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# Vapor Pressure Log Po = A + B/T + C log T + DT + ET2 komponen A B
C3H6 C3H8 C3H6O TBA TBHP
#
C
D
E
24.539
-1502.2
-6.48 -4.2845E-11
5.4982E-06
21.4469 38.5381 71.8181 11.5999
-1462.7 -2631 -4996.6 -2765.8
-5.261 3.282E-11 -11.104 4.2178E-10 -21.805 1.9238E-08 -2.12E-01 -4.20E-03
3.7349E-06 5.5025E-06 5.8247E-06 2.1416E-06
Cek fase pada arus keluar reaktor komponen
C3H6 C3H8 C3H6O TBA TBHP Σ
kmol/jam
xi
BM
1.00347371 0.0043880612
42
0.50425815 0.0022050559
44
99.3438976 0.4344180623 106.758128 0.4668395346 21.072948 0.092149286 228.682705 1
58 74 90
# T Bubble Point pada 40 atm P = 40 atm = = Trial T = 216.164447 oC
30400 mmHg 489.31444658 K
xi
Po
Ki
yi
C3H6 C3H8
0.00438806
2.27E+05
7.48E+00
3.28E-02
0.00220506
1.59E+05
5.24E+00
1.15E-02
C3H6O
0.43441806 0.46683953 0.09214929 1
4.11E+04 2.26E+04 6.87E+03
1.35E+00 7.45E-01 2.26E-01
5.87E-01 3.48E-01 2.08E-02 1.00E+00
komponen
TBA TBHP Σ
suhu dan tekanan keluar reaktor P T
#
= =
40 atm 78.8675 C
351.8675 K
Cek fase pada Flash Drum P operasi =
1.1 atm
836 mmHg
# T bubble point umpan separator pada 1.1 atm Trial T = 53.8366942 oC = 326.98669416 K Po komponen xi Ki
C3H6 C3H8 C3H6O TBA TBHP Σ
yi
0.00438806 17323.8143
20.7223
0.0909
0.00220506 13932.3114
16.6654
0.0367
0.43441806 1441.7438 0.46683953 214.1965
1.7246 0.2562
0.7492 0.1196
0.09214929 29.2171 1
0.0349
0.0032 0.9997
# T dew point umpan separator pada 1.1 atm Trial T = 89.0731276 oC = 362.2231276 K Po komponen yi Ki
xi
C3H6 C3H8
0.00438806 33967.3302
40.6308
0.000108
0.00220506 27292.7553
32.6468
6.75427E-05
C3H6O
0.43441806 3816.8817 0.46683953 976.3481 0.09214929 152.4717 1
4.5656 1.1679 0.1824
0.09514927 0.39973229 0.505253 1.0003101
TBA TBHP Σ
# Maka umpan akan berupa dua fase pada T=
57.7203461 oC = 330.870346 K
0.0003583
Throttling Valve Kode : Fungsi : Menurunkan tekanan produk keluar reaktor dari 40 atm menjadi 1.1 atm untuk diumpankan ke flash drum Tujuan
: 1. Menghitung neraca massa 2. Menghitung neraca panas 3. menentukan suhu sebelum masuk expansion valve
1. Neraca Massa
F, zi NM pada Flash Drum Umpan masuk Komp. BM 42 44 58 74 90
C3H6 C3H8 C3H6O TBA TBHP Total
2. Neraca Panas a. Menghitung Panas Penguapan (Hvap) T = 57.720346081 C Data A Tc 26.098 364.76 C3H6 26.89 369.82 C3H8 40.176 482.25 C3H6O 107.467 506.2 TBA TBHP Total
Q vap =
kg/hr 42.1459 22.1874 5761.9461 7900.1014 1896.5653 15622.9461
72.8
Komp. C3H6 C3H8 C3H6O TBA TBHP Total
zi 0.0044 0.0022 0.4344 0.4668 0.0921 1
330.8703460805 K n Hvap
611
0.358 0.365 0.366 0.813 0.204
kmol/hr 0.6327 0.2918 12.7610 2.4458
11.1474 11.8252 26.2902 45.3853 62.0927
0.0679 16.1312
Output Gas (Arus 6) kg/hr kmol/hr 26.5720 0.6327 12.8381 0.2918 740.1404 12.7610 180.9867 2.4458 6.1124 0.0679 966.6495 16.1992 15622.9461 228.6827
Qvap 7052.5715 3450.2933 335490.2383 111001.9154 4217.0411 456995.0184 kJ/hr
456995.01844 kJ/hr
b. Panas yang dibawa gas keluar T = 330.87 K Komp. A B C 31.298 7.2449E-02 C3H6 28.277 1.1600E-01 C3H8 29.501 9.2545E-02 C3H6O 8.866 4.2394E-01 TBA TBHP
Input (Arus 5) kmol/hr 1.0035 0.5043 99.3439 106.7581 21.0729 228.6827
2.1
kmol/hr 0.6327 0.2918 12.7610 2.4458 0.0679 16.1312
5.5655E-01
n.intCp.dT 1392.3204 740.6412 31538.5433 9611.7103 306.5341 43589.7492 kJ/jam
D
1.9481E-04 1.9597E-04 2.5626E-04 -2.4206E-04 -4.5845E-04
E
-2.1582E-07 -2.3271E-07 -2.9921E-07 6.1419E-08 2.1168E-07
Qtop =
6.2974E-11 6.8669E-11 9.0294E-11 -4.3829E-12 -4.2051E-11
int Cp dT 2200.7179 2538.3994 2471.4709 3929.9393 4513.4762
43589.7492 kJ/hr
c. Panas yang dibawa cairan keluar T= 330.87 K Tref = 298.15 K Komp. A B C D int Cp dT 54.7180 3.4512E-01 -1.6315E-03 3.8755E-06 C3H6 4012.4259 59.6420 3.2831E-01 -1.5377E-03 3.6539E-06 C3H8 4078.2311 53.3470 5.1543E-01 -1.8029E-03 2.7795E-06 C3H6O 4046.2745 -309.4150 4.4863E+00 -1.2958E-02 1.3642E-05 TBA 7990.6016 TBHP
Komp. C3H6 C3H8 C3H6O TBA TBHP Total
-18.8340
1.1762E+00
-3.2019E-03
kmol/hr n.Cp.dT 0.3708 1487.8366 0.2125 866.5559 86.5829 350337.9994 104.3124 833518.5229 21.0050 95726.0072 191.4785 ### kJ/jam
3.3722E-06
Qbot =
4557.2891
1281936.9220 kJ/hr
e. Menghitung panas yang harus dibawa produk masuk Panas yang harus dibawa oleh produk sebelum masuk expansion valve Qin = Qvap + Qtop + Qbot = 1782521.6897 kJ/hr Panas produk yang dibutuhkan sebelum masuk ekspansion valve Trial : T out = 67.9935 C 341.1435 K komp C3H6 C3H8 C3H6O TBA TBHP
komp C3H6 C3H8 C3H6O TBA TBHP Total
A
54.7180 59.6420 53.3470 -309.4150 -19.9540
kg/jam 42.1459 22.1874 5761.9461 7900.1014 1896.5653 15622.9461
B
C
3.4512E-01 3.2831E-01 5.1543E-01 4.4863E+00 1.1460E+00
-1.6315E-03 -1.5377E-03 -1.8029E-03 -1.2958E-02 -3.2019E-03
kmol/jam 1.0035 0.5043 99.3439 106.7581 21.0729 228.6827
n. Cp. dT 5402.8247 2755.7239 533052.3842 1130064.5444 111246.2126 1782521.6899
D
3.8755E-06 3.6539E-06 2.7795E-06 1.3642E-05 3.1722E-06
T 341.1435 341.1435 341.1435 341.1435
298.15 298.15 298.15
5464.9071 5365.7285 10585.2788
341.1435
298.15
5279.1006
Jadi umpan sebelum masuk ekspansion valve harus mempunyai suhu Neraca Panas total Input kJ/hr Output Qumpan 1782521.6899 Qtop Qbottom Qvap. Total 1782521.690 Total
kJ/hr 43589.7492 1281936.9220 456995.0184 1782521.690
Tref intgrl Cp. dT 298.15 5384.1218
selisih Qin-Qout=
67.9935 C
0.000
as (Arus 6)
Output Liquid (Arus 7) yi kg/hr kmol/hr xi 0.0391 15.5739 0.3708 0.0017 0.0180 9.3493 0.2125 0.0010 0.7878 5021.8056 86.5829 0.4075 0.1510 7719.1148 104.3124 0.4909 0.0042 1890.4529 21.0050 0.0989 1.0000 14656.2965 212.4835 1.0000
341.1435 K
PERANCANGAN DIMENSI DRUM
kondisi operasi
P = 1.1 atm T = 57.7203460805 °C
1.114575 bar =330.720346 K
=
16.17
Neraca Massa arus 5 Komponen
arus 6
Umpan kmol kg 1.0035 0.5043 99.3439 106.7581 21.0729 228.6827
C3H6 C3H8 C3H6O TBA TBHP
42.1459 22.1874 5761.9461 7900.1014 1896.5653 15622.9461
arus 7 fase uap fase cair kmol kg kmol kg 0.6327 26.5720 0.3708 0.2918 12.8381 0.2125 12.7610 740.1404 86.5829 2.4458 180.9867 104.3124 0.0679 6.1124 21.0050 16.1992 966.6495 212.4835 228.6827 15622.9461
15.5739 9.3493 5021.8056 7719.1148 1890.4529 14656.2965
1. Menentukan Densitas Uap dan Cairan # Densitas Uap rho uap = BM camp x P / (R T Z) BM camp
=
sigma BMi * yi
Operasi pada tekanan Rendah ( Smith Van Ness hal 89) , dan dari tabel 3.15 ( Smith Van Ness hal 90), digunakan persamaan generalized virial coeficient : Bº = 0,083 - 0,422/(Tr^1.6) (persm 3.50 Smith Van Ness ed 5) B´ = 0,139 - 0,172/(Tr^4,2) (persm 3.51 Smith Van Ness ed 5) BPc/RTc = Bº + ωB' (persm 3.48 Smith Van Ness ed 5) Z = 1 + (BPc/RTc x Pri/Tri) (persm 3.47 Smith Van Ness ed 5) dengan : T BMavg P R Tr Tc Pr
= = = = = = =
Komponen C3H6 C3H8 C3H6O TBA TBHP Σ
BM camp = Komponen C3H6 C3H8 C3H6O
Suhu separator = 57.7203 oC = 330.7203 K BM rata-rata = 59.6728 kg/kmol Tekanan = 1.1 atm = 836 mmHg Konstanta Gas = 0.08205 m3 atm/kmol K Suhu tereduksi (T/Tc) Suhu Kritis Tekanan tereduksi (P/Pc) BM 42 44 58 74 90
Kgmol/jam yi (fr. Mol) BM *yi 0.6327 0.0391 1.6403 0.2918 0.0180 0.7925 12.7610 0.7878 45.6900 2.4458 0.1510 11.1726 0.0679 0.0042 0.3773 16.1992 1 59.6728041
yi (fr.massa) 0.0274887547 0.0132810228 0.7656760706 0.1872308914 0.0063232605 1
59.67280406 Kg/Kmol ω 0.142 0.152 0.271
Pc (bar) Tc (K) Pc (atm) Tr 364.76 46.13 45.5267703 369.82 42.49 41.9343696 482.25 49.24 48.5961017
Pr 0.9067 0.8943 0.6858
0.0242 0.0262 0.0226
TBA
0.616
506.2
39.72 39.2005922
0.6533
0.0281
TBHP
0.668
576
43.4 42.8324698
0.5742
0.0257
Σ
Komp
Bo
B1
BPc/RTc
Zi
ρv (kg/m3)
yi*ρv (kg/m3)
C3H6
-0.4106
-0.1206
-0.4277
0.9886
2.4469
0.0673
C3H8
-0.4216
-0.1360
-0.4423
0.9870
2.4508
0.0325
C3H6O TBA TBHP Total
-0.6886 -0.7509 -0.9423
-0.6996 -0.8889 -1.6293
-0.8782 -1.2984 -2.0307
0.9710 0.9442 0.9092
2.4912 2.5618 2.6606
1.9074 0.4797 0.0168 2.5037
(1-T/Tc)^n Tc 428.25 0.647722765 506.2 0.7482933691 576 0.7835624114
r cairan 0.4778 0.7080 0.1514 1.3371
rho uap
=
2.503721666 kg/m3
=
0.1561619 lb/ft3
Densitas Cairan
^ T n ρ= A . B − 1 Tc
((
))
BM
Komponen C3H6O TBA TBHP Σ
#
58 74 90
Komponen C3H6O TBA TBHP Σ
Kg/jam 5021.8056 7719.1148 1890.4529 14631.3734
xi 0.3432 0.5276 0.1292 1.0000
B
n 0.29353 0.2737 0.2857
A 0.31226 0.26921 0.30445
r cairan =
0.27634 0.2565 0.26825
0.74786025 gr/ml
Laju Vol.trik Uap (qv)
=
=
747.86025 kg/m3
Massa Uap/ rho uap = 386.085058613 m3/j = 0.1072458496 m3/s
Laju Vol.trik cairan(q liq) =
=
46.645471318 lb/ft3
= 3.78735144 ft3/s
Massa Cairan/ rho Cairan = 19.5976407901 m3/j = 0.0054437891 m3/s =
0.1922456 ft3/s
menentukan kecepatan gas maksimum
Umax = K
(
ρ L - ρV ρV
0 .5
)
(Carl R Branan,1994)
dengan : U = kecepatan gas, ft/s K = system kostan merupakan fungsi (Wliq/Wvap)*(rho Vap/ rho liq)^0.5 W = flow rate, lb/s W liq = 14656.2965 Kg/jam = 8.9673865 lb/s W vap = 966.6495 Kg/jam = 0.59144 lb/s nilai (Wliq/Wvap)*(rho Vap/ rho liq)^0.5 = 0.88 dari figure 1 Carl R Branan, p.132 diperoleh nilai Kv = 0.15893029 untuk vessel horisontal nilai Kh = 0.19866287 sehingga U max = 3.427722901 ft/s = 1.04476994 m/s Av min = qv / U max = 1.1049176219 ft2
2.
menghitung Dmin Separator jika vessel pada keadaan volume penuh liquid maka : (A total) min =
(Av)min /0,2
(Carl R. Branan,1994)
X= Y=
D min
#
= =
5.52458811 ft2 2.65286484 ft
=
menghitung panjang Vessel Diambil waktu tinggal cairan dalam drum = V liq = q liq x waktu tinggal q liq = 0.005443789 m3/s over design = 20 % perancangan= 0.006532547 m3/s V liq = 3.919528158 m3 = D = D min + 6 in D = 37.83437808 in = D = D min = 31.83437808 in = Dstandar = 40 in = L = V liq/(0.25 pi D^2) L = 15.86944555 ft = L/D = 4.760833664
0.8085932 m
600 det
=
31.83437808 in
=
10 menit
138.41683 ft3 3.1528648 ft 2.6528648 ft 3.3333333 ft
= = =
0.9609932032 m 0.8085932032 m 1.016 m
4.837007 m
Menentukan Dimensi Flash Drum
#
menentukan tebal shell ts = (P.ri/(f.E-0.6*P)) + C Bahan kontruksi shell yang dipilih adalah karbon stell SA 283 grade C, karena : - tahan korosi - cukup kuat - harga murah - untuk pressure vessel dengan tebal shell lebih dari 5/8 inch - untuk vessel dengan tekanan moderat - untuk suhu operasi kurang dari 900 oF 482.22222 C Spesifikasi : Allowance stress (f) = 12650 lb/in2 (Tabel 13-1 brownell) Corrosion allowanse (C )= 0.1250 inch Efisiensi pengelasan (E) = 0.85 (tabel 13-2 brownell untuk single butt joint)
# P oerasi P design IDs ri t shell
= = = = =
1.1 1.3 40 20 0.127455424
digunakan tebal standar = Diameter Luar shell = diameter standar =
menentukan tebal head tabel 5.7 untuk OD = ts icr rc maka W W
atm atm in in in
=
over design = 1.016 m
0.1875 in = 3/16 in 40.375 in 42 in =
1.0668 m
42 in = 0.1875 in = 2.625 in = 42 in = 1/4 x (3+(r+icr)^0.5) = 2.420048652 in
t = ((P.r.W)/(2.f.E-0,2.P))+c t = 0.1279709402 in digunakan tebal standar = 0.1875 in = 3/16 in
#
20.00%
menentukan lebar head untuk tebal head 3/16 in, maka : sf = 1.5-2 dipilih sf = 2
(tabel 5.8 Brownell)
=
0.0047625
m
11.
#
Dari Tabel 5.6 Brownell &Young, sf = 1.5 - 2 Diambil nilai sf OD = ID + 2t a = ID/2 AB = a-icr BC = r-icr AC = ((BC^2)-(AB^2))^0.5 = b = r - AC OA = t + b + sf Maka lebar head = OA
= = = = = = = =
2 42 20 17.375 39.375 35.3341195 6.66588051 8.8534 8.8534 0.22487587 0.73778171
inch inch inch inch inch inch inch inch m ft
Dimensi Separator Keseluruhan lebar total separator = L + 2*OA = 5.2867587328 m = 17.3450089658 ft Tinggi Separator
=
42 in 3.5 ft 1.0668 m
= =
1.0668 m
Menentukan diameter pipa pemasukan dan pengeluaran Di,opt = 3.9*qf^(0.45)*rhof^(0.13) (pers.45 Peters) dengan Di,opt = diameter optimum, in qf = debit fluida, cuft/s ρf = densitas fluida, lbm/cuft Data densitas cairan ρ= A.B Komp C3H6 C3H8 C3H6O TBA TBHP
1 kg/m3 = 1 kg = 1 inch =
-(1-T/Tc)^n
A 0.23314 0.22151 0.31226 0.26921 0.30445
B 0.27517 0.27744 0.27634 0.2565 0.26825
Pipa pemasukan umpan ke Throttle Valve F = 15622.9461 kg/jam = 9.56740538 lbm/s Tumpan = 67.99353182 oC = 341.1435318 K wi Komp kg/jam C3H6 42.1459 0.0027 C3H8 22.1874 0.0014 C3H6O 5761.9461 0.3688 TBA 7900.1014 0.5057 TBHP 1896.5653 0.1214 Σ 15622.9461 1.0000 # 1 1.371837593 = 0.7289 gr/cm3 = 728.9493 kg/m3 = 45.5068 lbm/ft3
ρ=
qf = 9.56740538 lbm/s
n 0.30246 0.287 0.29353 0.2737 0.2857
ρi 0.4097 0.4099 0.6990 0.7327 0.8429
Tc 364.76 369.82 428.25 506.2 576
wi/ρi 0.0066 0.0035 0.5276 0.6902 0.1440 1.3718
0.06242796 2.20462 0.0254
45.5068 = 0.2102 Di,opt = Ukuran standar: D= ID = OD = SN =
lbm/ft3 ft3/s
3.1757 inch =
4 4.026 4.5 40
inch inch inch inch
0.0807 m
0.1016 m 0.1023 m 0.1143 m
Pipa pemasukan dari Throttle Valve ke drum F = 15622.9461 kg/jam = 9.5674 lbm/s o Tumpan = 57.7203 C = 330.8703 K wi ρi Komp kg/jam C3H6 42.1459 0.002697692 0.4372717 C3H8 22.1874 0.0014201776 0.4337706 C3H6O 5761.9461 0.3688130288 0.7180299 TBA 7900.1014 0.5056729635 0.7450128 TBHP 1896.5653 0.121396138 0.8535177 Σ 15622.9461 1
wi/ρi 0.00616937 0.00327403 0.51364578 0.67874404 0.14223038 1.3441
# 1 1.3441 = 0.7440 = 744.0124 = 46.4472
ρ=
gr/cm3 kg/m3 lbm/ft3
qf =
9.5674 46.4472 = 0.2060
Di,opt = Ukuran standar: D= ID = OD = SN =
lbm/s lbm/ft3 ft3/s
3.1550 inch =
4 4.026 4.5 40
inch inch inch inch
0.0801 m
0.1016 m 0.1023 m 0.1143 m
Pipa pengeluaran uap ρ=
0.1562 lbm/ft3
V= =
966.6495 kg/jam 0.5920 lbm/s
qf = = Di,opt = #
Ukuran standar: D= ID = OD = SN =
0.5920 lbm/s 0.1562 lbm/ft3 3.7908 ft3/s 5.5803 inch =
6 6.065 6.625 40
inch inch inch inch
0.1417 m
0.1524 m 0.1541 m 0.1683 m
Pipa pengeluaran cairan ρ=
46.6455 lbm/ft3
L= =
14656.2965 kg/jam 8.9754 lbm/s
qf = = Di,opt = #
Ukuran standar: D= ID = OD = SN =
8.9754 lbm/s 46.6455 lbm/ft3 0.1924 ft3/s 3.0614 inch =
3 3.068 3.5 40
inch inch inch inch
0 0.0778 m
0.0762 m 0.0779 m 0.0889 m
Resume
Tipe : Horisontal Jumlah : 1 buah 1.
2.
3
Kondisi operasi : Suhu : 57.72 oC Tekanan : 1.1 atm Spesifikasi : Drum/shell - Volume = 3.9195 m3 - ID = 1.0160 m - OD = 1.0668 m - L shell= 4.8370 m - panjang = 5.2868 m - Tebal = 0.1875 in = - material = carbon steel SA 283 grade C Head - Type = Torisperical dished head - Tebal = 0.1875 in = - Lebar = 8.8534 in = - material = carbon steel SA 283 grade C Diameter pipa pemasukan umpan ke trottle valve = Diameter pipa pemasukan umpan ke flash drum = Diameter pipa pengeluaran atas flash drum = Diameter pipa pengeluaran bawah flash drum = Lebar Drum= 5.2867587328 m
0.0048 m
0.0048 m 0.2249 m 4 4 6 3
in in in in
= = = =
psia
mith Van Ness ed 5) mith Van Ness ed 5) mith Van Ness ed 5) mith Van Ness ed 5)
-0.130929594 0.158930294
A B C D E F
-1.942936 -0.814894 -0.17939 -0.012379 0.000386 0.00026
nell untuk single butt joint)
lbm/ft3 lbm m
0.1016 0.1016 0.1524 0.0762
m m m m
Tabel 2 Kern hal 795 T isolasi = 42.92631
Asbestos rho [lb/ft3] 29.3 29.3 36 36 36 36 43.5 43.5
F
o
-328 32 32 212 392 752 -328 32
Sifat Fisis Udara
k [Btu/(hr.ft2.oF/ft)] 0.043 0.09 0.087 0.111 0.12 0.129 0.09 0.135
C
k=
o
-200 0 0 100 200 400 -200 0
0.097302
T2= k=
#DIV/0!
( Apendix A-5 Holman )
T ρ K kg/m3 250 1.4128 300 1.1774 309.613156 1.1429 2.2538 350 0.998
cp kJ/kg.oC 1.0053 1.0057 1.0063 0.9859 1.009
μ x 105 kg/m.s 1.599 0.8462 1.0825 -6.5266 2.075
ν x 106 m2/s 11.31 15.69 16.66 -14.7300 20.76
k α x 104 W/m.oC m2/s 0.02227 0.15675 0.02624 0.2216 0.02697 0.2363 0.0035 -0.2386 0.03003 0.2983
Pr 0.722 0.708 0.706 0.7740 0.697
Sifat-sifat logam (Appendix A-2 Holman hal 581) Baja karbon T, oC k, W/m.oC
T= k=
0 55
57.72035 oC (suhu T1) 53.26839 W/m.oC
100 52
200 48
300 45
400 42
600 35
30.874359 Btu/hr ft2 (F/ft) 191766.2 J/m.h.C
800 31
80
C (T3)
o
C
o
36.46316 oC
1000 29
1200 31
isolator yang digunakan adalah jenis asbestos dengan data-data sebagai berikut : Є= 0.96 k= 0.12 Btu/hr ft F (kern,1988)
(alasan dipilih asbestos
suhu isolator bagian luar T3 = Suhu rata2 dalam FD(T1) = Suhu udara luar(Ta) =
42.926313 C 57.720346 C 30 C
= = =
109.2674 F 135.8966 F 86 F
= = =
569.2674 R 595.8966 R 546 R
= = =
suhu film Tf = (T3 + Ta)/2 = β = 1/ Tf = delta T = T3 - Tf =
36.463156 C 0.0017933 R^-1 11.633681 F
=
97.63368 F
=
557.6337 R
=
Sifat Fisis udara pada Tf : 309.61316 rho 1.1429 kg/m3 = cp 1.0063 J/kgC = k 0.0270 W/mC = miu = 1.08E-05 kg/ms = Asumsi :
K 0.071349415 0.000240514 0.015631045 0.026184153
lb/ft3 Btu/lb F Btu/hr ft2 (F/ft) 97.087238 J/m.h.C lb/ft hr
(Daftar A-5, J.P.Holman)
Sifat-sifat udara tetap di sekeliling reaktor terjadi konveksi bebas Gr
l3r
Pr =
2
g t 2
cp . μ k
Raf =Gr . Pr Bila : Raf = 104 - 109
Raf = 109 - 1012
Gr = Bilangan Grasshof
Pr = Bilangan Prandtl
Raf = Bilangan Rayleigh
Δt hc=0 . 29 L
ID
0 . 25
( )
= koefisien perpindahan panas konveksi = L = tinggi silinder reaktor : 1.06680 m = 2.7802E+09 = 0.00040 = 1.1201E+06 hc = 0.526995051 Btu/hr ft2 F
2π L ( T 1 −T a ) R R ln 2 ln 3 R1 R2 1 + + k1 k2 ( hc ) R 3
=
ID/L = 4.534127 ID/L > 35/(Gr)^0.25 3.5 ft
=
Penentuan isolasi dilakukan dengan cara trial R1 = Jari-jari dalam menara R2 = Jari-jari luar menara R3 = Jari-jari menara stlh diisolasi T1 = Suhu dinding dalam menara T1 T2 T3 T3 = Suhu dinding luar menara Ta = Suhu udara luar k1 = konduktivitas dinding Menara k2 = konduktivitas panas isolator
Q loss=
4.837007
cek nilai l : 35/(Gr)^0.25
hc=0 . 19 ( Δ t )1/ 3(Mc Adams,1958)
hc l Gr Pr Raf
=
= = = = = = =
0.50800 m 0.53340 m 135.897 109 86 30.874359 0.09730
= =
1.666667 1.75
F F F BTU/(hr.ft2) (F/ft) BTU/(hr.ft2) (F/ft)
Q'loss=
2π L ( T 1 −T 3 ) R R ln 2 ln 3 R1 R2 + k1 k2
Q loss=
2π L ( T 1 −T a ) R R ln 2 ln 3 R1 R2 1 + + k1 k2 ( hc ) R 3
pada keadaan ajeg :
perhitingan Q loss :
2π L ( T 1 −T 3 ) R R ln 2 ln 3 R1 R2 + k1 k2
Q'loss Qloss = 2 π L 2 π L T1 -Ta ln (R2/R1) /k1 ln (R3/R2) /k2 1/(hc ) R3 = Q loss
selisih =
Q'loss=
0.0001
= 49.897 F = 0.0015802811 = 1.1121378215 0.973103091 =
23.9103
perhitingan Q` loss : ln (R2/R1) /k1 ln (R3/R2) /k2
= =
(alasan dipilih asbestos??)
R3
316.07631 K 330.87035 K 303.15 K
R2 R1
309.61316 K
W=J/s
5, J.P.Holman)
q3 q2 q1 q
T4 = Ta
T1
m
=
T3T2
15.86944555 ft
0.1524
asumsi l= L dapat di pakai
ft ft
T 1 −T 3 ) R ln 3 R1 R2 + k2
ID reaktor OD reaktor
rc=kb/ha tebal isolasi r3=rc+r2 Qloss
0.184636 ft 0.2 ft 1.95 ft 23.9103
0.056277 m 0.06096 m 0.59436 m
5.627709 6.096 59.436
T 1 −T 3 ) R ln 3 R1 R2 + k2
T1 -T3 ln (R2/R1) /k1 ln (R3/R2) /k2
Q' loss
= 26.629 F = 0.0015802811 = 1.1121378215
=
23.9102
R3 R2 R1
cm cm cm
