SI No Calculation

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5Revision History

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10 Scale up of reactor from lab to plant scale

11121314 Pipe flow distribution15

Heating time using hot water (Non Isothermal Heating Medium)

Heating time using steam (Isothermal Heating Medium)Cooling time using non isothermal cooling medium

Heating time using external Heat exchanger & Non Isothermal Heating Medium

Heating time using external Heat exchanger & Isothermal Heating Medium

Cooling time using external Heat exchanger & Non Isothermal Cooling Medium

Cooling time using external Heat exchanger & Isothermal Cooling Medium

Calculate the Heat Transfer coefficient for reactors based on different agitator types & utility flow rate

Calculate the addition time required based on heat of the reaction, Type of agitator, MOC & utility flow rate in a semi batch reactor

Calculate the vacuum pump flow rate based on air leakage ratePower required for pumpingPump affinity law

Pressure drop across pipe lines

Revision History

Issued to Downloads section on 5/5/2011

Non Isothermal HeatingReactor ContentsMass M 72500 lbInitial temp t1 120 °FFinal temp t2 200 °FSp.Heat Cm 1.05 Btu/ lb °FArea A 150 ft2HTC U 180 BTU/hr ft2 °F

Heating FluidInlet temp T1 230 °FSp.Heat Cw 1.2 Btu/ lb °FFlow rate W 18000 lb/hr

Time required for heating

ln T1 - t1 = WCw x K-1 x tT1 - t2 MCm K

K = e(UA/WC)

K 3.4903Time reqd = 6.42 hrs

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Isothermal HeatingReactor ContentsMass M 50000 lbInitial temp t1 68 °FFinal temp t2 257 °FSp.Heat Cm 0.5 Btu/ lb °FArea A 100 ft2HTC U 150 BTU/hr ft2 °F

Heating FluidInlet temp T1 320 °F

Time required for heating

ln T1 - t1 = U A timeT1 - t2 MC

Time reqd = 2.31 hrs

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Non Isothermal coolingMass M 72500 lbInitial temp T1 200 °FFinal temp T2 82 °FSp.Heat Cm 1.05 Btu/ lb °FArea A 150 ft2HTC U 110 BTU/hr ft2 °F

Cooling FluidInlet temp t1 72 °FSp.Heat Cw 1 Btu/ lb °FFlow rate W 15500 lb/hr

Time required for cooling

ln T1 - t1 = WCw x K-1 x tT2 - t2 MCm K

K = e(UA/WC)

K 2.8994Time reqd = 19.11 hrs

= 1146.78 min

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Non Isothermal HeatingReactor ContentsMass M 72500 lbMass Flow rate w 20000 lb/hrInitial temp t1 82 °FFinal temp t2 200 °FSp.Heat c 1.05 Btu/ lb °FArea A 150 ft2HTC U 210 BTU/hr ft2 °F

Heating FluidInlet temp T1 230 °FSp.Heat C 1.2 Btu/ lb °FFlow rate W 15500 lb/hr

Time required for heating

ln T1 - t1 = wWC x K3-1 x tT1 - t2 (K3wc-WC) M

K3 = eUA*(1/WC-1/wc)

K3 1.213548Time reqd = 10.03 hrs

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Isothermal HeatingReactor ContentsMass M 72500 lbMass Flow rate w 20000 lb/hrInitial temp t1 82 °FFinal temp t2 200 °FSp.Heat c 1.05 Btu/ lb °FArea A 150 ft2HTC U 210 BTU/hr ft2 °F

Heating FluidInlet temp T1 230 °F

Time required for heating

ln T1 - t1 = wc x K2-1 x tT1 - t2 Mc K2

K2 = eUA/wc

K2 4.481689Time reqd = 7.45 hrs

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Non Isothermal coolingMass M 72500 lbBatch flow rate W 20000 lb/hrInitial temp T1 200 °FFinal temp T2 82 °FSp.Heat C 1.05 Btu/ lb °FArea A 150 ft2HTC U 180 BTU/hr ft2 °F

Cooling FluidInlet temp t1 72 °FSp.Heat c 1 Btu/ lb °FFlow rate w 15500 lb/hr

Time required for cooling

ln T1 - t1 = Wcw x K4-1T2 - t1 (K4wc-WC) M

K4 = eUA*(1/WC-1/wc)

K4 0.6337Time reqd = 18.19 hrs

= 1091.62 min

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x t

Isothermal coolingMass M 72500 lbBatch flow rate W 20000 lb/hrInitial temp T1 200 °FFinal temp T2 82 °FSp.Heat C 1.05 Btu/ lb °FArea A 150 ft2HTC U 180 BTU/hr ft2 °F

Cooling FluidInlet temp t1 72 °FSp.Heat c 1 Btu/ lb °FFlow rate w 15500 lb/hr

Time required for cooling

ln T1 - t1 = WC K1-1T2 - t1 MC K1

K1 = eUA/WC

K1 3.6173Time reqd = 12.77 hrs

= 766.37 min

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x t

Calculation of Heat transfer coefficient in reactor

Reactor data

Diameter of the tank T 108 in 9 ft

Impeller speed N 45 rpm 2700 rph

Impeller Dia 42 in 3.5 ft

Impeller power Hp 2 Hp

Agitator Type Anchor-1

Vessel jacket flow depth L 1 in 0.08 ft

Height of liquid in cylindrical section Z 96 in 8 ft

Properties of batch side fluid

Heat capacity 0.997 Btu/lb °F

Sp gravity 1.0028 1002.8 62.60 lb/ft3

Bulk Viscosity µ 0.536 cp 1.29712 lb/hr ft

Viscosity at wall 1.45 cp

Thermal conductivity k 0.368 Btu/hr ft °F

Properties of jacket side fluid

Heat capacity 1.01 Btu/lb °F

Sp gravity 65.2 lb/ft3

Bulk Viscosity µ 1.45 cp 3.509 lb/hr ft

Viscosity at wall 1.45 cp 3.509 lb/hr ft

Thermal conductivity k 0.335 Btu/hr ft °F

Liquid flow rate Q 0.22 cu ft / sec

Calculation

Batch Side calculation

Agitated batch liquid reynolds number

=

= 1596296

Agitated batch liquid Prandlt number

=

= 3.51

µ/µw = 0.370

Z/T = 0.889

= 0.389

a 1

b 0.67

M 0.18

= 709.84 Btu/hr ft2 °F

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Da

Cp

ℓb

µw

Cp

ℓb

µw

NRe Da2 x N x ℓb / µ

NPr Cp x µ / k

Da/T

hb T / k = a x Nreb x NPr 1/3 x (µ/µw)M

Agitated liquid Heat Transfer coefficient,hb

Jacket Side calculation

A10.0243

b0.4

0.0265 0.3

Nu = hj Dj / Kj

Nre = Dj V ℓj / µj

Acs = L x W = 0.667 ft2

Wetted = L x 2 + W x 2 = 16.167 ft

= 0.165 ft

Aequivalent = 0.021 ft2

Velocity (V = Q/A) = 10.300 ft/sec 37082 ft/hr

= 113650.5

= 10.58

For cooling

A1 = 0.0243

b = 0.3

= 235.89 Btu / hr ft2 °F

Overall heat transfer coefficient

= 0.001

= 0.0066

= 150.42 Btu / hr ft2 °F

Nu = A1 x NRe0.667 x NPr b x (u/uw)0.14

Dj = Equivalent cross flow diameter of the jacket

Dj - Jacket cross flow diamater = 4 x Acs / Wetted

Dj

NRe

NPr

Jacket side Heat transfer coefficient, hj

1/U overall = 1/hJ + 1/hB + 1/hDM

RDM = 1/hDM

1/U overall

Uoverall

A reactor 2.0 KL capacity is used to carryout an exothermic reaction at 80°CThe reaction calorimetry shows heat of the reaction of 15000 BTU/mole of reagent 'Z'.If 5 mole of reagent Z is used calculate the addition rateCooling water is available at 0°C

U 55 BTU / hr sq F °F 264 kcal/hr m2 °CA 7 m2T 80 °C

Q 147840 kcal / hr

Heat of rxn 1000 kcal/mole

Addn Rate 29.568 lit/hr

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Two assumptions - Power per unit volume is constant - Tip speed is constant

Scale up (Lab scale data)Dt1 0.108 mH1 0.108 mVol 0.001 m3 1.00 LitDa1 0.043 mN1 400 rpm 6.67 rpsDensity 1300 kg/m3Viscosity 1000 CP 1.0000 kg / m s

Scale up times 1000 times

Assumption, Power per unit volume is constantN2 = N1 x (1/R) power 2/3

Ratio 10 = V2V1

N2 1.44 rps 86.2 rpmNp 5 (Assume)

Da2 0.433614 mDt2 1.084035 mV2 1000.00 LNre 351.0687

P 295.2272 J/sec 0.30 KW 0.40 hp0.515162 hp (30 % Loss)

Assuming Tip speed is constantN2 = N1 / RN2 0.666667 rps 40 rpm

Nre 162.9517 J/sec

P 29.52272 J/sec 0.03 KW 0.04 hp0.051516 hp (30 % Loss)

- The assumption is also based on power per unit volume as constant

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1/3

Designing Based on pressure drop test

ExampleTotal System Volume V = 350 ft3System Evacuation to Pi = 2 in Hg AbsFinal Pressure Pf = 3 in Hg AbsDrop test period t = 10 minDesired operating pressure Po = 1 in Hg Abs

Pump capacity required Q = (Rise in pressure) x System volume (V)time (t) x Desired operating pressure (Po)

Q = 35 CFM

Designing Based on pump evacuation time

Total system volume V = 100 ft3Initial pressure Pi = 760 mm Hg absFinal pressure Pf = 50 mm Hg absEvacuation time t = 2.25 min

Pump capacity required Q = V x ln(Pi/Pf)t

Q = 120.9 ACFM

Once the pump size is selected, we must recalculate the evacuation time by using that pump's averagecapacity.

Designing Based on Air Leakage Rate

Assuming the inlet gas composition is only air at 75°F

Fall in vacuum 300 mm Hg/hrTotal System volume V = 5000 litTemperature T = 75 °FInitial Pressure Pi = 760 mm Hg AbsOperating Pressure Po = 100 mm Hg AbsAir Leakage rate 2.348 kg/hr

The capacity in ACFM can be calculated using

ACFM = ( m / Mwt ) x (379/60) x (Initial pressure / Operating Pressure) x (460+T/520)

ACFM S = 8.82 ACFM

Correction factor Fc = 0.85

Corrected capacity Sc = 10.37 ACFM

Based on this, from the pump manufacturer curve required CFM, hp & gpm of water for service can be obtained

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Designing for air & solvent vapour

Composition of vapours are Air + Methanol Vapour + water vapour

Fall in vacuum 300 mm Hg/hrTotal System volume V = 5000 litTemperature T = 104 °FInitial Pressure Pi = 760 mm Hg AbsOperating Pressure Po = 50 mm Hg AbsMethanol vapour = 15 kg/hrWater vapour = 5 kg/hrAir Leakage rate = 2.23 kg/hrTotal = 22.23 kg/hr

Compound Qty % weight MwtMethanol 15.0 67.48 32.0Water 5.0 22.49 18.0Air 2.2 10.02 29.0

Calculating average molecular weight = 27.00

ACFM S = 85.74 ACFM

Correction factor Fc = 0.85

Corrected capacity Sc = 100.87 ACFM

(Rise in pressure) x System volume (V)time (t) x Desired operating pressure (Po)

Once the pump size is selected, we must recalculate the evacuation time by using that pump's average

0.4 atm5 m3

296.9 K

5.18 lb/hr

Based on this, from the pump manufacturer curve required CFM, hp & gpm of water for service can be obtained

0.394737 atm5 m3

313.0 K

4.91 lb/hr

Power Required for pumping

Flow rate Q 300 m3/hr 0.0833 m3/sec 83.33 kg/sec

Density ℓ 1000 kg/m3Pump head required H 25 m

Pump efficiency η 0.75

Power Required P 2777.8 kg m / sec 37.04 hp

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Affinity LawsQ, Flow Proportional N

H, Head Proportional

P , Power Proportional

Example

DataInitial RPM N1 1500 rpmInitial flow Q1 400 m3/hrInitial Head H1 25 mPower P1 49.4 hp

By adjustingFinal RPM N2 1400 rpm

ResultFinal flow Q2 373 m3/hrFinal Head H2 21.8 mFinal Power P2 40.1 hp

By reducing rpm by 6.67 %Head reduced by 12.89 %Power reduced by 18.70 %

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N2

N3

Pipe flow distribution

Data

Density ℓ 1000

Viscosity µ 0.8275 cp 0.00083 kg/m sec

Flow rate Q 500 0.13889Pipe Dia D 10 in 0.254

Acceleration, g 9.8

Calculation

Area A 0.05Velocity V 2.74 m/s

Reynolds Number 841775.2

Friction Factor f 0.002988Dia (m) Length (m)

0.20 30

0.15 60

0.10 40

Flow rate A (m3/sec)0.1389

Calculation0

Pipe Dia (m) Length(m) f V,m/s hf Q (m3/sec) Q (m3/hr)AaB 0.20 30 0.00299 2.95363 0.7980 0.0927 333.88AbB 0.15 60 0.00299 1.80872 0.7980 0.0319 115.01AcB 0.10 40 0.00299 1.80872 0.7980 0.0142 51.11

500.00SolverCons-1 0.00000Cons-2 0.00000Cons-3 0.00000

* Head loss across the pipe line is equal

Equations Used

kg/m3

m3/hr m3/sec

m/s2

m2

NRe

Equation of Continuity:Q_A = Q_a + Q_b + Q_c

Equality of Pressure drops:DelP due to friction in Pipe AaB = DelP due to friction in Pipe AbB DelP due to friction in Pipe AaB = DelP due to friction in Pipe AcB

Using these 3 sets of equations are formed; These equations are of non-linear. To solve these, the solver add-in is being used.

Target CellThis cell is to have the value of Q, in such a way that the constraints 1,2, and 3 are zero.

These values are arrived from solving the constraints 1, 2 and 3 by Solver Add-In

This program will run with Macros.

Set your macros setting to low and then proceed

the calculation.

Target CellThis cell is to have the value of Q, in such a way that the constraints 1,2, and 3 are zero.

These values are arrived from solving the constraints 1, 2 and 3 by Solver Add-In

Conversion Factors1 feet 0.3048 m

1 lb 0.454 kg1 inch 0.0254 m

1 centipoise 0.001 kg/m.sec1 atm 14.7 psi

1 atm 1.01E+05

g 9.812

Data given: Converted data:Mass flow rate 23100 lb/hr = 2.913167 kg/sec

Density r 52 = 833.7087Viscosity m 3.4 cP = 0.0034 kg/m.secPipe OD 3.5 inch

Pipe thickness 0.216 inch

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A 35o API distillate is being transferred from a storage tank at 1 atm absolute pressure to a pressure vessel at 50 psig by means of the piping arrangements shown in figure.The liquid flows at the rate of 23100 lb/hr through 3 inch Schedule 40 steel pipe; the length of the straight pipe is 450 feet. Calculate the minimum horsepower input to the pump having an efficiency of 60 percent.

The properties of the distillate are: viscosity = 3.4 cP, density = 52 lb/ft3.

The following are the data for the pipe and fittings:For 3 inch Schedule 40 Nominal pipe, OD = 3.5 inch; Thickness = 0.216 inchFlow coefficients for the fittings (K) are: Gate valve = 0.25; 90o elbow = 0.9; Check valve = 10

Friction factor can be calculated from Blasius equation. Account for entry and exit losses also.

N/m2

m/sec2

lb/ft3 kg/m3

Pipe length L 450 feet = 137.16 m

Vertical height 70 feet = 21.336 mPump efficiency (in fraction) 0.6

Loss coefficient of Gate Valve 0.25Loss coefficient of elbow 0.9

coefficient of check valve Valve 10

Pipe ID D 3.068 inch = 0.077927 m

Pressure at 2 50 psig = 3.45E+053.401361 34013.61

Calculations:

Volumetric flow rate Q 0.00349Velocity v 0.7326 m/sec

Reynolds Number NRe 13999Friction factor f 0.00726

1.3985 m

0.02735 m

0.00684 m

0.04923 m

0.27351 m

0.01094 m

0.02735 m

Total frictional head 1.76642 m

Pump head 22.561 m

Minimum power for the pump 1074.81 Watt

z1-z2

P2 N/m2

m3/sec

hf of pipe

v2/2g

hf of Gate valve

hf of 2 number of elbows

hf of Check valve

hf of sudden contraction at inlet

hf of sudden expansion at outlet

L62

0.079(NRe)-0.25