Unformatted text preview: Solutions to Chemical and Engineering Thermodynamics, 3e 4
4.1 Using the Mollier diagram = (510  490) C F T I = F T I = H P K H P K c1.241 10  7.929 10 hPa
H H 7 6 = 4.463 10 6 C Pa = 4.463 C MPa S =
S S 7 (510  490) C F T I F T I = H P K H P K c1.069 10  9.515 10 hPa
6 aH S f aH Sf
4.2 = 1.702 10 5 C Pa = 17.02 C MPa
T P a f a f a f a f a f a f a f a f a T , H f a P , S f = = = 0.262 ( unitless) a P , H f a T , S f = H, T S,P H, T S, P = S ,T H, P H, P S ,T
S V (a) Start from eqn. 4.427 H ( T , P)  H IG (T , P ) = RT ( Z  1) + RT a( T )  2 ; V  b V + 2bV  b 2 V = z LMN FH
T dP dT I K  P dV
V OP Q P= F P I H T K V = R da dt  2 so V  b V + 2 bV  b 2 H ( T , P)  H IG (T , P )
V = RT ( Z  1) + V = = RT ( Z  1) + a + T F H z LMN RST
T R da dt RT a( T )  2  + 2 dV 2 V  b V + 2bV  b V  b V + 2bV  b 2 da dT I K V V = V z U V W OP Q dV
2 + 2bV  b2 From integral tables we have z dx a x2 + b x + c = 1 b 2  4 a c ln 2a x + b  b 2  4 a c 2a x + b + b 2  4 a c for 4ac  b 2 < 0 In our case a = 1, b = 2b , c = b2 ; so 4 a c  b 2 = 4 1 b 2  ( 2b )2 =  8b2 c h and (b)2  4a c = 8b2 = 2 2b . Solutions to Chemical and Engineering Thermodynamics, 3e H ( T , P)  H IG (T , P ) = RT ( Z  1) + aa  Tda dTf LMln 2V + 2b  2 2b 2 2b MN 2V + 2b + 2 2b aa  Tda dTf ln V + d1  2 ib = RT ( Z  1) + 2 2b V + d1 + 2 ib
or finally H ( T , P)  H IG (T , P ) = RT ( Z  1) +  ln
V 2V + 2b  2 2b 2V + 2b + 2 2b V = OP PQ aTda dT  af lnLM Z + d1 + 2 iB OP 2 2b MN Z + d1  2 iB PQ (b) This part is similar except that we start from eqn. (4.428) S (T , P)  S IG ( T , P) = R ln Z + = R ln Z +
V V = V V = d d da dT F Z + d1 + 2 i B I = R ln( Z  B) + lnG J 2 2 b H Z + d1  2 i B K z LMNFH z LMN dP dT I K 
V R dv V OP Q R da dT R   dV V  b V 2 + 2 bV  b 2 V OP Q V V b da dT V + 1 + 2 b = R ln Z + R ln + ln V V = 2 2b V + 1  2 b i i V V = 4.7 (a) Ideal gas PV = NRT N = (50 bar ) 100 m 3 ( 27315 + 150) K 8.314 102 bar m3 kmol K . c h = 142.1 kmol Energy balance, closed nonflow system U = Q  PdV = Q + W . However, for ideal gas U = 0 since T is constant (isothermal). Thus W =  Q =  PdV =  z z z NRT V P dV =  NRT ln 2 =  NRT ln 1 V V1 P2 = 142 .1 kmol 8.314 J mol K (27315 + 150 ) ln . = 895.9 10 3 kJ = 895.9 MJ Q = 895.9 MJ Also, by Ideal Gas Law at fixed T and N F 50 I H 300 K Solutions to Chemical and Engineering Thermodynamics, 3e PV1 = PV2 V2 = V1 1 2 (b) Corresponding states Tr initial state final state P 50 1 = 100 m 3 = 16.67 m3 P2 300 Pr 50 = 0.679 7376 . 300 = 4.067 73.76 Z 0.94 0.765 150 + 27315 . = 1.391 304 .2 1.391 H IG  H TC 0.7 4.5 cal mol K S IG  S 0.4 2.4 cal mol K Number of moles of gas = N = Final volume = V f = = ZNRT Pf PV 142.1 PV = = 151.2 kmol ( 142 .1 = from above) ZRT 0.94 RT 0.765 151.2 8.314 102 (273.15 + 150 ) =1356 m3 . 300 Energy balance on gas: U = Q + W Entropy balance on gas processes in gas are reversible: S = S = Q or Q = T S T Q + S gen S gen = 0 . Therefore T S = S f  Si = N S f  S i = N S f  S f + S f  S i
IG IG Q = T S = ( 27315 + 150 ) 151.2 kmol ( 2326) = 14885 MJ . . . W = U  Q = N U f  U i  Q = N H f  H i  N Pf V f  PV i  Q i = N TC P R U  ( 0.4 4.184 )V S P T W R8.368  8.314 ln 300 U = 23.26N J K = NS V 50 W T = N 2.4 4.184  8.314 ln
f i c h mc h c IG h  cS  S hr
i IG i LM d H NM f Hf TC IG i+ c H f IG H IG i h TC dH  H i  Z RT
i IG i Since process is isothermal. 0 TC f f + Zi RTi  Q OP QP L304 .2(4.5  (0.7)) 4.184  8.314 OP + 14885 MJ = 151.2 kmolM N (273.15 + 150) (0.765  0.94) Q .
= 850.3 MJ = 151.2 103 48365 + 615.7 J + 14885 MJ =  638.2 + 14885 MJ . . . (c) PengRobinson E.O.S. Using the program PR1 with T = 27315 , P = 1 bar as the reference state, we obtain . T = 150 C , P = 50 bar Z = 0.9202 ; V = 06475 103 m3 mol ; H = 470248 J mol ; S = 17.57 J mol K . . . T = 150 C , P = 300 bar Solutions to Chemical and Engineering Thermodynamics, 3e Z = 0.7842 ; V = 0.9197 104 m3 mol ; H = 60.09 J mol ; S = 4124 J mol . .
N= V 100 m 3 = = 154 .44 kmol V 0.6475 10 3 m 3 mol Q = TN S = (273.15 + 150 ) 154 .44 ( 4124  ( 17.57)) = 1546.9 MJ . W = U  Q = ( H  PV ) f  ( H  PV )i  Q = N ( H  PV ) f  ( H  PV )i  Q
5 3 = 885.25 MJ {Note that N, Q and W are close to values obtained from corresponding states.} 4.8 LM60.09  300 0.9197 10 = 154.44 M10 J bar m  4702.48 MNM+50 0.6475 10 10
3 5 4 OP PP 10 + 1546.9 10 QP
3 6 FG T IJ H PK
and a f a S Pf aV Tf V T = a S Tf = C T = C
S T P P P P = (T , S ) (T , S ) ( P , T ) ( S , T ) ( P, T ) = = ( P, S ) ( P , T ) P, S ( S , P ) ( T , P) 1 V V dP S = T 1 V V dP a fa a fa f f S T = (V , S ) ( P, S ) (V , S ) ( P, T ) = (V , T ) ( P, T ) (V , T ) ( P, S ) ( S ,V ) (T , P ) S = = (T , V ) ( S , P ) T FG IJ FG T IJ H K HS K
V =
P CV T C = V T CP CP 4.9 (a) FG H IJ = ( H, T) = (H ,T ) (P ,T ) = FG H IJ FG P IJ H V K (V ,T ) (P ,T ) (V , T) H P K H V K F P IJ 0 (except at the critical point) Since G H V K
T T T T (b) FG H IJ = 0 H PK FG S IJ = (S, P ) = (S, P) (T , P) = FG S IJ FG T IJ H V K (V , P) (T, P ) (V , P) H T K H V K C 1 F dT I C TV C F S I = V G H dV JK = a1 V faV Tf = TV GH V JK T V
= 0 if
T T P P P P P P P P FG H IJ H V K ~ 1
P 4.10 (a) We start by using the method of Jacobians to reduce the derivatives Solutions to Chemical and Engineering Thermodynamics, 3e FG T IJ H V K =
H (T , H ) (T , H ) (T , P ) T ,V = (V , H ) (T , P ) (T ,V ) (V , H ) ( H , T ) ( P, T ) (V , T ) H = ( P, T ) ( H ,V ) (T ,V ) P a f = = a H V f ad H Tf FG V IJ H T K FG IJ a P V f H K a H T f
T T V T V Now from Table 4.1 we have that FG H IJ H PK =V T
T and
P FG H IJ H T K = CP + V  T
V LM N FG V IJ H T K P OP FG P IJ QH T K V alternatively, since H = U + PV FG H IJ = FG U IJ + FG ( PV )IJ H T K H T K H T K
V V = CV + V
V F dP I H dT K
T V Thus a f F P IJ FG V IJ =  FG P IJ Note: I have used G H V K H T K H T K
H T P FG T IJ H V K =  P V T V  T V T CV + V P T V a f a f P =  V P V CV a f + Ta P Tf + V a P T f
V V .
V FG T IJ H V K =
S = FG S IJ FG T IJ H V K H S K
T (T , S ) ( T , S ) (V , T ) ( S , T ) (T ,V ) = = (V , S ) (V , T ) (V , S ) (V , T ) ( S ,V ) =
V T P CV T FG IJ H K
= V (b) For the van der Waals fluid FG P IJ H T K
Thus = V R P , V b V FG IJ H K T  RT 2a + (V  b )2 V 3 FG T IJ H V K =
H 2 2   RTV (V  b ) + 2 a V + RT (V  b ) n CV + V R V  b s after simplification we obtain FG T IJ H V K =
H  2 a(V  b )  RTV b
2 2 CC (V  b )2 V 2 + R(V  b )V 3 Solutions to Chemical and Engineering Thermodynamics, 3e and FG T IJ H V K =
S RT CV (V  b ) 4.11 There are a number of ways to solve this problem. The method I use is a little unusual, but the simplest that I know of. At the critical point all three roots of V are equal, and equal to V C . Mathematically this can be expressed as V  V C a f 3 = 0 which, on expansion, becomes (1) V 3  3V C V 2 + 3V 2 V  V 3 = 0 C C compare this with P= RT a RT a  =  2 V  b V (V + b ) + b (V  b) V  b V + 2 bV  b 2 which multiplying through by the denominators can be written as V +V
3 2 F b  RT I + F 3b H PK H
2 2  2bRT a RTb 2 ab + V + b3 +  =0 P P P P I FG K H IJ K (2) Comparing the coefficients of V in Eqns. (1) and (2) gives TC , P C V : b RTC =  3V C P C (3) V :  3b 2  2bRTC a 2 + = 3V C PC PC (4) V 0: b3 + From Eqn. (3) RTC 2 ab 3 b  = V C PC PC (5) PCb 3 PCVC 1= = 3 ZC RTC RTC For convenience, let y = 1  3ZC or ZC = or PCb = 1  3ZC RTC (6) 1 y . Then 3 PCb =y RTC From Eqn. (4) Solutions to Chemical and Engineering Thermodynamics, 3e 3 FG P b IJ  2FG P b IJ + aP H RT K H RT K aRT f
C 2 C C C C C 2 2 = 3 ZC 3 y 2 + 2 y + or expanding and rearranging aRT f
C aP C 2 = 3(1  y)2 9 a RT f
C aP C 2 = 1 10 y 2 + 4 y + 1 3 c h (7) Finally from eqn. (5) FG P b IJ + FG bP IJ  FG P b IJ FG aP IJ =  Z H RT K H RT K H RT K H aRT f K 1 1 (1  y) y + y  y c10 y + 4 y + 1h =  3 27
3 2 C C C C C C C C 2 3 2 2 3 C 3 or 64 y3 + 6y 2 + 12y  1 = 0
This equation has the solution y = 0077796074 . b = 0.077796074 a = 0.457235529 RTC (from Eqn. (6)) PC
C 2 (8) aRT f
PC (from Eqn. (7)) 1 y = 0.307401309 . 3 Note that we have equated a and b to TC and P only at the critical point. Therefore these functions C Also ZC = could have other values away from the critical point. However, as we have equated functions of V , we have assumed a and b would only be functions of T. Therefore, to be completely general we could have aRT f FG T IJ HT K P RT F T I b = 0.077796074 G J HT K P T T FTI F TI with G J 1 as 1 and G J 1 as 1. HT K T HT K T
a = 0.457235529
C 2 C C C C C C C C C In fact, Peng and Robinson (and others) have set = 1 at all temperatures and adjusted a s a function of temperature to give the correct vapor pressure (see chapter 5). 4.12 (also available as a Mathcad worksheet) Solutions to Chemical and Engineering Thermodynamics, 3e . N1 N2 . M.B. . E.B. Q dN & & & & = N1 + N2 = 0 N 2 =  N1 dt & dU Q & & & = N1 H 1 + N 2 H 2 + Q = 0 = H 2  H1 dt N1 Also, now using the program PR1 with T = 27315 , P = 1 bar reference state we obtain . T = 100 C P = 30 bar Z = 0.9032 V = 09340 103 m3 mol .
H = 3609.72 J mol S = 1584 J mol K . T = 150 C P = 20 bar Z = 0.9583 V = 01686 102 m3 mol .
H = 679606 J mol . S = 4.68 J mol & Q . . & = 679606  3609.72 = 318634 J mol N 4.13 (also available as a Mathcad worksheet) Since process is adiabatic and reversible S = 0 or Si = S f , i.e., S (310 K, 14 bar ) = S (T = ?, 345 bar) . Using the program PR1 with the T = 27315 K and P = 1 . bar reference state we obtain T = 310 K , P = 14 bar , Z = 0.9733 , V = 01792 102 m3 mol , . H = 10908.3 J mol and S = 1575 J mol K . . By trial and error (knowing P and S , guessing T) we obtain T = 34191 K , P = 345 bar , . Z = 0.9717 , V = 08007 104 m3 mol , . Tf = 34191 K . .
System = contents of compressor dN & & & & M.B.: = 0 = N1 + N 2 N 2 =  N1 dt H = 188609 J mol , . S = 1575 J mol K . volume of compressor constant 0 dU dV & & & = 0 = N1 H 1 + N 2 H 2 + Q& 0 +Ws  P E.B.: dt dt & W & & & WS =  N1 H1 + N 2 H 2 or &S = H 2  H 1 = 188609  109083 = 7952.6 J mol . . N adiabatic 4.14 (a) FG P + a IJ (V  b) = RT PV = V  a H VK RT V  b RTV PV F V  a IJ = 1 i) lim = lim G RT H V  b RTV K PV RV a U I ii) B = lim V F H RT  1K = lim V SV  b  RTV  1V T W RV  (V  b)  a U = lim R bV  a U = b  a = lim V S T (V  b) RTV V S (V  b) RT V RT W T W
2 P 0 V V P 0 V V V V Solutions to Chemical and Engineering Thermodynamics, 3e iii) C = lim V 2
P 0 V FG PV  1  B IJ = lim V RbV  b(V  b) U = lim b V = b S V b V V b H RT V K T W
2 V V P 0 2 C = b2 (b) At the Boyle temperature: lim V F PV  1I = 0 B = 0 H RT K a a 9V c RTc V 0 =b , TB = but a = , b = c (Eqns. 4.63a) RTB Rb 8 3 TB = 9 8V c RTc 27 = Tc = 3.375Tc RVc 3 8 4.20 Mass balance (system = both tanks): N1i = N1f + N 2f
i energy balance (system = both tanks): N1i U 1 = N1f U 1f + N2f U 2f entropy balance (system = portion of initial contents of tank 1, also in there finally): S i1 = S 1f Also, P1 f = P2 f = P f ; N1i = V1
i V1 ; N1f = = Pf 1 T1 f V1
f V1 and N2f = V2 V2
f (a) Ideal gas solution: obtain Pi 1 T1i + P2f T2 f from mass balance and Pi = P f + P2 f = 2 P f P f = 250 bar = 25 107 Pa from energy balance . 1 1 Solutions to Chemical and Engineering Thermodynamics, 3e T1 =
f T1i = 23.9 C from entropy balance and 1 T2 Also N1f P f V RT i 2.5 10 7 293.15 = 1 f1 i 1 = = 0.588 i N1 RT1 P V1 249 .3 5.00 107 1 and N2f N1f = 1 i N1 N1i (b) Corresponding States Solution: Initial conditions Tr = 29315 . = 1538 ; . 190 .7 P = r 5 107 = 10.77 ; 4.64 106 Z = 1.22 ;
f FG P IJ HPK
f 1 i 1 R CP T1 f = (20 + 273.15) FH 1IK 2 8.314 35.565 = 249 .3 K = 2 T1i  1 T1
f T2 f = 355.9 K = 82.7 C FG IJ = 0.412 H K H IG  H = 18.0 J mol K ; SIG  S = 96 J mol K . . TC Mass balance: Pi 1 1 1 = Pf + f f i i f f Z1 T1 Z1 T1 Z2 T2 Entropy balance:
i IG S 1f  S 1 = 0 = S 1  S 1 R S T U = 5.0 10 = 1.398 10 V 122 293.15 W .
7 IG, f 1 5 (1) d i + dS f  S IG, i  S 1  S IG 1 1 i d i i or dS  S i
1 IG f 1 + CP ln T1 f Pf  R ln = 9 .6 29315 5.0 10 7 . (2) Energy balance:
i i N1i U 1 = N1f U 1f + N2f U 2f but N1i = N1f + N 2f N1f U 1f  U 1 + N2f U 2f  U i2 = 0 d i d i or Solutions to Chemical and Engineering Thermodynamics, 3e V ndH  P V i  dH  P V is + ZP RT ndH  P V i  dH  P V is = 0 1 ndH  H i + dH  H i  dH  H i  Z RT + Z RT s Z T 1 + ndH  H i + d H  H i  d H  H i  Z RT + Z RT s = 0 Z T
f 1 f 1 f 1 i 1 i 1 i 1 2 f 2 f f 2 f 2 f 2 i 1 i 1 i 1 2 f f 1 1 f 1 f , IG 1 f , IG 1 i , IG 1 i 1 i , IG 1 f 1 f 1 i 1 i 1 f f 2 2 f 2 f , IG 2 f , IG 2 i , IG 1 i 1 i , IG 1 f 2 f 2 i 1 i 1 P f V1 1 Z1f RT1 f f 2 Substituting in the known values gives . . ndH  H i + 35565cT  29315h  8.314 Z T + 6,406.5s 1 + . . ndH  H i + 35565cT  29315h  8.314 Z T + 6,406 .5s = 0 (3) Z T
f 1 Z1f T1 f f 1 f , IG 1 1 f f 1 1 f f 2 2 f 2 f , IG 2 f 2 f f 2 2 Eqns. (13) now must be solved. One possible procedure is i) Guess P f ii) Use Eqn. (2) to find T f 1 iii) Use Eqn. (1) to find T2 f iv) Use Eqn. (3), together with T f and T2 f to see if guessed P f is correct. If not, go back to 1 step i. After many iterations, I found the following solution
i i T2 f = 259.4 K ; N1f N1 = 0.645 ; N 2f N1 = 0.355 . P f = 9787 bar ; T1 f = 2216 K ; . . (c) PengRobinson equation of state Here we use the equations N1i = N1f + N 2f
i N1i U 1 (4) (5) (6) = N1f U 1f + N2f U 2f S i1 =S
f with U = H  PV P1 f = P2 f = P f
i f and N1i = V1 V 1 ; N1f = V1 V 1f ; N2f = V2 V 2f = V1 V 2 since V1 = V2 (value of V1 cancels out of problem, so any convenient value may be used). Procedure I used to solve problem was as follows. From PR1 we know V i1 N1i and S i1 given initial conditions. Then c h 1. 2. 3. i Guess value of T , find P1 = P that satisfies S 1f = S 1 1 f f f Use T f , P f and V 1f to get N1f ; then N 2f = N1i  N1f so V 2f is known. 1 From P f and V 2f find (trialanderror with PR1) T2 f 4. See if eqn. (5) energy balance is satisfied; if not go back to step 1. After a number of i iterations I find Pf =103.6 bar ; T1 f = 222.3 K ; T2 f = 2555 K ; N1f N1 = 0.619 ; .
i N2f N1 = 0381 . . Solutions to Chemical and Engineering Thermodynamics, 3e Summary ideal gas (part a) 250 bar 249.3 K 355.9 K 0.588 0.412 Corresponding states (part b) 97.87 221.6 K 259.4 K 0.645 0.355 PR E.O.S. (part c) 103.6 222.3 K 255.5 K 0.619 0.381 Pf T1 f T2 f
i N1f N1 N2f i N1 Clearly, the ideal gas assumption is seriously in error! ...
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 Spring '07
 Rethwisch

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