practice problems

23
Xa Ya Xc Yc Xe Ye 0.00 0.00 0.00 0.00 0.00 0.00 0.10 0.20 0.10 0.20 0.10 0.06 0.20 0.36 0.20 0.30 0.20 0.15 0.40 0.61 0.40 0.45 0.42 0.42 0.55 0.75 0.55 0.55 0.55 0.60 0.60 0.79 0.60 0.58 0.60 0.68 0.80 0.90 0.80 0.72 0.80 0.89 1.00 1.00 1.00 1.00 1.00 1.00 PT 101.3 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 Xe Ye Ya

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Page 1: Practice Problems

Xa Ya Xc Yc Xe Ye0.00 0.00 0.00 0.00 0.00 0.000.10 0.20 0.10 0.20 0.10 0.060.20 0.36 0.20 0.30 0.20 0.150.40 0.61 0.40 0.45 0.42 0.420.55 0.75 0.55 0.55 0.55 0.600.60 0.79 0.60 0.58 0.60 0.680.80 0.90 0.80 0.72 0.80 0.891.00 1.00 1.00 1.00 1.00 1.00

PT 101.3

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.000.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

Column CColumn TLinear (Column T)

Xa

Ya

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.000.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

Xe

Ye

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.000.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

Column IColumn TLinear (Column T)

XcYa

Page 2: Practice Problems

0.00 0.000.13 0.130.25 0.250.38 0.380.50 0.500.63 0.630.75 0.750.88 0.881.00 1.00

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.000.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

Column CColumn TLinear (Column T)

Xa

Ya

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.000.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

Column IColumn TLinear (Column T)

Xc

Ya

Page 3: Practice Problems
Page 4: Practice Problems

Tank 1 0.45 0.50 0.46Tank 2 0.70 0.80 0.71

kr 0.119

0.20 0.505 2

2.2360679775 1.4142135623731 1.82510.0894427191 0.353553390593274 0.1644901909296670.9105572809 0.646446609406726 0.835509809070333

(CAo-CA)/CAoExpected conversion using hydraulic residence time

Expected conversion using mean residence time

min-1

Page 5: Practice Problems

X 0.7Tau 6.934171 <-- Goal seek to find this valueRHS 0.699811

HRT Tank 2X 0.8Tau 10.37375 <-- Goal seek to find this valueRHS 0.799715

MRT Tank 2X 0.71Tau 7.199785 <-- Goal seek to find this valueRHS 0.709944

Page 6: Practice Problems

a) Reynolds Calc Bernoulli Eqn b)D 0.1016 m u2 12.33u 12.33 m/s u1 0rho 1020 kg/m3 alpha 1mu 1.90E-03 Pa_s KE term 76.01445Reynolds # 672517.1

g 9.81epsilon/D (Cast Iron Diagram) z2 20eps 2.00E-04 z1 0D 0.1016 m height term 196.2 c)eps/D 1.97E-03 m/mf from moody 0.006 Pressure Term

P2 345 kPa gfriction loss from pipes P1 0 kPa g d)f 0.006 rho 1020 kg/m3deltaL 25 m p term -338.2353D 0.1016 mu 12.33 m/s RHS 948.4432Ff 448.9042 m2/s2 LHS -338.2353

Ws -1286.679 J/kgFriction loss from fittingssum of kf 1.7 Convert Wsu 12.33 Ws -1286.679 J/kghf 129.2246 m2/s2 rho 1020

Vol. Flow 0.1 m3/sFf+hf 578.1288 m2/s2 Ws -131.2412 kWHX 10 mg 9.81 m/s2Ff+hf+HX 676.2288 m2/s2

Page 7: Practice Problems

Pmax calcWs -1286.679 J/kgrho 1020 kg/m3Pmax 1312412 PaPmax 1312.412 kPa

accounting for efficiency, pump A or B

increasing roughnessincreases friction factorincreases Ws (check efficiency equation)

if pump goes below vapour pressure of fluidbubbles can form and collapse creating shockwavescausing your pumps damage

Page 8: Practice Problems

2

9.9858D 15 cm

3 deg C

20 deg C

5 deg Calpha 1.04E-07 m2/sr 0.075 m

a) Unaccomplished Temp Change

-2

-17Y 0.117647

t 1.53E+04 st 4.262684 hours

b) New Ambient Temperature

-5 deg C

-10

-25Y 0.4

t 8.72E+03 st 2.421464 hours

Temperature gradient is the driving forceLarger temp gradient means faster heat transfer

c) Dough would freeze on the surface

d) Keep the dough in the freezer until it reaches the temp of the fridge at the surfaceThe overall delta T between freezer and dough is larger so it will reach fridge temperature fasterThen shove it in the fridge.

e) increase surface area

asp

Ksp

Tfridge

Ti

Tideal

Ta-T

Ta-Ti

Tfreezer

Ta-T

Ta-Ti

Page 9: Practice Problems

a) Jetting vs Non-Jetting Regime c)

Increasing flow rate decreases bubble sizeIncreases the number of bubblesResidence time increases (smaller bubbles are less buoyant)More mass transferHTU decreases

b) 2 Mass transfers

High concentration of BA in centre

Low concentration at surface

Concentration gradient from centre to surface

Mass Transfer happens tau

Then concentration of BA at surface is high

Low concentration in water Volume of Kerosene in ColumnConcentration gradient from surface to water tau

Mass Transfer happens

JETTING REGIME NON-JETTING REGIMEVolume of Droplet in Column

r

Number of Droplets#

Mass Transfer AreaD

Vk

Vw

Dp

Kc

Vk

Volk

Voldrop

Voldrop

AMT 1 drop

Total AMT

Page 10: Practice Problems

0.05 L/min

2 L/min 0.01 m/s

5 mm 0.03 m2

30 s 4.79 unit conversion

0.01 m/s X 0.03 mol/L 30 mol/m3

0 mol/L 0 mol/m3

Volume of Kerosene in Column 0.04311 mol/s0.5 min

0.05 L/min Convert flow rates

0.025 L rhomW 54.89 kmol/m3rhomK 4.75 kmol/m3

Volume of Droplet in Column rhomBA 8.92 kmol/m3 unit conversion

0.0025 m 0.05 L/min 0.00005 m3/min

6.54E-08 m3 2 L/min 0.002 m3/min

6.54E-05 L 0.2375 mol/min

109.78 mol/minNumber of Droplets

381.97 droplets

Mass Transfer Area0.005 m

7.853975E-05 m2

0.03 m2

NOTE, mk should be mBA

mBA calculation

Kc

Total AMT

Kp

Yi

MBA

Vk

Vw

Mk

Mw

Page 11: Practice Problems

Xi Calculation

2.5866 mol/min

109.78

0.2375

X 0.03

0

10.92095 mol/LY 0.023562 mol/L

MBA

Mw

Mk

Yi

Xi

Page 12: Practice Problems

RunHot Inlet Inlet Cold

(°C) (°C) (GPM) (°C) (°C)1 1 68.9 18.7 3 7 15.83 2 36.1 15.7 3 5.8 13.15 3 29.7 15.6 3 5.4 13.17 3 31.2 26.3 1 5 19.49 3 31 24.3 2 5 17.9

11 3 31.2 22.8 3 5 15.7GPM to m3s 0.0000631

997Di 2.5 cmDo 3.5 cm

pipe thickness 0.5 cm

CoefficientscP waterkJ/(kg·C˚)

4.19

Hot Side: Use a different correlationRun Tm mu (Pa*s) rho (kg/m^3) v (m/s) k (W/m*K) Re

1 27.370306807932 5.66E-04 9.88E+02 1.30E-01 0.64 2.27E+03

3 15.540486394002 1.02E-03 9.98E+02 2.60E-01 0.609 2.54E+03

5 13.141279688679 1.20E-03 9.99E+02 3.90E-01 0.603 3.25E+03

7 16.085131557911 1.55E-03 1.00E+03 3.90E-01 0.595 2.52E+03

9 16.000294582818 1.47E-03 1.00E+03 3.90E-01 0.596 2.65E+03

11 16.623489757232 1.40E-03 1.00E+03 3.90E-01 0.598 2.79E+03

Cold side:Run Tm mu (Pa*s) rho (kg/m^3) v (m/s) k (W/m*K) Re

1 27.370306807932 1.39E-03 1.00E+03 4.00E-01 0.598 7.19E+03

3 15.540486394002 1.45E-03 1.00E+03 4.00E-01 0.597 6.90E+03

5 13.141279688679 1.43E-03 1.00E+03 4.00E-01 0.598 6.99E+03

7 16.085131557911 1.19E-03 9.99E+02 1.30E-01 0.603 2.73E+03

9 16.000294582818 1.24E-03 9.99E+02 2.70E-01 0.602 5.44E+03

11 16.623489757232 1.32E-03 9.99E+02 4.00E-01 0.6 7.57E+03

Overall:U

645.858316281085

734.256574641391

Hot Flow rate (GPM)

Hot Outlet

Cold Flow Rate

Outlet Cold

Page 13: Practice Problems

833.640979093152

485.47559423851

692.786340932194

812.752899881096

Heat Duty

Run Q (kW) Cold Side

1 17.677340271068 17.6773403

3 11.410704307921 11.4107043

5 10.955109266208 10.9551093

7 7.8089388014813 7.8089388

9 11.084785537868 11.0847855

11 13.510789506334 13.5107895

Q (kW) Hot side

Page 14: Practice Problems

Use a different correlationPr Nu hi (W/m^2*K)

3.71E+00 1.65E+01 1055.04272

7.02E+00 2.19E+01 1332.31578

8.34E+00 2.80E+01 1688.60323

1.09E+01 2.48E+01 1473.12553

1.03E+01 2.54E+01 1514.45873

9.81E+00 2.60E+01 1555.50188

Pr Nu ho (W/m^2*K)

9.74E+00 6.96E+01 1665.28368

1.02E+01 6.85E+01 1635.72704

1.00E+01 6.88E+01 1646.49241

8.27E+00 3.00E+01 724.109289

8.63E+00 5.30E+01 1276.90347

9.22E+00 7.09E+01 1702.10758

Page 15: Practice Problems

Bernoulli Equation Friction TermsDn 0.02664 mZt 5.76 m eps/D calculationZdiff 4 m eps 0.00015 mZ2 9.76 m D 0.0525 m

eps/D 0.002857Velocity CalcVdot 11 m3/h SystemDpipe 0.0525 m Re 57003.04CSA 0.002165 m2 fric factor 0.006975u 5081.414 m/h pipe L 6.5

1.411504 m/s Extra H 5.76Ffsystem 6.490617

Reynolds Number Calckin visc 1.30E-06 m2/s Nozzleu 1.411504 m/s eps 0.00015D 0.0525 D 0.02664Re 57003.04 eps/D 0.005631

CSA 0.000557 m2g 9.81 m/s2 u 19734.89 m/h

5.481914 m/sRe 1.12E+05fric factor 8.06E-03L 1 mFfnozzle 18.19447

FittingsKf1 1.1 m2/s2Kf2 0.7 m2/s2Kf3 2.1 m2/s2u 1.411504 m/shf 3.885069 m2/s2

sum of F 28.57016

Page 16: Practice Problems

Ws CalculationsPressure Term 0Height Term 95.7456KE Term 15.02569Friction 28.57016Ws 139.3415 J/kg

Ws Conversionrho 999 kg/m3Vdot 11 m3/hWs 425.3398 W

Page 17: Practice Problems
Page 18: Practice Problems

Particles Unaccomplished Temp Change

8.5 m/s Ta - TK 0.79 W/m2C Ta - Ti

1.8 kJ/kgC Yrho 652 kg/m3

350 deg C Finite CylinderT 160 deg C rcylalpha 6.731425E-07 m2/s rslab

Air

flow rate 2000 cfm

1000 J/kgC

rho 1 kg/m3

2 deg Ct

uterminal

CP

Ti

Ysc = Ys x Yc

acyl

CP kcyl

as

Ta ks

Page 19: Practice Problems

Unaccomplished Temp Change

-158-348

0.454023

Finite Cylinder0.01 m

0.0125 m

1.5811

5.7565

1.2911

2.486122.37146 s

Ysc = Ys x Yc

Page 20: Practice Problems