| Name: |
Twoeq9 - Pipe flow velocity
calculation for a specified pressure drop |
| Source: |
Wilkes, J. O., Fluid mechanics
for chemical engineers, Prentice-Hall, 1998 |
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Engineering with Numerical
Methods, Prentice Hall Inc., 1999 |
| Reference/s |
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| Model: |
2 implicit equations, indep.
variables fF and u |
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Higher difficulty level |
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Constraints: fF>0, u>0 |
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Discontinuities: The function
f(fF) undefined for u<=0 and/or fF<=0 |
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Initial estimates: 1. fF =
0.1, u=10; 2. fF = 1, u=10; 3. fF = 0.1, u=1; 4. fF = 0.1, u=0.1; |
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| Solved by |
Shacham, M., POLYMATH 5.1,
build 19, April 18, 2001 |
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| Model Eqs. |
Pipe flow velocity calculation
for a specified pressure drop |POLVER05_3 |
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EXCEL
FILE |
f(fF) = fF-1/(2.28-4*log(eps/D+4.67/(Re*fF^0.5)))^2
# |
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TEXT
FILE |
f(u) = 133.7-(2*fF*rho*u*u*L/D+rho*g*200)/(g*144)
# |
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POLYMATH
FILE |
L=6000 # |
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D=0.505 # |
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rho=53 # |
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g=32.2 # |
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eps=.00015 # |
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Re=rho*D*u/(13.2*0.000672)
# |
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fF(0)=.1 |
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u(0)=10 |
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| Variable/function values |
Variable |
Value |
f(x) |
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fF |
0.1 |
9.5371E-02 |
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u |
10 |
-2.6560E+03 |
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L |
6000 |
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D |
0.505 |
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rho |
53 |
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g |
32.2 |
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eps |
0.00015 |
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Re |
30173.39 |
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| Solution |
Variable |
Value |
f(x) |
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fF |
0.006874616348157 |
-1.7347E-18 |
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u |
5.6728221306731 |
0 |
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L |
6000 |
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D |
0.505 |
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rho |
53 |
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g |
32.2 |
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eps |
0.00015 |
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Re |
17116.824982804 |
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| Additional Information |
Only a constrained method
which keeps u and fF positive throughout the |
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iterations converges from
the 4th initial guess. |