Perhit. HE Tipe HorisontalPerhitungan Shell and
TubeTipe:HorizontalShell sideID:104in8.66666632ftpass:1Tube
sideOD:1in (dalam BWG 14 )
=0.0833333333ftL:402in33.5ftpass:4passjumlah:680buahA:210m22260.4211ft2Baffle:7in0.5833333333ftPitch:1.25inFluida
Panas:Long ResiduLaju alir fluida
W1=2021.60ton/day=185702.15lb/hrT1=642.56oFT2=592.16oFFluida
Dingin:Crude OilLaju alir fluida
W2=4040.20ton/day=371128.73lb/hrt1=317.12oFt2=340.88oF1LMTDFluida
PanasFluida DinginSelisih642.56Higher TempSuhu
tinggi340.88301.68592.16Higher TempSuhu
rendah317.12275.04Selisih26.6LMTD=DTm=288.152Menghitung faktor
koreksi dengan terlebih dahulu menghitung R dan
SR==2.12S==0.0730088496Dari grafik 18 kern atau fig 10.31 A.
Ludwig, makaFT =0.97jadi True Temperature Different,DT=DTm x
FT=279.5101458533oF3Menghitung caloric temperature dan data
physical fluida==0.9116945107T1 - T2=50.40oFSpesific Gravity long
residu pada 60 0F/60 0F =0.90930API==24.114208732Dari fig. 17, Kern
atau fig. 10-32, Ludwig, didapatKc=0.35Fc=0.465Tc=T2 +
Fc(T1-T2)=615.596oFtc=t1 + Fc(t2-t1)=328.1684oFsehingga diperoleh
data2 sebagai berikut :Crude oiltc=328.1684oF0API=35.5t=0.35cpt
=0.847lb/ft j (p.165 maxwell)VABP=572oFB=0.03Cpt + B=0.68(p.93
maxwell)Cpt=0.65Btu/(lb 0F)kt=0.0725Btu/(hr.ft.0F)Long
Residutc=615.596oF0API=24.114208732s=0.4s =0.968lb/ft j (p.165
maxwell)VABP=834.8oFB=0.02Cps + B=0.65(p.93 maxwell)Cps=0.63Btu/(lb
0F)ks=0.0615Btu/(hr.ft.0F) (fig. 1 Kern)4Menentukan Heat DutyQt=Mt
x Cpt x (t2-t1) =5731712.1339192Btu / hrQs=Ms x Cps x (T1-T2)
=5896414.8065088Btu / hr5Menghitung
UDA=2260.4211ft2UD==9.0718830129Btu/j ft2 0F6Menghitung Flow Area
(at & as)a. Tube1 in OD BWG 14Flow area per tube
(at')=0.546in2(lihat tabel 10 kern)at=Nt x at'=0.6445833333ft2144 x
passb. Shellas=ID x C' x B144 x PTC'=PT -
ODt=0.25inB=7inPT=1.25inas=1.011ft27Menghitung Mass Velocity (Gt)a.
TubeGt=Wt575765.32405947lb/(ft2j)atb.
ShellGs=Ws183661.471384615lb/(ft2j)at8Menghitung Reynold Number
(Re) dan Heat Transfer Factor (JH)a. Tubetc=328.1684oFt=0.35cp
=0.8466747744lb/ft j (p.165 maxwell)IDt=0.834in(lihat kern tabel
10, untuk OD 1" BWG 10)0.06950000ftb. Shelltc=615.596oFs=0.4cp
=0.9676283136lb/ft j (p.165 maxwell)De=0.99in0.0825ftRet=D x
Gt47262.1734248436Res =De x
Gs15658.9789449819tsL/D=482.0143884892Dari fig 24 Kern atau fig
10-36 LudwigDari fig 28 kern atau fig 10-44 LudwigJH =250JH
=1169Menghitung Thermal Funtion, ka. Tubetc=t1 +
Fc(t2-t1)=328.17oFt=0.35t =0.8466747744lb/ft j (p.165 maxwell)Lihat
Fig. 16 Kernk((Cp x ) / k )1/3 =0.1834462011Btu/j.ft2.0Ffta.
Shelltc=t1 + Fc(t2-t1)=615.60oFs=0.4s =0.968lb/ft j (p.165
maxwell)Lihat Fig. 16 Kernk((Cp x ) / k )1/3
=0.20328Btu/j.ft2.0Fft10Menghitung Tube-Wall Temperature, twa.
Tubehi=JH x k/D (Cp . /k)1/3=659.878421295Btu/j.ft2.0Fthio=(hi/t) x
(ID/OD)=550.33860336Btu/j.ft2.0Ftb. Shellh0=JH x k/De (Cp .
/k)1/3=285.824Btu/j.ft2.0Fs=328.5102282507oF11Koreksi ho dan hioa.
Tubetw=328.5102282507oFw=0.325cp =0.7861980048lb/ft j (p.165
maxwell)t=0.8466747744lb/ft
jt==1.0104291242hio==556.078153027Btu/j.ft2.0Fb.
Shelltw=328.5102282507oFw=0.9cp =2.1771637056lb/ft j (p.165
maxwell)s=0.968lb/ft
js==0.8927252077ho==255.1622897534Btu/j.ft2.0F12Menghitung Clean
Overall Coefficient, UcUc==174.905203594Btu/j.ft2.0F13Menghitung
Dirt Factor, Rd (fouling Factor)Rd==0.104513315614Friction Factor,
fa. Tubeb. Shelldari figure 26
kernRet=23907.3909Res=23343.9436f=0.00022f=0.001815Specific
Gravity, sa. Tubetc=328.1684oFs=0.765(fig. 39, lampiran)b.
ShellTc=615.596oFs=0.78(fig. 39, lampiran)16Banyak lintasan yang
melintang (Number of Croses)N + 1=12 x L /
B=57.4285714286Ds=IDs=8.66666632ft17Pressure Drop,Pa.
TubePt==3.4849447076PsiPr (Pressure Drop
Return)Pr==0.4810457516dari fig.27 kernv2=0.0232 x g'Pr=Pt +
Pr=3.9659904593Psib. ShellPs==10.0774780389Psi
Perhitungan HE Tipe VertikalPerhitungan Shell and
TubeTipe:VertikalShell sideID:73.5in6.124999755ftpass:1Tube
sideOD:2in (dalam BWG 14 )
=0.0833333333ftL:300in25ftpass:4passjumlah:498buahA:90m2968.7519ft2Baffle:5in0.4166666667ftPitch:1.25inFluida
Panas:Long ResiduLaju alir fluida
W1=2021.60ton/day=185702.15lb/hrT1=642.56oFT2=592.16oFFluida
Dingin:Crude OilLaju alir fluida
W2=4040.20ton/day=371128.73lb/hrt1=317.12oFt2=340.88oF1LMTDFluida
PanasFluida DinginSelisih642.56Higher TempSuhu
tinggi340.88301.68592.16Higher TempSuhu
rendah317.12275.04Selisih26.6LMTD=DTm=288.152Menghitung faktor
koreksi dengan terlebih dahulu menghitung R dan
SR==2.12S==0.0730088496Dari grafik 18 kern atau fig 10.31 A.
Ludwig, makaFT =0.97jadi True Temperature Different,DT=DTm x
FT=279.5101458533oF3Menghitung caloric temperature dan data
physical fluida==0.9116945107T1 - T2=50.40oFSpesific Gravity long
residu pada 60 0F/60 0F =0.90930API==24.114208732Dari fig. 17, Kern
atau fig. 10-32, Ludwig, didapatKc=0.35Fc=0.465Tc=T2 +
Fc(T1-T2)=615.596oFtc=t1 + Fc(t2-t1)=328.1684oFsehingga diperoleh
data2 sebagai berikut :Crude oiltc=328.1684oF0API=35.5t=0.35cpt
=0.847lb/ft j (p.165 maxwell)VABP=572oFB=0.03Cpt + B=0.68(p.93
maxwell)Cpt=0.65Btu/(lb 0F)kt=0.0725Btu/(hr.ft.0F)Long
Residutc=615.596oF0API=24.114208732s=0.4s =0.968lb/ft j (p.165
maxwell)VABP=834.8oFB=0.02Cps + B=0.65(p.93 maxwell)Cps=0.63Btu/(lb
0F)ks=0.0615Btu/(hr.ft.0F) (fig. 1 Kern)4Menentukan Heat DutyQt=Mt
x Cpt x (t2-t1) =5731712.1339192Btu / hrQs=Ms x Cps x (T1-T2)
=5896414.8065088Btu / hr5Menghitung
UDA=968.7519ft2UD==21.1677270301Btu/j ft2 0F6Menghitung Flow Area
(at & as)a. Tube1 in OD BWG 14Flow area per tube
(at')=0.546in2(lihat tabel 10 kern)at=Nt x at'=0.4720625ft2144 x
passb. Shellas=ID x C' x B144 x PTC'=PT -
ODt=-0.75inB=5inPT=1.25inas=-1.53125ft27Menghitung Mass Velocity
(Gt)a. TubeGt=Wt786185.583053091lb/(ft2j)atb.
ShellGs=Ws-121274.876342857lb/(ft2j)at8Menghitung Reynold Number
(Re) dan Heat Transfer Factor (JH)a. Tubetc=328.1684oFt=0.35cp
=0.8466747744lb/ft j (p.165 maxwell)IDt=0.834in(lihat kern tabel
10, untuk OD 1" BWG 10)0.06950000ftb. Shelltc=615.596oFs=0.4cp
=0.9676283136lb/ft j (p.165 maxwell)De=0.99in0.0825ftRet=D x
Gt64534.6946363327Res =De x
Gs-10339.897208115tsL/D=359.7122302158Dari fig 24 Kern atau fig
10-36 LudwigDari fig 28 kern atau fig 10-44 LudwigJH =250JH
=1169Menghitung Thermal Funtion, ka. Tubetc=t1 +
Fc(t2-t1)=328.17oFt=0.35t =0.8466747744lb/ft j (p.165 maxwell)Lihat
Fig. 16 Kernk((Cp x ) / k )1/3 =0.1834462011Btu/j.ft2.0Ffta.
Shelltc=t1 + Fc(t2-t1)=615.60oFs=0.4s =0.968lb/ft j (p.165
maxwell)Lihat Fig. 16 Kernk((Cp x ) / k )1/3
=0.20328Btu/j.ft2.0Fft10Menghitung Tube-Wall Temperature, twa.
Tubehi=JH x k/D (Cp . /k)1/3=659.878421295Btu/j.ft2.0Fthio=(hi/t) x
(ID/OD)=550.33860336Btu/j.ft2.0Ftb. Shellh0=JH x k/De (Cp .
/k)1/3=285.824Btu/j.ft2.0Fs=328.5102282507oF11Koreksi ho dan hioa.
Tubetw=328.5102282507oFw=0.325cp =0.7861980048lb/ft j (p.165
maxwell)t=0.8466747744lb/ft
jt==1.0104291242hio==556.078153027Btu/j.ft2.0Fb.
Shelltw=328.5102282507oFw=0.9cp =2.1771637056lb/ft j (p.165
maxwell)s=0.968lb/ft
js==0.8927252077ho==255.1622897534Btu/j.ft2.0F12Menghitung Clean
Overall Coefficient, UcUc==174.905203594Btu/j.ft2.0F13Menghitung
Dirt Factor, Rd (fouling Factor)Rd==0.041524345114Friction Factor,
fa. Tubeb. Shelldari figure 26
kernRet=23907.3909Res=23343.9436f=0.00022f=0.001815Specific
Gravity, sa. Tubetc=328.1684oFs=0.765(fig. 39, lampiran)b.
ShellTc=615.596oFs=0.78(fig. 39, lampiran)16Banyak lintasan yang
melintang (Number of Croses)N + 1=12 x L /
B=60Ds=IDs=6.124999755ft17Pressure Drop,Pa.
TubePt==4.8489782206PsiPr (Pressure Drop
Return)Pr==0.4810457516dari fig.27 kernv2=0.0232 x g'Pr=Pt +
Pr=5.3300239723Psib. ShellPs==3.2444019955Psi
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