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# Smith e Van Ness Introdução a engenharia Química Termodinâmica 7ª ed Soluções

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P 3000atm\ufffd D 0.17in\ufffd A S 4 D 2\ufffd A 0.023 in2 F P A\ufffd g 32.174 ft sec 2 mass F g \ufffd mass 1000.7 lbm Ans. 1.7 Pabs U g h Patm\ufffd= U 13.535 gm cm 3 \ufffd g 9.832 m s 2 \ufffd h 56.38cm\ufffd Patm 101.78kPa\ufffd Pabs U g h Patm\ufffd\ufffd Pabs 176.808kPa Ans. 1.8 U 13.535 gm cm 3 \ufffd g 32.243 ft s 2 \ufffd h 25.62in\ufffd Patm 29.86in_Hg\ufffd Pabs U g h Patm\ufffd\ufffd Pabs 27.22psia Ans. Chapter 1 - Section A - Mathcad Solutions 1.4 The equation that relates deg F to deg C is: t(F) = 1.8 t(C) + 32. Solve this equation by setting t(F) = t(C). Guess solution: t 0\ufffd Given t 1.8t 32\ufffd= Find t( ) 40\ufffd Ans. 1.5 By definition: P F A = F mass g= Note: Pressures are in gauge pressure. P 3000bar\ufffd D 4mm\ufffd A S 4 D 2\ufffd A 12.566mm2 F P A\ufffd g 9.807 m s 2 mass F g \ufffd mass 384.4kg Ans. 1.6 By definition: P F A = F mass g= 1 FMars K x\ufffd FMars 4 10 3\ufffdu mK gMars FMars mass \ufffd gMars 0.01 mK kg Ans. 1.12 Given: z P d d U\ufffd g= and: U M P R T= Substituting: zP d d M P R T\ufffd g= Separating variables and integrating: Psea PDenver P 1 P µ´ µ¶ d 0 zDenver z M g R T §¨ © · ¹\ufffd µ´ µ¶ d= After integrating: ln PDenver Psea §¨ © · ¹ M\ufffd g R T zDenver= Taking the exponential of both sides and rearranging: PDenver Psea e M\ufffd g R T zDenver §¨ © · ¹= Psea 1atm\ufffd M 29 gm mol \ufffd g 9.8 m s 2 \ufffd 1.10 Assume the following: U 13.5 gm cm 3 \ufffd g 9.8 m s 2 \ufffd P 400bar\ufffd h PU g\ufffd h 302.3m Ans. 1.11 The force on a spring is described by: F = Ks x where Ks is the spring constant. First calculate K based on the earth measurement then gMars based on spring measurement on Mars. On Earth: F mass g= K x= mass 0.40kg\ufffd g 9.81 m s 2 \ufffd x 1.08cm\ufffd F mass g\ufffd F 3.924N Ks F x \ufffd Ks 363.333 N m On Mars: x 0.40cm\ufffd 2 Ans. wmoon M gmoon\ufffd wmoon 18.767 lbf Ans. 1.14 costbulb 5.00dollars 1000hr 10 hr day \ufffd costelec 0.1dollars kW hr 10 hr day 70 W\ufffd costbulb 18.262 dollars yr costelec 25.567 dollars yr costtotal costbulb costelec\ufffd\ufffd costtotal 43.829 dollars yr Ans. 1.15 D 1.25ft\ufffd mass 250lbm\ufffd g 32.169 ft s 2 \ufffd R 82.06 cm 3 atm mol K\ufffd T 10 273.15\ufffd( )K\ufffd zDenver 1 mi\ufffd M g R T zDenver 0.194 PDenver Psea e M\ufffd g R T zDenver §¨ © · ¹\ufffd PDenver 0.823atm Ans. PDenver 0.834bar Ans. 1.13 The same proportionality applies as in Pb. 1.11. gearth 32.186 ft s 2 \ufffd gmoon 5.32 ft s 2 \ufffd 'lmoon 18.76\ufffd 'learth 'lmoon gearth gmoon \ufffd 'learth 113.498 M 'learth lbm\ufffd M 113.498 lbm 3 Ans. (b) Pabs F A \ufffd Pabs 110.054kPa Ans. (c) 'l 0.83m\ufffd Work F 'l\ufffd Work 15.848kJ Ans. 'EP mass g' l\ufffd 'EP 1.222kJ Ans. 1.18 mass 1250kg\ufffd u 40 m s \ufffd EK 1 2 mass u 2\ufffd EK 1000kJ Ans. Work EK\ufffd Work 1000kJ Ans. 1.19 Wdot mass g' h time 0.91 0.92= Wdot 200W\ufffd g 9.8 m s 2 \ufffd 'h 50m\ufffd Patm 30.12in_Hg\ufffd A S 4 D 2\ufffd A 1.227 ft2 (a) F Patm A mass g\ufffd\ufffd F 2.8642 103u lbf Ans. (b) Pabs F A \ufffd Pabs 16.208psia Ans. (c) 'l 1.7ft\ufffd Work F 'l\ufffd Work 4.8691 103u ft lbf Ans. 'PE mass g' l\ufffd 'PE 424.9 ft lbf Ans. 1.16 D 0.47m\ufffd mass 150kg\ufffd g 9.813 m s 2 \ufffd Patm 101.57kPa\ufffd A S 4 D 2\ufffd A 0.173m2 (a) F Patm A mass g\ufffd\ufffd F 1.909 104u N 4 mdot Wdot g 'h 0.91 0.92\ufffd mdot 0.488 kg s Ans. 1.22 a) cost_coal 25.00 ton 29 MJ kg \ufffd cost_coal 0.95GJ 1\ufffd cost_gasoline 2.00 gal 37 GJ m 3 \ufffd cost_gasoline 14.28GJ 1\ufffd cost_electricity 0.1000 kW hr\ufffd cost_electricity 27.778GJ 1\ufffd b)The electrical energy can directly be converted to other forms of energy whereas the coal and gasoline would typically need to be converted to heat and then into some other form of energy before being useful. The obvious advantage of coal is that it is cheap if it is used as a heat source. Otherwise it is messy to handle and bulky for tranport and storage. Gasoline is an important transportation fuel. It is more convenient to transport and store than coal. It can be used to generate electricity by burning it but the efficiency is limited. However, fuel cells are currently being developed which will allow for the conversion of gasoline to electricity by chemical means, a more efficient process. Electricity has the most uses though it is expensive. It is easy to transport but expensive to store. As a transportation fuel it is clean but batteries to store it on-board have limited capacity and are heavy. 5 1.24 Use the Matcad genfit function to fit the data to Antoine's equation. The genfit function requires the first derivatives of the function with respect to the parameters being fitted. Function being fit: f T A\ufffd B\ufffd C\ufffd( ) e A B T C\ufffd\ufffd §¨ © · ¹\ufffd First derivative of the function with respect to parameter A A f T A\ufffd B\ufffd C\ufffd( )d d exp A B T C\ufffd\ufffd §¨ © · ¹o First derivative of the function with respect to parameter B B f T A\ufffd B\ufffd C\ufffd( )d d 1\ufffd T C\ufffd exp A B T C\ufffd\ufffd §¨ © · ¹o First derivative of the function with respect to parameter C C f T A\ufffd B\ufffd C\ufffd( )d d B T C\ufffd( )2 exp A B T C\ufffd\ufffd §¨ © · ¹o t 18.5\ufffd 9.5\ufffd 0.2 11.8 23.1 32.7 44.4 52.1 63.3 75.5 §¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ©¨ · ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¹ \ufffd Psat 3.18 5.48 9.45 16.9 28.2 41.9 66.6 89.5 129 187 §¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ©¨ · ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¹ \ufffd 6 T t 273.15\ufffd\ufffd lnPsat ln Psat( )\ufffd Array of functions used by Mathcad. In this case, a0 = A, a1 = B and a2 = C. Guess values of parameters F T a\ufffd( ) exp a0 a1 T a2\ufffd \ufffd§¨© · ¹ exp a0 a1 T a2\ufffd \ufffd§¨© · ¹ 1\ufffd T a2\ufffd exp a0 a1 T a2\ufffd \ufffd§¨© · ¹ a1 T a2\ufffd\ufffd\ufffd 2 exp a0 a1 T a2\ufffd \ufffd§¨© · ¹ ª« « « « « « « « « «¬ º» » » » » » » » » »¼ \ufffd guess 15 3000 50\ufffd §¨ ¨ ©¨ · ¸ ¹ \ufffd Apply the genfit function A B C §¨ ¨ ©¨ · ¸ ¹ genfit T Psat\ufffd guess\ufffd F\ufffd( )\ufffd A B C §¨ ¨ ©¨ · ¸ ¹ 13.421 2.29 10 3u 69.053\ufffd §¨ ¨ ©¨ · ¸ ¹ Ans. Compare fit with data. 240 260 280 300 320 340 360 0 50 100 150 200 Psat f T A\ufffd B\ufffd C\ufffd( ) T To find the normal boiling point, find the value of T for which Psat = 1 atm. 7 This is an open-ended problem. The strategy depends on age of the child, and on such unpredictable items as possible financial aid, monies earned by the child, and length of time spent in earning a degree. c) The salary of a Ph. D. engineer over this period increased at a rate of 5.5%, slightly higher than the rate of inflation. i 5.511% i Find i( )\ufffd C2 C1 1 i\ufffd( )t2 t1\ufffd=Given C2 80000 dollars yr \ufffd C1 16000 dollars yr \ufffd t2 2000\ufffd t1 1970\ufffd b) The increase in price of gasoline over this period kept pace with the rate of inflation. C2 1.513 dollars gal C2 C1 1 i\ufffd( )t2 t1\ufffd\ufffd i 5%\ufffd C1 0.35 dollars gal \ufffd t2 2000\ufffd t1 1970\ufffd a) 1.25 Tnb 273.15K\ufffd 56.004degC Ans. Tnb 329.154K Tnb B A ln Psat kPa §¨ © · ¹\ufffd C\ufffd§¨ ©¨ · ¹ K\ufffd Psat 1atm\ufffd 8 t1 20 degC\ufffd CP 4.18 kJ kg degC\ufffd MH2O 30 kg\ufffd t2 t1 'Utotal MH2O CP \ufffd\ufffd t2 20.014degC Ans. (d) For the restoration process, the change in internal energy is equal but of opposite sign to that of the initial process. Thus Q 'Utotal\ufffd\ufffd Q 1.715\ufffd kJ Ans.Ans. (e) In all cases the total internal energy change of the universe is zero. 2.2 Similar to Pb. 2.1 with mass of water = 30 kg. Answers are: (a) W = 1.715 kJ (b) Internal energy change of the water = 1.429 kJ (c) Final temp. = 20.014