shell side tube side shell side tube side shell side tube side shell side tube side Shell side

1 1 1 2 t T t t S − − = = 0.37 178 199 77 104 = − − Dari Fig. 18, Kern, 1988 didapat Ft = 0.97 CMTD Δt = 88.32 × 0.97 = 85.67 Caloric Temperature T F c dan t c 188.5 2 178 199 2 T T T 2 1 c = + = + = 5 . 99 2 104 77 2 t t t 2 1 c = + = + = F Menghitung jumlah tubes yang digunakan F Dari Tabel 8. Kern, 1965, kondensor untuk fluida panas light organic dan fluida dingin air, diperoleh U D = 75 – 150, faktor pengotor R d Diambil U = 0,003 D = 90 Btujam ⋅ft 2 a. Luas permukaan untuk perpindahan panas, ⋅°F 2 D ft 915 , 165 85.67 90 1279251,64 Δt U Q A = × = × = Luas permukaan luar a ″ = 0.1963 ft 2 Jumlah tube, ft Tabel 10. Kern, 1965 43 , 70 ft ft 0.1963 ft 12 ft 915 , 165 a L A N 2 2 t = × = × = buah Nilai terdekat adalah 70 buah dengan ID shell = 10 in Tabel 9. Kern, 1965 b. Koreksi U D Koefisien menyeluruh kotor t A Q U D ∆ ⋅ = A = 0.1963 × 12 × 70 = 164,892 ft 2 558 , 90 85,67 164,892 1279251,64 = ⋅ = D U Btu h ft 2 Penentuan R F D

1. Flow Area a

design:

a. shell side

Pt 144 B C ID a s × × × = Kern, 1965 Universitas Sumatera Utara Keterangan: C’ = 1 – 0.75 = 0.25 in B = 2.67 in 0.046 1 144 67 . 2 25 . 10 = × × × = s a ft

b. tube side

2 n 144 a Nt a t t × × = a ’t = 0.302 Tabel 10, Kern, 1965 0.037 4 144 0.302 70 = × × = t a ft

2. Mass Velocity G

2

a. shell side

s a W Gs = Kern, 1965 217 , 500812 0.046 23037,362 = = Gs lbh ft G” = 2 3 2 t N L W ⋅ Kern, 1965 G” = h 2 3 2 lbft 579 , 83 70 16 23037,362 = ⋅

b. tube side

t a W Gt = Kern, 1965 568 , 767958 0.037 28414,467 = = Gt lbh ft V = 2 ρ 3600 Gt V = fps 311 , 3 64,428 3600 568 , 767958 = ⋅

3. Koefisien Perpindahan Panas

a. shell side

asumsi awal h o = 200 Btuhr ft2 F Universitas Sumatera Utara

b. tube side

untuk V = 3,311 fps 99.5 F, h i OD ID h h i io × = = 850 Btuhr ft2 F Fig 25, Kern, 1965 950 , 702 75 . 0.62 850 = × = io h Btuhr ft2 F Temperatur dinding T w T w c c c t T ho hio ho t − + + = T w 119,169 5 . 99 188.5 200 950 , 702 200 5 . 99 = − + + = o Temperatur film t F f 159,084 2 119,169 199 2 1 = + = + = w f T T t untuk t F f μf = 1.2 lbft h didapat data sebagai berikut: kf = 0.1 Btu ft h ºF sf = 0.5 kgL dari nilai G” = 739 , 93 lbh ft 2 dan data-data pada t f h didapat, o sebenarnya = 180 Btuft 2

4. Koefisien perpindahan panas menyeluruh bersih Uc

h fig 12.9, Kern, 1965 o io o io h h h h Uc + × = 143,305 180 702,950 180 702,950 = + × = Uc Btu h ft 2

5. Faktor Pengotor R

F D D C D C D U U U U R ⋅ − = 0.037 221 , 93 143,305 221 , 93 143,305 = ⋅ − = D R R D ≥ hitung R D ketentuan, maka spesifikasi dapat diterima. Universitas Sumatera Utara

6. Bilangan Reynold N

Re

a. shell side

f s Gs De Re µ × = in 0.08 75 . 12 4 75 . 1 4 De 2 2 = ⋅ π ⋅ ⋅ π × = 481 , 33387 1.2 217 , 500812 08 . Re = × = s

b. tube side

µ × = Gt D Re t D = ID tube = 0.62 in Tabel 10. Kern, 1965 279 , 24120 1,645 568 , 767958 12 62 , Re = × = t Perhitungan Pressure Drop :

a. Shell side

s e 10 2 s s s D 10 22 . 5 1 N D G f 2 1 P φ ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ = ∆ Kern, 1965 untuk R e 481 , 33387 = , f = 0.0015 ft 2 in 2 Fig.29, Kern, 1965 N+1 = LB Kern, 1965 = 144 2.67 = 53,93 ΔP s ≤ yang diperbolehkan adalah 10 psi, maka ΔP s

b. Tube side