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Burner (Repaired)

The document provides calculations to determine the composition and heat of combustion of natural gas. It first calculates the molar composition of the outlet gas as 8.6% CO2, 16.15% H2O, and 75.25% N2. It then calculates the heat of combustion for the main components of natural gas (CH4, C2H6, C3H8, C4H10) based on their chemical reactions and heats of formation. The heat of combustion is calculated to be -799083.6326 kJ/mol for CH4, -1428279.217 kJ/mol for C2H6, -2044633.472 kJ/mol for C3H

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Hengky Fernando
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51 views21 pages

Burner (Repaired)

The document provides calculations to determine the composition and heat of combustion of natural gas. It first calculates the molar composition of the outlet gas as 8.6% CO2, 16.15% H2O, and 75.25% N2. It then calculates the heat of combustion for the main components of natural gas (CH4, C2H6, C3H8, C4H10) based on their chemical reactions and heats of formation. The heat of combustion is calculated to be -799083.6326 kJ/mol for CH4, -1428279.217 kJ/mol for C2H6, -2044633.472 kJ/mol for C3H

Uploaded by

Hengky Fernando
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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CH4 90%

CO2
C2H6 7.5%
H2O
C3H8 1.25%
N2
C4H10 1.25%

N2 79%

O2 21%

Asumsi : - basis perhitungan 1 jam

- Steady state

-excess O2 20%

- mol input 5689.0768 mol/jam

Persamaan reaksi stokhiometri :

CH4 + 2O2 CO2 + 2H2O

5120.1691 mol 10240.3382 mol 5120.1691 mol 10240.3382 mol

C2H6 + 3.5O2 2CO2 + 3H2O

420.6808 mol 1493.3827mol 853.3615 mol 1280.0423 mol

C3H8 + 5O2 3CO2 + 4H2O


71.1135 mol 355.5673 mol 213.3404 mol 284.4538 mol

C4H10 + 6.5O2 4CO2 + 5H2O

71.1135 mol 462.2375 mol 284.4538 mol 355.5673 mol

Menghitung komposisi gas hasil keluaran (udara panas)

n=( 5120.1691+853.3615+ 213.3404+284.4538 ) mol CO 2+ ( 10240.3382+1280.0423+284.4538+355.5673 ) mo

n=6471.3248 mol CO 2+12160.4016 mol H 2 O+56661.1730 mol N 2

n=75292.8995mol /jam

komposisi udara panas

6471.3248 mol CO 2
CO 2= =8.6
75292.8995 mol total

12160.4016 mol H 2O
H 2 O= =16.15
75292.8995 mol total

56661.1730 mol N 2
N 2= =75.25
75292.8995 mol total

Excess O2 20%

mol oksigen input=( 10240.3382+1493.3827+355.5673+ 462.2375 ) x 1.2 mol

mol oksigen input=15061.8308 mol / jam

Menghitung laju alir udara


100
mol udara= x 15061.8308 mol/ jam
21

mol udara=71723.0038 mol/ jam

Mr komponen
CH4 16 g/mol
C2H6 30 g/mol
C3H8 44 g/mol
C4H10 58 g/mol
O2 32 g/mol
N2 28 g/mol
H2O 18 g/mol
CO2 44 g/mol

massa
mol=
Mr

massa=mol x Mr

Perhitungan neraca massa

Massa input

- CH4

massa=mol x Mr CH 4

g
massa=5120.1691 mol x 16
mol

massa=81922.7058 g

- C2H6

massa=mol x Mr C 2 H 6
g
massa=420.6808 mol x 30
mol

massa=12800.4228 g

- C3H8

massa=mol x Mr C 3 H 8

g
massa=71.1135 mol x 44
mol

massa=3128.9922 g

- C4H10

massa=mol x Mr C 4 H 10

g
massa=71.1135 mol x 58
mol

massa=4124.5807 g

- N2

massa=mol x Mr N 2

g
massa=56661.1730 mol x 28
mol

massa=1586512.8445 g

- O2
massa=mol x Mr O 2

massa=7326.907 mol x 32 g /mol

massa=481978.5857 g

Massa output

- N2

massa=mol x Mr N 2

g
massa=56661.1730 mol x 28
mol

massa=1586512.8445 g

- O2

massa=mol x Mr O 2

g
massa=2510.3051mol x 32
mol

massa=80329.7623 g

- CO2

massa=mol x Mr CO 2
g
massa=6471.3248mol x 44
mol

massa=284738.2934 g

- H2O

massa=mol x Mr H 2O

g
massa=12160.4016 mol x 18
mol

massa=21887.2295 g

Tabel neraca massa

komponen input output


81922.705
CH4 8 gram -
12800.422
C2H6 8 gram -
C3H8 3128.9922 gram -
C4H10 4124.5807 gram -
481978.58 80329.764
O2 57 gram 3 gram
1586512.8 1586512.8
N2 445 gram 445 gram
284738.29
CO2 - 34 gram
218887.22
H20 - 95 gram
2170468.1 2170468.1
akumulasi 316 gram 316 gram

Kapasitas Panas Gas, Cpg = a + bT + cT2 + dT3 + eT4 (J/mol.K)


Komponen A B C D E
29.
O2 9 -0.0114 4.3E-05 -4E-08 1.01E-11
29.
N2 4 -0.003 5.5E-05 5.1E-09 -4.25E-12
CO2 19 0.0796 -7E-05 3.8E-08 -8.13E-12
H2O 3.4 -0.0097 3.3E-05 -2E-08 4.3E-12
38.
CH4 4 -0.0237 0.00029 -3E-07 8.01E-11
33.
C2H6 8 -0.0155 0.00038 -4E-07 1.39E-10
47.
C3H8 3 -0.131 0.00117 -2E-06 8.19E-10
66.
C4H10 7 -0.186 0.00153 2.2E-06 1.05E-09

Komponen Hr (J/mol)
CH4 -78451.68
C2H6 -84684.0665
C3H8 -103846.765
C4H10 -126147.461
CO2 -393504.766
H2O -241834.933
O2 0

a. Menghitung panas reaksi pembakaran gas alam (Hr)

Reaksi pembakaran gas alam

CH4 + 2O2 CO2 + 2H2O

C2H6 + 3.5O2 2CO2 + 3H2O

C3H8 + 5O2 3CO2 + 4H2O

C4H10 + 6.5O2 4CO2 + 5H2O


1. Panas reaksi pembakaran CH4

CpgCO2
( +2 Cpl H 2 O Cpg CH 4 2CpgO 2 ) dT
303

Hr( 1) (30 C )= Hr +

(1)
298

n CH4 = 5120.1691 mol


Hr(1)= HrCO 2+ 2 Hr H 2 O HrCH 42 Hr O 2


Hr(1)=393504.7656+2 x (241834.933 )(78451.68 )2 x (0)


Hr(1)=798722.9516 J /mol

95.99
(+2 x 16.88191.722 x 149.36)dT
303
Hr (1) ( 30 C )=798722.9516+
298

J
Hr (1) ( 30 C )=799083.6326
mol

2. Panas reaksi pembakaran C2H6

2Cpg CO 2
( +3 Cpl H 2O Cpg C 2 H 63.5 Cpg O 2 )dT
303

Hr( 2) (30 C)= Hr +



(2)
298

n C2H6 = 426.6808 mol



Hr(2)=2 Hr CO2 +3 Hr H 2 O HrC 2 H 6 3.5 HrO 2


Hr(2)=2 x (393504.7656)+3 x (241834.933 ) 84684.06653.5 x(0)

J
Hr(2)=1427830.264
mol

2 x 95.99
(+3 x 16.88168.823.5 x 149.36) dT
303
Hr (2) ( 30 C )=1427839.264+
298

J
Hr (2) ( 30 C )=1428279.217
mol

3. Panas reaksi pembakaran C3H8

3 Cpg CO2
( +4 Cpl H 2O CpgC 3 H 85 CpgO 2 )dT
303

Hr (3) (30 C)= Hr +



(3 )
298

n C3H8 = 71.1135 mol


Hr(3)=3 HrCO 2+ 4 Hr H 2 O HrC 3 H 85 HrO 2


Hr(3)=3 x (393504.7656 )+ 4 x (241834.933 ) 103846.76545 x (0)

J
Hr(3)=2044007.263
mol
3 26307504.7656

3 x 95.99
(+ 4 x 16.88234.915 x 149.36)dT
303
Hr (+33 ranC 3 H 8 ropana ) ( 30 C )=2044007.263+
298

J
Hr (3) ( 30 C )=2044633.472
mol

4. Panas reaksi pembakaran C4H10


4 CpgCO 2
( +5 Cpl H 2O Cpg C 4 H10 6.5CpgO 2)dT
303

Hr( 4 )( 30 C)= Hr +

(4)
298

n C4H10 = 71.1135 mol

Hr(4 )=4 HrCO 2 +5 HrH 2O Hr C 4 H 10 6.5 HrO 2


Hr(4 )=4 x (393504.7656 ) +5 x (241834.933 ) 126147.46076.5 x (0)

J
Hr(4 )=2657046.267
mol

4 26307504.7656

4 x 95.99
(+5 x 16.88331.246.5 x 149.36)dT
303
Hr (+33 ranC 3 H 8 ropana ) ( 30 C )=2657046.267+
298

J
Hr (4 ) ( 30 C )=2657879.79
mol
5. Panas reaksi total

4 4

r i Hr i ( T )= r i Hr i ( 30 C )
i=4 i=4

n CH 4 x Hr (1 ) ( 30 C )+ n C 2 H 6 x Hr( 2) ( 30 C ) +n C 3 H 8 x Hr (3 ) ( 30 C ) +n C 4 H 10 x Hr( 4 ) ( 30 C )

( (
5120.1691 mol X 799083.6326
J
mol )) ( (
+ 426.6808 mol X 1428279.217
J
mol )) (
+ 71.1135 mol X 2044633 (
5034296125 J

b. Menghitung panas input


- CH4

303
QCH 4 =n CH 4 x CpgCH 4 dT
298

J
QCH 4 =5120.1691mol x 191.72
mol

QCH 4 =981617.4904 J

- C2H6

303
QC 2 H 6=n C 2 H 6 x CpgC 2 H 6 dT
298
J
QC 2 H 6=426.6808 mol x 168.82
mol

QC 2 H 6=72033.0539 J

- C3H8

303
QC 3 H 8=nC 3 H 8 x CpgC 3 H 8 dT
298

J
QC 3 H 8=71.1135 mol x 234.91
mol

QC 3 H 8=16705.3329 J

- C4H10

303
QC 4 H 10=nC 4 H 10 x CpgC 4 H 10 dT
298

J
QC 4 H 10=71.1135 mol x 331.24
mol

QC 4 H 10=23555.5579 J

- N2
303
QN 2=n N 2 x Cpg N 2dT
298

J
QN 2=56661.1730 mol x 146.96
mol

QN 2=8327189.2703 J

- O2

303
QO 2=n O 2 x Cpg O 2dT
298

J
QO 2=15061.8308 mol x 149.36
mol

QO 2=2249624.5439 J

c. Menghitung panas keluar


- N2

1334
QN 2=n N 2 x Cpg N 2dT
298

J
QN 2=56661.1730 mol x 70987.507
mol

QN 2=4020445685.8126 J
- O2

1334
QO 2=n O 2 x Cpg O 2dT
298

J
QO 2=2510.3051 mol x 34597.5352
mol

QO 2=86826735.3381 J

- CO2

1334
QCO2 =n CO 2 x Cpg CO 2 dT
298

J
QCO2 =6471.3248 mol x 52040.0127
mol

QCO2 =336671035.7809 J

- H2O

1334 373
Q H 2 O=n H 2 O x
(298
Cpg H 2 O dT + H VL + Cpl H 2 O dT
298
)
J
QH 2 O=12160.4016mol x ( 8781.9531+40631.4 +16.88 )
mol

QH 2 O=602023393.0686 J
Tabel neraca energi

komponen input (J) output(J)


CH4 981617.4904 -
C2H6 72033.0539 -
C3H8 16705.3329 -
C4H10 23555.5579 -
8327189.270 4020445685.8
N2 3 126
2249624.543 86826735.338
O2 9 1
336671035.78
CO2 - 09
602023393.06
H2O - 86
sub total 5034296125 -
jumlah 5045966850 5045966850

Menghitung daya blower

mol udara=71723.0038 mol/ jam


PV =n R T

Keterangan :

P = tekanan, 1 atm

R= nilai tetapan gas ideal, 0.082 atm L/mol K

T = suhu, K

n = mol gas input

n RT
V=
P

L
71723 mol x 0.082 atm x 303 K
mol K
V=
1atm

V = 1782030 L/jam x 1 jam/ 3600 detik

V = 495 L/s

P1 V1 = P2 V2

1 atm x 495 L = 2 atm x V2

V2 = 495 L atm/2 atm

V2 = 247.5 L/s

V2 = 247.5 L/s x 1 m3/1000 L

V2 = 0.2475 m3/s
V2 = 0.2475 m3/s x 1 ft3/(0.3048 m)3 x 60 detik/1 menit

V2 = 519 ft3/menit

Menentukan

Daya blower

Daya(P) = 1.57 x 10-4 . Qu . Pop (Perrys 7ed)

Keterangan :

Qu = laju alir volumetric udara, 519 ft3/menit

Pop = 5 in H2O

Maka daya teoritis blower adalah :

Pteoritis = 1.57.10-4 x 519 x 5

= 0.41 hp

Efisiensi blower = 40% - 80% (Perrys 7ed)

Diambil nilai efisiensi 80%, maka daya actual blower adalah :

Paktual = Pteoritis/

= 0.41/0.8

= 0.51 hp

Alat Blower
Fungsi Mengalirkan udara
Tipe Centrifugal blower
Kapasitas 519 ft3/menit
Power motor 0.51 hp
Harga $210-260

Menghitung kecepatan dalam pipa pada blower

V = 519 ft3/menit

Q = v/A

Q = v/(1/4 D2)

519 ft3/menit = v/

Storage vessel ( tempat penyimpanan gas alam)

Asumsi :

- gas ideal

PV = nRT

Keterangan :

R= nilai tetapan gas ideal, 0.082 atm L/mol K

T = suhu, K

n = mol input gas alam, 5689.0768 mol/jam

n RT
V=
P
atm
5689.0768 mol x 0.082 L x 303 K
mol K
V=
197 atm

V = 717.517 L

Lama penyimpanan pada pressure vessel adalah 30 hari ( 720 jam ) :

V = 717.517 L/jam x 720 jam

V = 516612.1 L

Volume safety 20% = 516612.1 L*1.2

= 619934.5 L x 1 m3/1000 L

=619.934 m3

Volume = 4/3 x x R3

4
619.934 m3= x 3.14 x R3
3

3
R 3= x 619.934 m3
4 x 3.14

R=5.29 m

kecepatan gas alam dalam pipa

P1V2 = P1V2

197 atm x 619.934 L = 2 atm x V2

V2 = 61063.5 L/jam x 1 jam/3600 detik


V2 = 16.96 L/s (digunakan regulator untuk mengatur tekanan pada 2 atm)

Tebal pressure vessel

Spec no. yang digunakan :


SA-283 Grade A

Allowable stress : 110 MPa

D
t=P
4

101 kPa
197 atm x x 5.29 m x 2
1 atm
t=
1000 kPa
4 x 110 MPa x
1 MPa

t = 0.478 m

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