Heat Balance Calculations Problem Statement: Operating temperatures, flow rates, flue gas analysis and feed properties are provided for Refinery A. Perform heat balance and estimate the following: Argon corrected flue gas analysis Dry air to Regenerator Flue gas flow rate Coke make H2 in coke Air to coke ratio Catalyst circulation rate Cat / Oil ratio Delta coke Heat of Reaction 6-1
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Operating Data Base Data: Parameter
Value
UOM
Combined Feed Temperature
375
°F
Lift Gas Temperature
100
°F
Steam Temperature
380
°F
Reactor Temperature
970
°F
Regenerated Catalyst Temperature
1371
°F
Ave. Hottest Regen Temperature
1375
°F
399
°F
Ambient Temperature
62
°F
Ambient Relative Humidity
97
%
Air to Regenerator (MAB Discharge)
6-2
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Operating Data (continued)
Parameter
Value
UOM
412,923
lb/hr
Fresh Feed Gravity at base temperature
21.3
°API
Fresh Feed UOP K
11.8
Fresh Feed Rate (30,565 BPD)
MCB Recycle Flow to Riser
0
lb/hr
Lift Gas Rate (Sponge Gas)
3,250
lb/hr
12,900
lb/hr
Feed Steam Flow Rate
1,800
lb/hr
Stripping Steam Flow Rate
5,000
lb/hr
380,200
lb/hr
Lift Steam Flow Rate
Air to Regenerator Flow Rate
6-3
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Flue Gas Analysis
Flue Gas Composition
(Mole % by GC)
CO
0
CO2
15.50
O2 + Ar
3.47
N2
6-4
81.03
SOx
0
NOx
0
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Step 1 : Argon correction Adjust Flue Gas Composition & correct for Argon, if necessary: Component CO CO2 O2 + Ar N2
GC, Mol%
Argon correction
Ar Corrected GC, mol%
0
0
15.50
15.50
3.47
[O2 - (0.012)x(N2)]
2.50
81.03
[N2 + O2 - O2Corrected]
82.00
SOx
0
0
NOx
0
0
100.00
100.00
Total
Note: Correction required for GC flue gas analyses, flue gas analyses by Orsat are correct, i.e. N2+Ar. 6-5
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Step 2: Correct Air to dry basis Estimate Moisture content of Air - Antoine’s equation a = Ambient Temperature
= 62 °F
b = Relative Humidity
= 97 %
c
= Saturated Vapor Pressure = 10 =
10
3165 .36 6 .40375 - 62 + 392 .595
3165.36 6 .40375- a + 392 . 595
= 0.2756 psia
[ 0.00622 × b × c ] Moisture content of ambient air = (14.696 − [0.01 × b × c]) = 6-6
[0.00622 × 97 × 0.2756] (14.696 − [0.01× 97 × 0.2756]) UOP - CONFIDENTIAL
= 0.0115 lbwater/lbdry air
Step 2: Correct Air to dry basis (continued)
Total Air to Regenerator = Dry Air to Regenerator =
380,200 lb/hr
Total Airlb/hr 1 + (lb H 2O lbdry air )
[
380,200 = [1 + 0.0115]
]
= 375,870 lb/hr
H2O vapor in Air (Total Air - Dry Air) = (380,200-375,870) = 4,330 lb/hr
6-7
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Step 3: Calculate flue gas rate • Flue gas rate is not measured by a flow meter, so we need to calculate it • Nitrogen stays constant, so we use it to find the flue gas rate N2 + Ar (in air) = N2 + Ar (in flue gas) lb 375,870 hr Dry Air, mol/hr = 28.966 (MW air)
6-8
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= 12,976 lbmol/hr
Step 3: Calculate flue gas rate (continued)
Estimate number of moles of N2 + Ar in Air N2 + Ar (in air) = 12,976 lbmol/hrdry air x 79%(N2 + Ar in air) = 10,251 lbmol/hr
The moles of N2 + Ar in Air and flue gas stay constant N2 + Ar (in flue gas) = n lbmol/hrflue gas x 82%(N2 + Ar in flue gas) ∴ Flue gas rate, n lbmol/hrflue gas = 10,251 / 82% = 12,501 lbmol/hr
6-9
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Step 4 : Estimate Carbon Content of Coke C + O2 + H2 + S + N CO + CO2 + H2O + O2 + SOX + NOX If 1 mole of C burns for each mole of CO or CO2 produced, then: No. of moles of carbon burnt,
(Mol%CO + Mol%CO2 ) Clbmol/hr = lbmol hr flue gas × 100 (0 + 15.50) Clbmol/hr = 12,501 × 100
6-10
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= 1,938 lbmol/hr
Step 5: Perform Oxygen balance C + O2 + H2 + S + N CO + CO2 + H2O + O2 + SOX + NOX Calculate H2O formed due to combustion by O2 balance O2 in air = (O2 in flue gas) + (O2 in CO) + (O2 in CO2) + (O2 in H2O) O2 in air =
12,976 lbmol/hrdry airx 21% = 2,725 lbmol/hr
O2 in flue gas = O2 in CO =
12,501 lbmol/hrflue gas x 2.5% =
312 lbmol/hr
12,501 x 0% x (0.5 lbmol O2/lbmol CO) =
0 lbmol/hr
O2 in CO2 =
12,501 x 15.5% = 1,938 lbmol/hr
O2 in SO2 =
12,501 x 0% =
0 lbmol/hr
O2 in NO2 =
12,501 x 0% =
0 lbmol/hr
O2 in H2O =
2,725 - 312 - 0 - 1,938 - 0 - 0 =
475 lbmol/hr
Calculate the H2 burnt to H2O H2 burned to H2O = 6-11
2 lbmolH2/lbmolO2 x 475 lbmol/hrO2 = UOP - CONFIDENTIAL
950 lbmol/hr
Step 6: Estimate Coke Yield
Calculate Coke from C and H2 Clb/hr = lbmol hrC × MWC = 1,938 x 12.01
= 23,275 lb/hr
H 2 (lb/hr) = lbmol hrH 2 × MWH 2 = 950 x 2.016
= 1,915 lb/hr
Coke Ratelb/hr = C + H2 = 23,275 + 1,915
= 25,190 lb/hr
Coke hrlb Coke Yieldwt% FF = × 100 lb Fresh Feed hr = 25,190 / 412,923 x 100
6-12
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= 6.10 wt% of Fresh Feed
Step 7: Calculate Hydrogen in Coke
H2 in Cokewt% =
H 2 hrlb × 100 lb Coke hr = 1,915 / 25,190 x 100
Air to Cokelb/lb =
Dry Air hrlb Coke hrlb
= 375,870 / 25,190
6-13
= 7.60 wt%
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= 14.92 lbair/lbcoke
Step 8: Correct ∆HC for Regen temperature Basis: 1 lb of Coke Average hottest regenerator temperature
= 1,375 °F
Correct ∆HC for operating temperature: ∆HC (CO) = 46,216 + (1.47 x 1375°F)
= 48,237 Btu/lbmol
∆HC (CO2) = 169,135 + (0.500 x 1375°F)
= 169,822 Btu/lbmol
∆HC (H2O) = 104,546 + (1.586 x 1375°F)
= 106,725 Btu/lbmol
Heat of combustion, BTU/lbmol @ 25°C
Note: The ∆HC for various compounds in reference handbooks is provided at 25°C (77°F). These heat of combustions must be corrected for the operating temperature of regenerator. 6-14
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Step 9: Convert ∆HC of compounds to Btu/hr Since: C + O2 + H2 CO + CO2 + H2O + O2 Convert ∆HC of compounds to Btu/hr ∆HC (CO2) = ∆HC (CO2)Btu/lbmol x C in Cokelbmol/hr = 169,822 x 1,938
=
329,100,000 Btu/hr
∆HC (H2O) = ∆HC (H2O)Btu/lbmol x H2O in Cokelbmol/hr = 106,725 x 950
6-15
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=
101,400,000 Btu/hr
Step 10: Calculate gross heat of combustion The gross heat of combustion of coke in Btu/lb of coke is sum of heat of combustions of CO, CO2 and H2O divided by total coke make in lb/hr. i.e. ∆HComb
= [∆HC (CO2) + ∆HC (H2O)]Btu/hr/Cokelb/hr) = [329,100,000 + 101,400,000]/25,190 = 17,090 Btu/lb Coke
6-16
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Step 11: Correction factor for H2 in Coke Calculate Correction Factor for Hydrogen in Coke H2 Correction Factor = 1133 - (134.64 x wt% H2 in Coke) = 1133 - (134.64 x 7.60) == + 110 Btu/lb Coke
Corrected Gross Heat of Coke Combustion (∆H (∆HCombCoke) ∆HCombCoke = ∆HComb + H2 Corr Factor = 17,090 + 110 = 17,200 Btu/lb Coke Note: The base heats of combustion are for the burning of H2, Carbon (graphite), or CO to the combustion products H2O + CO2. Since Coke is not an ideal compound, it does not behave necessarily like these ideal compounds. Therefore, an adjustment must be made for the effect of the Hydrogen-Carbon interactions in the coke. 6-17
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Step 12: Heat consumed by streams Regenerator Heat Balance: Basis: 1 lb of Coke
Heat consumed to heat up catalyst, QNet = ∆HCombCoke - ∆HCoke - ∆HAir - ∆HH2O - ∆HRadLoss - ∆HHtRemoval Also, QNet = mass x CP x ∆T Heat consumed to heat up coke,
∆HCoke = (0.4 Btu/lb°F) x (TFG-TRx)
= 0.4 x (1375 - 970) = 162 Btu/lb Coke
Heat consumed to heat up Air,
∆HAir
6-18
= Air/Cokelb/lb x 0.26 Btu/lb°F x (TFG-TMAB) = 14.92 x 0.26 x (1375 - 399) = 3,786 Btu/lb Coke UOP - CONFIDENTIAL
Step 12: Heat consumed by streams (contd..) Heat consumed to heat up water,
∆HH2O = (H2O in Air)/Cokelb/lbx 0.485 Btu/lb°F x (TFG-TMAB) = (4,330/25,190) x 0.485 x (1375 - 399) = 81 Btu/lb Coke Radiation losses,
=
∆HRadiation Loss
250 Btu/lb Coke
Heat consumed to generate steam in catalyst cooler,
∆HHt Removal
6-19
= Cooler Duty, calc separately = 58,750,000 Btu/hr = (Cooler DutyBtu/hr)/Cokelb/hr) = 58,750,000/25,190 = 2,332 Btu/lb Coke
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Step 13: Net Heat of Combustion Heat consumed to heat up catalyst, QNet = ∆HCombCoke - ∆HCoke - ∆HAir - ∆HH2O - ∆HRadLoss ∆HHtRemoval = 17,200 - 162 - 3786 - 81 - 250 - 2,332 = 10,589 Btu/lb Coke
6-20
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Step 14: Catalyst circulation rate Also, QNet = mass x Cp x ∆T then mass = QNet / (Cp x ∆T) Therefore, Catalyst Circulation Rate, CCR = (Cokelb/hr x Net Regen HeatBtu/lb Coke) [0.275 Btu/lb°F x (TCat-TRx)] = (25,190 x 10,589)/[0.275 x (1371 - 970)] = 2,419,000 lb/hr = 2,419,000lb/hr/ (60 min/hr x 2,000 lb/ton) = 20.16 ton/min 6-21
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Step 15: Catalyst to oil ratio & Delta coke Cat/Oil Ratio
= Cat Circ lb/hr/FFlb/hr = 2,419,000 / 412,923 = 5.86 lb/lb
∆Cokewt%
= Cokelb/hr / Cat Circ lb/hr x 100 = 25,190 / 2,419,000 x 100 = 1.04 wt%
6-22
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Reactor Heat Balance
Reactor Heat Balance:
Basis: 1.0 lb of Fresh Feed
Reactor Temp (TRx)
970 °F
Combined Feed Temp (TCF)
375 °F
Fresh Feed Rate (FF)
412,923 lb/hr
Recycle Rate (Recy)
6-23
0 lb/hr
FF Enthalpy at TCF (HFFCFT)
252 Btu/lb
FF Enthalpy at TRx (HFFRxT)
760 Btu/lb
Recycle Enthalpy at TCF (HRecyCFT)
0 Btu/lb
Recycle Enthalpy at TRx (HRecyRxT)
0 Btu/lb
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Reactor Heat Balance
Inerts from Regen (assumed)
0.007 lb/lb FF
Inerts Specific Heat (Cp)
0.275 Btu/lb°F
Lift Gas Temp (TLG)
100 °F
Lift Gas Rate (Sponge Gas)
3,250 lb/hr
Lift Gas Specific Heat (CpLG)
0.5 Btu/lb°F
Steam Temp (TStm)
380 °F
Steam Specific Heat (CpStm)
0.485 Btu/lb°F
Feed Dispersion Steam Rate
1,800 lb/hr
Lift Steam Rate
12,900 lb/hr
Stripping Steam Rate
6-24
5,000 lb/hr
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Entrained Inerts
Inertsmass time
1 1 Pemulsion ⋅ MWgas ⋅ = Cat Circ Rate ⋅ − ρ emulsion ρ skeletal R ⋅ Tcatalyst
- MWgas ≅ 29 for Regen standpipe, 18 for spent standpipe - ρemulsion ≅ 40 lb/ft³ The standpipe overall density should not be used, as it’s not indicative
of the gas that is carried with the catalyst) ρskeletal ≅ 160 lb/ft³ (use the catalyst vendor’s value if available)
- Pemulsion ≅ abs. pres. at the lowest point in the standpipe, usually just upstream of the slide valve
- All catalyst temperatures in R or K 6-25
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Entrained Inerts
Inertsmass time
1 1 Pemulsion ⋅ MWgas ⋅ = Cat Circ Rate ⋅ − ρ emulsion ρ skeletal R ⋅ Tcatalyst
Regenerated Catalyst
Inertslbs/hr
Cat Circ Rate lb/hr ⋅ PRCSV inlet = ⋅ 29 572.352 ⋅ TRegen cat
Spent Catalyst
Inertslbs/hr =
6-26
Cat Circ Ratelb/hr ⋅ PSCSV inlet ⋅ 18 572.352 ⋅ Tspent cat UOP - CONFIDENTIAL
Reactor Heat Demand Reactor Heat Balance: Reactor heat demand QRx by rearranging formula (2): QRx = ∆HFF + ∆HRecy + ∆HLG + ∆HStm + ∆HInerts + ∆HRad Loss + ∆HRx Also, QRx = mass x CP x ∆T
6-27
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Step 16: Heat consumed by streams Heat Consumed by Feed ∆HFF
= HFFRxT-HFFCFT = 760 - 252
= 508 Btu/lb FF
Heat Consumed by Recycle ∆HRecy = Recy x (HRecyRxT-HRecyCFT) /FFlb/hr
= 0 Btu/lb FF
Heat Consumed by Steam ∆HStm = Total Stm x CpStm x (TRx-TStm)/FFlb/hr = (5,000 + 1,800 + 12,900)x 0.485 x (970 - 380) /412,923
= 13.6 Btu/lb FF 6-28
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Step 16: Heat consumed by streams (contd) Heat Consumed by Lift Gas ∆HLG
= LG x CpLG x (TRx-TLG)/FFlb/hr = 3,250 x 0.500 x (970 - 100) /412,923
= 3.4 Btu/lb FF
Heat Consumed by Inerts ∆HInerts = Inerts Rate lb/lb FF x CpInerts x (TRx-TRegen Cat) = 0.007lb/lb FF x 0.275 x (970 - 1371) = -0.8 Btu/lb FF
Reactor Radiation Losses (typically 2 Btu/lb FF)
= 2 Btu/lb FF
∆HRad Loss
6-29
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Step 17: Calculate Reactor heat demand Reactor Heat Demand, QRx= ∆HFF + ∆HRecy + ∆HLG + ∆HStm + ∆HInerts + ∆HRadLoss + ∆HRxn = (508 + 0 + 3.4 + 13.6 - 0.8 + 2) + ∆HRxn = 526.2 + ∆HRxn Btu/lbFF
6-30
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Step 18 : Overall Heat Balance Overall Balance: QNet = QRx
Q net
( BTU
lb Coke = Q Rx ( BTU lb FF ) lb Coke ) × lb Fresh Feed = 526.2 Btu/lbFF + ∆HRxn
10,589 Btu/lbCoke x (25,190 lb Coke / 412,923 lb FF) = 646.0 Btu/lbFF
Heat of Reaction, ∆HRxn = 646.0 - 526.2 = 120 Btu/lb Fresh Feed Usually 150 to 200 btu/lb
6-31
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