Process Fluiid Duty

GIVEN DATA: FLUID HANDLED: 3

Total flow rate (m /hr): 19.5 3

Oil Flow rate (m /hr):7.0 3

Water flow rate (m /hr):12.5

Temperature: 60

PROBLEM STATEMENT: Calculate: 

Process fluid duty.



Heat Transfer Area,Diameter and length of firetube.

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Fired duty

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Fuel LHV

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Adiabatic fuel Temperature

PROCESS FLUIID DUTY: Fluid temperature=60 degree C Outlet temperature=70 degree C ∆T=10 degree C Density of oil (Ww):

3

897.6 kg/m 3

Density of water (W o): 980.6 kg/m

3

Q o=7 m /h 3

Q w=12.5 m /h Specific gravity of oil:

(Density of oil/Density of water) = (897.6/980.6)=0.915

Specific gravity of water:

1

Specific heat of oil (Co): 0.479 kcal/kg ˚C

=2004 J/kg degree C

Specific heat of water (Cw):1.035 kcal/kg ˚C =4330.44 J/kg degree C

Q o=WoCo∆T = [Wo*SGo*Q o]*(Co)*(∆T) =(980.6*0.915*7)*(2004)*(10) =125866089.72 J/hr 1 J/hr=2.778*10**-7 kW

Q o=35 kW Q w=Ww*Cw*∆T =(980.6*1*12.5)*4330.44*10

147.5 kW Q lost=(0.1)q

q=Q o+Q w+Q lost 0.9(q)=147.5+35 kW Q=202.778 kW Total heat required=202.778 kW=730000.8 kJ/h = Fluid duty =691907.25 Btu/hr

FIRED DUTY Efficiency=70% Therefore : Fire duty=Process fluid duty/Efficiency Fire duty=691907.25/0.7=988439 Btu/h

SURFACE AREA REQUIRED: 2

Assume heat flux=10000 Btu/ (hr.ft ) Area=Q/flux =691907.25/10000 =69.2 ft

2

DIAMETER AND LENGTH From table

OD=18’’ Therefore, Surface area=4*pi*(D/2)*L=69.2 ft2 OR A=9964.8 sq inches=4*pi*(D/2)*L Or L=88”

For calculating LHV: We list down the reactions: 1. 2. 3. 4. 5. 6. 7. 8.

CH4+2.4O2-CO2+2H2O+0.4O2 C2H6+4.2O22CO2+3H2O+0.7O2 C3H8+6O23CO2+4H2O+O2 C4H10+7.8O24CO2+5H2O+1.3O2 C5H12+12O25CO2+10H2O+2O2 C6H14+11.4O26CO2+7H2O+1.9O2 C7H16+13.2O27CO2+7H2O+2.2O2 N2+2O22NO2 Heat of formation data:

Compound CO2 H2O CH4 C2H6 C3H8 C4H10 (iso) C4H10 (n) C5H12 (iso) C5H12 (n) C6H14 C7H16 NO2

Heat of combustion for individual reactions:

r1=((393.5+2*(241.82))-(74.87)) =802.27 kJ/mol r2=(2*(393.5)+3*(241.82))-(84.68);=1427.78 kJ/mol r3=(3*(393.5)+4*(241.82))-(103.85);=2043.93 kJ/mol r4i=(4*(393.5)+5*(241.82))-(134.3);=2648.8 kJ/mol r4n=(4*(393.5)+5*(241.82))-(125.5);=2657.6 kJ/mol

Heat or formation (kJ/mol) 393.5 241.82 74.87 84.68 103.85 134.3 125.5 154.4 146.9 167.4 187.9 33.2

r5i=(5*(393.5)+10*(241.82))-(154.4);=4231.3 kJ/mol r5n=(5*(393.5)+10*(241.82))-(146.9);=4238.8 kJ/mol r6=(6*(393.5)+7*(241.82))-(167.4);=3886.34 kJ/mol r7=(7*(393.5)+7*(241.82))-(187.9);=4259.34 kJ/mol

MASS FRACTIONS:

Compound CH4 C2H6 C3H8 C4H10 (iso) C4H10 (n) C5H12 (iso) C5H12 (n) C6H14 (n) C7H16 CO2 N2

MASS FRACTION 0.9345 0.0105 0.0053 0.0028 0.0000628 0.0003896 0.0002597 0.0001034 0.0003 0.0355 0.01

TOTAL heat of combustion: ∑(mass frac)*(heat of comb)/(Mol wt) *(1000*0.948/2.2) BTU/Lb =20549.32 BTU/lb

FUEL QUANTITY REQUIRED: Fire duty/LHV=Fuel required Fire duty=988439 BTU/h LHV=20549.32 BTU/lb Therefore : Fuel required=48 lb