048 m or 48 mm

hr× 3600 s/hr = ,

2

=

c

m KE g

2 V

2 2

2500 lbm

88 300 600 ft-lbf

2 32 2 ft-lbm/lbf-s ,

×

or 300 600 ft-lbf 386 Btu

778 ft-lbf/Btu, = The English unit on energy is most often the Btu (some authors use BTU)

2 + =

KE PE mV2+ mgh = 1 5000 802 5000 9.81 1000 65 10 N m6 65 MJ

1.59 At 10 000 m, g=9.81 3.32 10− × −6×10 000=9.777 m/s2

6 surface

6

10 km

140 000 9.81 1.373 10 N

140 000 9.777 1.369 10 N

6

140 000(9.81 3.32 10− )

2

10

10 000

140 000 9.81 10 000 3.32 10 1.373 10 0.0023 10

2 1.371 10 N m or 13.71 GJ

−