Redesign dan Recalculation High Pressure Heater (HPH) 7 PT. PJB UP Paiton pada Zona Subcooled Menggunakan Analisa Termodinamika dan Perpindahan Panas

Redesign dan Recalculation High Pressure Heater (HPH) 7 PT. PJB UP Paiton pada Zona Subcooled Menggunakan Analisa Termodinamika dan Perpindahan Panas

Dhany Ahmad Barkah 2107 100 143

Dosen Pembimbing

Dr. Eng. Ir. Prabowo, M.Eng

WTP

Latar Belakang

• Kerusakan pada tube.

Latar Belakang

• Kerusakan pada baffle. • Kerusakan pada shell.

Latar Belakang

• Replugging hampir 10% dari total tube.

Latar Belakang

Kerusakan pada HPH 7

Turunnya efisiensi termal pembangkit

Konsumsi batubara semakin meningkat

Semakin besarnya biaya operasi

Perlu redesign dan recalculation

Tujuan Perancangan

• Mendapatkan dimensi zona subcooled.

• Mendapatkan

karakteristik tekanan dan kecepatan yang diizinkan.

• Mendapatkan performa dari HPH 7

Dimensi Subcooled

PENELITIAN TERDAHULU

Irfan S. Husaini, Syed M. Zubair, M. A.

Antar (2005)

Irfan S. Husaini, Syed M. Zubair,

M. A. Antar (2005)

Mohammed A. Antar, Syed M. Zubair

(2006)

Simin Wang, Jian Wen, Yanzhong Li (2008)

METODOLOGI

FLOWCHART PERANCANGAN

HPH 7

A

B

C

Asumsi Overall heat transfer coefficient Subcooling (Uas,sub)

1300 ≤ Uas,sub ≤ 2500

Asumsi Overall heat transfer coefficient Condensing (Uas,cond)

1000 ≤ Uas,cond ≤ 4000 Menghitung Luas Perpan Subcooling (Asub) Menghitung Luas Perpan Condensing (Acond)

Menghitung Panjang tube Subcooling Menghitung Panjang tube Condensing

Menghitung Diameter Dalam Shell (IDshell)

Menghitung Crossflow Area pada Subcooling

Zone Menghitung Crossflow Area pada Condensing

Zone

Menghitung Shell Side Mass Velocity

Subcooling Zone Menghitung Shell Side Mass Velocity Condensing Zone

Menghitung Kecepatan Maksimum Fluida sisi Shell, Subcooling

Menghitung Reynolds Max sisi Shell, Subcooling

Menghitung Nusselt number Subcooling

Menghitung Koefisien Konveksi sisi Shell, Subcooling

Menghitung Koefisien Konveksi sisi Shell, Condensing sub

sub as

sub

sub U LMTD

A Q



 

, ascond cond

cond

cond U LMTD

A Q



 

,

Nt do L A

tube sub

sub 

Nt do L A

tube cond

cond  

 ²  ²

637 .

0 do N PR

CTP

ID_shell CL  _tube  _t

  1

  Nb

L

As _sub ID ^{shell cond}

cond cond

cond

cond As

Vs m



  

sub sub

sub

sub As

Vs m



 

sub tube

sub Vs

do ST

V ST 

 





  max

sub tube sub sub sub

do d V



  

 max

max Re

Re max  Pr ³¹ 13

.

1 sub

m sub sub

sub C d

Nu    

sub sub sub

sub do

k ho  Nu 

 

 

4 1

, ,

3 ' 729

.

0 













 

tube cold mean hot sat f

fg l v l l

cond N T T do

h k ho g









 1

  Nb

L

As _cond ID^{shell cond}

START

INPUT :

1. Feedwater : - Temp. Inlet feedwater - Temp. outlet feedwater - Laju alir massa feedwater - Tekanan kerja feedwater 2. Heating Steam : - Temp. Inlet steam

- Laju alir massa steam - Tekanan kerja steam - Temp. outlet steam

Dapatkan :

 Properti feedwater sisi tube

 Properti steam sisi shell

Hitung Tx1 dan Tx2

Input Dimensi :

- Bahan tube K tube dari asumsi - Jumlah laluan Ni dari asumsi - Layout tube dari asumsi

- Tube Pitch

- Asumsi kecepatan dalam tube

A

Menghitung Laju Perpindahan Panas :

Menghitung ∆LMTD Masing-masing Zona

E

2 . . 2

4 4 1

di v Nt m

Nt di

m v













) (

) (

) (

) (

sup

liquid f

steam sub

f g steam

konden

g steam steam

de

ci co c

c

h h m

q

h h m

q

h h

m q

T T Cp m

q



































   

 

 

   

 

 ^f_o ¹_i

i o 1

subcooling f

1 f

2 g

1 f 2

g condensasi

Tc Th

Tx ln Th

Tc Th Tx

ΔTLMTD Th

Tx Th

Tx ln Th

Tx Th Tx

ΔTLMTD Th









 









 

1 (1 ²)^1/2 2 Cr Cr

 

 

^{ } _^







  ₂² ₁¹_/^/₂² ) 1 ( exp 1

) 1 ( exp

1

Cr NTU

Cr NTU

1 (1 ²)^1/2 2 Cr Cr

 

 

^{ } _^







  ₂² ₁¹_/^/₂² ) 1 ( exp

1

) 1 ( exp

1

Cr NTU

Cr NTU

D

Hitung :

- Kapasitas panas fluida panas (Ch) - Kapasitas panas fluida dingin (Cc)

Cc< Ch

h c

r C

C C

Hitung :

- Rasio kapasitas panas

- NTU

- Effektiveness

Cc

NTU UA

c h

r C

C C

Hitung :

- Rasio kapasitas panas

- NTU

- Effektiveness

Ch

NTU UA

YES NO

∆P ≤ 1.5 E

YES NO

Menghitung pressure drop sisi tube Pressure drop major

Pressure drop minor

mnor

major p

p

p    



di v L f

p_major ²

2 1





180 , ,

,entr mnorext mnorreturn mnor

mnor p p p

p   



, 2 2

1 v k p_mnorentr  



, 2 2

1 v k p_mnorext  



2 180

, 2

1 v k p_mnorreturn  



04 . 0

 k

1

 k

2.

0

 k

Finish

FLOWCHART PERHITUNGAN

VARIASI KECEPATAN DALAM TUBE

START

Variasi kecepatan Uas,sub

V1 = 1.8 m/s Uas,cond

V2 = 2.1 m/s ρ

V3 = 2.44 m/s di

V4 = 3 m/s

Vi = V1

2

4 di v Nt m

i



 

Menghitung Luas Perpan Subcooling (Asub) Menghitung Luas Perpan Condensing (Acond)

Menghitung Panjang tube Subcooling Menghitung Panjang tube Condensing

Menghitung Reynolds Number Sisi Tube

Menghitung Nusselt number Sisi Tube

Menghitung Koefisien Konveksi Sisi Tube

Menghitung pressure drop sisi tube Pressure drop major

Pressure drop minor

mnor

major p

p

p    



di v L f

p_major ²

2 1 





180 , ,

,entr mnor ext mnor return

mnor

mnor p p p

p    



, 2 2

1 v k pmnor entr  



, 2

2 1 v k pmnor ext  



2 180

, 2

1 v k

p_{mnor return}  



04 . 0

 k

1 k 

2 . 0 k

AA

1

 _i

i V

V AB

sub sub

as

sub

sub U LMTD

A Q



 

,

Nt do

L A

tube sub

sub  

cond cond

as

cond

cond U LMTD

A Q



 

,

Nt do

L A

tube cond

cond   

tube tube tube tube tube

di d V



  

Re 

4 . 0 5

/

4 Pr

Re 023 ,

0 tube tube

Nutube  

tube tube tube

tube di

k

hi Nu 



m

AA

Vi < V4 AB

Grafik hi & ∆p vs Re

FINISH

Ya

Tidak

FLOWCHART PERHITUNGAN

VARIASI JUMLAH BAFFLE

START

- Kecepatan optimal,Vtube - Lsubcooled

- Nb1 = 1 - Nb3 = 4 - Nb2 = 2

Menghitung Diameter Dalam Shell (IDshell)

Nbi = Nb1

) 1

( 

 Nbi B L^subcooled

Menghitung Crossflow Area pada Subcooling Zone

Menghitung Shell Side Mass Velocity Subcooling Zone

Menghitung Kecepatan Maksimum Fluida sisi Shell, Subcooling

Menghitung Reynolds Max sisi Shell, Subcooling

Menghitung Nusselt number Subcooling

Menghitung Koefisien Konveksi sisi Shell, Subcooling

- Menghitung nilai pressure drop sisi shell zona condensing

- Menghitung nilai pressure drop sisi shell zona subcooled

 

s De N Ds Gs pshell cond f

. 10 22 . 5

1 .

2 1

10 2

, 

 



v f N

p_{shellsub L} 

 



 

 2

max2 ,

 

Menghitung Crossflow Area pada Condensing Zone

Menghitung Shell Side Mass Velocity Condensing Zone

Menghitung Koefisien Konveksi sisi Shell, Condensing

BB

1 BA

 _i

i Nb

Nb

 

 ^{2 2}

637 .

0 do N PR

CTP

IDshell  CL  tube  t

T shell

sub S

C B

As ID  



sub sub

sub

sub As

Vs m



 

cond shell

cond ID L

As 0.5 

cond cond

cond

cond As

Vs m



  

sub tube

sub Vs

do ST

V ST 







  max

sub tube sub sub

sub

do d V



  

 max

max Re

Re max  Pr ³¹ 13

.

1 sub

m sub sub

sub C d

Nu    

sub sub sub

sub do

k

ho Nu 



 

 

4 1

, ,

3 ' 729

.

0 













 

tube cold mean hot sat f

fg l v l l

cond N T T do

h k ho g









FLOWCHART PERHITUNGAN NTU &

ε

ANALISA TERMODINAMIKA

Control Volume HPH 7

m

𝑚 =125.280 Kg/hr T = 349 °C P =41,9 Kg/cm²

𝑚 = 1.351,2 Kg/hr

T = 206 °C 𝑚 = 1.351,2 Kg/hr

T = 250 °C

𝑚 = 125.280 Kg/hr

T = 215 °C

properties of steam

T inlet (C) 349 T outlet (C) 215 Pressure (bar) 41.58

h inlet (J/kg) 3085200 h outlet (J/kg) 920940

34.8 T sat 252.63 hg(J/kg) 2798400 hf (J/kg) 1098200 𝑚 h(kg/s)

properties of water

T inlet (C) 206 T rata-rata (C) 228 T outlet (C) 250 ρ (kg/m3) 843.1 Pressure (bar) 193.19 k (W/m.K) 0.6607

h inlet (J/kg) 886680 µ 1.21E-04 h outlet (J/kg) 1087400 Pr 0.835727

375.32 Tx1 (C) 209.6256293 Cp (J/kg.K) 4533.2 Tx2 (C) 244.1338684

𝑚 (kg/s)

𝑄_𝑕 = 𝑚 _{𝑠𝑡𝑒𝑎𝑚} 𝑕_{𝑕𝑜𝑡,𝑖} − 𝑕_{𝑕𝑜𝑡,𝑜}

𝑄_𝑕 = 34.8^𝑘𝑔_𝑠 (3085200 − 920940)_𝑘𝑔^𝐽 𝑄_𝑕 = 75316248^𝐽_𝑠

𝑄_𝑐 = 𝑚 _{𝑐𝑜𝑙𝑑}𝐶𝑝_{𝑐𝑜𝑙𝑑} 𝑇_{𝑐𝑜𝑙𝑑,𝑜} − 𝑇_{𝑐𝑜𝑙𝑑,𝑖} 𝑄_𝑐 = 375.32^𝑘𝑔_𝑠 𝑥 4533.2_{𝑘𝑔 𝐶}^𝐽 𝑥 (250 − 206) 𝐶^𝑜

𝑄_𝑐 = 74861627.46^𝐽_𝑠

ANALISA TERMODINAMIKA

Feedwater in Feedwater out Steam in

Steam out Zona desuperheating

Zona condensing

Zona subcooled

𝑄_{𝑑𝑒𝑠𝑢𝑝} = 𝑚 _{𝑠𝑡𝑒𝑎𝑚} 𝑕_{𝑕𝑜𝑡,𝑖} − 𝑕_𝑔

𝑄_{𝑑𝑒𝑠𝑢𝑝} = 34.8^𝑘𝑔_𝑠 3085200 − 2798400 _𝑘𝑔^𝐽 𝑄_{𝑑𝑒𝑠𝑢𝑝} = 9980640 J/s

𝑄_{𝑐𝑜𝑛𝑑} = 𝑚 _{𝑠𝑡𝑒𝑎𝑚} 𝑕_𝑔 − 𝑕_𝑓

𝑄_{𝑐𝑜𝑛𝑑} = 34.8^𝑘𝑔_𝑠 2798400 − 1098200 _𝑘𝑔^𝐽 𝑄_{𝑐𝑜𝑛𝑑} = 59166960 J/s

𝑄_𝑠𝑢𝑏 = 𝑚 _{𝑠𝑡𝑒𝑎𝑚} 𝑕_𝑓 − 𝑕_{𝑕𝑜𝑡,𝑜}

𝑄_𝑠𝑢𝑏 = 34.8^𝑘𝑔_𝑠 1098200 − 920940 _𝑘𝑔^𝐽 𝑄_𝑠𝑢𝑏 = 6168648 𝐽/𝑠

𝑄_{𝑠𝑢𝑏,𝑕𝑜𝑡} = 𝑄_{𝑠𝑢𝑏,𝑐𝑜𝑙𝑑}

6168648 𝐽/𝑠 = 𝑚 _{𝑐𝑜𝑙𝑑}𝐶𝑝_{𝑐𝑜𝑙𝑑}(𝑇_𝑥1 − 𝑇_{𝑐𝑜𝑙𝑑,𝑖}) 𝑇_𝑥1 = _𝑚 ^{𝑄𝑠𝑢𝑏,𝑠𝑡𝑒𝑎𝑚}

𝑓𝑒𝑒𝑑𝑤𝑎𝑡𝑒𝑟𝐶𝑝 + 𝑇_{𝑐𝑜𝑙𝑑,𝑖} 𝑇_𝑥1 = 209.6 𝐶^𝑜

𝑄_{𝑑𝑒𝑠𝑢𝑝,𝑕𝑜𝑡} = 𝑄_{𝑑𝑒𝑠𝑢𝑝,𝑐𝑜𝑙𝑑}

9980640 J/s = 𝑚 _{𝑐𝑜𝑙𝑑}𝐶𝑝_{𝑐𝑜𝑙𝑑}(𝑇_{𝑐𝑜𝑙𝑑,𝑜} − 𝑇_𝑥2) 𝑇_𝑥2 = 𝑇_{𝑐𝑜𝑙𝑑,𝑜} − ( ^𝑄𝑑𝑒𝑠𝑢𝑝,𝑠𝑡𝑒𝑎𝑚

𝑚 _{𝑓𝑒𝑒𝑑𝑤𝑎𝑡𝑒𝑟}𝐶𝑝_{𝑐𝑜𝑙𝑑}) 𝑇_𝑥2 = 244.13 𝐶^𝑜

ANALISA TERMODINAMIKA

Tho

Tsat1

Tx2 Tco

Tsat2

349

Desuperheating Thi Subcooling

215

Condensing T (^oC)

206 252.63

D.C.A

T.T.D 250

Tx1 Tci

x

ANALISA PERPINDAHAN PANAS

METODE LMTD

∆𝑇𝐿𝑀𝑇𝐷_𝑠𝑢𝑏= ^{𝑇𝑕𝑠𝑎𝑡;𝑇𝑥2 ; 𝑇𝑕𝑠𝑎𝑡;𝑇𝑥1}

ln ^{𝑇𝑕𝑠𝑎𝑡−𝑇𝑥2}

𝑇𝑕𝑠𝑎𝑡−𝑇𝑥1

∆𝑇𝐿𝑀𝑇𝐷_𝑠𝑢𝑏= 252.63;244.13 ; 252.63;209.6 ln 252.63−244.13

252.63−209.6

∆𝑇𝐿𝑀𝑇𝐷_𝑠𝑢𝑏= 19.151

∆𝑇𝐿𝑀𝑇𝐷_{𝑐𝑜𝑛𝑑}= ^{𝑇𝑕𝑠𝑎𝑡;𝑇𝑥1 ; 𝑇𝑕𝑜;𝑇𝑐𝑖}

ln ^{𝑇𝑕𝑠𝑎𝑡−𝑇𝑥1}

𝑇𝑕𝑜−𝑇𝑐𝑖

∆𝑇𝐿𝑀𝑇𝐷_{𝑐𝑜𝑛𝑑}= 252.63;209.6 ; 215;206 ln 252.63−209.6

215−206

∆𝑇𝐿𝑀𝑇𝐷_{𝑐𝑜𝑛𝑑}= 19.5668

Analisa Perpan Bagian Internal

variasi kecepatan 1.8

2.1 2.44

3

𝑁𝑡 = _𝜌 ^4𝑚𝑐

𝑓𝑒𝑒𝑑𝑤𝑎𝑡𝑒𝑟𝑣_{𝑡𝑢𝑏𝑒}𝜋𝑑𝑖² =

4 𝑥 34.8^𝑘𝑔_𝑠

843.1_𝑚3^𝑘𝑔 𝑥 2.44^𝑚_𝑠 𝑥 𝜋 𝑥 0.011659𝑚 ² 𝑁𝑡 = 1709.032 ≈ 1710

METODE LMTD Analisa Perpan Bagian Internal

𝐴_𝑠𝑢𝑏 = _𝑈 ^{𝑄𝑠𝑢𝑏}

𝑠𝑢𝑏 𝑥 ∆𝐿𝑀𝑇𝐷_𝑠𝑢𝑏

𝐴_{𝑐𝑜𝑛𝑑} = _𝑈 ^{𝑄𝑐𝑜𝑛𝑑}

𝑐𝑜𝑛𝑑 𝑥 ∆𝐿𝑀𝑇𝐷_{𝑐𝑜𝑛𝑑} 𝐴_𝑠𝑢𝑏 = 4226.338 𝑥 19.57^{6168648 J}

𝐴_{𝑐𝑜𝑛𝑑} = 2942.43 𝑥 19.15^{59166960 J} 𝐴_𝑠𝑢𝑏 = 74.594 𝑚²

𝐴_{𝑐𝑜𝑛𝑑} = 1049.968 𝑚²

𝐿_𝑠𝑢𝑏 = _𝜋𝑑𝑜^{𝐴𝑠𝑢𝑏}

𝑡𝑢𝑏𝑒𝑁𝑡

𝐿_{𝑐𝑜𝑛𝑑} = _𝜋𝑑𝑜^{𝐴𝑐𝑜𝑛𝑑}

𝑡𝑢𝑏𝑒𝑁𝑡 𝐿_𝑠𝑢𝑏 = 𝜋 𝑥 0.01588 𝑚 𝑥 1710^{74.594 𝑚2}

𝐿_{𝑐𝑜𝑛𝑑} = 𝜋 𝑥 0.01588 𝑚 𝑥 1710^{1049.968 𝑚2} 𝐿_𝑠𝑢𝑏 = 0.874 𝑚

𝐿_{𝑐𝑜𝑛𝑑} = 12.308 𝑚 𝑅𝑒𝑑_{𝑡𝑢𝑏𝑒} = ^{𝜌𝑡𝑢𝑏𝑒 𝑥 𝑉}_𝜇^{𝑡𝑢𝑏𝑒 𝑥 𝑑𝑖𝑡𝑢𝑏𝑒}

𝑡𝑢𝑏𝑒

𝑅𝑒𝑑_{𝑡𝑢𝑏𝑒} = ^843.1

𝑘𝑔

𝑚3 𝑥 2.44^𝑚_𝑠 𝑥 0.011659 𝑚 1.21𝐸;04 𝑃𝑎.𝑠 𝑅𝑒𝑑_{𝑡𝑢𝑏𝑒} = 198218.802

METODE LMTD Analisa Perpan Bagian Internal

𝑁𝑢_{𝑡𝑢𝑏𝑒} = 0.023 𝑥 𝑅𝑒_{𝑡𝑢𝑏𝑒}^4/5 𝑥 𝑃𝑟_{𝑡𝑢𝑏𝑒}^0.4

𝑁𝑢_{𝑡𝑢𝑏𝑒} = 0.023 𝑥 198218.802 ^4/5 𝑥 0.835727^0.4 𝑁𝑢_{𝑡𝑢𝑏𝑒} = 369.1302

𝑕𝑖_{𝑡𝑢𝑏𝑒} = ^{𝑁𝑢𝑡𝑢𝑏𝑒}_𝑑𝑖 ^{𝑥 𝑘𝑡𝑢𝑏𝑒}

𝑡𝑢𝑏𝑒

𝑕𝑖_{𝑡𝑢𝑏𝑒} = 369.1302 𝑥 0.6607_𝑚.𝐾^𝑊 0.011659 𝑚

𝑕𝑖_{𝑡𝑢𝑏𝑒} = 20918.1163_𝑚^𝑊_2.𝐾

variasi kecepatan Ltot hi Re

1.8 12.81 16462.41 146469.08

2.1 14.82 18623.05 170880.60

2.44 17.14 20998.45 198546.98

3 21.13 24772.65 244115.14

METODE LMTD Analisa Perpan Bagian Internal

Pressure Drop

Pressure Drop Major Pressure Drop Minor

∆𝑝_{𝑚𝑎𝑦𝑜𝑟}= 𝑓1

2𝜌𝑣² 𝐿_𝑡𝑜𝑡 𝑑𝑖

∆𝑝_{𝑚𝑎𝑦𝑜𝑟}= 0.029 𝑥 ¹₂ 𝑥 843.1_𝑚^𝑘𝑔₃ 𝑥 (2.44 𝑚/𝑠)² 𝑥 _{0.011659 𝑚}^{17.418 𝑚}

∆𝑝_{𝑚𝑎𝑦𝑜𝑟}= 111797.183 𝑃𝑎 = 1.139 𝑘𝑔/𝑐𝑚²

∆𝑝_{𝑚𝑖𝑛𝑜𝑟}= ∆𝑝_{𝑚𝑖𝑛𝑜𝑟,𝑒𝑛𝑡𝑟} + ∆𝑝_{𝑚𝑖𝑛𝑜𝑟,𝑒𝑥𝑡} + ∆𝑝𝑚𝑖𝑛𝑜𝑟,𝑟𝑡𝑟𝑛 𝑏𝑒𝑛𝑑

METODE LMTD Analisa Perpan Bagian Internal

Pressure Drop Minor

∆𝑝_{𝑚𝑖𝑛𝑜𝑟,𝑒𝑛𝑡𝑟}= ^𝑘𝜌𝑣₂ ²

∆𝑝_{𝑚𝑖𝑛𝑜𝑟,𝑒𝑛𝑡𝑟}= 0.04 𝑥 866.83 _𝑚3^𝑘𝑔 𝑥 2.44^𝑚_𝑠 ²

2

∆𝑝_{𝑚𝑖𝑛𝑜𝑟,𝑒𝑛𝑡𝑟}= 103.215 𝑃𝑎 = 0.001052 _𝑐𝑚^𝑘𝑔₂

∆𝑝_{𝑚𝑖𝑛𝑜𝑟,𝑒𝑥𝑡}= 𝑘𝜌𝑣² 2

∆𝑝_{𝑚𝑖𝑛𝑜𝑟,𝑒𝑥𝑡}= 1 𝑥 819.69 _𝑚3^𝑘𝑔 𝑥 2.44^𝑚_𝑠 ²

2

∆𝑝_{𝑚𝑖𝑛𝑜𝑟,𝑒𝑥𝑡}= 2440.0532 𝑃𝑎 = 0.0249 𝑘𝑔/𝑐𝑚²

∆𝑝𝑚𝑖𝑛𝑜𝑟,𝑟𝑡𝑟𝑛 𝑏𝑒𝑛𝑑= 𝑘𝜌𝑣² 2

∆𝑝𝑚𝑖𝑛𝑜𝑟,𝑟𝑡𝑟𝑛 𝑏𝑒𝑛𝑑= 0.2 𝑥 844.43 ^𝑘𝑔

𝑚3 𝑥 2.44^𝑚_𝑠 ²

2

∆𝑝𝑚𝑖𝑛𝑜𝑟,𝑟𝑡𝑟𝑛 𝑏𝑒𝑛𝑑= 502.7398 𝑃𝑎 = 0.00512 𝑘𝑔/𝑐𝑚²

METODE LMTD Analisa Perpan Bagian Internal

Pressure Drop

∆𝑝_𝑡𝑜𝑡= ∆𝑝_{𝑚𝑎𝑦𝑜𝑟} + ∆𝑝_{𝑚𝑖𝑛𝑜𝑟,𝑒𝑛𝑡𝑟} + ∆𝑝_{𝑚𝑖𝑛𝑜𝑟,𝑒𝑥𝑡} + ∆𝑝𝑚𝑖𝑛𝑜𝑟,𝑟𝑡𝑟𝑛 𝑏𝑒𝑛𝑑

∆𝑝_𝑡𝑜𝑡= 1.139 𝑘𝑔/𝑐𝑚² + 0.001052 _𝑐𝑚^𝑘𝑔₂ + 0.0249 𝑘𝑔/𝑐𝑚² + 0.00512 𝑘𝑔/𝑐𝑚²

∆𝑝_𝑡𝑜𝑡= 1.17 𝑘𝑔/𝑐𝑚²

variasi kecepatan ∆p mjr tube (kg/cm2) ∆p mnr (kg/cm2) ∆p total (kg/cm2) 1.8 0.455882183 0.016891593 0.472773776 2.1 0.717783678 0.022991335 0.740775013 2.44 1.120755218 0.031038824 1.151794042

3 2.05300897 0.046921092 2.099930062

METODE LMTD Analisa Perpan Bagian Internal

𝑣 = ^𝑅𝑒𝜇_𝜌𝐷

𝑣 = 198546.98 𝑥 1.21𝑥10⁻⁴ 𝑃𝑎.𝑠 843.1^𝑘𝑔

𝑚3𝑥 0.011659 𝑚 𝑣 = 2.44 𝑚/𝑠

METODE LMTD Analisa Perpan Bagian Eksternal

𝐼𝐷_{𝑠𝑕𝑒𝑙𝑙} = 0.637 _𝐶𝑇𝑃^𝐶𝐿 𝜋𝑑𝑜²𝑁𝑡(𝑃𝑅²)

𝐼𝐷_{𝑠𝑕𝑒𝑙𝑙} = 0.637 ^0.87_0.9 𝜋 𝑥 (0.01588 𝑚)² 𝑥 1710 𝑥 1.385² 𝐼𝐷_{𝑠𝑕𝑒𝑙𝑙} = 1.0099 𝑚

𝐴𝑠_𝑠𝑢𝑏 = ^{𝐼𝐷𝑠𝑕𝑒𝑙𝑙}₂ ^{𝑥 𝐿𝑠𝑢𝑏}

𝐴𝑠_𝑠𝑢𝑏 = 1.0099 𝑚 𝑥 0.8731 𝑚

2 𝑚

𝐴𝑠_𝑠𝑢𝑏 = 0.44086 𝑚²

𝐴𝑆_{𝑐𝑜𝑛𝑑} = 0.5 𝑥 𝐼𝐷_{𝑠𝑕𝑒𝑙𝑙} 𝑥 𝐿_{𝑐𝑜𝑛𝑑}

𝐴𝑠_{𝑐𝑜𝑛𝑑} = 0.5 𝑥 1.0099 𝑚 𝑥 12.308 𝑚 𝐴𝑠_{𝑐𝑜𝑛𝑑} = 6.215 𝑚²

𝑉𝑠_𝑠𝑢𝑏 = _𝐴𝑠 ^{𝑚 𝑠𝑢𝑏}

𝑠𝑢𝑏 𝑥 𝜌_𝑠𝑢𝑏 𝑉𝑠_{𝑐𝑜𝑛𝑑} = _𝐴𝑠 ^{𝑚 𝑐𝑜𝑛𝑑}

𝑐𝑜𝑛𝑑 𝑥 𝜌_{𝑐𝑜𝑛𝑑}

𝑉𝑠_𝑠𝑢𝑏 = ^{34.8 𝑘𝑔/𝑠}

0.44086 𝑚² 𝑥 851.52 𝑘𝑔/𝑚³ 𝑉𝑠_{𝑐𝑜𝑛𝑑} = ^{34.8 𝑘𝑔/𝑠}

6.215 𝑚² 𝑥 818.53 𝑘𝑔/𝑚³

𝑉𝑠_𝑠𝑢𝑏 = 0.0926 𝑚/𝑠 𝑉𝑠_{𝑐𝑜𝑛𝑑} = 0.0068 𝑚/𝑠

METODE LMTD Analisa Perpan Bagian Eksternal

𝑉𝑚𝑎𝑥 = _{(𝑆𝑇;𝑑𝑜)}^𝑆𝑇 𝑉𝑠_𝑠𝑢𝑏

𝑉𝑚𝑎𝑥 = (0.022 𝑚;0.01588 𝑚)^{0.022 𝑚} 0.0926 𝑚/𝑠 𝑉𝑚𝑎𝑥 = 0.333 𝑚/𝑠

𝑅𝑒𝑑, 𝑚𝑎𝑥 = 𝜌_𝑠𝑢𝑏 𝑥 𝑉𝑚𝑎𝑥_𝑠𝑢𝑏 𝑥 𝑑𝑜 𝜇_𝑠𝑢𝑏

𝑅𝑒𝑑, 𝑚𝑎𝑥 = ^825.58

𝑘𝑔

𝑚3𝑥0.333^𝑚_𝑠𝑥 0.01588 𝑚 0.000114 𝑃𝑎.𝑠 𝑅𝑒𝑑, 𝑚𝑎𝑥 = 38265.433

𝑁𝑢_𝑠𝑢𝑏 = 1.13 𝑥 𝐶_𝑠𝑢𝑏 𝑥 𝑅𝑒𝑑, 𝑚𝑎𝑥_𝑠𝑢𝑏 ^𝑚 𝑥 𝑃𝑟_𝑠𝑢𝑏¹³

𝑁𝑢_𝑠𝑢𝑏 = 1.13 𝑥 0.470501 𝑥 38265.433 ^0.563917 𝑥 0.844001^1/3 𝑁𝑢_𝑠𝑢𝑏 = 193.043

METODE LMTD Analisa Perpan Bagian Eksternal

𝑕𝑜_𝑠𝑢𝑏 = 𝑁𝑢_𝑠𝑢𝑏 𝑥 𝑘_𝑠𝑢𝑏 𝑑𝑜_𝑠𝑢𝑏

𝑕𝑜_𝑠𝑢𝑏 = 193.043 𝑥 0.6367 _𝑚.𝐾^𝑊 0.01588 𝑚

𝑕𝑜_𝑠𝑢𝑏 = 8102.227^𝑊 _𝑚2.𝐾

𝑕𝑜_{𝑐𝑜𝑛𝑑} = 0.729 𝑔𝜌_𝑙(𝜌_𝑙 − 𝜌_𝑔)𝑘_𝑙³𝑕′_𝑓𝑔

𝑁𝑙 𝑥 𝜇_𝑙(𝑇_{𝑚𝑒𝑎𝑛,𝑐𝑜𝑙𝑑}− 𝑇_{𝑠𝑎𝑡,𝑕𝑜𝑡})𝑑𝑜_{𝑡𝑢𝑏𝑒}

1/4

𝑕𝑜_{𝑐𝑜𝑛𝑑} = 0.729 ^9.81

𝑚

𝑠2 𝑥 818.53 𝑘𝑔/𝑚³(818.53;20.904)𝑘𝑔/𝑚³ 𝑥 (0.6303_𝑚.𝐾^𝑊 )³ 𝑥 1743615.77 𝐽/𝑘𝑔 88 𝑥 0.0001114 𝑃𝑎.𝑠 239.19 𝐶⁰ ;252.63 𝐶⁰ 𝑥 0.01588 𝑚

1/4

𝑕𝑜_{𝑐𝑜𝑛𝑑} = 4411.767 ^𝑊 _𝑚2.𝐾

METODE LMTD Analisa Perpan Bagian Eksternal

∆𝑝_𝑠𝑢𝑏= 𝑁_𝐿𝑋 𝜌𝑣_𝑚𝑎𝑥² 2 𝑓

∆𝑝_𝑠𝑢𝑏= 44 𝑥 1.04 𝑥 825.58 𝑘𝑔/𝑚³ 0.86 𝑚/𝑠 ²

2 𝑥 0.45

∆𝑝_𝑠𝑢𝑏= 6197.0733 𝑃𝑎 = 0.06315 𝑘𝑔/𝑐𝑚²

∆𝑝

_{𝑐𝑜𝑛𝑑}

=

¹

2

𝑓𝐺𝑠²𝐼𝐷𝑠𝑕𝑒𝑙𝑙 𝑁:1 5.22 𝑥 10¹⁰𝐷𝑒 𝑠

∆𝑝

_{𝑐𝑜𝑛𝑑}

=

1 2

0.0011^𝑓𝑡2_𝑖𝑛2 𝑥 (151372.134 _{𝑕𝑟.𝑓𝑡2}^𝑙𝑏 )² 𝑥 3.31 𝑓𝑡 𝑥 261 5.22 𝑥 10¹⁰ 𝑥 0.075 𝑓𝑡 𝑥 62.5

∆𝑝

_{𝑐𝑜𝑛𝑑}

= 0.044 𝑝𝑠𝑖 = 0.00309

_𝑐𝑚^𝑘𝑔₂

METODE LMTD Analisa Perpan Bagian Eksternal

Nb ho ∆p (kg/cm²)

4 13473.74 0.06

2 9916.97 0.03

1 7775.43 0.01

1 17291.83 0.01

METODE LMTD Pemilihan Design Optimal

0 0,5 1 1,5 2 2,5

0 5000 10000 15000 20000 25000 30000

140000 160000 180000 200000 220000 240000 260000

hi

Re

hi & ∆p vs Re

hi vs Re pressure drop

0 0,01 0,02 0,03 0,04 0,05 0,06 0,07

0 5000 10000 15000

0 1 2 3 4

ho

Nb

ho & ∆p vs Nb

ho sv Nb dlta (p) vs Nb

Asub Acond Adesup Lsub Lcond Ldesup IDshell kecepatan Jumlah tube Jumlah baffle sub

74.5945 1049.968 361.397 0.8744 12.30777 4.23632 1.009889 2.44 1710 4

METODE NTU

𝐶𝑐 = 𝑚 _𝑐𝐶𝑝_𝑐

𝐶𝑐 = 375.2^𝑘𝑔_𝑠 𝑥 4.533_{𝑘𝑔.𝐶}^𝑘𝐽 𝐶𝑐 = 1700.78_𝑠.𝐶^𝑘𝐽

𝐶𝑕 = 𝑄𝑐/(∆𝑇_𝑕) 𝐶𝑕 = 74834.39 𝑘𝐽/𝑠

(349;215)⁰𝐶 𝐶𝑕 = 558.46_𝑠.𝐶^𝑘𝐽

>

Tci 206C Thi 349C Nt 1710

Tco 250C Tho 215 do 0.01588

Tmean 228C Tmean 282C di 0.011659

Pc,i 197kg/cm2 Phs,i 42.4 kg/cm2 Atube 0.000107

mc 375.2_{kg/s mh} 34.8 _{kg/s ST} 0.022

cp,c 4.533kj/kgC cp,h 16.04786207kj/kgC C 0.470501

Cc 1700.782 Ch 558.4656Cmin m 0.563917

ρc 843.1 kg/m3 _ρh 18.755 kg/m3 ID 1.009889

µc 0.000121 Pa.s µh 0.00001943 Pa.s L 17.41849

Pr 0.835727 Pr 1.182744 As 17.59074

kc 0.6607 W/mC kh 0.04851 _{W/mC Aselim(m²) 0.9}

cold hot Tube &Shell dimension

METODE NTU

𝐶𝑟 = 𝐶𝑚𝑖𝑛

𝐶𝑚𝑎𝑥 = 𝐶𝑕

𝐶𝑐 = 0.328 𝑈_𝑡𝑜𝑡 = _𝐴 ^{𝑄𝑡𝑜𝑡}

𝑡𝑜𝑡∆𝑇𝐿𝑀𝑇𝐷 𝑈_𝑡𝑜𝑡 = 74861627 𝐽/𝑠

1450.152 𝑚² 𝑥 37.533 𝐶 𝑈_𝑡𝑜𝑡 = 1375.414_𝑚^𝑊_20𝐶

𝑈_𝑡𝑜𝑡 = _𝑑𝑜 ¹

𝑑𝑖 1

𝑕𝑖:𝑅_{𝑘𝑜𝑛𝑑}:_𝑕𝑜¹

𝑕𝑜 = ₁ ¹

𝑄𝑡𝑜𝑡;𝑅_{𝑘𝑜𝑛𝑑};^𝑑𝑜_{𝑑𝑖𝑕𝑖}¹

𝑕𝑜 = ₁ ¹

74861627 𝐽/𝑠 ; 4.80758𝐸;05 ^𝑚2.𝐾_𝑊 ; _{0.011659 𝑚}^{0.01588 𝑚 1}

20918.1163 𝑊 𝑚2𝐶

𝑕𝑜 = 1582.735_𝑚^𝑊_2𝐶

Atot ∆LMTD qtotal Utot Rkond di do hi ho

1485.959 37.53292 74861627 1342.271 4.81E-05 0.011659 0.01588 20918.1163 1582.73531

METODE NTU

𝐴_{𝑝𝑒𝑟𝑝𝑎𝑛} = 𝜋. 𝑑𝑜. 𝑁𝑡_{𝑝𝑙𝑢𝑔}. 𝐿_𝑡𝑜𝑡

𝐴_{𝑝𝑒𝑟𝑝𝑎𝑛} = 𝜋 𝑥 0.011659 𝑚 𝑥 1710 𝑥 17.418 𝑚 𝐴_{𝑝𝑒𝑟𝑝𝑎𝑛} = 1485.96 𝑚²

𝑁𝑇𝑈 = ^𝑈𝐴_𝐶^{𝑝𝑒𝑟𝑝𝑎𝑛}

𝑚𝑖𝑛 = ^1.376

𝑘𝑊

𝑚2𝐶 𝑥 1485.96 𝑚² 558.46^𝑘𝑊_𝐶 𝑁𝑇𝑈 = 3.57

𝜀 = 2 1 + 𝐶𝑟 + (1 + 𝐶𝑟²)^1/2 𝑥 1:𝑒𝑥𝑝 ;𝑁𝑇𝑈(1:𝐶𝑟²)^1/2 1;𝑒𝑥𝑝 ;𝑁𝑇𝑈(1:𝐶𝑟²)^1/2

;1

𝜀 = 2 1 + 0.328 + (1 + 0.328²)^1/2 𝑥 1:𝑒𝑥𝑝 ;3.57 (1:0.328²)^1/2

1;𝑒𝑥𝑝 ;3.57 (1:0.328²)^1/2

;1

𝜀 = 0.822

METODE NTU _{Grafik ε vs NTU}

METODE NTU _Grafik hi & ∆p vs % plugging

KESIMPULAN

• Panjang tube dan jumlah tube HPH 7adalah 17.1363 m dan 1710 buah.

• Diameter dalam shell adalah 1.099 m.

• Pada zona subcooled, dimensi panjangnya adalah 0.874 m dengan jumlah baffle yang digunakan adalah empat buah.

• Panjang zona condensing adalah 12.308 m.

• Pressure drop pada tube sebesar 1.1702 kg/cm

²

, pada shell zona subcooled 0.0631 kg/cm

²

, dan pada sisi shell zona

condensing 0.00309 kg/cm

²

.

• Pada Cr = 0.328 diperoleh effectiveness maks HPH 7 adalah

0.8227 dan NTU-nya 3.573.