GIVEN DATA:

Fluid Handled:

Total flow rate (m3/hr): 19.5 Oil Flow rate (m3/hr):7.0 Water flow rate (m3/hr):12.5 Temperature: 60 ˚C

S No. Properties Oil Water1. Density (kg/m3) 897.6 980.62 Viscosity (cP) at 60 ˚C 65.3 0.53 Viscosity (cP) at 70 ˚C 43 0.44 Specific Heat (kcal/kg ˚C) 0.479 1.0355 Thermal conductivity (kcal/m hr ˚C) 0.136 0.5626 Surface Tension (dynes/cm) 28.6 66

Over capacity factor on flow 110% of normal

Operating conditions:

Outlet temperature=70 ˚C Allowable pressure drop=0.5 kg/cm2

Water Content in oil at outlet(% v/v)= 1 (max) Oil Content in water at outlet(ppm)= 125 (max) Allowable liquid carry over in gas (L/mm/m3)=13.4 (max)

Fuel Data:

COMPOUND

MOLE% MASS%

CH4 97.15 93.45C2H6 0.6 1.05C3H8 0.205 0.53

C4H10 (iso) 0.081 0.3C4H10 (n) 0.002 0.01

C5H12 (iso) 0.009 0.04C5H12 (n) 0.006 0.03C6H14 (n) 0.002 0.01

C7H16 0.005 0.03CO2 1.34 3.55N2 0.6 1.0

PROBLEM STATEMENT:

Calculate:

Process fluid duty. Assume flux=10000 BTU/hr /ft2

Heat Transfer Area, Diameter and length of fire tube. Fired duty Fuel LHV Adiabatic fuel Temperature

PROCESS FLUIID DUTY:

Fluid temperature: 60˚ C

Outlet temperature: 70 ˚ C

∆T: 10 ˚ C

Density of oil (Ww): 897.6 kg/m3

Density of water (Wo): 980.6 kg/m3

Qo: 7 m3/h

Qw: 12.5 m3/h

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

Specific heat of water (Cw): 1.035 kcal/kg ˚C

Qo=WoCo∆T

= [Density of oil*Qo]*(Co)*(∆T)

=897.6*7*0.479*10

=30096 kcal/h

Qw=Ww*Cw*∆T

= (980.6*12.5)*1.035*10

=126865.125 kcal/h

Qlost= (0.1) q

q=Qo+Qw+Qlost

0.9(q) =30096+126865.125

q=174401 kcal/h

Total heat required=174401 kcal/h = Fluid duty

=692079.1 Btu/hr

Over capacity factor=110%

Therefore:

Fluid duty:

=761287 BTU/h

=0.76 MMBTU/hr

FIRED DUTY

Efficiency=70%

Therefore:

Fired duty=Process fluid duty/Efficiency

Fired duty=761287/0.7=1087552.85 Btu/h

SURFACE AREA REQUIRED:

Assume heat flux=10000 Btu/ (hr.ft2)

Area=Q/flux

=761287/10000

=76.13 ft2

DIAMETER AND LENGTH

The maximum allowable Heat Density should be 15000 BTU/hr. /sq. in

Therefore, the required fire tube should have:

NPS=14” O.D=350 mm Length=64.3 m

CALCULATION OF L.H.V:

We list down the reactions:

1. CH4+(1.2)(2)(O2+3.76N2)CO2+2H2O+0.4O2

2. C2H6+(3.5)(1.2)(O2+3.76N2)2CO2+3H2O+0.7O2

3. C3H8+(5)(1.2)(O2+3.76N2)3CO2+4H2O+O2

4. C4H10+(6.5)(1.2)(O2+3.76N2)4CO2+5H2O+1.3O2

5. C5H12+(1.2)(10)(O2+3.76N2)5CO2+10H2O+2O2

6. C6H14+(1.2)(9.5)(O2+3.76N2)6CO2+7H2O+1.9O2

7. C7H16+(1.2)(11)(O2+3.76N2)7CO2+8H2O+2.2O2

Heat of formation data:

Compound Heat or formation (kJ/mol)CO2 393.5H2O 241.82CH4 74.87C2H6 84.68C3H8 103.85

C4H10 (iso) 134.3C4H10 (n) 125.5

C5H12 (iso) 154.4C5H12 (n) 146.9

C6H14 167.4C7H16 187.9NO2 33.2

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

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 of fuel gas:

COMPOUND MOLE% MASS%

CH4 97.15 93.45C2H6 0.6 1.05C3H8 0.205 0.53

C4H10 (iso) 0.081 0.3C4H10 (n) 0.002 0.01

C5H12 (iso) 0.009 0.04C5H12 (n) 0.006 0.03C6H14 (n) 0.002 0.01

C7H16 0.005 0.03CO2 1.34 3.55N2 0.6 1.0

Total heat of combustion:

∑ (mass frac)*(heat of comb)/ (Mol wt.) *(1000*0.948/2.2) BTU/Lb

=20549.4 BTU/lb

FUEL QUANTITY REQUIRED:

FuelQuantity Req=Fire DutyL . H .V

Fire duty=1087552.85 BTU/h

LHV=20549.4 BTU/lb

Therefore:

Fuel required=53 lb/h

Product flue gases component:

Flue gas Molecular weight =∑ (M .W ) i∗¿¿

Therefore, molecular wt. of Flue gas: (0.081*44) + (0.161*18) + (0.032*32) + (0.726*28)

Molecular weight=27.8

Amount of Flue gas per lb of fuel:

Basis: 453.6 g=1 lb

Fuel data:

Component Mass(g) No. of molesCH4 423.8 26.5C2H6 4.75 0.158C3H8 2.4 0.05C4H10 (i) 1.27 0.02C4H10 (n) 0.028 0.0004C5H12 (i) 0.17 0.0023C5H12 (n) 0.11 0.0015C6H14 0.04 0.0005C7H16 1.36 0.013N2 4.51 0.16

No. of moles of O 2 :-

26.5(0.4)+(0.7)(0.158)+0.054+(0.02)(1.3)+(0.0004)(1.3)+(2)(0.0023)+(2)(0.0015)

+ (1.9) (0.0005) + (2.1) (0.013)

=10.8 moles

Moles of N2:-

26.5*9.02+15.79*0.158+0.05*22.8+0.02*29.33+ (0.0004)*29.33+45.12*0.0023

+0.0015*45.12+42.86*(0.0005) +49.632*0.013+4.53

=248.6 moles

S No. Product component Mole % Mass %

1 CO2 8.1 12.82 H2O 16.1 10.4

3 O2 3.2 3.74 N2 72.6 73.1

Moles of CO2:-

26.5+2*0.158+0.05*3+0.02*4+ (0.0004)*4+5*0.0023+0.0015*5+6*(0.0005) +7*0.013

=27.1 moles

Moles of H2O:-

26.5*2+3*0.158+4*0.05+0.02*5+ (0.0004)*5+10*0.0023+0.0016*10

+7*(0.0005) +7*0.013

=52.9 moles

Product Flue gas:

Therefore, Amount of flue gas per pound of fuel=20.8 lb

In this case, Amount of fuel=53 lbs

Therefore Amount of flue gas produced=53*20.8 lbs

=1102.4 lbs

Amount of Air per lb of fuel:

From Stoichiometry and No. of moles of fuel component,

No.of moles of air:-

=(26.5)(2)(1.2)+(3.5)(1.2)(0.158)+(5)(1.2)(0.05)+(6.5)(1.2)(0.02)+

(6.5)(1.2)(0.0004)+(1.2)(10)(0.002+(1.2)(10)(0.0016)+(1.2)(9.5)(0.0005)+(1.2)(11)(0.013)

=65.1 moles of air

Now,

Mass (g) =Molecular wt.*No. of moles

= (2(16) +3.76(28))*66.8

=8937 g

=19.78 lb

S No. Product component No. of moles Mass(g)

1 CO2 27.1 1192.42 H2O 52.9 952.23 O2 10.8 345.64 N2 248.6 6960.8TOTAL: 9451 g=20.8 lb

Therefore, Air per pound of fuel is 20.78 lb

Air required in this case= Amount of fuel (lb)*Air per pound of fuel

=53*20.78

=1101.3 lbs

ADIABATIC FLAME TEMPERATURE:

Heat of Combustion( kJmol )=∫TrTf

∑ ¿ (Cp ) i

ni=No.of moles of component ‘i’

Cpi=Specific heat capacity of component ‘i’

Cp=n*(a+b*T)

(Cp)O2=n*(7.16+10-3T) (Cp) CO2=n*(10.55+2.16*10-3T)

(Cp)H2O=n*(7.17+2.56*10-3T) (Cp) N2=n*(6.83+9*10-4T)

Heat of combustion=795 kJ/mol

Therefore,

795= ∫ [0.396(7.16*T+10-3*T2/2) +1.974(7.17T+2.56*10-3*T2/2)

+0.9938(10.55+2.16*10-3T2/2) +8.946(6.83T+9*10-4T2/2)]

On solving, we get:

T=2028.7˚C