Tugas 1_papp_rahman Maulana_1610814210022_.docx

Nama = Rahman Maulana NIM = 1610814210022 Matkul = PAPP

86 oF 30 oC

203 oF 95 oC

122 oF 50 oC

Cooler COOLER

113 oF 45 oC Data : m etanol OD Cp etanol

= 10022 kg/jam x 2,205 lb/kg = 22098,51 lb/jam = ¾” 16 BWG, 1 in. square pitch = 0,75 Btu/lb. oF (fig 2 kern page 804)

Ditanya : evaluasi HE tersebut “ Prosedur Evaluasi” 1. Neraca Energi Qetanol =m.Cp.∆t

= 22098,51 lb/jam x 0,75 Btu/lb.oF x (203-122)oF = 1342484 Btu/jam

Qetanol

= Qair

= 1342484 Btu/jam Btu

Massa air =

Qair Cp.∆t

=

1342484 jam Btu

1,0lb.℉ .(113−86)℉

= 49721,65 lb/jam

2. Menentukan Suhu Kalorik /Suhu Rata-Rata Th = T1+T2/2 = 203 oF + 122 oF = 162,5 oF Tc = t1 + t2/2 = 113 oF + 86 oF = 99,5 oF 3. ∆t LMTD Hot Fluid 203 122 81

Higher Temp. Lower Temp. Differences

Cold Fluid 113 86 27

Differences 90 36 54

Nama = Rahman Maulana NIM = 1610814210022 Matkul = PAPP

LMTD = 81

54 90 ln( ) 36

s=

27

= 0,23 27 (203−86) FT=0,87 (fig 18 kern hal 828) ∆t LMTD = FT.LMTD = 0,87 x 58,93 oF = 51,27194 oF R=

=3

= 58,93326 oF

4. Surface Area UD = 75-150 (Kern, Page 840) Trial = 80 Q 1376092,2570 Btu/jam A= = = 327,2951 ft2 ≥ 120 ft2 UD .∆t 80 x 51,8613 ℉ 5. Number Tube A Nt = ′′ (Kern, Page 843) 𝑎 .𝐿 331,6761 = 0,1963 𝑥 8 = 208,4151 buah Nt L ID HE

= 220 buah (Kern, Page 841) = 8 in = 19 ¼ in =1–2

6. Standarisasi UD Nt UD Standar = x UD trial (Kern, Page 840) 𝑁𝑡 𝑠𝑡𝑎𝑛𝑑𝑎𝑟 211,2048 = x 80 220 = 75,78733 Hot fluid: Shell side, metanol 1) Flow Area, as IDs .C.B 19,25 .0,25.7,7 as = = = 0,257335 ft2 𝑛 .144 1 .144 2) Laju Alir Massa, Gs

Nama = Rahman Maulana NIM = 1610814210022 Matkul = PAPP

w

22098,51 lb/jam

= 85874,46 lb/jam.ft2 𝑎𝑠 0,257335 ft2 3) Reynold Number, Re (Kern, Page 838) De .Gs 0,95/12 .104289,7072 Re = = = 5852,613 (Turbulen) 𝜇 0,48 .2,42 4) Coefficient Heat Transfer Gs =

=

ho = 𝑗𝐻. (

𝑘

).(

𝐶𝑝.𝜇 1

)3 . (

𝜇

𝐷𝑒 𝑘 𝜇.𝑤 20 K=0,105 btu/jam.ft . F/ft Jh = 24 (fig 28 kern page 838)

)0,14

0,105 1,16 1 ho= 24. ( ).( )3 . (1)0,14 0,079 0,105 =117,3827 btu/jam.ft2.0F

Cold fluid: Tube side, air 1) Flow Area, at (Kern, Page 843) Nt .a′t 220.0,302 at = = = 0,230694 ft2 𝑛 .144 2 .144 2) Laju Alir Massa, Gt w 49721,65 lb/jam Gt = = = 215530,3 lb/jam.ft2 𝑎𝑡 0,230694 ft2 3) V=Gt/3600.ρ 215530,3 =0,9579 ft/s 3600.62,5 4) Reynold Number, Re (Kern, Page 823 & 843) D .Gt 0,05165 .220920,5891 Re = = = 6135,39 (Turbulen) 𝜇 0,75 .2,42 5) Coefficient Heat Transfer (Kern, Page 164) hi = 710 btu/jam.ft2 oF hio = hi.ID/OD = 10250,625 btu/jam.ft2 oF 6) Clean overall coefficient hi0.h0 Uc = ℎ𝑖0+ℎ𝑜 10250,625.117,38 = 10250,625+117,38

Nama = Rahman Maulana NIM = 1610814210022 Matkul = PAPP

= 117,3814 7) Menentukan Dirt factor 117,3814−Ud Dirt factor = 117,3814.𝑈𝑑 117,3814−75,78733 = 117,3814.75,78733 = 0,004676 Menentukan Pressure Drop Hot fluid: Shell side, metanol 1) Untuk Re= 5852,613 2) N+1 = 12 L/B = 12.8/6,5

f = 0,0024 (fig 29 kern page 839)

= 14,76 3) Ds =19 ¼ in = 1,603 ft f.G2 .Ds.(N+1) 4) ∆PS = 5.22.1010 .𝐷𝑒.𝑠𝑔 0,0024.104289,70722 .1,603.14,76 = 5.22.1010 .0,079.0,79.1 = 0,010842 psi

Cold fluid: Tube side, air 1) Untuk Re = 6135,39 f = 0,0025 (fig 26 kern page 836) f.G2 .L.n 2) ∆Pt = 5.22.1010 .𝐷𝑒.𝑠𝑔 0,0025.220920,58912 .8.2 = 5.22.1010 .0,05165.1.1 = 0,7240 psi v^2 3) Gt = 220920,5891 = 0,006 (fig 27 kern page 837) 𝑔 4n v^2 ∆Pn= . 𝑠 𝑔 4.2 = .0,006 1

Nama = Rahman Maulana NIM = 1610814210022 Matkul = PAPP

= 0,048 psi ∆Pt = 0,7240+0,048 = 0,772 psi

Nama = Rahman Maulana NIM = 1610814210022 Matkul = PAPP