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for Engineers 1st Edition by Kroos and Potter

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

2 /s

3

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 8000×10−6m

3

v = V = 1

= 1 = 0.0008 m3 /kg

m ρ 1250 kg/m3

SG = ρHg 1250 kg/m3 = 1.25

=

ρ 3

water 1000 kg/m

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) P = Fn = 36cos30° kN = 1559

kN/m2 or 1560 kPa

A 200 cm2 ×10

−4 m

2/cm

2

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

p = ρ gH = (13.6 × 1000 kg/m3 ) × 9.81 m/s

2 × 0.42 m

= 56 030 kg/m ⋅ s2 or 56.03×10

3

N/m2

or

56.03 kPa

1.14 (C) ∑ F = 0 PA + Kx = mg

P × π × 0.052 + 400× 0.2 = 40× 9.81. ∴ P = 39 780 N/m

2 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,

12 m V 2 = 12 Kx2 . (The velocity must be expressed in m/s.)

1 80×1000 2 1 K = 98.8×106 N/m or 100

× 2000 × = × K × 0.12. 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.

1.20 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.

1.27

1.28 The number of

molecules in a cubic meter

of air at sea level is (3 ×

1016

) × 10 9 = 3 ×10

25 .

V

π

3 r 3 =3×10102512

molecules/mmolecules 3 . ∴ r = 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:

lbf = slug × sft2 is simpler than lbf = lbm ×

32.232.2 ft-lbm/lbf ft/s2 -s2

The system and

c.v. are identical

The system

is the air

inside plus

that which

has exited.

The c.v.

extends to

the exit of

the balloon

nozzle.

Before

Control surface

After

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.

0.000998 − 0.001008

1.33 % change = × 100 = − 0.992% or

−1 % 0.001008

1.34 ρ ice = = v 0.01747

1 1 = 57.2 lbm/ft3 . ρwater = 62.4 lbm/ft

3 . 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.

1.35 SGHg = = 13.6

1000

kg m

W = γ V = 13 600 3 × 9.81 2 × 2 m3 =

266 800 N m s

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

N

1.36 a ) ρ = = 1 = 0.2

kg/m

1 3, 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 m

3 /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/m3, W = mg

= 1000× 9.8 = 9800 N m 1000

d ) m = W

= 1000 = 102 kg, ρ = m = 102

= 51

kg/m3 , v = 1

= 1 = 0.0196 m3 /kg g 9.8

V 2 ρ 51

1 1 = 0.02

lbm/ft3, m = ρV = 0.02× 20 = 0.4 lbm,

1.37 a ) ρ = =

v 50

W = m g

= 0.4 lbm × 32.2

ft/s2

= 0.4 lbf

gc 32.2 ft-lbm/lbf-s2

b ) v = 1

= 1

= 50 ft

3 /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/s2 2 = 1000 lbf

gc 32.2 ft-lbm/lbf-s

v = V = 20

= 0.02 ft3 /lbm, ρ = 1 = 1 = 50 lbm/ft3

m 1000 v 0.02

d ) m = W g g c = 500 lbf × 32.2 32.2 ft-lbm/lbf ft/s2 -s

2 = 500 lbm

v = V

= 20

= 0.04 ft

3 /lbm, ρ = 1 = 1 = 25 lbm/ft

3

m 500 v 0.04

3

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/m

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

m3

= 2 kg

v 4 ρwater 1000 v 4 m /kg

m

2.)

W = mg = 2 kg × 9.81s 2 = 19.62 N . (We used N = kg·m/s

1 1 ρ

1.39 v = ρ = 0.2 = 5 ft3 /lbm , SG = ρwaterx =62.40.2 lbm/ft

lbm/ft33 = 0.00321

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

ft/s2 = 4 lbf

m

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

0.6012×1.225 kg/m3

ii) SG = = 0.000 736

1000 kg/m3

0.3376×1.225 kg/m3

iii) SG = = 0.000 414

1000 kg/m3

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

atmospheric pressure. First, cm

2 × 9.81 m = 206 000 N/m

2 or 206 kPa

gage P = 2.1 kg × 1002

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 334 000 N/m

2

gage cm2 m2 s2 m2

P 50 m or 2500

mm

2 1 2 m3 s2 m

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

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.

− P = ρ g h . 334 000 N = ( 1000 × 13 . 6 ) kg × 9 . 81 m × h . ∴ h = 2 .

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 P =P +P

1.48 a) abs gage atm = 5 + 0.371×

14.7 = 10.45 psia or 1505 psfa

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

b) R = R e β ( T0 −T

)/T0T = 3000e4220(25 −120)/ 298 ×393 = 97.8 Ω

0

c) R = R e β ( T0 −T

)/T0T = 3000e4220(25 −180)/ 298 ×453 = 23.6 Ω

0

1.56 Use V = β V T .

0.00018 2 4 3

H = 0.00018× 3 π × 0.003 × 20. ∴ a) π 4

H = 0.016 m or 16 mm

b) π 4 H = 0.00018× 3 π × 0. 32 mm

0.000182 4

c) π 4 H =

0.00018× 3 π × 0.

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

hr 3600 s/hr 2gc

= 2500 lbm × 88

2 = 300,600 ft-lbf

2 × 32.2 ft-lbm/lbf-s2

6 © 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.

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 = 12 m V 2 + mgh = 12 × 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

−6 10 0002 10 − 0.0023×1010

= 1.371× 1010 N

m or 13.71 GJ

Solutions to Final Exam

1. A

2. D W = N ⋅ m/s = ( kg ⋅ m/s 2 )m/s = kg ⋅ m

2 /s

3

3. C Sum forces in the vertical direction (be sure and use Pa and not kPa):

∑ F = 0. PA − W − Kx = 0

x 254 m or 25.4 cm

If absolute pressure is used, the atmospheric pressure acting on the top of

the cylinder must be included.

4. A

5. B Since the temperature is below the boiling point (120.2°C from Table

C-2) of water at 0.2 MPa, refer to Table C-1 and use hf = 461 kJ/kg.

6. D The specific volume at 160°C is vg = 0.307 m3/kg (Table C-1). From

Table C-3 we search at 800°C and

observe at 1.6 MPa that v = 0.309 m3/kg. So, P2 = 1.59 MPa. (No

careful interpolation is needed.)

7. C The volume is assumed to be constant (it doesn’t blow up like a

balloon!). The ideal gas law is used: P1V1 = mRT1 and P2 V 2 =

mRT2 so P1 / P2 = T1 / T2 . Then, using absolute temperatures and

pressures (assume atmosphere pressure of 100 kPa since it is not

specified),

539 kPaor 439 kPa gage

P2 433

8. B If the pressure is constant, the work is mP(v2 – v1). The result is

W = mP ( v 2Units: kg − v1 ) = 8⋅× kN800 ⋅× m( 03. =2608 kN −⋅ 0m

. 2404) = 131 kJ

= kJ m2 kg

9. A First, find the height H of the piston above the cylinder bottom:

m = ρ AH . 688 m

The temperature when the piston hits the stops in this constant-pressure

process is

T = T V2 = 473× 1.188A = 333 K or 59.9°C

2 1 V 1.688A

1

10. A For this constant pressure process, the heat transfer is

Q = mC p T = 2 × 1.0 × ( 400 − 20) = 760 kJ

11. B The temperature at the final state must be known. It is T = T P2 =

673× = 168 K. So

2 1 P 400

1

Q = mC v T = 2 × 0. 717 × ( 673 − 168) = 724 kJ

12. C For an isothermal process Q = W = mRT ln V 2 / V1 so the heat

transfer is

Q = mRT ln V2 = 1. 0 × 0. 287 × 473× ln 1

= −94.1 kJ

V 2

1

13. D The heat from the copper enters the water:

mc C p , c Tc = mw C p , w Tw . 10 × 0.39 × (200 − T2 ) = 50 × 4.18× (T2 −

10). T2 = 13.5°C

14. A Only A does not have a fluid flowing in and/or out.

15. C The density is found from Table C-3 using ρ = 1/ v :

m = ρ AV = 1 AV = 1

× (π × 0. 142 ) × 2

= 1.89 kg/s v 0.06525

16. C The power required by a pump involving a liquid is

P2 − P1 6000 − 20

W = m = 5 ×

= 29.9 kW or 40.1 hp ρ 1000

17. D Equate the expression for the efficiency of a Carnot engine to the

efficiency in general:

TL Wout 10

η = 1− = . Qin = = 29.6 kJ/s TH Qin 1 − 313 /

473

Qout = Qin − W = 29. 6 − 10 = 19.6 kJ/s

18. A The maximum possible efficiency would be ηmax = 1 − TL = 1 −

283 = 0.237 . Consequently,

TH 371

Wout 2000

= 0.237. Wout, max

= 0.237 × + 10 = 10.2 kW. The engine is improbable.

Qin 60

B If COP > 1, A is violated. If COP = 1, C occurs. The condition that

W > QL is very possible.

20. D The heat transfer from the high

temperature reservoir is Q H = Q L + W = 20 + 8 = 28 kJ/s.

The

efficiency of the engine is

thus η = W =

8

= 0.286. The Carnot efficiency

yields

QH 28

η T = 410 K or 137°C

H

TH

Obviously the temperatures must be absolute.

21. C The entropy change of the copper is mCplnT2/T1 and that of the

water is Q/T. Hence,

mcu C p ,cu T

T 2 + Q = m C ln T 2 +

S = m C ln

net cu p ,cu T1 Twater cu p,cu T1 Twater

= 10 × 0.39ln 293

+ = 0.123 kJ/K

373 293

The copper loses entropy and the water gains entropy. Make sure the

heat transfer to the water is positive.

22. C The maximum work occurs with an isentropic process: s1 = s2 =

7.168 kJ/kg·K. The enthalpy at the turbine exit and the work are found

as follows (use Tables C-3 and C-2):

s2 = s1 = 7. 168 = 0. 6491+ x 2 ( 7.502). ∴ x2 = 0.869

∴ h2 = 192 + 0.869 × 2393 = 2272 kJ/kg and wT = h = 3658 − 2272

= 1386 kJ/kg

The copper loses entropy and the water gains entropy. Make sure the

heat transfer to the water is positive.

23. A The minimum work occurs with an isentropic process for which

T2 = T1 ( P2 / P1)k −1/ k = 300× 60.2587 = 477 K

∴ wC = C p (T2 − T1) = 1.0 × ( 477 − 300) = 177 kJ/kg

24. D The maximum possible turbine work occurs if the entropy is constant,

as in Problem 22 which is 1386 kJ/kg. The actual work is wT = h1 − h2

= 3658 − 2585 = 1073 kJ/kg . The efficiency is then

η = w = 10731386 = 0.774 or

77.4% T w

T , s

25. D The heat that leaves the condenser enters the water:

mw C p , w Tw = ms ( h1 − h2 ). mw × 4.18× 20 = 4 ×

( 2609.7 − 251). ∴ mw = 113 kg/s

26. B Heat transfer across a large temperature difference, which occurs in

the combustion process, is highly irreversible. The losses in A, C¸ and D

are relatively small.

27. B The combustion process is not reversible.

28. A The rate at which heat is added to the boiler is

Q B = m( h3 − h2 ) = 2 × ( 3434 − 251) = 6366 kJ/s

29. C The turbine power output is

WT = m( h3 − h4 ) = 2 × ( 3434 − 2609. 7) = 1648 kW

30. D The rate of heat transfer from the condenser is

QC = m( h4 − h1) = 2 × ( 2610 − 251) = 4718 kJ/s

31. B The required pump horsepower follows:

P 5000 − 20 WP = m = 2 ×= 9.96 kW or 13.35

hp ρ 1000

k−1

32. C First, find the temperature at state 2: T2 =

T

1 V1

= 293× 80.4 =

673 K. Then

V2

q 2 -3 = C v (T3 − T2 ) = 0. 717 × (1473 − 673) = 574 kJ/kg

33. A The efficiency of the Otto cycle is

1 1

η = 1− k−1 = 1− = 0.435 or 43.5%

r 8

34. A The work output can be found using the efficiency and the heat

input:

w net = ηqin = 1 − 1

q

2-3 = 1 − 1 × 574

= 250

kJ/kg

rk−1 80 . 4

35. D The heat input is provided by the combustor:

qin = C p (T3 − T2 ) = 1.0 × (1073 − 513) = 560 kJ/kg

36. B The pressure ratio for the assumed isentropic process is

P 2 =T 2 k /( k−1) = 513 3 .5 = 7.1

T1

293

P1

37. A The back-work ratio is found to be w comp C p (T2

− T1) = 240 − 20 = 0.44 or 44%

BWR = =

w C (T − T ) 800 − 300

T p 3 4

38. C The efficiency is the net work divided by the heat input:

T

η = w −wcomp = C p (T

3 − T4 ) − C p (T2 − T1) = 800 − 300 − (

240 − 20)

= 0.5

qin

C p (T3 − T2 ) 800 − 240

39. D The vapor pressure, using Pg = 5.628 kPa from Table C-1, is Pv =

Pgφ = 5.628× 0.9 = 5.065 kPa.

Then we have

Pv = 0 . 622 ×

5.065 ω = 0.622

Pa 100 − 5.065

40. C Locate state 1 at 10°C and 60% humidity. Move at constant ω (it is

assumed that no moisture is added. If it were, the amount of water would

be needed) to the right until T2 = 25°C. There the chart is read to provide

the humidity as φ = 24%.

= 0 . kg 0332 w /kg a