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458.401 Process & Product Design

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11

Jong Min Lee

School of Chemical and Biological Engineering Seoul National University

Heuristics (“Rules of Thumb”)

Heuristics 11

Use of Experienced-Based Principles to:

Check new process designs

Provide equipment size and performance estimates

Help with troubleshooting problems with existing systems Verify computer simulations and results

Provide initial estimates for simulator inputs Use as a starting place for initial designs Develop preliminary process layouts

However, keep in mind that a heuristic:

Doesn’t guarantee a solution May contradict other heuristics

Can reduce the time to solve problems Depends on the context in which it is used

(2)

Examples of Heuristics Usage

What is the density of water in SI units? In English units?

What does a gallon of milk weigh? What about 4 gallons of milk?

What does a cubic meter of air weigh?

The ratio of is approximately _______?

3

L

/⇢

v

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Heuristics 11

Creation of Heuristics / Shortcuts

1. Predict. (assumptions, heuristics, back of the envelope calculations) 2. Authenticate/Analyze (equations, simulations, operational data) 3. Reevaluate/Rethink (compare, revise)

Example-11.1

Suppose you calculated a turbulent heat transfer coefficient inside a 1.5” pipe as:

5250 W/[m2·℃], for water at 21℃ and a velocity of 1.83 m/s.

Estimate a heat transfer coefficient when the water temperature is 93℃ and a velocity of 3.05 m/s.

(3)

5

1. Predict: assume the velocity and temp. have no effect. Predicted Heat Transfer Coef. = 5250 W/[m2·℃]

2. Authenticate: Find appropriate equation for calculation of Nusselt number.

Solution-11.1

Re = u⇢Dpipe/µ= (1.83)(997.4)(1.5)(0.0254)/(9.8⇥10 4) = 71⇥103

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For highly turbulent flow, use the _______________ equation hD/k = (0.023)(Du⇢/µ)0.8(Cpµ/k)1/3

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h0/h = (D0/D)0.2(u0/u)0.8(⇢0/⇢)0.80/µ)0.47(Cp0/Cp)0.33(k0/k)0.67

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= (1)(1.50)(0.973)(1.73)(1.00)(1.08) = 2.725<latexit sha1_base64="/maFZZlv1jxJRD5H8GXwqJHIhTA=">AAACFnicbVDLSgMxFM3UV62vUZdugkWYLhxmppZWQSi4cVnBPqAdSibNtKGZB0lGKEO/wo2/4saFIm7FnX9jpu1CWw/kcnLuuST3eDGjQlrWt5ZbW9/Y3MpvF3Z29/YP9MOjlogSjkkTRyziHQ8JwmhImpJKRjoxJyjwGGl745us334gXNAovJeTmLgBGobUpxhJJfX1c3gNDbtk2GbFKhmWeVktZ5d5taxZrZWUyTGrTqWvFy2lZoCrxF6QIlig0de/eoMIJwEJJWZIiK5txdJNEZcUMzIt9BJBYoTHaEi6ioYoIMJNZ2tN4ZlSBtCPuDqhhDP190SKAiEmgaecAZIjsdzLxP963UT6NTelYZxIEuL5Q37CoIxglhEcUE6wZBNFEOZU/RXiEeIIS5VkQYVgL6+8SlqOaZdN5+6iWL9axJEHJ+AUGMAGVVAHt6ABmgCDR/AMXsGb9qS9aO/ax9ya0xYzx+APtM8fD+CWjA==</latexit>

h0 = (2.725)(5250) W/m2 C = 14,300W/m2 C

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(11.2)

Heuristics 11

3. Reevaluate/Rethink (compare, revise) Solution-11.1

• The temperature effect on viscosity must be evaluated.

• The effects of temperature on Cp, ⍴, and k are negligible.

• Pipe diameter has a small effect on h (all other things being equal)

• Results are limited to the range where the Sieder-Tate equation is valid

With these assumption, the values for water at 21℃ are substituted into Eq. (11.2). This creates a useful heuristic for evaluating the heat transfer coefficients for water

(4)

Follow the logic carefully…

7

From Rules 9 and 12

u = 0.0305

r850

8 1 = 0.313 m/s

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uact = (0.75)(0.313) = 0.23 m/s

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Mass flow rate of vapor = 9200 kg/h = 9200/3600 (kg/s) = 2.56 (kg/s)

uactv⇡D2/4 = 2.56 ! D = 1.33 m

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From Rule 5, the volume of the liquid =

0.5L⇡D2/4 = 0.726L m3

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5 min of liquid flow 


= (5 min)(60 sec/min)(11,570 kg/hr)/(850 kg/m

3

)/3600 (hr/sec) = 1.13 m

3

Equating the two results above, L = 1.56 m

Heuristics 11

Rule 1: Set F = 0.9

Rule 6: min 𝛥 T = 10℃; but E-105 has 8℃ (ignore for now) Rule 7: Water enters at 30℃ and leaves at 40℃

Rule 8: U = 850 W/m

2

T

lm

= [(112 40) (38 30)]/ ln[(112 40)/(38 30)] = 29.1 C

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Q = 1085MJ/h = 301kW from Table 1.7

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A = Q

F U T

lm

= 301, 000

0.9 ⇥ 850 ⇥ 29.1 = 13.52 m

2

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From Rule 9, this heat exchanger should be double-pipe or multiple- pipe design

In Table 1.7, E-105 is multiple-pipe design with the area of 12 m

2

(5)

9

Rule 2:

Fs = u⇢0.5v = 1.2 1.5 m/s(kg/m3)0.5

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Rule 3:

Ptray = 0.007 bar

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Rule 4:

"tray = 60 90%

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x<latexit sha1_base64="AUHygUqOawO2OsD/95c1w9r8X30=">AAACDHicbVDLSgMxFM34rPVVdekmWAQXUmZa0RYUCm5cVrAPaIeSyWTa0EwyJJliGerejb/ixoUibv0Ad/6NaTsLbT0QOJxzLjf3eBGjStv2t7W0vLK6tp7ZyG5ube/s5vb2G0rEEpM6FkzIlocUYZSTuqaakVYkCQo9Rpre4HriN4dEKir4nR5FxA1Rj9OAYqSN1M3l77uJGPb9MbyCdqFSOS+ePhjJE3qm2CW7bFKGTAEXiZOSPEhR6+a+Or7AcUi4xgwp1XbsSLsJkppiRsbZTqxIhPAA9UjbUI5CotxkeswYHhvFh4GQ5nENp+rviQSFSo1CzyRDpPtq3puI/3ntWAdlN6E8ijXheLYoiBnUAk6agT6VBGs2MgRhSc1fIe4jibA2/WVNCc78yYukUSw4pULx9ixfvUzryIBDcAROgAMuQBXcgBqoAwwewTN4BW/Wk/VivVsfs+iSlc4cgD+wPn8Aa3WYoA==</latexit> ovhd = 0.9962, xbot = 0.0308

ovhd = 2.44, ↵bot = 2.13, ↵geom ave =p

2.44⇥2.13 = 2.28

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Nmin = {ln[0.9962/(1 0.9962)]/[0.0308/(1 0.0308)]}/ln(2.28) = 10.9

<latexit sha1_base64="8oNSSofXdqd1lpfcWqv6LV2+Pog=">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</latexit>

R<latexit sha1_base64="SVMEVoJcedBDisFohnO4jPMdMDE=">AAACEXicbZDLTgIxFIY7eEO8oS7dNBITWDhMR1AWmpC4cYlGLgkQ0ikFGjqdSdsxIRNewY2v4saFxrh15863sVwWCp6kyZf/Pyen5/dCzpR2nG8rsbK6tr6R3Extbe/s7qX3D2oqiCShVRLwQDY8rChnglY105w2Qkmx73Fa94bXE7/+QKVigbjXo5C2fdwXrMcI1kbqpLN3nbjlMzGGVzCLCq7t5pFTtM9z+azh0inKGQPZTrGTzji2My24DGgOGTCvSif91eoGJPKp0IRjpZrICXU7xlIzwuk41YoUDTEZ4j5tGhTYp6odTy8awxOjdGEvkOYJDafq74kY+0qNfM90+lgP1KI3Ef/zmpHuldoxE2GkqSCzRb2IQx3ASTywyyQlmo8MYCKZ+SskAywx0SbElAkBLZ68DDXXRme2e1vIlC/ncSTBETgGWYDABSiDG1ABVUDAI3gGr+DNerJerHfrY9aasOYzh+BPWZ8/8suXWQ==</latexit> min = (142.2/105.6)/(2.28 1) = 1.05

stream 15

1.26 ≤ R ≤ 1.58

N

theory

= 2 x 10.9 = 21.8 N

actual

= (21.8 / ε (=0.6) ) x 1.1 = 43 trays Vapor flowrate (Str. 13) = 22,700 kg/h

v = 6.1 kg/m3

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u = (1.2 1.5)/6.10.5 = (0.49 0.60) m/s

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Heuristics 11

Vapor flowrate (Str. 13) = 22,700 kg/h

22,700kg

h ⇥ 1 h

3,600 s ⇥ m3

6.1 kg = 1.03m3

s =v

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Dtower = [4v/⇡u]0.5 = [(4)(1.03)/(3.142)/(0.49 ⇠ 0.60)]0.5 = 1.64 ⇠ 1.48 m

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Ptower = (Nactual)( Ptray) = (43)(0.007) = 0.30 bar

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Tables 1.5 and 1.7 and F 1.5 From Heuristics

Tower diameter 1.5 m 1.48 - 1.64 m

Reflux ratio, R 1.75 1.26 - 1.58

Number of trays 42 43

Referensi

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