Process Calcs

DesignSI NoCalculation1Heating time using hot water (Non Isothermal Heating Medium)2Heating time using steam (Isothermal Heating Medium)3Cooling time using non isothermal cooling medium4Heating time using external Heat exchanger & Non Isothermal Heating Medium5Heating time using external Heat exchanger & Isothermal Heating MediumRevision History6Cooling time using external Heat exchanger & Non Isothermal Cooling MediumRev. 0 Issued to Downloads section on 5/5/20117Cooling time using external Heat exchanger & Isothermal Cooling Medium8Calculate the Heat Transfer coefficient for reactors based on different agitator types & utility flow rate9Calculate the addition time required based on heat of the reaction, Type of agitator, MOC & utility flow rate in a semi batch reactor10Scale up of reactor from lab to plant scale11Calculate the vacuum pump flow rate based on air leakage rate12Power required for pumping13Pump affinity law14Pipe flow distribution15Pressure drop across pipe lines

1HomeNon Isothermal HeatingReactor ContentsMassM72500lbInitial tempt1120FFinal tempt2200FSp.HeatCm1.05Btu/ lb FAreaA150ft2HTCU180BTU/hr ft2 F

Heating FluidInlet tempT1230FSp.HeatCw1.2Btu/ lb FFlow rateW18000lb/hr

Time required for heating

lnT1 - t1=WCw xK-1 x tT1 - t2MCmK

K=e(UA/WC)

K3.4903429575Time reqd=6.42hrs

2HomeIsothermal HeatingReactor ContentsMassM50000lbInitial tempt168FFinal tempt2257FSp.HeatCm0.5Btu/ lb FAreaA100ft2HTCU150BTU/hr ft2 F

Heating FluidInlet tempT1320F


lnT1 - t1=U A timeT1 - t2MC

Time reqd=2.31hrs

3HomeNon Isothermal coolingMassM72500lbInitial tempT1200FFinal tempT282FSp.HeatCm1.05Btu/ lb FAreaA150ft2HTCU110BTU/hr ft2 F

Cooling FluidInlet tempt172FSp.HeatCw1Btu/ lb FFlow rateW15500lb/hr

Time required for cooling

lnT1 - t1=WCw xK-1 x tT2 - t2MCmK

K=e(UA/WC)

K2.8994356918Time reqd=19.11hrs=1146.78min

4HomeNon Isothermal HeatingReactor ContentsMassM72500lbMass Flow ratew20000lb/hrInitial tempt182FFinal tempt2200FSp.Heatc1.05Btu/ lb FAreaA150ft2HTCU210BTU/hr ft2 F

Heating FluidInlet tempT1230FSp.HeatC1.2Btu/ lb FFlow rateW15500lb/hr


lnT1 - t1=wWC xK3-1 x tT1 - t2(K3wc-WC)M

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

K31.2135481052Time reqd=10.03hrs

5HomeIsothermal HeatingReactor ContentsMassM72500lbMass Flow ratew20000lb/hrInitial tempt182FFinal tempt2200FSp.Heatc1.05Btu/ lb FAreaA150ft2HTCU210BTU/hr ft2 F

Heating FluidInlet tempT1230F


lnT1 - t1= wc xK2-1 x tT1 - t2McK2

K2=eUA/wc

K24.4816890703Time reqd=7.45hrs

6HomeNon Isothermal coolingMassM72500lbBatch flow rateW20000lb/hrInitial tempT1200FFinal tempT282FSp.HeatC1.05Btu/ lb FAreaA150ft2HTCU180BTU/hr ft2 F

Cooling FluidInlet tempt172FSp.Heatc1Btu/ lb FFlow ratew15500lb/hr


lnT1 - t1=Wcw xK4-1 x tT2 - t1(K4wc-WC)M

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


7HomeIsothermal coolingMassM72500lbBatch flow rateW20000lb/hrInitial tempT1200FFinal tempT282FSp.HeatC1.05Btu/ lb FAreaA150ft2HTCU180BTU/hr ft2 F

Cooling FluidInlet tempt172FSp.Heatc1Btu/ lb FFlow ratew15500lb/hr


lnT1 - t1=WCK1-1 x tT2 - t1MCK1

K1=eUA/WC


8HomeCalculation of Heat transfer coefficient in reactorReactor dataDiameter of the tankT108in9ftImpeller speedN45rpm2700rphImpeller DiaDa42in3.5ftImpeller powerHp2HpAgitator TypeAnchor-1Vessel jacket flow depthL1in0.08ftHeight of liquid in cylindrical sectionZ96in8ft

Properties of batch side fluidHeat capacityCp0.997Btu/lb FSp gravityb1.00281002.862.60lb/ft3Bulk Viscosity0.536cp1.29712lb/hr ftViscosity at wallw1.45cpThermal conductivityk0.368Btu/hr ft FProperties of jacket side fluidHeat capacityCp1.01Btu/lb FSp gravityb65.2lb/ft3Bulk Viscosity1.45cp3.509lb/hr ftViscosity at wallw1.45cp3.509lb/hr ftThermal conductivityk0.335Btu/hr ft FLiquid flow rateQ0.22cu ft / secCalculationBatch Side calculationAgitated batch liquid reynolds numberNRe=Da2 x N x b / =1596296.06904527Agitated batch liquid Prandlt numberNPr=Cp x / k=3.51/w=0.370Z/T=0.889Da/T=0.389a1b0.67M0.18hb T / k = a x Nreb x NPr 1/3 x (/w)M
rmadhankumar: Heat Transfer CoefficientsHeat Transfer in an agitated vessel is analogous to that in a double-pipe heat exchanger. The liquid agitation affects the fluid velocity and Reynolds Number. Since the coefficient of heat transfer depends very little on agitator speed, it is not practical to size an agitator for a specific heat transfer coefficient. The best approach is to size the agitator for the best mixing and size the heat transfer area as required.

The general correlation for a jacketed agitated vessel is:

(hjT/k) = 0.85(D2Np/u)0.66(Cpu/k)0.33(Z/T)-0.56(D/T)0.13(u/uw)0.14

A general correlation for heat transfer in an agitated tank with an internal coil was developed by Oldshue and Getton and includes the diameter of the coil as follows.

(hcdt/k) = 0.17(D2Np/u)0.67(Cpu/k)0.37(D/T)0.1(dt/T)0.5

hj = heat transfer coefficient of vessel jacketT = tank diameterk = thermal conductivity of liquidD = impeller diameterN = agitator speedp = densityu = viscosityCp = heat capacityZ = height of liquiduw = viscosity at walldt = diameter of tube

Agitated liquid Heat Transfer coefficient,hb=709.84Btu/hr ft2 FhbT / K = 0.85 x NRe0.66 x NPr0.33 x (Z/T)-0.56 x (D/T)0.13 x (u/uw)0.14Agitated liquid heat transfer coefficient, hb537.0521553916Btu/hr ft2 FJacket Side calculation

Nu = A1 x NRe0.667 x NPr b x (u/uw)0.14A10.0243b0.4
rmadhankumar: Heating

0.02650.3
rmadhankumar: Cooling

Nu = hj Dj / Kj

Dj = Equivalent cross flow diameter of the jacketNre = Dj V j / jDj - Jacket cross flow diamater = 4 x Acs / WettedAcs = L x W=0.667ft2Wetted = L x 2 + W x 2=16.167ftDj=0.165ftAequivalent=0.021ft2Velocity (V = Q/A)=10.300ft/sec37081.648089172ft/hr

NRe=113650.540103428

NPr=10.58

For coolingA1=0.0243b=0.3

Jacket side Heat transfer coefficient, hj=235.89Btu / hr ft2 F

Overall heat transfer coefficient

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

RDM = 1/hDM=0.001

1/U overall =0.0066

Uoverall =150.42Btu / hr ft2 FAgitator TypeabmRange of Reynolds NumberAnchor-11.000.500.1810-300Anchor-20.360.670.18300-40000Disk,flat blade turbine0.540.670.1440-300000Helical Ribbon0.6330.500.188-100000Paddle0.360.670.21300-300000Pitched blade turbine0.530.670.2480-200Propeller0.540.670.142000

9HomeA reactor 2.0 KL capacity is used to carryout an exothermic reaction at 80CThe 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 0C

U55BTU / hr sq F F264kcal/hr m2 CA7m2T80C

Q147840kcal / hr

Heat of rxn1000kcal/mole

Addn Rate29.568lit/hr

10HomeTwo assumptions - Power per unit volume is constant - Tip speed is constantScale up (Lab scale data)Dt10.108mH10.108mVol0.0009999995m31.00LitDa10.043mN1400rpm6.67rpsDensity1300kg/m3Viscosity1000CP1.0000kg / m s

Scale up times1000timesAssumption, Power per unit volume is constantN2 =N1 x (1/R) power 2/3Ratio10=V21/3V1N21.44rps86.2rpmNp5(Assume)

Da20.4336140728mDt21.0840351821mV21000.00LNre351.0687427296

P295.2272375749J/sec0.30KW0.40hp0.5151616226hp(30 % Loss)Assuming Tip speed is constantN2 =N1 / RN20.6666666667rps40rpm

Nre162.9516756084J/sec

P29.5227237575J/sec0.03KW0.04hp0.0515161623hp(30 % Loss)

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

11HomeDesigning Based on pressure drop test

ExampleTotal System VolumeV=350ft3System Evacuation to Pi=2in Hg AbsFinal PressurePf=3in Hg AbsDrop test periodt=10minDesired operating pressurePo=1in Hg Abs

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

Q=35CFMDesigning Based on pump evacuation time

Total system volumeV=100ft3Initial pressurePi=760mm Hg absFinal pressurePf=50mm Hg absEvacuation timet=2.25min

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

Q=120.9ACFM

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 75F

Fall in vacuum300mm Hg/hr0.4atmTotal System volumeV=5000lit5m3TemperatureT=75F296.9KInitial PressurePi=760mm Hg AbsOperating PressurePo=100mm Hg AbsAir Leakage rate2.348kg/hr5.18lb/hrThe capacity in ACFM can be calculated using

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

ACFMS=8.82ACFM

Correction factorFc=0.85

Corrected capacitySc=10.37ACFM

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

Composition of vapours are Air + Methanol Vapour + water vapour

Fall in vacuum300mm Hg/hr0.3947368421atmTotal System volumeV=5000lit5m3TemperatureT=104F313.0KInitial PressurePi=760mm Hg AbsOperating PressurePo=50mm Hg AbsMethanol vapour=15kg/hrWater vapour=5kg/hrAir Leakage rate=2.23kg/hr4.91lb/hrTotal=22.23kg/hr

CompoundQty% weightMwtMethanol15.067.4832.0Water5.022.4918.0Air2.210.0229.0Calculating average molecular weight=27.00ACFMS=85.74ACFM

Correction factorFc=0.85

Corrected capacitySc=100.87ACFM

12Home

Power Required for pumping

Flow rateQ300m3/hr0.0833m3/sec83.33kg/secDensity1000kg/m3Pump head requiredH25mPump efficiency0.75

Power RequiredP2777.8kg m / sec37.04hp

13Home

Affinity LawsQ, FlowProportionalNH, HeadProportionalN2P , PowerProportionalN3

Example

DataInitial RPMN11500rpmInitial flowQ1400m3/hrInitial HeadH125mPowerP149.4hp

By adjustingFinal RPMN21400rpm

ResultFinal flowQ2373m3/hrFinal HeadH221.8mFinal PowerP240.1hp

By reducing rpm by 6.67%Head reduced by12.89%Power reduced by18.70%

14

Pipe flow distribution

DataDensity1000kg/m3This program will run with Macros.Viscosity0.8275cp0.00083kg/m secSet your macros setting to low and then proceedFlow rateQ500m3/hr0.13889m3/secthe calculation.Pipe DiaD10in0.254Acceleration,g9.8m/s2

CalculationAreaA0.05m2VelocityV2.74m/sReynolds NumberNRe841775.2

Friction Factorf0.0029879881Dia (m)Length (m)0.2030

0.1560

0.1040

Flow rate A (m3/sec)0.1389Calculation0PipeDia (m)Length(m)fV,m/shfQ (m3/sec)Q (m3/hr)AaB0.20300.00298798812.9536278148
rmadhankumar: These values are arrived from solving the constraints 1, 2 and 3 by Solver Add-In0.79800.0927333.88AbB0.15600.00298798811.80872025240.79800.0319115.01AcB0.10400.00298798811.80872026060.79800.014251.11500.00
rmadhankumar: Target CellThis cell is to have the value of Q, in such a way that the constraints 1,2, and 3 are zero.

SolverCons-10.00000
rmadhankumar: From Equation of continuity

Cons-20.00000
rmadhankumar: From Equality of pressure drop

Cons-3-0.00000* Head loss across the pipe line is equalEquations UsedEquation 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.

15HomeA 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.

Conversion Factors1 feet 0.3048m1 lb 0.454kg1 inch 0.0254m1 centipoise0.001kg/m.sec1 atm 14.7psi1 atm 1.01E+05N/m2g9.812m/sec2

Data given:Converted data:Mass flow rate 23100lb/hr=2.9131666667kg/secDensity r 52lb/ft3=833.7086519609kg/m3Viscosity m 3.4cP=0.0034kg/m.secPipe OD 3.5inchPipe thickness 0.216inchPipe length L 450feet=137.16mVertical height z1-z2 70feet=21.336mPump efficiency (in fraction)0.6

Loss coefficient of Gate Valve 0.25Loss coefficient of elbow 0.9Loss coefficient of check valve Valve 10

Pipe ID D 3.068
chem: OD - 2(thickness)inch=0.0779272mPressure at 2 P2 50psig=3.45E+05N/m23.401360544234013.6054421769Calculations:Volumetric flow rateQ 0.00349
chem: mass flow rate / densitym3/secVelocity v 0.7326
chem: volumetric flow rate/aream/secReynolds Number NRe 13999
chem: Dvr/m

Friction factor f 0.00726
chem: 0.079(NRe)-0.25

chem: 0.079(NRe)-0.25hf of pipe 1.3985
chem: hf = Dp/(rg)where Dp= 2fLrv2/Dm

v2/2g 0.02735mhf of Gate valve 0.00684
chem: hf = Kv2/(2g)mhf of 2 number of elbows 0.04923mhf of Check valve 0.27351m

hf of sudden contraction at inlet 0.01094
chem: hf = 0.4(1 S2/S1)v2/2gwhere S2 is the area of pipe; and S1 is the area of tank 1mhf of sudden expansion at outlet 0.02735
chem: hf= (1 - S2/S3)2.v2/2gwhere S2 is the area of pipe; S3 is the area of tank 2.m

Total frictional head 1.76642
chem: = frictional lossesdue to skin friction in pipe + losses in (entry, gatevalve, 2 nos of elbow, checkvalve, exit)m

Pump head 22.561
chem: = (p2/rg) + (z2 - z1) + total frictional headm

Minimum power for the pump 1074.81
chem: = mass flow rate X g X pump head / efficiencyWatt

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