(Solution Manual) Introduction to Chemical Engineering Thermodynamics

Smith,Van Ness, Abbott, 6

Engineering

Solution Manual

6th Edition

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Solution Manual for Introduction to Chemical Engineering Thermodynamics, The downloadable file consists of 1242 pages, and the total size of the file is 9.71 MB. Purchase now only for highly discounted price.

Author(s): Smith, Van Ness, Abbott

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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 Given t = 1.8t + 32 Find(t) = −40 Ans. 1.5 By definition: P = F A F = mass g Note: Pressures are in gauge pressure. P = 3000bar D = 4mm A =  D2 4 F = PA g = 9.807 m 2 s mass = F g Ans. 1.6 By definition: P = F A F = mass g P = 3000atm D = 0.17in A =  D2 4 F = PA g = 32.174 ft 2 sec mass = F g Ans. 1.7 Pabs = gh + Patm Patm = 101.78kPa Pabs = gh + Patm Pabs = 176.808 kPa Ans. 1.8 A = 0.023 in  = 13.535  gm cm g = 32.243  ft h = 25.62in mass = 1000.7 lbm mass = 384.4 kg A = 12.566 mm2  = 13.535  gm cm h = 56.38cm Patm = 29.86in_Hg Pabs = gh + Patm Pabs = 27.22 psia Ans. 1 1.10 Assume the following: P = 400bar h = P g h = 302.3 m 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 F = mass g On Mars: x = 0.40cm g= FMars Mars mass mass = 0.40kg F = 3.924 N FMars = Kx Ks = F x Ans. x = 1.08cm 1.12 Given: d P = −g dz and:  = MP RT Substituting: d P = − MP g dz RT Separating variables and integrating: PDenver   Psea zDenver 1 dP =  P  0 − Mg dz  RT   PDenver  After integrating: ln = −Mg zDenver  Psea  RT Taking the exponential of both sides  − Mg z  and rearranging: PDenver = Pseae RT Denver gMars = 0.01 m K kg  = 13.5 gm 3 cm g = 9.8 m 2 g = 9.81 m 2 FMars = 4  10 m K − 3 Ks = 363.333 N m Psea = 1atm g = 9.8 m 2 s 2 M = 29 gm mol zDenver = 1mi  − Mg  z  PDenver = Pseae RT Denver PDenver = 0.823 atm Ans. PDenver = 0.834 bar Ans. 1.13 The same proportionality applies as in Pb. 1.11. learth = lmoon  gearth gmoon M = learthlbm M = 113.498 lbm Ans. wmoon = Mgmoon wmoon = 18.767 lbf Ans. 1.14 costbulb = 5.00dollars 10 1000hr hr day costelec = 0.1 dollars 10 kWhr hr day 70W costtotal = costbulb + costelec 1.15 D = 1.25ft mass = 250lb m Ans. costbulb = 18.262 dollars yr costelec = 25.567 dollars yr costtotal = 43.829 dollars yr R = 82.06 cm atm molK 3 T = (10 + 273.15)K learth = 113.498 Mg zDenver = 0.194 RT gearth = 32.186  ft 2 gmoon = 5.32 ft 2 lmoon = 18.76 3 g = 32.169 ft 2 10 N u Patm = 30.12in_Hg (a) F = PatmA + mass g A =  D2 4 Ans. (b) Pabs = F A Ans. (c) l = 1.7ft Work = Fl PE = mass g l PE = 424.9 ftlbf Ans. Ans. 1.16 D = 0.47m mass = 150kg Patm = 101.57kPa A =  D2 4 (a) F = PatmA + mass g F = 1.909  4 Ans. (b) Pabs = F A Ans. (c) l = 0.83m Work = Fl EP = mass g l Work = 15.848 kJ EP = 1.222 kJ Ans. Ans. 1.18 mass = 1250kg EK = 1 mass  2 2 Ans. Work = EK Work = 1000 kJ Ans. 1.19 Wdot = mass g time h 0.910.92 A = 1.227 ft2 F = 2.8642  10 lbf 3 Work = 4.8691  10 ftlbf 3 2 A = 0.173 m g = 9.813 m 2 EK = 1000 kJ u = 40 m s Pabs = 110.054 kPa Pabs = 16.208 psia Wdot = 200W 4 g = 9.8 m 2 s h = 50m 0.95 GJ 14.28 GJ 27.778 GJ 1.22 a) mdot = Wdot gh0.910.92 cost_coal = − 1 Ans. cost_gasoline = − 1 cost_electricity = − 1 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 mdot = 0.488 kg 25.00 cost_coal = ton 29 MJ kg 2.00 cost_gasoline = gal 37 GJ 3 m cost_electricity = 0.1000 kWhr 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. A− B  Function being fit: f(T  A  B  C) = e T+C First derivative of the function with respect to parameter A d f(T  A  B  C) → expA − B  dA  T + C  First derivative of the function with respect to parameter B d f(T  A  B  C) → −1 expA − B  dB T + C  T + C  First derivative of the function with respect to parameter C d f(T  A  B  C) → B expA − B  dC (T + C)2  T + C  −18.5   3.18   −9.5   5.48   0.2   9.45       11.8   23.1   16.9   28.2  t =   Psat =    32.7   44.4   41.9   66.6       52.1   63.3    75.5   89.5   129    187  6 7   T = t + 273.15 lnPsat = ln(Psat) Array of functions used by Mathcad. In this case, a0 = A, a1 = B and a2 = C.    exp a0 − a1   T + a  Guess values of parameters      exp a0   2   − a1   T + a2    15  F(T  a) = guess = 3000   −1  T + a  exp a0 − a1   T + a   −50   2  2    a1  − a1   exp a0 2 T + a   ()T + a2  Apply the genfit function A  2   B  = genfit(T  Psat  guess  F) C  Compare fit with data. Ans. 200 150 Psat f( T  A  B C) 100 50 0 240 260 280 300 320 340 360 T 13.421 B  =  2.29  10 C 3  −69.053 8 To find the normal boiling point, find the value of T for which Psat = 1 atm. 9  Psat = 1atm Tnb =  B  Psat  − CK  A − ln   kPa   1.25 a) t1 = 1970 C2 = C1(1 + i) Tnb − 273.15K = 56.004 degC t2 = 2000 t2−t1 Ans. The increase in price of gasoline over this period kept pace with the rate of inflation. b) t1 = 1970 Given t2 = 2000 C2 = (1 + i)t2−t1 C1 i = Find(i) i = 5.511 % The salary of a Ph. D. engineer over this period increased at a rate of 5.5%, slightly higher than the rate of inflation. c) 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. C2 = 80000 dollars yr C1 = 16000 dollars yr Tnb = 329.154 K C1 = 0.35 dollars gal C2 = 1.513 dollars gal i = 5% 10 Chapter 2 - Section A - Mathcad Solutions 2.1 (a) Mwt = 35kg Work = Mwtg z (b) Utotal = Work z = 5m Work = 1.715 kJ Utotal = 1.715 kJ Ans. Ans. (c) By Eqs. (2.14) and (2.21): dU + d(PV) = CPdT Since P is constant, this can be written: MH2OCPdT = MH2OdU + MH2OPdV Take Cp and V constant and integrate: MH2OCP()t2 − t1 = kJ Utotal t1 = 20degC Utotal CP = 4.18 kgdegC MH2O = 30kg t2 = t1 + MH2OCP t2 = 20.014 degC 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 Q = −1.715 kJ 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 g = 9.8 m 2 11 (c) Final temp. = 20.014 deg C (d) Q = -1.715 kJ 2.4 The electric power supplied to the motor must equal the work done by the motor plus the heat generated by the motor. i = 9.7amp Wdot elect = iE E = 110V Wdot mech = 1.25hp Qdot = Wdot elect − Wdot mech 2.5 Eq. (2.3): Ut = Q + W Step 1 to 2: Ut12 = −200J Q12 = Ut12 − W12 Qdot = 134.875 W W12 = −6000J Ans. Ans. Step 3 to 4: Q34 = −800J Ut34 = Q34 + W34 W34 = 300J Ut34 = −500 J Ans. Step 1 to 2 to 3 to 4 to 1: Since Ut is a state function, Ut for a series of steps that leads back to the initial state must be zero. Therefore, the sum of the Ut values for all of the steps must sum to zero. Ut41 = 4700J Ut23 = −Ut12 − Ut34 − Ut41 Ut23 = −4000 J Ans. Step 2 to 3: Ut23 = −4  103 J W23 = Ut23 − Q23 Q23 = −3800J W23 = −200 J Ans. For a series of steps, the total work done is the sum of the work done for each step. W12341 = −1400J Q12 = 5.8  10 J 3 Wdot elect = 1.067  10 W 3 10 W41 = W12341 − W12 − W23 − W34 Step 4 to 1: Ut41 = 4700J Q41 = Ut41 − W41 Note: Q12341 = −W12341 Q41 = 200 J Ans. Ans. 2.11 The enthalpy change of the water = work done. M = 20kg MCP t Wdot = 0.25kW  =  = 0.929 hr Wdot Ans. 2.12 Q = 7.5kJ U = −12kJ U = −12kJ Q = U W = U − Q W = −19.5 kJ Q = −12 kJ Ans. Ans. 2.13 Subscripts: c, casting; w, water; t, tank. Then mcUc + mwUw + mtUt = 0 Let C represent specific heat, C = CP = CV Then by Eq. (2.18) mcCc tc + mwCw tw + mtCt tt = 0 mc = 2kg mw = 40kg mt = 5kg Cw = 4.18 kJ kgdegC tc = 500degC t1 = 25degC t2 = 30degC (guess) Given −mcCc()t2 − tc = ()mwCw + mtCt ()t2 − t1 Cc = 0.50 kJ kgdegC Ct = 0.5 kJ kgdegC W41 = 4.5  10 J 3 W41 = 4.5  10 J 3 CP = 4.18 kJ kgdegC t = 10degC t2 = Find()t2 t2 = 27.78 degC Ans. 11 U = 2022.4 kJ kg 10 kW 2.15 mass = 1kg (a) T = 1K Ut = mass CV T Ut = 4.18 kJ Ans. (b) g = 9.8 m 2 s EP = Ut z = EP mass g z = 426.531 m Ans. (c) EK = Ut u = Ans. 2.17 z = 50m D = 2m mdot = uA A =  D2 4 2.18 (a) (b) Wdot = mdotg z H1 = U1 + P1V1 Wdot = 7.697  3 Ans. Ans. H2 = U2 + P2V2 Ans. U = U2 − U1 H = H2 − H1 Ans. A = 3.142 m 2 mdot = 1.571  104 kg H1 = 763.131 kJ kg U2 = 2784.4  kJ kg V2 = 169.7  cm 3 gm H = 2275.8 kJ kg CV = 4.18 kJ kgK EK 1 mass 2 u = 91.433 m s  = 1000 kg 3 m u = 5 m s P2 = 1500 kPa P1 = 1002.7 kPa U1 = 762.0  kJ kg V1 = 1.128  cm 3 gm 12 mdot 2 = 0.8 kg 1 2 2.22 D1 = 2.5cm (a) For an incompressible fluid, =constant. By a mass balance, mdot = constant = u1A1 = u2A2  D1 2 u2 = u1  D2  Ans. (b) E = u 2 − 1 u12 Ans. K 2 2 2 2.23 Energy balance: mdot 3H3 − ()mdot 1H1 + mdot 2H2 Mass balance: mdot 3 − mdot 1 − mdot 2 = 0 = Qdot Therefore: mdot 1()H3 − H1 + mdot 2()H3 − H2 = Qdot or mdot Cp()T3 − T1 + mdot 2CP()T3 − T2 = Qdot T3CP()mdot 1 + mdot 2 = Qdot + mdot 1CPT1 + mdot 2CPT2 T1 = 25degC T2 = 75degC Qdot + mdot 1CPT1 + mdot 2CPT2 T3 = T3 = 43.235 degC Ans. ()mdot 1 + mdot 2 CP 2.25 By Eq. (2.32a): H + u = 0 H = CPT By continuity, u = u 2  A1 incompressibility 2 1 A2 mdot 1 = 1.0 kg u1 = 2 m u2 = 0.5 m kg E = −1.875 J D2 = 5cm C = 4.18 kJ P kgdegC Qdot = −30 kJ s CP = 4.18 kJ kgK 13