Kasus shell and tube 200 000 lb/jam air suling memasuki HE pada suhu 93 oF dan keluar pada suhu 85oF. Panas akan diserap oleh air 320 000 lb/jam air pendingin dengan suhu masuk 75 oF dan suhu keluar 80 oF. Pressure drop diperoleh sebesar 10 psi pada kedua arus dengan fouling factor sebesar 0.0005 untuk air suling dan 0.0015 untuk air pendingin ketika kecepatan melebihi 6 fps. Spesifikasinya adalah ShellTubeID = 15 1/4 inchNumber and length = 160, 1600Baffle space = 12 inchOD, BWG, pitch = in, 10 bwg,1 in. sqPasses = 1 Passes = 1Apakah HE ini sudah layak?Jawab:Heat balances:Distiied water: 200000 1 (93-85)= 1.600.000Btu/hrRaw water: 320000 1 (80-75)= 1.600.000Btu/hrLMTD = = = 11,4 0F R= = =1,6S= = =0,278Ft=0,945 didapat dari Fig.18t=0,945 11,4=10,750FPada temp rata-rata:Ta= =890F dan ta= =77,50F

Hot fluid; Shell side, distilled water

as = IDCB/144PT = = 0.254 ft2

Gs = w/as = 200000/0,254 = 787401,575 lb/hr ft2

Pada temp rata-rata: Ta= 890F, = 0,817,42= 1,96 lb/hr ft2De = = 0,0458 ftRe = DeGa/ = (0.0458 x 787401,575)/1,96 = 18.399,486JH=78

Pada temp rata-rata: Ta= 890F, c= 1,0 Btu/(lb)(0F)k = 0.36 Btu/hr ft2 = = 1,76

ho = jH a ho = 78 =1079,057

Cold fluid; tube sid, raw water

Hot fluid; Shell side, distilled water

Rea = 18.399,486, f=0,0017 ft2/in2No of cross, (N+1)= 12L/B = (1216)/12= 16

P = = = 8,958 Psi

Pa = 8,958 Psi

Allowable Pa = 10 psi Cold fluid; tube sid, raw water

Kasus Double Pipe 1000 lb,jam minyak pelumas 26 oAPI akan didinginkan dari 450 ke 350 oF dengan 10598.3 lb/jam minyak mentah 34o API. Minyak mentah akan dipanaskan dari 300 hingga 310 oF. Setiap arus mempunyai foul factor sebesar 0.003 dan pressure drop sebesar 10 psi. Beberapa HE double pipe dengan panjang 15 ft ukuran 4x3 in telah disediakan. Berapa yang dibutuhkan dan bagaimanakah penyusunannya? VIskositas minyak mentah dapat diperoleh dari fig 14. Viskositas minyak pelumas sebesar 1.4 cp pada 500 oF, 3 cp pada 400 oF dan 7.7 cp pada 300 oF. Viskositas tersebut dapat menghasilkan kesalahan yang cukup besar jika diasumsikan (/w)0.14 = 1 Neraca Panas Minyak pelumas = 1000 x 0.62(450 350) = 62 000 btu/jam Minyak mentah = 10598.3 x 0.585(310-300) = 62 000 Btu/jam Hot fluidCold FluidDifference

450Higher temperature310140t2

350Lower Temperature30050t1

90t2 - t1

LMTD = = = 87.41 0F

Suhu rerata = = 0.357Dari interpolasi grafik diperolehKc = 0.43Fc = 0.395Tc = 350 x 0.395(450 -350) = 389.5 0F tc = 300 x 0.395(310 -300) = 304 0F Hot fluid; annulus minyak pelumas

D2 = = 0.3335 ft D1 = = 0.292 ft

aa = (D22 D12)/4 = (0.33552 0.2922)/4 = 0.0214 ft2

Diameter ekuivalen De = (D22 D12)/D1 = (0.33552 0.2922)/0.292 = 0.0935 ft

Kecepatan umpan, Ga = w/aa = 1000/0.0214 = 46729.97 lb/hr ft2Pada 389.5 0F , = 3 cp = 3 x 2.42 = 7.25 lb/ft hr

Re = DeGa/ = (0.0935 x 46729.97)/7.25 = 838.795

L/D = 2 x (15x2)/0.0935 = 641.711

Dari figure 24 diperoleh jH = 1.9

Pada Tc = 389.5 0F Dari figure 4: C = 0.615 Btu/lb 0FDari figure 1: K = 8.067 Btu/hr ft2

= = 4.05

ho = jH a = = 5.514 Btu/hr ft2 0F

tw = tc + =304 + = 305.558 0F

w = 6.6 x 2.42 = 16 lb/ft hr a = (/w)0.14 = 0.9

ho = x a = 5.514 x 0.9 = 4.9626

Cold fluid; inner pipe minyak mentah

D = 3.068/12 = 0.25567 ft

ap = D2/4 = (0.25567)2/D = 0.0513 ft2

Gp = w/ap = 10598.8/0.0513 = 206541.34 lb/hr ft2

Pada suhu 304 0F = 0.83 cp = 0.83 x 2.42 = 2.01 lb/ft hr

Rep = DGp/M = = 26271.852

jH = 90

pada t2 = 304 0F c = 0.585 Btu/lb 0Fk = 0.073 Btu/hr ft2

= = 2.52

hi = jH p

= = 64.756 Btu/hr ft2 0F = x = = 56.763

w = 0.77 x 2.42 = 1.86 p = (/w)0.14 = (2.01/1.86)0.14 = 1

hio = x p = 297 x 1 = 297

Uc = = = 4.88 Btu/hr ft2 0F = + Rd Rd = 0.003 + 0.003 = 0.006 UD = = 4.74A = = = 149.641 ft2 External surface/linear ft, as = 0.917 L = = = 163.185 ftJumlah hairpin yang dibutuhkan = = 5.44 6 Hot fluid; annulus minyak pelumas

De = (D2 D1) = (0.3355 0.292) = 0.0435 ft

Rea = DeGa/ = (0.0435 x 46728.97)/7.25=2032.71

f = = = 0.00787s =0.75 = 62.5 x 0.75 = 46.875 lb/ft3

Fa = = = 7.259 ft

v = = = 0.2769 fps

F1 = 8 = 8 = 0.00899 ft

Pa = = 2.366 psi

Allowable Pa = 10 psi Cold fluid; inner pipe minyak mentah

Rep = 26271.852 f = 0.0035 + = 0.007176

s = 0.76 = 62.5 x 0.76 = 47.5

Fp = = = 0.296 ftPp = = 0.098 psi

Allowable Pp = 10 psi