FAKHRI BURHANUDDIN
TEKNIK MESIN
41313120068
PERPINDAHAN PANAS-QUIZ 10
/diketahui :
Aliran air
Tci = 200C = 293K
Tco = 360C=309K
Cpc = 4176,61 J/kg-K
Aliran oli
mh = 4,5 kg/s
Thi = 1600C=433K
Th0=1000C=373K
Cph =2131 J/kg-K
Dimensi Tube
Diameter : 3/4in(OD = 19 mm;0,019m ID = 16 mm;0,016m)
Panjang = 4,8m
Jumlah lintasan
1 pass
α=450
CL = 1
CTP = 0,93
PR = 1,25
dalam keadaan clean
Uc=325 W/m2K
os = 20% = 1,2
Ditanya = berapa diameter shell dan jumlah tube yang diperlukan?
Dijawab :
- Balans energi pada aliran di sisi shell
)( hohihphh TTcmQ 
= (4,5kg/s)( 2131 J/kg-K)(433K-373K)
=958950 J/s
- Laju Aliran massa di sisi tube (Qh=Qc)
)( cicopc
c
c
TTc
Q
m


= 958950 J/s
4176,61 J/kg-K ( 309K-293K)
=14,3 kg/s
- Beda temperatur rata-rata logaritmatik
CFmcSTm TFT ,, 










2
1
21
,
ln
T
T
TT
T CFm ....(1)
ciho TTT  1 =373K- 293K=80K
cohi TTT  2 =433K-309K=124K










2
1
21
,
ln
T
T
TT
T CFm = = ,
= 100,46
- Mencari uf dalam keadaan foul
2,1
f
c
c
f
U
U
A
A
OS
)( cicopccc TTcmQ 
Uf = Uc/1,2
=(325 W/m2K)/1,2
=271 W/m2K
- Luas permukaan perpindahan panas total



STmf
tot
TU
Q
A
,
=
/
/ ( , )
= 35,25m2
-menentukan diameter shell
5,025,0
637,0 












L
dPRA
CTP
CL
D oo
s
=0,637 (1/0,93)0,5((35,25m2)(1,25)2(0,019m)/4,8m)0,5
=0,637 (1.04)(0,47m)
=0,310m
-menentukan jumlah tube dalam shell












 22
2
785,0
o
s
t
dPR
D
CL
CTP
N
=0,785 ((0,93/1))((0,310m)2/(1,25)2(0,019m)2)
=0,785(0.93)(549,58)
=401buah

10

  • 1.
    FAKHRI BURHANUDDIN TEKNIK MESIN 41313120068 PERPINDAHANPANAS-QUIZ 10 /diketahui : Aliran air Tci = 200C = 293K Tco = 360C=309K Cpc = 4176,61 J/kg-K Aliran oli mh = 4,5 kg/s Thi = 1600C=433K Th0=1000C=373K Cph =2131 J/kg-K Dimensi Tube Diameter : 3/4in(OD = 19 mm;0,019m ID = 16 mm;0,016m) Panjang = 4,8m Jumlah lintasan 1 pass α=450 CL = 1 CTP = 0,93 PR = 1,25 dalam keadaan clean Uc=325 W/m2K os = 20% = 1,2
  • 2.
    Ditanya = berapadiameter shell dan jumlah tube yang diperlukan? Dijawab : - Balans energi pada aliran di sisi shell )( hohihphh TTcmQ  = (4,5kg/s)( 2131 J/kg-K)(433K-373K) =958950 J/s - Laju Aliran massa di sisi tube (Qh=Qc) )( cicopc c c TTc Q m   = 958950 J/s 4176,61 J/kg-K ( 309K-293K) =14,3 kg/s - Beda temperatur rata-rata logaritmatik CFmcSTm TFT ,,            2 1 21 , ln T T TT T CFm ....(1) ciho TTT  1 =373K- 293K=80K cohi TTT  2 =433K-309K=124K           2 1 21 , ln T T TT T CFm = = , = 100,46 - Mencari uf dalam keadaan foul 2,1 f c c f U U A A OS )( cicopccc TTcmQ 
  • 3.
    Uf = Uc/1,2 =(325W/m2K)/1,2 =271 W/m2K - Luas permukaan perpindahan panas total    STmf tot TU Q A , = / / ( , ) = 35,25m2 -menentukan diameter shell 5,025,0 637,0              L dPRA CTP CL D oo s =0,637 (1/0,93)0,5((35,25m2)(1,25)2(0,019m)/4,8m)0,5 =0,637 (1.04)(0,47m) =0,310m -menentukan jumlah tube dalam shell              22 2 785,0 o s t dPR D CL CTP N =0,785 ((0,93/1))((0,310m)2/(1,25)2(0,019m)2) =0,785(0.93)(549,58) =401buah