Download Solution Manual for Thermodynamics for Engineers 1st Edition by Kroos and Potter

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Download "Download Solution Manual for Thermodynamics for Engineers 1st Edition by Kroos and Potter"

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1 Download Solution Manual for Thermodynamics for Engineers 1st Edition by Kroos and Potter Link download full: 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/s (D) W = J/s = N m/s = (kg m/s 2 ) m/s = kg m 2 /s 3

2 1.9 (A) N = N = 34 GN (or MN.) 1.10 (A) ρ = m = 10 kg = 1250 kg/m3 V m 3 = 1 = m3 /kg m ρ 1250 kg/m 3 v = V = 1 = 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 = F n = 36cos30 kn = 1559 kn/m 2 or 1560 kpa A 200 cm m 2 /cm (A) Use Eq to convert to pascals: p = ρ gh = ( kg/m 3 ) 9.81 m/s m = kg/m s 2 or N/m 2 or kpa 1.14 (C) F = 0 PA + Kx = mg

3 P π = P = N/m 2 gage or 39.8 kpa The atmospheric pressure acts down on the top and up on the bottom of the cylinder and hence cancels out (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 (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 Kx 2. (The velocity must be expressed in m/s.) K = N/m or = K MN/m 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 True. Thermodynamics presents how energy is transferred, stored, and transformed from one form to another. If you use it to dry your hair,

4 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 Consult the Internet A large number of engineers were required when the industrial revolution occurred 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 It s CO2 and it keeps things very cold. Check it out on the Internet.

5 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. Before 1.27 Control surface After 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 The number of molecules in a cubic meter of air at sea level is ( ) 10 9 = V π 3 r 3 = molecules/mmol 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 From Wikipedia, 1 stone = 6.35 kg (= 14 lbm). 6.3 stones = = 40 kg 1.31 The units using Newton s 2 nd law are simpler: lbf = slug sft2 is simpler than lbf = lbm ft-lbm/lbf ft/s2 -s2

6 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 Volume is extensive since it increases when the mass is increased, other properties remaining constant % change = 100 = 0.992% or 1 % ρ ice = = v = 57.2 lbm/ft 3. 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 SGHg = = kg m W = γ V = m 3 = N m s W = N lbf = 59,980 lbf N 1.36 a ) ρ = = 1 = 0.2 kg/m 1 3, m = ρv = = 0.4 kg, W = mg = = 3.92 N v 5 b ) v = 1 = 1 = 0.5 m 3 /kg, m = ρv = 2 2 = 4 kg, W = mg = = 39.2 N ρ 2 c ) v = V = 2 = m3 /kg, ρ = 1 = 500 kg/m3, W = mg = = 9800 N m 1000

7 d ) m = W = 1000 = 102 kg, ρ = m = 102 = 51 kg/m 3, v = 1 = 1 = m 3 /kg g 9.8 V 2 ρ = 0.02 lbm/ft3, m = ρv = = 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-s 2 b ) v = 1 = 1 = 50 ft 3 /lbm, m = ρv = = 0.4 lbm, ρ 0.02 g 32.2 ft/s2 0.4 lbf gc W = m = 0.4 lbm = 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 ft-lbm/lbf ft/s 2 -s 2 = 500 lbm v = V = 20 = 0.04 ft 3 /lbm, ρ = 1 = 1 = 25 lbm/ft 3 m 500 v 0.04

8 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/s 2, which will be true for problems of interest in our study. In space travel, g is not 32.2 ft/s 2. v ρ = 1 = 1 = 0.25 kg/m 3, SG = ρ x = 0.25 = , m = V = 8 m3 = 2 kg ρ water 1000 v 4 m /kg m W = mg = 2 kg 9.81s 2 = N. (We used N = kg m/s 3 2.)

9 1 1 ρ 1.39 v = ρ = 0.2 = 5 ft3 /lbm, SG = ρwaterx = lbm/ft lbm/ft33 = = ρv = = 4 lbm, W = m g = 4 lbm 32.2 ft/s2 = 4 lbf m gc 32.2 ft-lbm/lbf-s 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 From Table B-1 in the Appendix, we observe that. So, i) SG = kg/m 3 = kg/m kg/m 3 ii) SG = = kg/m kg/m 3 iii) SG = = kg/m From Table B-1 in the Appendix, we observe find the local atmospheric pressure. First, cm m = N/m 2 or 206 kpa gage P = 2.1 kg 1002 g cm 2 m 2 s 2 (We used N = kg m/s 2.)

10 i) P = 206 kpa kpa = 307 kpa ii) P = 206 kpa kpa = 296 kpa iii) P = 206 kpa kpa = 260 kpa (We could have used Patm = kpa or even 100 kpa since extreme accuracy is not of interest) 1.43 Refer to Fig. 1.6 and Eq The pressure in the tire would be P2 and P1 would be open to the atmosphere: P N/m 2 gage cm2 m2 s2 m2 m P = ρ g h N = ( ) kg m h. h = 2. P 50 m or 2500 mm m3 s2 N kg m P = ρ gh m2 = 786m 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 = ρ Hg g h Hg = ( ) 9.81 h Hg. h Hg = m or in A measured pressure is a gage pressure.

11 a) P = ρ gh = = Pa or kpa gage b) P = ρ gh = = Pa or kpa gage P =P +P P =P +P 1.48 a) abs gage atm = = psia or 1505 psfa b) abs gage atm = = psia or 3665 psfa 1.49 Consult the Internet 1.50 Consult the Internet 1.51 T ( R ) = T( F) = = 580 R 1.52 T ( C) = T(K) 273 = = 270 C 1.53 T ( R) = T( F) = = 860 R 1.54 T (K) = = 310 K 1.55 Use Eq. 1.20: a) R = R e β ( T0 T )/T0T = 3000e4220(25 60)/ = 677 Ω 0 b) R = R e β ( T0 T )/T0T = 3000e4220(25 120)/ = 97.8 Ω 0 c) R = R e β ( T0 T )/T0T = 3000e4220(25 180)/ = 23.6 Ω Use V = β V T

12 a) π 4 H = π H = m or 16 mm b) π 4 H = π mm c) π 4 H = π 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 ft-lbm/lbf-s 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) KE + PE = 12 m V 2 + mgh = = N m = 65 MJ 1.59 At m, g = = m/s 2 Wsurface = mg = = N

13 W10 km = mg = = N PE = mgdh = ( h ) dh = N m or 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.

14 6. D The specific volume at 160 C is vg = m 3 /kg (Table C-1). From Table C-3 we search at 800 C and observe at 1.6 MPa that v = m 3 /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 P B If the pressure is constant, the work is mp(v2 v1). The result is W = mp ( v 2Units: kg v1 ) = 8 kn800 m( 0 3. =2608 kn 0 m. 2404) = 131 kj = kj m 2 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 = A = 333 K or 59.9 C 2 1 V 1.688A

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16 10. A For this constant pressure process, the heat transfer is Q = mc p T = ( ) = 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 Q = mc v T = ( ) = 724 kj 12. C For an isothermal process Q = W = mrt ln V 2 / V 1 so the heat transfer is Q = mrt ln V2 = ln 1 = 94.1 kj V D The heat from the copper enters the water: mc C p, c Tc = mw C p, w Tw (200 T2 ) = (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 ) 2 = 1.89 kg/s v

17 16. C The power required by a pump involving a liquid is P2 P W = m = 5 = 29.9 kw or 40.1 hp ρ 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 Q in / 473 Qout = Qin W = = 19.6 kj/s 18. A The maximum possible efficiency would be ηmax = 1 TL = = Consequently, TH 371 Wout 2000 = W out, max = 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 efficiency of the engine is thus η = W = 8 = Q L + W = = 28 kj/s. The = The Carnot efficiency

18 yields Q H 28 η T = 410 K or 137 C TH Obviously the temperatures must be absolute. H 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 = ln = kj/k 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 = 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 = = x 2 ( 7.502). x2 = h2 = = 2272 kj/kg and wt = h = = 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

19 T2 = T1 ( P2 / P1) k 1 / k = = 477 K wc = C p (T2 T1) = 1.0 ( ) = 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 = = 1073 kj/kg. The efficiency is then η = w = = 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 ( ). 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 ( ) = 6366 kj/s 29. C The turbine power output is WT = m( h3 h4 ) = 2 ( ) = 1648 kw

20 30. D The rate of heat transfer from the condenser is QC = m( h4 h1) = 2 ( ) = 4718 kj/s 31. B The required pump horsepower follows: P W P = m = 2 = 9.96 kw or hp ρ 1000 k C First, find the temperature at state 2: T2 673 K. Then V2 = T1 V1 = = q 2-3 = C v (T3 T2 ) = ( ) = 574 kj/kg 33. A The efficiency of the Otto cycle is 1 1 η = 1 k 1 = 1 = or 43.5% r A The work output can be found using the efficiency and the heat input: kj/kg w net = ηqin = 1 1 q 2-3 = = 250 rk

21 35. D The heat input is provided by the combustor: qin = C p (T3 T2 ) = 1.0 ( ) = 560 kj/kg 36. B The pressure ratio for the assumed isentropic process is P 2 =T 2 k /( k 1) = = 7.1 T1 293 P1 37. A The back-work ratio is found to be w comp C p (T2 T1) = = 0.44 or 44% BWR = = w C (T T ) T p C The efficiency is the net work divided by the heat input: T η = w w comp = C p (T 3 T4 ) C p (T2 T1) = ( ) = 0.5 qin C p (T3 T2 ) D The vapor pressure, using Pg = kpa from Table C-1, is Pv = Pgφ = = kpa. Then we have

22 Pv = = kg w /kg a ω = Pa 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%.

CONCEPTS AND DEFINITIONS. Prepared by Engr. John Paul Timola

CONCEPTS AND DEFINITIONS. Prepared by Engr. John Paul Timola CONCEPTS AND DEFINITIONS Prepared by Engr. John Paul Timola ENGINEERING THERMODYNAMICS Science that involves design and analysis of devices and systems for energy conversion Deals with heat and work and