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- .MCD 20000 60
- .CMD PLOTFORMAT logs=0,0 subdivs=1,1 size=5,15 type=l
- .CMD FORMAT rd=d ct=10 im=i et=5 zt=15 pr=3 mass length time charge
- .CMD SET ORIGIN 0
- .CMD SET TOL 0.001000
- .CMD MARGIN 0
- .CMD LINELENGTH 78
- .CMD SET PRNCOLWIDTH 8
- .CMD SET PRNPRECISION 4
- .TXT 1 0 1 41
- a1,40,78,39
- PIPE SIZING PROBLEM FOR TURBULENT FLOW
- .TXT 2 10 5 44
- a5,44,41,188
- This example comes from "Fundamentals of
- Fluid Mechanics" by Philip M. Gerhart and
- Richard J. Gross, Addison-Wesley, 1985.
- See p. 437. [Method based on F.M. White,
- "Fluid Mechanics."]
- .TXT 6 -10 1 15
- a1,15,78,14
- Define units:
- .EQN 2 10 1 10
- ft:1L
- .EQN 0 20 1 11
- sec:1T
- .EQN 0 20 1 10
- lb:1M
- .EQN 2 -40 3 10
- in:ft/12
- .EQN 0 20 1 15
- min:60*sec
- .TXT 0 20 2 20
- a2,20,31,34
- (use lb for mass,
- lbf for force)
- .EQN 2 -20 4 18
- g:32.174*ft/sec^2
- .EQN 1 20 1 13
- lbf:lb*g
- .EQN 1 -40 2 16
- gal:231*in^3
- .EQN 1 40 4 12
- psi:lbf/in^2
- .TXT 5 -50 2 64
- a2,64,78,82
- Problem: find required pipe size for the following parameters
- for flow of water:
- .EQN 3 10 3 14
- Q:175*gal/min
- .TXT 1 30 1 23
- a1,23,51,22
- ... maximum flow rate
- .EQN 3 -30 1 15
- Dp:1.2*psi
- .TXT 0 30 1 27
- a1,27,38,26
- ... maximum pressure drop
- .EQN 2 -30 1 13
- L:100*ft
- .TXT 0 30 1 17
- a1,17,38,16
- ... pipe length
- .EQN 2 -30 4 15
- ▌:62.4*lb/ft^3
- .TXT 1 30 1 22
- a1,22,38,21
- ... density of water
- .EQN 4 -30 1 17
- ε:0.00015*ft
- .TXT 0 30 1 20
- a1,20,38,19
- ... pipe roughness
- .EQN 2 -30 4 20
- v:1.22*10^-5*ft^2/sec
- .TXT 2 30 1 25
- a1,25,38,24
- ... kinematic viscosity
- .TXT 4 -40 1 1
- x1,1,0,0
- .TXT 2 0 1 51
- a1,51,78,50
- Now solve problem by computing with these values:
- .EQN 2 10 3 12
- gh.L:Dp/▌
- .EQN 0 13 5 20
- gh.L=?ft^2/sec^2
- .TXT 2 27 1 17
- a1,17,38,16
- ... energy loss
- .EQN 3 -40 6 19
- f':(128*gh.L*Q^3)/(π^3*L*v^5)
- .EQN 2 21 2 19
- f'=?
- .TXT 1 19 2 25
- a2,25,38,34
- ... dimensionless flow
- parameter
- .EQN 4 -40 3 10
- e:(ε*v)/Q
- .EQN 0 21 2 18
- e=?
- .TXT 1 19 2 25
- a2,25,28,34
- ... dimensionless flow
- parameter
- .TXT 3 -50 1 63
- a1,63,78,62
- Now try three different approximations for Reynolds number R:
- .TXT 2 0 1 17
- a1,17,78,16
- First estimate:
- .EQN 1 11 2 21
- R:1.43*(f')^0.208
- .EQN 0 22 2 17
- R=?
- .EQN 3 -22 3 12
- D:(4*Q)/(π*v*R)
- .EQN 1 22 1 14
- D=?in
- .TXT 3 -33 1 18
- a1,18,78,17
- Second estimate:
- .EQN 1 11 5 41
- R:(-2*\f'*log((π*e*R)/14.8+2.51/\f'*R^1.5))^0.4
- .EQN 6 22 2 17
- R=?
- .EQN 3 -22 3 12
- D:(4*Q)/(π*v*R)
- .EQN 1 22 1 14
- D=?in
- .TXT 3 -33 1 51
- a1,51,78,50
- Colebrook's implicit formula, solved iteratively:
- .EQN 2 11 1 7
- Given
- .EQN 0 9 5 40
- R≈(-2*\f'*log((π*e*R)/14.8+2.51/\f'*R^1.5))^0.4
- .EQN 6 -9 1 14
- R:Find(R)
- .EQN 2 21 2 17
- R=?
- .EQN 2 -21 3 12
- D:(4*Q)/(π*v*R)
- .EQN 1 21 1 14
- D=?in
-