Solution manual for thermodynamics for engineers 1st edition by kroos

Chapter 1 SolutionsDownload FULL Solution Manual for Thermodynamics for Engineers 1st Edition by Kroos at https://getbooksolutions.com/download/solution-manual-for-thermodynamics-for-engineers-1st-edition-by-kroos

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 8000×10−6m3

v = V = 1 = 1 = 0 . 0008 m 3 /kgρ 1250 kg/m3m

SG =ρ

Hg = 1250 kg/m 3 = 1.25ρ

water 1000 kg/m3

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 =

F

n = 36cos30° kN = 1559 kN/m2 or 1560 kPa200 cm2 ×10−4 m2/cm2A

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

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= 56 030 kg/m ⋅ s2or 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,

12 m V 2 =

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

1

× 2000

80 × 1000 2 1

× K × 0.12.

∴ K = 98.8×106 N/m or 100

MN/m2

×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

© 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.

1.26 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 .

V =4

π r 3 =1012 molecules

. ∴ r = 0.00002 m or 0.02 mm3 3×1025 molecules/m3

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 × ft is simpler than lbf = lbm × 32.2 ft/s2

s2 32.2 ft-lbm/lbf-s2

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 % 0.001008

1.34 ρ ice = 1

= 1 = 57.2 lbm/ft3 . ρwater = 62.4 lbm/ft3 . So, ice is lighter than water at0.01747v

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

= 13.61000

kg mW = γ V = 13 600 × 9.81 × 2 m3 = 266 800 N

m3

s2

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

N

1.36

a ) ρ = 1

= 1 = 0.2 kg/m 3 ,m = ρV = 0.2× 2 = 0.4 kg, W = mg = 0.4× 9.8 = 3.92

Nv 5

b ) v =

1=

1 = 0.5 m 3 /kg, m = ρV = 2× 2 = 4 kg, W = mg = 4× 9.8 = 39.2 N

2ρc ) v =

V=

2= 0.002 m3 /kg, ρ =

1= 500 kg/m3,

W = mg = 1000× 9.8 = 9800 N

m 1000 0.002d ) 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

1.37a ) ρ =

1

=1

= 0.02 lbm/ft 3 ,m = ρV = 0.02× 20 = 0.4 lbm,

50v32.2 ft/s2

W = m g = 0.4 lbm × = 0.4 lbfg

c 32.2 ft-lbm/lbf-s2

b ) v = 1 = 1 = 50 ft 3 /lbm,

m = ρV = 0.02× 20 = 0.4 lbm,

0.02ρW = m

g= 0.4 lbm ×

32.2 ft/s2

= 0.4 lbf32.2 ft-lbm/lbf-s2gc

c ) W = m g = 1000 lbm × 32.2 ft/s2

= 1000 lbfgc 32.2 ft-lbm/lbf-s2

v =V = 20 = 0.02 ft 3 /lbm, ρ = 1

= 1 = 50 lbm/ft 3 1000 0.02m v

d ) m = W g c = 500 lbf × 32.2 ft-lbm/lbf-s 2 = 500 lbm

g 32.2 ft/s2

v = V = 20 = 0.04 ft 3 /lbm, ρ = 1

= 1 = 25 lbm/ft 3 500 0.04m 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.

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 m3 /kg

W = mg = 2 kg × 9.81sm

2 = 19.62 N . (We used N = kg·m/s2.)

4


1.39 v =1

=1

= 5 ft 3 /lbm ,SG =

ρx=

0.2 lbm/ft3= 0.00321ρ 0.2

ρwater 62.4 lbm/ft3

m = ρV = 0.2× 20 = 4 lbm, W = m

g= 4 lbm ×

32.2 ft/s2

= 4 lbfg

c 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/m 3 = 0.0012251000 kg/m3

ii) SG = 0.6012×1.225 kg/m3

= 0.000 7361000 kg/m3

iii) SG = 0.3376 × 1.225 kg/m 3 = 0.000 4141000 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 gagecm2 s2

g m2

(We used N = kg·m/s2.)

i) P = 206 kPa + 101 kPa = 307 kPaii) P = 206 kPa + 101× 0.887 kPa = 296 kPaiii) 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 × 9.81m = 334 000 kg ⋅ m/s2= 334 000 N/m2

gage cm2 m2 s2 m2

P − P = ρ g h. 334 000 N = (1000 × 13.6 ) kg × 9.81m × h. ∴ h = 2.50 m or 2500 mm2 1

m2 m3 s2

N kg m 2

1.44 P = ρ gh. 100 000 = 786 × 9.81 × h. ∴ h = 13 . 0 m (We used N = kg·m/s .)m2 m3 s2

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 gagem3 s2

water

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

Pabs

= P

gage+

P

atm1.48 a) = 5 + 0.371× 14.7 = 10.45 psia or 1505 psfa

b)P

abs=

P

gage+

P

atm = 20 + 0.371× 14.7 = 25.45 psia or 3665 psfa1.49 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

(

T

0 −T

)/T

0T = 3000e4220(25 − 60)/ 298 ×333 = 677 Ω

0

b) R = R e β

( T

0 −T

)/T

0T = 3000e4220(25 −120)/ 298 ×393 = 97.8 Ω

0

c) R = R e β

( T

0 −T

)/T

0T = 3000e4220(25 −180)/ 298 ×453 = 23.6 Ω

0

1.56 UseV = β V T .

0.00018 2 H = 0.00018×

4 3 ∴ H = 0 . 016 m or 16 mma) π 3 π × 0.003 × 20.4

0.000182 H = 0.00018×

4 3 ∴ H = 0 . 032 m or 32 mmb) π π × 0.003 × 40.4 3

0 . 00018 2 H =

4 3 ∴ H = 0 . 048 m or 48 mmc) π 4 0.00018× 3 π × 0.003 × 60.

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 × 32.2 ft-lbm/lbf-s22


or 300,600 ft-lbf = 386 Btu778 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 NW10 km = mg = 140 000 × 9.777 = 1.369 ×106 N

10 000 10 000

140 000(9.81− 3.32 ×10−6 h ) dhPE = ∫ mgdh = ∫0 0

= 140 000 9.81× 10 000 − 3.32× 10 −6 × 10 000 2 = 1.373 10 10× 10 − 0.0023×10

2

= 1.371× 1010 N ⋅ m or 13.71 GJ

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Solution manual for thermodynamics for engineers 1st edition by kroos