solution manual for thermodynamics for engineers 1st ... · pdf filesolution manual for...

Solution Manual for Thermodynamics for

Engineers 1st Edition by Kroos and Potter

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thermodynamics-for-engineers-1st-edition-by-kroos-and-potter

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 2

) ⋅ m/s = kg ⋅ m 2

/s3

1.9 (A) 34 000 000 000 N = 34 × 109

N = 34 GN (or 34 000 MN.)

1.10 (A) ρ = m

= 10 kg

= 1250 kg/m3

V 8000×10

−6 m

3

v = V = 1

= 1

= 0.0008 m3 /kg

m ρ 1250 kg/m3

SG =

ρHg

= 1250 kg/m

3

= 1.25

ρ 1000 kg/m

3 water

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.)

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1.12 (C) P = Fn

= 36cos30° kN

= 1559 kN/m2 or 1560 kPa

A 200 cm2

×10−4

m2

/cm2

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

p = ρ gH = (13.6 × 1000 kg/m3

) × 9.81 m/s2

× 0.42 m

= 56 030 kg/m ⋅ s2

or 56.03×103

N/m2

or 56.03 kPa

1 © 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1.14 (C) ∑ F = 0 PA + Kx = mg

P × π × 0.052

+ 400× 0.2 = 40× 9.81. ∴ P = 39 780 N/m2 or 39.8 kPa gage

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 ) / x .

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 2 1 2 2 m V = 2 Kx . (The velocity must be expressed in m/s.)

1 80×1000 2 1

∴ K = 98.8×106 N/m or 100

= × K × 0.12

. × 2000 × MN/m

2 3600 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.

2

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.

1.27 Before

The system and c.v. are identical

Control surface

After

The system is the air inside plus that which has exited. The c.v. extends to the exit of the balloon nozzle.

1.28 The number of molecules in a cubic meter of air at sea level is (3 × 1016

) × 10 9

= 3 ×1025

.

4 π r

3 =

1012

molecules

V =

. ∴ r = 0.00002 m or 0.02 mm

3×10 25

molecules/m 3

3 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:

32.2 ft/s2

lbf = slug × ft

is simpler than lbf = lbm ×

s 2 32.2 ft-lbm/lbf-s 2

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 % change =

0.000998

−

0.001008 × 100 = − 0.992% or

1

−1 % 0.001008

1 = 57.2 lbm/ft

3 . ρwater = 62.4 lbm/ft

3 . So, ice is lighter than water at 1.34 ρ ice =

=

v 0.01747

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.


1.35 SGHg = 13 600 = 13.6

1000 kg

m

W = γ V = 13 600 × 9.81

× 2 m

3 = 266 800 N 3 2

m s W = 266 800 N × 0.2248 lbf = 59,980 lbf

N

1.36 a

) ρ

= 1

= 1 = 0.2 kg/m3

, m = ρV = 0.2× 2 = 0.4 kg, W = mg = 0.4× 9.8 = 3.92 N

v 5

b ) v = 1 = 1 = 0.5 m3 /kg, m = ρV = 2× 2 = 4 kg, W = mg = 4× 9.8 = 39.2 N

ρ 2

c ) v = V = 2 = 0.002 m3 /kg, ρ = 1 = 500 kg/m

3, W = mg = 1000× 9.8 = 9800 N

m 1000

0.002

d ) m = W = 1000 = 102 kg, ρ = m = 102 = 51 kg/m3 , v = 1 = 1 = 0.0196 m 3 /kg

g 9.8

V 2

ρ 51

1.37 a ) ρ = 1

= 1 = 0.02 lbm/ft3

,

m = ρV = 0.02× 20 = 0.4 lbm,

v 50

32.2 ft/s2

W = m g = 0.4 lbm × = 0.4 lbf

gc

32.2 ft-lbm/lbf-s2

b ) v = 1 = 1 = 50 ft3

/lbm , m = ρV = 0.02× 20 = 0.4 lbm,

ρ 0.02

g 32.2 ft/s2

W = m = 0.4 lbm × = 0.4 lbf

gc

c ) W = m g = 1000 lbm × 32.2 ft/s

2 = 1000 lbf

gc

32.2 ft-lbm/lbf-s2

v = V = 20

= 0.02 ft3

/lbm , ρ = 1

=

1

= 50 lbm/ft3

m 1000 v 0.02

d ) m = W g

c = 500 lbf × 32.2 ft-lbm/lbf-s

2 = 500 lbm

32.2 ft/s2

g

v = V = 20 = 0.04 ft3

/lbm, ρ = 1

=

1

= 25 lbm/ft3

m 500 v 0.04

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 ρ = 1

= 1 = 0.25 kg/m3

, SG = ρx = 0.25 = 0.00025 , m = V = 8 m3

= 2 kg

ρ

3

v 4

1000

v 4 m /kg m2 =

water

W = mg = 2 kg × 9.81

19.62 N . (We used N = kg·m/s2

.)

s

4

1 1 5 ft

3 ρx 0.2 lbm/ft3

1.39 v =

=

=

/lbm , SG =

ρwater =

= 0.00321

ρ 0.2

62.4 lbm/ft3

m = ρV = 0.2× 20 = 4 lbm, W = m g = 4 lbm × 32.2 ft/s2

= 4 lbf

gc 32.2 ft-lbm/lbf-s2

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,

i) SG = 1.225 kg/m3

=

0.001225 1000 kg/m3

ii) SG = 0.6012×1.225 kg/m3 = 0.000

736 1000 kg/m3

iii) SG = 0.3376×1.225 kg/m3 = 0.000

414 1000 kg/m3

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

P = 2.1 kg × 1002

cm2 × 9.81 m = 206 000 N/m2 or 206 kPa gage

g cm2 m

2 s

2

(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:

P = 3. 4 kg × 1002 cm

2 × 9.81 m = 334 000 kg ⋅ m/s

2 = 334 000 N/m

2

gage cm 2 m 2 s 2 m 2

P − P = ρ g h. 334 000 N = (1000 × 13.6 ) kg × 9.81 m × h. ∴ h = 2.50 m or 2500 mm

2 1

m2

m3

s2

N kg m 2

1.44 P = ρ gh. 100 000 m2 = 786m3 × 9.81 s2 × h. ∴ h = 13.0 m (We used N = kg·m/s .)

1.45 P = 10 atm ⋅ 100 kPa

= 1000 kPa.

∴ P = 1000 kPa − 100 kPa = 900 kPa

atm

g

5


1.46 P = ρ g h = 1000 kg × 9.81 m × 0.25 m = 2453 Pa = 2.45 kPa gage

water m3

s2

P = ρHg g hHg . 2453 = (1000× 13.6)× 9.81× hHg . ∴ hHg = 0.0184m or 0.724 in.

1.47 A measured pressure is a gage pressure.

a) P = ρ gh = 13 600 × 9. 81× 0.10 = 13 340 Pa or 13.34 kPa gage

b) P = ρ gh = 13 600 × 9. 81× 0.28 = 37 360 Pa

or 37.36 kPa gage

P = P +P 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

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) R = R e β ( T0 −T )/T0T = 3000e4220(25 − 60)/ 298 ×333 = 677 Ω

0

= 97.8 Ω b) R = R e β ( T

0 −T

)/T

0

T = 3000e4220(25 −120)/ 298 ×393

0

c)

= 23.6 Ω R = R e β ( T0 −T )/T0T = 3000e4220(25 −180)/ 298 ×453 0

1.56 Use V = β V T . 0.00018

2 4 3

π

H = 0.00018× 3 π × 0.003 × 20.

∴ H = 0.016 m or 16 mm a) 4

0.000182 4 3

b) π 4 H = 0.00018× 3 π × 0.003 × 40. ∴ H = 0.032 m or 32 mm

0.000182

4 3

c) π 4 H = 0.00018× 3 π × 0.003 × 60. ∴ H = 0.048 m or 48 mm

1.57 V = 60 mi ×5280 ft/mi = 88 ft/s , KE = m V 2

hr 3600 s/hr 2gc

= 2500 lbm × 882 = 300,600 ft-lbf

2 × 32.2 ft-lbm/lbf-s2


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).

1.58 KE + PE = 1

2 m V 2

+ mgh = 1

2 × 5000 × 802

+ 5000 × 9.81× 1000 = 65× 106

N ⋅ m = 65 MJ

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

× 10 000 = 9.777 m/s2

Wsurface = mg = 140 000 × 9.81 = 1.373×106

N

W10 km = mg = 140 000 × 9.777 = 1.369 ×106

N

10 000 10 000

PE = ∫ mgdh = ∫ 140 000(9.81− 3.32 ×10−6

h ) dh 0 0

10 0002

= 140 000 9.81× 10 000 − 3.32 × 10 −6

× = 1.373

2 = 1.371× 10

10 N ⋅

m or 13.71 GJ

10 10

× 10 − 0.0023×10


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