Production of
Chlorinated Polyvinyl
Chloride
1
5
3
2
7
4
6
6
7
8
12.4. 탄소, 수소, 염소의 질량분율
PVC
(C2H3Cl)n
67% PVC
(C9H11Cl7)n
24g C
3g H
35.45g Cl
(total 62.45g)
108g C
11 g H
248.15 g Cl
(total 367.5g)
0.384g C/g
0.048g H/g
0.568g H/g
0.294g C/g
0.030g H/g
0.676g Cl/g
12.5.
67% 염소 3.5x10^6 kg/yr 생성 ,
10wt%PVC 슬러리와 Cl2(g)를 흡수기로 보내야 할 공급량(kg/h)=?
염소처리 반응기속 염화수소 생산량(kg/h)=?
ṁ1(kg/h) slurry
0.1kg PVC/kg
0.90 kg H2O/kg
ṁ2(kg Cl2/h)
ṁ4(kg CPVC/h)
absorber
ṁ3(kg HCl/h)
ṁ4 =
C:
3.5×106 kg CPVC
yr
1 yr
1 day
300days
= 486.1 kg CPVC/h
24 h
0.10ṁ1(kg PVC) 0.384kg C
kg
=
486.1kg CPVC 0.294kg C
h
kg
⇒ ṁ1 = 3722kg/h slurry
Cl :
0.5 ṁ2 + 0.10 ṁ1(0.568) = 0.676 ṁ4
0.5 ṁ2 + 0.10(3722)(0.568) = (0.676)(486.1)
⇒ ṁ2 = 234.4kg Cl2/h
(Polymer)n-Cl + Cl2
ṁ3 =
234.4kg Cl2 fed
h
(Polymer)(n+1)-Cl + HCl
1kmol Cl2
1kmol HCl
36.45kg HCl
70.9kg Cl2
1kmol Cl2
1kmol HCl
⇒ ṁ3 = 120.5kg HCl/h
12.6. 염소기화기
T(℃)
-20.0
-10.0
0.0
10.0
20.0
30.0
40.0
p*(atm) 1.78
2.59
3.64
4.99
6.69
8.78
11.32 17.95 27.04 39.06
(a) 염소기화열
Clausius-Clapeyron Equation :
lnp* = -∆Ĥv/RT+B
p* VS 1/T
∆Ĥv/R=2426K and B=10.17
∆Ĥv/R=2426 ⇒ ∆Ĥv = 2426(K)8.314(J/mol •K)(kJ/1000J)
= 20.17kJ/mol
lnp*=-2426/T+10.17
p*Cl2=2.611×104 e-2426/T
60.0
80.0
100.0
(b) 운전압력=?
Cl2 vaporizer
T=5℃=278.2K
⇒ P=p*Cl2 = (4.262 atm)(760torr/1atm) = 3240 torr
(C) 염소 기화기 속 가해야 할 열(kW)
Q=
234.4kg Cl2
hr
1kmol
1000mol
70.9kg
1kmol
20.17kJ
1hr
mol
3600s
= 18.5 kW
(d) 백분율 차이
Table B.1 : 20.4kJ/mol
Calculated : 20.17kJ/mol
0.23/20.4×100=1.1%error
12.8~12.12 deal with the chlorine absorber
pcl2(atm)
pH2O(atm)
25℃, P(atm)
ṁ0 (kg/h)
0.10kg PVC(s)/kg
0.90kg H2O(1)/kg
Vapor
Slurry
ṁcl2[kg Cl2(g)/h]
ṁp(kg polymer/h)
xp(kg Cl2/kg)
(1-xp)(kg PVC/kg)
ṁa(kg liquid/h)
xa(kg Cl2/kg)
(1-xa)(kg H2O(l)/kg)
25℃, P(atm)
Head space
pcl2(atm), pH2O(atm)
25℃, P(atm)
Polymer phase(solid)
xp (kg Cl2/kg polymer)
(1-xp)(kg PVC/kg polymer)
25℃
Aqueous phase(liquid)
xa (kg Cl2/kg liquid)
(1-xa) (kg H2O(l)/kg liquid)
25℃
• Henry's law
xp(kg Cl2/kg PVC phase) = αppcl2 xa(kg Cl2/kg aqueous phase) = αapcl2
pcl2(atm)
0.75
1.18
1.23
2.95
3.03
3.87
4.82
Wt% Cl2
2.9%
4.6%
5.1%
11.9%
12.3%
15.4%
19.8%
12.8
αp
pcl2
αp = 0.0405 mass fraction/atm
αa = αp/2.68
=0.0151 mass fraction/atm
12.9.
xp = 0.0405 pcl2 (1)
xa = 0.0151 pcl2 (2)
pH2O=(1-xa) p*H2O(25℃) = 23.756/760(1-xa) from [Table B.3]
PVC : 0.1 ṁ0=372.2=ṁp(1-xp) (4)
H2O : 0.9 ṁ0=3350= ṁa(1-xa) (5)
Cl2 : ṁcl2=234.4= ṁpxp+ṁaxa (6)
P= pcl2+pH2O (7)
(2) → pcl2=66.225 xa (8)
(1)+(2) → xp=2.68 xa (9)
(5) → ṁa=3350/(1- xa) (10)
(4) → ṁp=372.2/(1-2.68 xa) (11)
(6) → ṁpxp+ṁaxa=f(xa)=234.4 →xa=0.0501
⇒ xp=0.1343, ṁa=3527kg/h, ṁp=430kg/h, pcl2=3.315atm
pH2O=0.030atm, P=3.34atm
pcl2(atm)=3.315
pH2O(atm)=0.030
25℃, P(atm)=3.34
ṁ0 (kg/h)=3722
0.10kg PVC(s)/kg
0.90kg H2O(1)/kg
Vapor
Slurry
ṁcl2[kg Cl2(g)/h] =234.4
ṁp(kg polymer/h) =430
xp(kg Cl2/kg)=0.1343
(1-xp)(kg PVC/kg)=0.8657
ṁa(kg liquid/h)=3527
xa(kg Cl2/kg)=0.0502
(1-xa)(kg H2O(l)/kg)=0.9498
25℃, P(atm)=3.34
12.10. 오류
Using the equation from 6(a),
p*Cl2=2.611×104 e-2426/T
T=25℃ = 298.2K 대입 → p*Cl2(25℃)=7.63atm
Using Raoult’s law : pcl2=xa·p*Cl2=(0.0501)(7.63) = 0.382atm
% error = (0.382-3.315)/3.315×100 = -88.5%
12.11.
탱크 : 슬러리 70%, 상부공간 30%
슬러리 체류시간 : 12분
슬러리부피, 상부공간부피, 탱크부피 =?
Vapor
sluurry
429.79(kg polymer/h)
0.134(kg Cl2/kg)
0.866 (kg PVC/kg)
3526.7(kg liquid/h)
0.0502 (kg Cl2/kg)
0.9498 (kg H2O/kg)
Specific volume of PVC = 0.709L/kg
Specific volume of liquid water = 1.00L/kg
Specific volume of dissolved chlorine (in either PVC or water) = 0.690L/(kg
dissolved)
=
430kg (1-0.1343)kg PVC
0.709L
hr
kg
kg
+ (430)(0.1343)(0.690)
PVC
Cl2
+ (3527)(1-0.0501)(1.00) + (3527)(0.0501)(0.690)
H2O
Cl2(l)
= 3776 L/hr = 3.776 ㎥/h
t = Vslurry /
→ Vslurry = (12min) (3.766㎥/hr) (60min) = 0.755 ㎥
Tank volume : 0.755 ㎥ / 0.70 = 1.08 ㎥
Head Space volume : 1.08 – 0.755 = 0.325 ㎥
12.12. 염소 총 질량
nhead space = PV/RT =
(3.34atm)(0.325㎥)(103L/㎥)
(0.08206L·atm/mol·K)(25+273.2K)
(mcl2)head space = 44.36·ycl2
= 44.36mol 3.315 mol Cl2 70.9g
3.34
mol
mol
= 44.36mol
kg
1000g
= 3.12kg Cl2 in head space
Basis :
430kg PVC phase
3527kg aqueous phase
3776㎥
(mcl2)actual = ṁcl2(Vslurry/V) = (234.5)(0.755/3.776) = 46.9kg Cl2 in slurry
(mcl2)total = (mcl2)head space + (mcl2)slurry = 3.12 + 46.9 = 50.0kg Cl2
12.14.
Too short – not enough time in the reactor to obtain the desired extent of
chlorination
Too long – too much chlorination (resin above 70% Cl is too hard to
process)
– paying for reactor you don’t need.
12.15. 액상PH
Sol)
3722kg/hr Slurry
0.1kg PVC/kg
0.90 kg H2O/kg
234.4 kg Cl2/h
486.1kg PVC/h
Absorber
and
Reactor
nHCl = 120.5kg HCl 1kmol
hr
36.45kg
Va = (3350+120.5)kg
hr
103mol
kmol
Va(L/h)
3350kg H2O/h
120.5kg HCl/h
= 3306 mol/hr
L
= 3453 L/hr
1.005kg
[HCl] = [H+] = 3306/3453 = 0.9573 → pH=-log(0.9573)
⇒ pH = 0.019, highly acidic
12.16. wet cake 질량유속, 폐수의 질량유속 및 질량
486.1kg CPVC/hr
Reactor
Centifuge
3470.5kg HCl(aq)/hr
ṁ1 kg wetcake/hr
0.90kg CPVC/kg
0.10 kg acid/kg
ṁ2 kg acid/hr
CPVC balance : 486.1 = 0.90ṁ1 ⇒ ṁ1 = 540.1kg wetcake/hr
Mass balance : 486.1 + 3470.5 = 540.1 + ṁ2
⇒ ṁ2 = 3416.5kg waste liquor/hr [3.47wt% HCl, 96.5wt% H2O]
CPVC balance: (540.1)(0.90) = ṁ4(0.15) ⇒ ṁ4=3240.6kg slurry/h
kmol HCl reacting: (540.1)(0.10)(0.0347)kg HCl
h
kmol = 0.05142kmol HCl/h
36.45kg
NaHCO3 balance: 0.10(ṁ1)= 0.0512kmol HCl kmol NaHCO3 84.0kg
h
kmol HCl
kmol NaHCO3
⇒ ṁ1=43.2kg 10% soln/h
CO2 balance: ṁ3= 0.05142kmol HCl kmolCO2 44.0 kg ⇒ ṁ3=2.26kg CO2/h
h
kmol HCl kmol CO2
NaCl balance: 3240.6(0.85)x= 0.05142kmol HCl kmol NaCl 58.45kg
h
kmol HCl
kmol
⇒ x= 0.0011kg NaCl/kg
H2O balance: 540.1(0.10)(0.965) + 43.2(0.90) + (0.05142kmol H2O/h)(18kg/kmol)
+ ṁ2 = 3240.6(0.85)(1-0.0011)
⇒ ṁ2 = 2660kg make up water/h
12.17.
540.1kg wetcake/h
0.90kg CPVC/kg
0.10 kg acid/kg
0.0347 HCl
0.965 H2O
ṁ1 (kg NaHCO3(aq)/h)
0.10 kg NaHCO3/kg
0.90 kg H2O/kg
ṁ3 (kg CO2(g)/h)
Resin
Neutralizer
ṁ4 (kg slurry/h)
0.15kg CPVC/kg
0.85kg solution/kg
x kg NaCl/kg
(1-x) kg H2O/kg
ṁ2 (kg H2O(l)/h)
HCl(aq) + NaHCO3(aq) → NaCl(aq) + CO2(g) + H2O(l)
12.18. wetcake를 왜 건조기로 직접 보내지 않는지?
중화제로 sodium 사용?
중화기에 물 추가?
sol)
- If the wetcake were sent directly to the dryer, the entrained HCl would
degrade the resin.
- NaHCO3 is a weak base and so it does not degrade CPVC.
- The makeup water lowers the solids content of the slurry enough to enable
easy pumping.
12.19. (a)
ṁ1(kg NaHCO3/h)
ṁ2(kg H2O/h)
NaHCO3
dissolve
Feed
drum
43.2kg/h
0.10 kgNaHCO3/kg
0.90 kgH2O/kg
ṁ1=4.32kg NaHCO3/h
ṁ2=38.88kg H2O/h
L
24h 2.5days = Vfd=2400L
Feed drum volume : Vsd= 43.2kg
h
1.08kg 1day
Volume of solution/batch :
2x50lbm NaHCO3(s) 0.4536kg kg sol
L
Vsoln=
lbm
0.1kg NaHCO3 1.08kg
Vsoln = 420L solution
Volume of dissolving tank : V= 420 = 525L
0.80
(b) 최소한의 용해탱크 부피?
Vmin = 43.2 kg soln
h
4h
L
1.08kg
1Liter Tank
0.80 liter soln
= 200L
Advantages of a larger tank include : - down time for maintance
- frees operator for other tasks
- provides extra capacity, if production
rate increases or if dissolver goes
down for a long time
(c)
When the agitator breaks down, the Feed drum contains :
43.2kg 48h
h
= 2074kg
Continued
After 32 hrs, drum contains 2074-43.2(32) = 692kg
Working at maximum capacity, the operators can produce :
420L soln 1.08kg
4h
L
= 113.4kg/h or 453.6kg/batch
After each 4hrs, the drum is replenished by : (453.6 - 4(43.2)) = 281kg
t 0
4
8
12
16
20
m 692 973 1254 1535 1816 2097 > 2074
If the minimum tank volume had been used, we would never be able to
Replenish the feed drum
(d)
Basis : 100kg solution
ΔTm=
RTm²
ΔHm
10kg NaHCO3
kmol
84.01kg
= 0.119 kmol
NaHCO3
90kg H2O Kmol
18kg = 5.00 kmol H2O
= (8.314 J/mol k)(273.2)²k²
6010 J/mol
X
X = 0.119/(5+0.119)=0.232 , ΔTm = 2.4k or 2.4℃
= (TH2O-T’)
Tfreeze = -2.4 ℃
X
12.20. 원심분리기
3240.6 kg/h
0.15CPVC
0.85soln
0.011 NaCl
0.999 H2O
ṁ1kg wetcake/h
0.90 CPVC
0.10 soln
Product
centrifuge 0.011 NaCl
0.999 H2O
ṁ2 kg soln
0.011 NaCl
0.999 H2O
CPVC balance : 3240.6(0.15) = 0.90ṁ1
Mass balance : 3240.6 = 540.1 + ṁ2
ṁ1= 540.1 kg/h wetcake
ṁ2= 2700.5 kg liquor/h
12.21. (a)
2700.5 kg/h
Product
Centrifuge 0.0011 NaCl
Reactor centrifuge
3416.5 kg/h
0.0347 HCl(aq)
0.965H2O
2.26 kgCO2/h
0.999 H2O
Waste
ṁ2 kg NaCl/h
Liquor
ṁ3 kg
Neutrallizer
Na2CO3/h
ṁ4 kg H2O/h
ṁ5 kg NaCl/h
ṁ3 kg
Na2CO3/h
ṁ6 kg H2O/h
Waste
Treatment
Resin neutralizer
ṁ1 kg/h
0.10 kg NaOH/kg
x1a kg Na2CO3/kg
(0.9-x1a-x1b) kgH2O/kg
NaOH
Feed
drum
rṁ5 kg NaCl/h
Rṁ3 kg Na2CO3/h
Rṁ6 kg H2O/h
NaOH
Makeup
tank
ṁ3 kg NaOH/h
(b)
Speculate on the reason that the HCl must be nuutralized.
Minimize the difficulty of treating waste.
(c)
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O
CO2 + 2NaOH
Na2CO3 + H2O
kmol NaOH 40kg NaOH
0.1ṁ1 = (3416.5)(0.0347)kg HCl kmol HCl
h
36.45kg HCl kmol HCl
kmol NaOH
+
2.26kg CO2 kmol CO2 2kmol NaOH 40kg NaOH
h
44kg CO2 kmol CO2
kmol NaOH
ṁ1 = 1342 kg soln/hr
ṁ2= 3416.5(0.0347)kg HCl
Kmol HCl Kmol NaCl 58.45kg NaCl
h
36.45kg HCl kmol HCl
kmol NaCl
+ 1342(x1a)
= (190.1+1342x1a) kg NaCl/h
ṁ3= 2.26kg CO2 kmol CO2 kmol Na2CO3 106kg NaCO3 + 1342x1b
h
44kg CO2
kmol CO2 kmol Na2CO3
= (5.444+1342x1b) kg NaCl/h
ṁ4 = 3416(0.965) + 1342(0.9-x1a-x1b)
= (4505 – 1342 x1b) kg H2O/h
ṁ5 = ṁ2 + 2700.5(0.0011)
ṁ5 = (193.1+1342 x1a) kg NaCl/h
ṁ6 = ṁ4 + 2700.5(0.999)
ṁ6= (7203-1342(x1a+x1b)) kg H2O/h
NaOH balance : ṁ7 = 1342(0.1) = 134.2 kg NaOH(s)/h
(1) NaCl balance : rṁ5 = 1342(x1a)
r(193.1+1342 x1a) = 1342 x1a
(2) Na2CO3 balance : rṁ3 = 1342(x1b)
r(5.444+1342 x1b) = 1342 x1b
(3) H2O balance : rṁ6 = 1342(0.9-x1a-x1b)
r(7203-1342(x1a+x1b) = (0.9-x1a-x1b)(1342)
(1) x1a = 193.1r/(1342(1-r))
(2) x1b = 5.444r/(1342(1-r))
(3) r[ 7203 – 193.1r/(1-r) – 5.44r/(1-r) ] = 1208 – 193.1r/(1-r) 5.44r/(1-r)
r=0.163, x1a=0.0281 x1b=0.00079
r=0.9998, x1a=doesn’t make sense
ṁ2 = (190.1+1342(0.0281)) = 227.8 kg NaCl/h
ṁ3 = (5.444+1342(0.00079)) = 6.504 kg Na2CO3/h
ṁ4 = (4505-1342(0.0281+0.00079)) = 4466.2 kg H2O/h
Total mass flow rate leaving neutralizer = ṁ2+ṁ3+ṁ4
= 4700.5 kg/h soln
0.048 kg NaCl/kg
0.0014 kg Na2CO3/kg
0.950 kg H2O/kg
From of combined salt solution sent to makeup tank
ṁ5= (193.1+1342x1a) = 230.8 kg NaCl/h
ṁ6= (7203-1342(x1a+x1b)) = 7164 kg H2O/h
ṁ3= 6.504 kg Na2CO3/h
Total = ṁ5+ṁ6+ṁ3 = 7401 kg soln/h
rṁ5= 0.163(230.8) = 37.62 kg NaCl/h to makeup tank
rṁ3= 0.163(6.504) = 1.06 kg Na2CO3/h to makeup tank
rṁ6= 0.163(7164) = 1167.7 kg H2O/h to makeup tank
Totoal to makeup tank = rṁ5+rṁ3+rṁ6 = 1206.4 kg soln/hr
1206.4/7401 = 0.163 : fraction of combined salt solution sent to the
makeup tank
(d) 필요한 고체량, tank 크기
i) 1 batch contains : 8(1342kg NaOH/h) = 1074kg NaOH/batch
ii) 8(0.163)[ṁ5+ṁ3+ṁ6] + 1074
= 8(0.163)[ 193.1+1342(0.0281) + 5.444+1342(0.00079)
+ 7203-1342(0.0281-0.00079) ] + 1074
= 10730 kg
batch
= 16,000 liters
L
1
1.11kg 0.60
(e) 처리해야 할 NaCl, Na2CO3 / yr
ṁ5= 230.8kg NaCl/h , ṁ3= 6.504kg Na2CO3/h
230.8kg NaCl (1-0.613) 1metric ton 24h
h
1000kg
day
300days
= 1390 metric tons
yr
NaCl/yr
6.504kg Na2CO3 (1-0.163) 1metric ton 24h 300days
h
1000kg
day
yr
= 39 metric tons Na2CO3/yr
12.22. 기화기
(a)
234.4kg Cl2(l)/h
Cl2
vaporizer
T=22℃
234.4kg Cl2(v)/h
T=5 ℃
P=2bar
Path : Cl2(l)
T=22℃
Cl2(v)
T=22 ℃
234.4kg Cl2(l) 290kJ +
hr
kg
So, Ó= ΔH= 66063kJ/h
Cl2(v)
T=5℃
: use Q=ΔH
234.2kg Cl2(v) 0.48kJ (5-22)℃
hr
kg ℃
ṁs kg H2O(v)/hr
Steam coil
ṁs H2O(l)/hr
Satd, P=2bar
Use Q= ΔH
Ó= ΔH= -66063kJ/hr= (ṁs kg/hr)(-2201.6kJ/kg)
ṁs= 30kg/hr steam
(b) 단열공정, 출구의 염산증기 온도?
1kg Cl2(l)
T=22℃
Cl2
vaporizer
1kg Cl2(v)
Tad(℃)
ΔH=0 = m(ΔHv)22℃ = mCp(22-Tad)
ΔHv/Cp= 22- Tad
Tad= -582℃
Tad= 22 - 290kJ/kg
0.48kJ/kg ℃
12.23.
234.4kg Cl2(v)/h
T=5℃
T=3℃
Heat
Exchanger
Cp(Cl2) = 0.48 kJ/kg ℃
Ó hx= (234.4kg/h)(0.48kJ/kg ℃)(25-3)℃
= 2475kJ/h
234.4kg Cl2(v)/h
T=25 ℃
12.24.
Slurry from absorber
T=25℃
Slurry Heat
Exchanger
T=50℃
430kg polymer/h
0.1343 Cl2
0.8657 PVC
3527 kg liq/h
0.0501 Cl2
0.9499 H2O(l)
Ó = (∑miCpi)(50-25)℃
= 25[ 430(0.1343)(0.96) + 430(0.8657)(1.2) + 3527(0.0501)(0.96)
+ 3527(0.95)(4.2) ]
= 368,612kJ/h
12.25. 염소 반응기
ṁw kg H2O(l)/h
T=45 ℃
T=50 ℃
430kg polymer/h
0.1343 Cl2
0.8657 PVC
3527 kg liquid/h
0.0501 Cl2
0.9499 PVC
T=65 ℃
ṁw kg H2O(l)/h
T=15 ℃
486.1 kg CPVC/h
3470 kg HCl(l)/h
120.5 kg HCl/h
3350 kg H2O/h
(a)
(b)
∆H = Q
(cooling water)
+ Q (heat of mixture) + Q
Q (escape to outside)
(heat of reactor wall)
+
Neglect heat losses to outside + reactor wall
Refs : [PVC, Cl2(diss.), H2O(l), CPVC, HCl] at T=50℃
Qcw = ∑outmiĤi - ∑InmiĤi + mCl2 react • ∆Ĥr (kJ/kg Cl2react)
= [(486.1)(1.9)+(3470)(4.0)](65-50) + [(430)(0.1343)+3527(0.0501)] (-1770)
Qcw = -193000 kJ/h
Qcw =ṁw Cp(45-15) ⇒ ṁcw=
H2O/h
193000 kJ/h
(4.2kJ/kg℃)(30℃)
= 1532kg cooling
(c) 냉각수 차단, 생성물 온도?
Qcw = 0 ⇒ [(486.1)(1.9) + (3470)(4.0)] (Tad -50) – 414980 = 0
Tad = 78℃
Problems with adiabatic operation:
CPVC degrades.
Excessive vaporization, pressure buildup.
Unreacted polymer in core of resin pellets.
12.26 NaOH makeup tank
(a) q = -2.5×105 kJ/h
Ws=(20kJ/s)(0.65)(0.80)(60s/min)(60min/h) = 3.74×104 kJ/h
(b) 용해가 완료될때의 온도?
Batch contains 1074kg NaOH
1074kg NaOH
kmol
39.98kg
9342kg H2O
kmol
18kg
= 26.8NaOH
n=19kmol H2O/kmol NaOH
= 519kmol H2O
Using Table 8.5-1 with n=19 → ∆ĤNaOH(s) = -42.8kJ/mol NaOH
Energy balance : qt+wst = NNaOH(s) ∆Ĥ + MsolnCp(T-25)
(-2.5×105+3.74×104)kJ/h·4h=(26.8kmol NaOH)(1000mol/kmol)
(-42.8kJ/mol NaOH)+(10730kg soln)(3.8kJ/kg soln℃)(T-25℃)
Solve for T → 32.3℃
(c)
qt = MsolnCp(25-32.3)
→ t = (10730)(3.8)(-7.3)/(-2.5 ×105) ⇒ t=1.2h
ttotal = 4+1.2=5.2h
12.27.
(a)
- Tilt dryer, so solids move axially along cylinder length by gravity
- Heat required to vaporize water transferred from air, causing Q and
H to drop
- Preheater raises temperature of air fed to dryer
heat transfer to wetcake
evaporation of water
(b) Dryer
9n3
y3(kmol H2O/kmol)
(1-y3)(kmol DA/kmol)
T=110, P=1atm
ṅ1(kmol/h)
y1(kmolH2O/kmol)
(1-y1)kmol dryair/kmol
T=27℃, P=1atm
60% rh
10n3
y3, n=y3
T=110℃
ṅ3(kmol air/h)
Y3(kmol H2O/kmol
(1-y3)(kmol DA/kmol)
T=110℃, P=1.3atm
10% rh
ṅ2(kmol/h)
Y2(kmol H2O/kmol)
(1-y2)(kmol DA/kmol)
T2(℃)
Ó(kJ/h)
ṁp(kmol/h) c 120℃
0.001 kg H2O/kg
Xs kg Nacl/kg
(0.999-xs) kg CPVC/kg
0.1 kg W.cake/h
0.90 CPVC
0.10 soln
0.0011NaCl
0.999 H2O
T=80℃, P=1atm
(c)
10 unknowns ( ṅ1, y1, Ó, ṅ2, y2, T2, ṅ3, y3, ṁp, xs)
Z relative humidities : y1, y3
Overall CPVC balance
Overall NaCl balannce
ṁp, xs
Overall dry air balance
Overall H2O balance
ṅ1, ṅ3
Mixing point mal balance : ṅ2
Mixing point H2O balance : y2
Energy balance on dryer :T2
Energy balance on mixing preheater : Ó
(d)
y1= 0.60 P*w(27℃)/P = (0.60)(26.739)/(760) = 0.0211
y3= 0.20 P*w(110℃)/P = (0.20)(1075)/(1.3)(760) = 0.218
Overall CPVC balance : (540.1)(0.90) = ṁp(0.999-xs)
Overall NaCl balance : (540.1)(0.0011)(0.1) = ṁp(xs)
ṁp= 486.6 kg/h dried resin
xs= 0.00012 kg NaCl/kg
Overall dry air balance : ṅ1(1-0.0211) = ṅ3(1-0.218)
Overall H2O balance : ṅ1(0.0211)+(540.1)(0.10)(0.999)/18kg/kmol
= ṅ3(0.218) + 486.6(0.001)/18kg.kmol
ṅ1= 11.8 kmol/h fresh air
ṅ3= 14.8 kmol/h air to exhaust vent
Mixing point mol balance : ṅ1 + 9ṅ3 = ṅ2 = 145.0 kmol/h
Mixing point H2O balance : (11.8)(0.0211)+9(14.8)(0.218) = 145.0(y2)
y2= 0.202 kmol H2O/kmol in air senting dryer
Refs: CPVC(s, 80℃), H2O(l, 0℃), DA(110℃)
Subst.
ṅin
Ĥ in
ṅout
Ĥout
CPVC
486.1kg/h
0
486.1kg/h
1.88(40)
DA
115.7kmol/h
29.2(T2-110)
115.7kmol/h
0
H2O(l)
540.0/h
0.49kg/h
504
H2O(v)
29.29kmol/h
336
45028+34T2
32.26kmol/h
45028+34(110)
ΔH=0= ∑out mi Ĥi - ∑in mi Ĥi
= (486.1kg/h)(1.88kJ/kg ℃) + (0.49kg/h)(504kJ/kg)
+32.26kmol/h [ 45028+34kJ/kmol ℃(336kJ/kg) ]
- 115.7kmol/h [ 29.2(T2-110) ] – (54kg/h)(336kJ/kg)
- 29.29kmol/h [ 45028 + 34T2] = 0
T2= 147.4 ℃ (건조기로 들어오는 공기의 온도)
133.2 kmol/h
(e)
0.218 H2O(v)
0.782 DA
T=110℃
11.8 kmol/h
145 kmol/h
0.0211 H2O(v)
0.9789 DA
T=27℃ P=1atm
0.202 H2O(v)
0.798 DA
T=147.4℃
Q (kW)
Refs : H2O(v, 27℃), DA(27℃)
Substance
nin
Hin
nout
Hout
H2O(v) 27℃
0.249
0
29.29
4093.6
H2O(v) 110℃
29.04
2822.0
-
-
DA
27℃
11.55
0
115.7
3524.4
DA
110℃
104.16
2423.6
-
-
Ó = ∑ out nH - ∑ in nH = 193282kJ/h = 53.7KW
12.28.
Basis:
100mol/s
Fuel feed
92.5mol/s CH4
4.8mol/s C2H6
2.7mol/s C3H8
T=25℃
ṅ0 mol O2/s
3.76 ṅ0 mol N2/s
ṅ1 mol H2O/s
T=27 ℃, P=1atm rh=60%
Dry Air
furnace
ṅCO2
ṅH2O
ṅO2
ṅN2
T=250℃
CH4 + 2O2 → CO2 + 2H2O
C2H6 + 7/2O2 → 2CO2 + 3H2O
C3H8 + 5O2 → 3CO2 + 4H2O
ṅ0 = 1.2[2(92.5)+3.5(4.8)+5(2.7)] = 258.4mol O2/s
3.76ṅ0 = 971.4mol N2/s
From problem 27: yH2O = 0.0211mol H2O/mol = ṅ1/(258.4+971.4+ṅ1)
ṅ1 = 26.5mol H2O/s
ṅCO2 = [92.5+2(4.8)+3(2.7)] = 110.2mol CO2/s
ṅH2O = [26.5+2(92.5)+3(4.8)+4(2.7)] = 236.7mol H2O/s
ṅO2 = 258.4-[2(92.5)+3.5(4.8)+5(2.7)] = 43.1mol O2/s
Refs: C(s), H2(s), O2(s), N2(s) at 25℃
Subst.
ṅIn
ĤIn
ṅOut
ĤOut
CH4
92.5
-74.85
―
―
C2H6
4.8
-84.67
―
―
C3H8
2.7
-103.8
―
―
O2
258.4
0.058
43.1
6.87
N2
971.4
0.058
971.4
6.62
H 2O
26.5
-241.8
236.7
-234.05
CO2
―
―
110.2
-384.2
Q = ∑outṅiĤi -∑inṅiĤi = -77065 kJ/s
100 mol/s
From problem 12.27:
-53.7 kJ/s
So, n mol/s
f
Q = 53.7 kJ/s
= -77065 kJ/s
100 mol/s
= 0.70 mol/s = 0.252 kmol/h
(MW)fuel = 0.925(16) + 0.048(30) + 0.027(44)
=17.43 kg/kmol
So, ṅf = (0.252 kmol/h)(17.43kg/kmol) = 4.39kg fuel/h
12.29.
(3722)(0.1)(kg PVC/h) × (24h/day) × (300day/yr) = 2.68×10⁶ kg PVC/yr
67.6wt% CPVC : ṁcpvc = 2.68×10⁶kg PVC
y
38.4kg C
100kg PVC
100kg CPVC
29.9kg C
= 3.50×10⁶ kg PVC6.76/y
Cl2 consumption : (3.50×10⁶)(0.676) - (2.68×10⁶)(0.568)
Cl2 consumption(67% CPVC) = 843,800kg Cl2/yr
70.0wt% CPVC : ṁ =
2.868×10⁶kg PVC 38.4kg C
y
100kg PVC
100kg CPVC
27.4kg C
= 3.756×10⁶kg CPVC70/y
Cl2 consumption(70% CPVC) : (3.756×10⁶)(0.70) - (2.68×10⁶)(0.568)
= 1,107,000kg Cl2/y
(b)
PVC
$0.7/kg
Cl2
$0.15/kg
67%Cl CPVC
$2.5/kg
70%Cl CPVC
$2.90/kg
C67($/kg 67% Cl CPVC) = 1.80-(6.0×10^-8)M67
C70(%/kg 70% Cl CPVC) = 1.85-(5.0×10^-8)M70 + (4.0×10^-14)(M70) ²
P(X) = 판매세 – 원자재비용 - 제조비용
X : 67%Cl CPVC 생산 분율
1- X : 70%Cl CPVC 생산 분율
(b)
P=
X
$2.50
kgCPVC67
3.50×10⁶kg CPVC67
yr
+
$/yr revenue
(1-X)
-
$2.90
3.756×10⁶kg CPVC70
kg CPVC70
yr
(0.70)(2.68×10⁶kgPVC/y)
$/yr PVC
%/yr Cl2
-X
$0.15 8.438×10⁵kg Cl2
kg
yr
- (1-X) (0.15/kg)(1.107×10⁶)
- X(3.5×10⁶)[ 1.80-(6.0^-8)(3.5×10⁶)X ]
$/yr 67% CPVC
- (1-X)(3.756×10⁶)[ 1.85-(5.0×10^-8)(3.75×10⁶)(1-X)
+ (4.0×10^-14)[(3.756×10⁶)(1-x) ²]
$/yr
70%CPVC
( all×10⁶ )
X
Rev
PVC
Cl2
Cost 67% Cost 70% P
0
10.9
1.876
0.166
0
8.37
0.48
0.54
9.736
1.876
0.145
3.19
3.25
1.27306
0.55
9.714
1.876
0.144
3.24
3.18
1.27333
0.56
9.693
1.876
0.144
3.30
3.10
1.27330
1
8.75
1.876
0.127
5.57
0
1.182
% increase in P =
1.27333-1.182
1.182
×100 = 7.7%