Process Calcs
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Transcript of 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
Time required for heating
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
Time required for heating
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
Time required for heating
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
Time required for cooling
lnT1 - t1=Wcw xK4-1 x tT2 - t1(K4wc-WC)M
K4=eUA*(1/WC-1/wc)
K40.6336736542Time reqd=18.19hrs=1091.62min
7HomeIsothermal coolingMassM72500lbBatch flow rateW20000lb/hrInitial tempT1200FFinal tempT282FSp.HeatC1.05Btu/ lb FAreaA150ft2HTCU180BTU/hr ft2 F
Cooling FluidInlet tempt172FSp.Heatc1Btu/ lb FFlow ratew15500lb/hr
Time required for cooling
lnT1 - t1=WCK1-1 x tT2 - t1MCK1
K1=eUA/WC
K13.6172507852Time reqd=12.77hrs=766.37min
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