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1. A hygrometer, which measures the amount of moisture in a gas stream, is to be calibrated Steam and dry air are fed a

##### Citation preview

1. A hygrometer, which measures the amount of moisture in a gas stream, is to be calibrated

Steam and dry air are fed at known flow rates and mixed to form a gas stream with a known water con recorded; the flow rate of either the water or the air is changed to produce a stream with a different w recorded and so on. The following data are taken: Mass Fraction of Water, y 0.011 0.044 0.083 0.126 0.17

Hygrometer Reading, R. 5 20 40 60 80

(a) Draw a calibration curve and determine an equation for y( R ).

Calibration Curve

90 80 70 60

Calibration Curve

50

Linear (Calibration Curve)

40 30 20 10 0 0

0.02 0.04 0.06

0.08

0.1

0.12 0.14 0.16

0.18

(b) Suppose a sample of a stack gas is inserted in the sample chamber of the hygrometer and a readin flow rate of the stack gas is 1200 kg/h, what is the mass flow rate of water vapor in the gas? y = 0.0021 R If R = 43 what is y? y = 0.0021*43 y = 0.0903 mass flow rate of vapor in the gas is: 108.36

tream, is to be calibrated

as stream with a known water content and the hygrometer reading is roduce a stream with a different water content and the new reading is

y = 0.0021 R

e

Calibration Curve Linear (Calibration Curve)

ber of the hygrometer and a reading of R=43 is obtained. If the mass water vapor in the gas?

SUMMARY OUTPUT Regression Statistics Multiple R 0.9997912573 R Square 0.9995825581 Adjusted R Square 0.9994434108 Standard Error 0.0014944649 Observations 5 ANOVA df Regression Residual Total

SS MS F 1 0.0160440997 0.0160440997 7183.62906617 3 6.700276E-006 2.233425E-006 4 0.0160508

Coefficients Standard Error t Stat P-value 0.0004848066 0.0012181169 0.3979967891 0.7172630512 0.0021052486 2.483885E-005 84.7562921922 3.620239E-006

Intercept X Variable 1

RESIDUAL OUTPUT

PROBABILITY OUTPUT

Observation 1 2 3 4 5

Predicted Y 0.0110110497 0.042589779 0.0846947514 0.1267997238 0.1689046961

Residuals -1.10497E-005 0.001410221 -0.0016947514 -0.0007997238 0.0010953039

Percentile 10 30 50 70 90

X Variable 1 0.002 0.001 Residuals

0 -0.001 0

10

20

-0.002

Significance F 3.620239E-006

X Variable 1 0.2

Lower 95% Upper 95% Lower 95.0% Upper 95.0% -0.0033917851 0.0043613984 -0.0033917851 0.0043613984 0.0020262003 0.0021842969 0.0020262003 0.0021842969

Y

0.1 0 0

10 20 30 40 50 60 X Variable 1

BILITY OUTPUT

Normal Prob

Y 0.011 0.044 0.083 0.126 0.17

0.2 Y

0.1 0 0

10

20

30

40

Sample

X Variable 1 Residual Plot 0.002 0.001

als

0 -0.001 0

10

20

30

40

50

60

70

80

90

-0.002 X Variable 1

X Variable 1 Line Fit Plot Y Predicted Y Linear (Predicted Y) 10 20 30 40 50 60 70 80 90 X Variable 1

Normal Probability Plot

10

20

30

40

50

60

Sample Percentile

70

80

90

100

2. A process instrument reading, Z(volts), is thought to be related to a process stream flow rate V(L/s) and pressure P(kPa) by the following expression: Z = aVbPc Process data have been obtained in two sets of runs---one with V held constant, the other with P held constant. The data are as follows: Point 1 2 3 4

V(L/s) 0.65 1.02 1.75 3.43

P(kPa) 11.2 11.2 11.2 11.2

Z(volts) 2.27 2.58 3.72 5.21

lnZ lnV lnP 0.81978 -0.430783 2.415914 0.947789 0.019803 2.415914 1.313724 0.559616 2.415914 1.65058 1.23256 2.415914

5 6 7

1.02 1.02 1.02

9.1 7.6 5.4

3.5 4.19 5.89

1.252763 0.019803 2.208274 1.432701 0.019803 2.028148 1.773256 0.019803 1.686399

(b) Now use a graphical method and all the data to calculate a, b, and c. Comment on why you would have more confidence in this result than in that of part (a). (Hint: You will need at least two plots.)

s stream flow rate

nt, the other with P

ment on why you will need at least

ln Z = ln a + blnV + clnP ln Z = 3.5371 + 0.5165^lnV - 1.0466^lnP 34.36711 Z = 34.36711V0.5165P-1.0466

SUMMARY OUTPUT Regression Statistics Multiple R 0.9952054236 R Square 0.9904338352 Adjusted R Square 0.9856507528 Standard Error 0.0415073647 Observations 7 ANOVA df Regression Residual Total

Intercept X Variable 1 X Variable 2

2 4 6

SS 0.7135064805 0.0068914453 0.7203979258

MS 0.3567532402 0.0017228613

Coefficients 3.5371210232 0.5165475686 -1.0466377554

Standard Error 0.1385889391 0.0328641812 0.0625996286

t Stat 25.522390502 15.717646062 -16.7195521478

Predicted Y 0.7860145812 1.0187634481 1.2976026237 1.6452104551 1.2360866467 1.4246134931 1.7823012067

Residuals 0.0337652503 -0.0709740491 0.0161210446 0.0053694006 0.0166763218 0.0080872409 -0.0090452091

RESIDUAL OUTPUT Observation 1 2 3 4 5 6 7

F Significance F 207.0702006454 9.15115095E-005

P-value 0.000013997 9.57131179E-005 7.49835179E-005

Lower 95% 3.1523364416 0.4253019735 -1.2204421879

PROBABILITY OUTPUT Percentile 7.1428571429 21.4285714286 35.7142857143 50 64.2857142857 78.5714285714 92.8571428571

Y 0.8197798315 0.9477893989 1.2527629685 1.3137236683 1.4327007339 1.6505798558 1.7732559977

Upper 95% 3.9219056048 0.6077931637 -0.8728333229

Lower 95.0% 3.1523364416 0.4253019735 -1.2204421879

X Variable 1 Residual Plot 0.05 0 Residuals -0.6 -0.4 -0.2 0 -0.05

0.2

0.4

0.6

0.8

1

1.2

1.4

-0.1 X Variable 1

X Variable 2 Residual Plot Upper 95.0% 3.9219056048 0.6077931637 -0.8728333229

0.05 Residuals

0 1.6 -0.05

1.7

1.8

1.9

2

2.1

2.2

2.3

2.4

2.5

90

100

-0.1 X Variable 2

Normal Probability Plot 2 Y

1 0 0

10

20

30

40

50

60

Sample Percentile

70

80

lot

X Variable 1 Line Fit Plot 2

1

1.2

1.4

1

Y -1

Linear (Predicted Y)

-0.5

0

0.5

1

1.5

X Variable 1

X Variable 2 Line Fit Plot 2 Y

2.4

2.5

Y

1

Predicted Y

0 1.6 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 X Variable 2

ot

0

Predicted Y

0

lot

2.3

Y

90

100

Linear (Predicted Y)

3. A solution containing hazardous waste is charged into a storage tank and subjected to a chemical treatment that decomposes the waste to harmless products. The concentration of the decomposing waste, C, has been reported to vary with time according to the formula. C = 1/(a+bt) When sufficient time has elapsed for the concentration to drop to 0.01 g/L, the contents of the tank are discharged into a river that passes by the plant. The following data are taken for C and t: t(h) 1 2 3 4 5

C(g/L) 1.43 1.02 0.73 0.53 0.38

a. If the given formula is correct, what plot would yield a straight line that would enable you to determine the parameters a and b?

Calibration Curve 6 5 4

Calibration Curve Linear (Calibration Curve)

3 2 1 0 0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

b. Estimate a and b using the method of least squares (Appendix A.1). Check the goodness of fit by generating a plot of C versus t that shows both the measured and predicted values of C. a = 1.595 b = -0.259 c. Using the results of part (b), estimate the initial concentration of the waste in the tank and the time required for C to reach its discharge level.

When t = 0 C = 1/(a+bt) C = 1/(1.595 - 0.259*0) C= 0.6269592476 when C = 0.01 g/L, t = ? t = (a-(1/C))/b t= (1.595-(1/0.01))/0.259 t= -379.9420849 d. You should have very little confidence in the time estimated in part ( c ). Explain why The value we have for t is -379.942085 so we have little confidence in part c.

cted to a chemical the decomposing

ntents of the tank

nable you to

urve

ration Curve)

goodness of fit by s of C.

he tank and the

ence in part c.

SUMMARY OUTPUT Regression Statistics Multiple R 0.9804137374 R Square 0.9612110965 Adjusted R Square 0.948281462 Standard Error 0.0949912277 Observations 5 ANOVA df Regression Residual Total

1 3 4

SS MS F 0.67081 0.67081 74.341706686 0.02707 0.0090233333 0.69788

Coefficients Standard Error t Stat P-value 1.595 0.0996276401 16.009613385 0.0005299827 -0.259 0.0300388637 -8.6221636894 0.0032808106

Intercept X Variable 1

RESIDUAL OUTPUT

PROBABILITY OUTPUT

Observation 1 2 3 4 5

Predicted Y 1.336 1.077 0.818 0.559 0.3

Residuals 0.094 -0.057 -0.088 -0.029 0.08

Percentile 10 30 50 70 90

X Variable 1 R 0.2 0.1

Residuals

0 0.5 -0.1

1

1.5

2

X

Significance F 0.0032808106

X Variable 1 L 2

Lower 95% Upper 95% Lower 95.0% Upper 95.0% 1.2779403849 1.9120596151 1.2779403849 1.9120596151 -0.3545970708 -0.1634029292 -0.3545970708 -0.1634029292

Y

1 0 0.5

1

1.5

2

2.5

3

3.5

X Variable 1

ABILITY OUTPUT

Normal Prob

Y 0.38 0.53 0.73 1.02 1.43

2 Y

1 0 0

10

20

30

40

Sample P

X Variable 1 Residual Plot 0.2 0.1

uals

5

0

0 0.5 -0.1

1

1.5

2

2.5

3

3.5

4

4.5

5

5.5

X Variable 1

X Variable 1 Line Fit Plot Y Predicted Y 1

1.5

2

2.5

3

3.5

4

4.5

5

5.5

X Variable 1

Normal Probability Plot

10

20

30

40

50

60

Sample Percentile

70

80

90

100

4. A published study of chemical reaction. A-----P. indicates that if the reactor initially contains A at a concentration CAO (g/L) and the reaction temperature, T, is kept constant, then the concentration of P in the reactor increases with time according to the formula CP(g/L) = CAO(1-e-kt) The rate constant, k(s-1), is reportedly a function only of the reaction temperature. To test this finding, the reaction is run in four different laboratories. The reported experimental results are given below. Lab 1

Lab 2

Lab 3

Lab 4

T = 275 C T = 275 C T = 275 C T = 275oC CAO = 4.83 CAO = 12.2 CAO = 5.14 CAO = 3.69 o

t(s) 0 10 20 30 60 120 240 360 480 600

o

o

Cp (g/L) 0 0.287 0.594 0.871 1.51 2.62 3.91 4.3 4.62 4.68

0 1.21 2.43 3.38 5.89 8.9 11.2 12.1 12.1 12.2

0 0.31 0.614 0.885 1.64 2.66 3.87 4.61 4.89 5.03

k 0 0.245 0.465 0.67 1.2 2.06 3.03 3.32 3.54 3.59

#DIV/0! 0.006126 0.006561 0.006628 0.006248 0.006515 0.006909 0.006138 0.006532 0.005787

a. What plot would yield a straight line if the given equation is correct? b. Enter the given data into a spreadsheet. For each data set (Cp versus t), generate the plot of part (a) and determine the corresponding value of k. (Your spreadsheet program probably has a built-in function to perform a linear regression on the data in two specified columns.) c. Use the results in part (b) to come up with a good estimate of the value of k at 275 oC. Explain how you did it. d. If you did the calculation in part (b) correctly, one of the calculated values of k should be considerably out of line with the others. Think of as may possible explanations for this result as you can (up to 10)

initially contains A then the

ted experimental

ln Cp = ln CAO + ln(1-ekt) k = (1/t)[ln(Ca) - ln(Ca-Cp)] k #DIV/0! #DIV/0! 0.010445 0.006221 0.011106 0.006361 0.010814 0.006299 0.010988 0.006405 0.010896 0.006073 0.010423 0.005825 0.013345 0.006311 0.010008 0.006299 Err:502 0.006407

enerate the plot of am probably has a lumns.)

k at 275 oC. Explain

of k should be s for this result as

#DIV/0! 0.00687 0.006735 0.006679 0.006556 0.006809 0.007171 0.006389 0.006672 0.006014

SUMMARY OUTPUT Regression Statistics Multiple R 0.9270327498 R Square 0.8593897192 Adjusted R Square 0.8418134341 Standard Error 0.7574775216 Observations 10 ANOVA df Regression Residual Total

Intercept X Variable 1

1 8 9

SS 28.0545020344 4.5901775656 32.6446796

MS 28.0545020344 0.5737721957

Coefficients 0.7853894855 0.0080927631

Standard Error 0.3267340946 0.0011573515

t Stat 2.4037573629 6.9924848273

Predicted Y 0.7853894855 0.8663171164 0.9472447474 1.0281723784 1.2709552713 1.7565210571 2.7276526286 3.6987842002 4.6699157718 5.6410473434

Residuals -0.7853894855 -0.5793171164 -0.3532447474 -0.1571723784 0.2390447287 0.8634789429 1.1823473714 0.6012157998 -0.0499157718 -0.9610473434

RESIDUAL OUTPUT Observation 1 2 3 4 5 6 7 8 9 10

F 48.8948440595

Significance F 0.000113487

P-value 0.0429243054 0.000113487

Lower 95% 0.0319393123 0.0054239057

PROBABILITY OUTPUT Percentile

Y 5 15 25 35 45 55 65 75 85 95

0 0.287 0.594 0.871 1.51 2.62 3.91 4.3 4.62 4.68

Upper 95% 1.5388396586 0.0107616205

Lower 95.0% 0.0319393123 0.0054239057

X Variable 1 Residual Plot 2 Residuals

0 0

100

200

300

400

500

600

700

-2 X Variable 1

X Variable 1 Line Fit Plot 10

Upper 95.0% 1.5388396586 0.0107616205

Y

Y

5

Predicted Y Linear (Predicted Y)

0 0

100 200 300 400 500 600 700 X Variable 1

Normal Probability Plot 5 Y 0 0

10

20

30

40

50

60

Sample Percentile

70

80

90

100

lot

600

700

ot

ted Y (Predicted Y)

t

0

90

100

SUMMARY OUTPUT Regression Statistics Multiple R 0.8721906137 R Square 0.7607164667 Adjusted R Square 0.730806025 Standard Error 2.553643501 Observations 10 ANOVA df Regression Residual Total

Intercept X Variable 1

1 8 9

SS 165.8519289579 52.1687610421 218.02069

MS 165.8519289579 6.5210951303

Coefficients 3.1630427678 0.0196768606

Standard Error 1.1015011975 0.0039017174

t Stat 2.8715745157 5.0431280517

Predicted Y 3.1630427678 3.3598113736 3.5565799795 3.7533485853 4.3436544028 5.5242660379 7.8854893081 10.2467125782 12.6079358484 14.9691591185

Residuals -3.1630427678 -2.1498113736 -1.1265799795 -0.3733485853 1.5463455972 3.3757339621 3.3145106919 1.8532874218 -0.5079358484 -2.7691591185

RESIDUAL OUTPUT Observation 1 2 3 4 5 6 7 8 9 10

F 25.4331405454

Significance F 0.0009977367

P-value 0.0207786193 0.0009977367

Lower 95% 0.6229764514 0.0106794841

PROBABILITY OUTPUT Percentile

Y 5 15 25 35 45 55 65 75 85 95

0 1.21 2.43 3.38 5.89 8.9 11.2 12.1 12.1 12.2

Upper 95% 5.7031090842 0.0286742371

Lower 95.0% 0.6229764514 0.0106794841

X Variable 1 Residual Plot 5 Residuals

0 0

100

200

300

400

500

600

700

-5 X Variable 1

X Variable 1 Line Fit Plot 20

Upper 95.0% 5.7031090842 0.0286742371

Y

Y

10

Predicted Y Linear (Predicted Y)

0 0

100 200 300 400 500 600 700 X Variable 1

Normal Probability Plot 20 Y

10 0 0

10

20

30

40

50

60

Sample Percentile

70

80

90

100

lot

600

700

ot

ted Y (Predicted Y)

ot

0

90

100

SUMMARY OUTPUT Regression Statistics Multiple R 0.9389616743 R Square 0.8816490258 Adjusted R Square 0.866855154 Standard Error 0.7337671823 Observations 10 ANOVA df Regression Residual Total

SS 32.0870986778 4.3073142222 36.3944129

MS 32.0870986778 0.5384142778

Coefficients Standard Error 0.7891641143 0.3165067597 0.0086548744 0.0011211245

t Stat 2.4933562715 7.7198158533

1 8 9

Intercept X Variable 1

RESIDUAL OUTPUT Observation 1 2 3 4 5 6 7 8 9 10

Predicted Y 0.7891641143 0.8757128583 0.9622616024 1.0488103464 1.3084565786 1.8277490429 2.8663339714 3.9049189 4.9435038286 5.9820887571

Residuals -0.7891641143 -0.5657128583 -0.3482616024 -0.1638103464 0.3315434214 0.8322509571 1.0036660286 0.7050811 -0.0535038286 -0.9520887571

F 59.5955568081

Significance F 0.000056392

P-value 0.0373264813 0.000056392

Lower 95% 0.0592982176 0.0060695568

PROBABILITY OUTPUT Percentile

Y 5 15 25 35 45 55 65 75 85 95

0 0.31 0.614 0.885 1.64 2.66 3.87 4.61 4.89 5.03

Upper 95% 1.519030011 0.0112401921

Lower 95.0% 0.0592982176 0.0060695568

Upper 95.0% 1.519030011 0.0112401921

X Variable 1 Residual Plot 2 Residuals

0 0

100

200

300

400

500

600

700

-2 X Variable 1

X Variable 1 Line Fit Plot 10 Y

Y

5

Predicted Y Linear (Predicted Y)

0 0

100 200 300 400 500 600 700 X Variable 1

Normal Probability Plot 10 Y

5 0 0

10

20

30

40

50

60

Sample Percentile

70

80

90

100

SUMMARY OUTPUT Regression Statistics Multiple R 0.9232473245 R Square 0.8523856222 Adjusted R Square 0.833933825 Standard Error 0.5951834366 Observations 10 ANOVA df Regression Residual Total

Intercept X Variable 1

1 8 9

SS 16.3643634142 2.8339465858 19.19831

MS 16.3643634142 0.3542433232

Coefficients 0.6252852741 0.0061808059

Standard Error 0.2567293625 0.0009093821

t Stat 2.4355814538 6.7967097045

Predicted Y 0.6252852741 0.6870933327 0.7489013914 0.81070945 0.9961336259 1.3669819778 2.1086786815 2.8503753852 3.5920720889 4.3337687926

Residuals -0.6252852741 -0.4420933327 -0.2839013914 -0.14070945 0.2038663741 0.6930180222 0.9213213185 0.4696246148 -0.0520720889 -0.7437687926

RESIDUAL OUTPUT Observation 1 2 3 4 5 6 7 8 9 10

F 46.1952628073

Significance F 0.0001382865

P-value 0.0408450575 0.0001382865

Lower 95% 0.0332663024 0.0040837671

PROBABILITY OUTPUT Percentile

Y 5 15 25 35 45 55 65 75 85 95

0 0.245 0.465 0.67 1.2 2.06 3.03 3.32 3.54 3.59

Upper 95% 1.2173042457 0.0082778447

Lower 95.0% 0.0332663024 0.0040837671

X Variable 1 Residual Plot 2 Residuals

1 0 -1

0

100

200

300

400

500

600

700

X Variable 1

X Variable 1 Line Fit Plot Upper 95.0% 1.2173042457 0.0082778447

5

Y Predicted Y

Y

Linear (Predicted Y)

0 0

100 200 300 400 500 600 700 X Variable 1

Normal Probability Plot 4 Y

2 0 0

10

20

30

40

50

60

Sample Percentile

70

80

90

100

lot

600

700

lot

cted Y

r (Predicted Y)

ot

0

90

100