Perhitungan Co Current.docx

Perhitungan co current a)kalor yang dilepas fluida panas Laju alir (L/min ) 0,5

c

T1 k

c

T3 k

44, 4

317, 4

36, 1

309, 1

1

45, 8

318, 8

39, 3

312, 3

1,5

46, 3

319, 3

41, 1

314, 1

∆t T∆V (k) k c

ρ∆V (kg/m3 )

wH (kg/jam )

Cp (kj/kg˚ k)

QH (kg/jam )

8,3 31 3,2 5 6,8 31 5,5 5 5,2 31 6,7

40,2 5

992,17

29,765 1

4,179

1032,4 32

42,5 5

991,49

59,489 4

4,179

1690,5 22

43,7

991

89,19

4,179

1938,1 70

∆t T∆V (k) k c

ρ∆V (kg/m3 )

wC (kg/jam )

Cp (kj/kg˚ k)

QC (kg/jam )

8,3 31 3,2 5 6,8 31 5,5 5 5,2 31 6,7

22,4

996,50

59,79

4,181

5599,5 9

24,6

996,15

59,769

4,181

6147,3 9

26,4

996

59,76

4,181

659621

b)kalor yang diserap oleh fluida dingin Laju alir (L/min ) 1

c

T4 k

c

T6 k

10, 8

283, 8

33, 2

306, 2

1

11, 9

284, 9

36, 5

309, 5

1

10, 8

283, 8

37, 2

310, 2

Data Q2 terkecil  laju alir hot fluid = 1 L/min 3) menghitung flow area a). inner pipe ∆p = 1,9895 x 10-4 m2  0,0021414 ft2 D1 = 0,01592 m b). annulus D2 = 0,0286 De = 0,0382 m2 ∆p = 4,5857 x 10-4 m2

4) menghitung mass velocity (Gp) 𝑤𝑐

59,769

Gp = ∆𝑝 = 4,5857 𝑥 10−4 𝑚2 = 299016,83

5)menghitung viscositas cairan a) inner pipe T∆v = 42,55 ˚c T˚c

µ(kg/ms) 7,96 x 10-4 7,20 x 10-4

30 35 T∆v−T1

µ∆v = µ1 + ( 𝑇2−𝑇1 ) (µ2- µ1) 42,55−30

µ∆v = 7,96 x 10-4 +(

35−30

) (7,20 x 10-4-7,96 x 10-4)

= 6,05 x 10-4 kg/ms b)annulus T∆v = 27 ˚c T˚c

µ(kg/ms) 1,00 x 10-3 7,96 x 10-4

20 30 T∆v−T1

µ∆v = µ1 + ( 𝑇2−𝑇1 ) (µ2- µ1) 27−20

µ∆v = 1,00 x 10-3 +(30−20) (7,96 x 10-4-1,00 x 10-3) = 8,572 x 10-4 kg/ms

6)menghitung bilangan Re a) inner pipe (fluida panas) Re =

𝐷.𝐺𝑝 µ

=

0,01592 𝑚.299016,83 𝑘𝑔 .3600𝑠/𝑗𝑎𝑚 𝑚𝑠

6,05 𝑥 10−4

= 2185,65

b)annulus (fluida dingin) Re =

𝐷.𝐺𝑝 µ

=

0,0286 𝑚.130337,78 𝑘𝑔 .3600𝑠/𝑗𝑎𝑚 𝑚𝑠

8,572 𝑥 10−4

= 1207,65 7)Menghitung JH dari grafik a)inner pipe Re = 2185,65 JH = 6 b)annulus Re = 1207,65 JH =20 8)menghitung konduktivitas termal air (K) a)inner pipe T∆v = 33,45 ˚c = 92,21 ˚f T˚f 86 140

K(Btu/hr.ft˚f ) 0,381 0,398

T∆v−T1

K = K1+ ( 𝑇2−𝑇1 ) (K2-K1) 92,21−86

= 0,381+ ( 140−86 ) (0,398 - 0,381) = 0,388 Btu/hr.ft˚f b)annulus T∆v = 27 ˚c = 80,6 ˚f Dari Ekstrapolasi maka didapatkan nilai K yaitu = 0,379 btu/hr.ft ˚f 9)menghitung bilangan ( Cµ/K)1/3 a)inner pipe ∆Tv = 108,9 ˚f C = 0,999 Btu/lb˚f

µ = 1,46354 lb/ft jam K= 0,38295 Btu/hr ft ˚f (C

𝑏𝑡𝑢 𝑙𝑏 𝑓.1,79873 𝑗𝑎𝑚 𝑙𝑏 𝑓𝑡 0,388𝑏𝑡𝑢 𝑓𝑡 𝑓 ℎ𝑟

0,997

µ/K 1/3

)

=

= 1,5666 b)annulus ∆Tv = 80,6 ˚f C = 0,998 Btu/lb˚f K = 0,379 Btu/hr ft˚f µ = 2,0736 lb/ft jam (C

𝑏𝑡𝑢 𝑙𝑏 𝑓.2,0736 𝑗𝑎𝑚 𝑙𝑏 𝑓𝑡 0,379𝑏𝑡𝑢 𝑓𝑡 𝑓 ℎ𝑟

0,998

µ/K 1/3

)

=

= 1,76091 10)menghitung h1 (fluida panas) dan h0 (fluigda dingin) a) h1 inner pipe (fluida panas) D1 inner pipe = 0,03937 ft h1 = JH x

𝐾

( Cµ/K)1/3

𝐷

𝑏𝑡𝑢 𝑗𝑎𝑚 𝑓𝑡2

0,388

=6x

0,03937 𝑓𝑡

(1,5666)

= 92,65 btu/ft2 jam ˚f b)annulus De annulus = 0,12532 ft 𝐾

h0= JH x

𝐷𝑒

( Cµ/K)1/3 𝑏𝑡𝑢 𝑗𝑎𝑚 𝑓𝑡2

0,379

= 20 x

0,12532

𝐹

(1,76091)

= 106,508 btu/ft2 jam ˚f

11)menghitung h10

ID = 0,0159 m = 0,05216 ft OD = 0,0286 m = 0,0938 ft 𝐼𝐷

h10 = h1 x 𝑂𝐷 = 92,65 x

0,05216 ft 0,0938 ft

= 45,15 12) menghitung Uc Uc =

ℎ10 𝑥ℎ0 (ℎ10+ℎ0) 51,5205 𝑥 106,508

= (51,5205+106,508) = 34,7237 13)menghitung Ud Rd = 0,001 1

1

Rd = 𝑈𝐷 - 𝑈𝑐 0,001 =

1 𝑈𝐷

-

1 34,7237

1

0,001 = 𝑈𝐷 − 0,0287987 UD = 33,670 14) menghitung efisiensi alat a) Qin = Qout Qshell =Qtube  Qdingin = Q panas n.Cp.Dt = A.Ud.LMTD n.Cp.LMTD = A.Ud.LMTD 

Qin = n.Cp.Dt LMTD = 12,47 ˚C = 54,46˚F 1 𝑙𝑏

n = 18 𝑙𝑏/𝑙𝑏𝑚𝑜𝑙 = 0,055 lbmol Cp = 0,999 Qin = n.Cp.LMTD = 0,055 lbmol . 0,999 . 54,46˚F





= 2,7202 Qout = Ud.Atube.LMTD A = 0,0021414 Ud = 33,670 LMTD = 54,46˚F Qout = Ud.Atube.LMTD = 28,12 . 0,0021414 . 54,46˚F = 3,5870 EFISIENSI =

𝑄𝑖𝑛 𝑄𝑜𝑢𝑡

x 100%

2,7202

= 3,9266 x 100% = 69,27%