1 Civil Engineering Hydraulics Pipes in Parallel Calvin: You know, Hobbes, some days even your lucky rocket ship underpants don’t help. Parallel Pipes There are often cases where we find pipes laid in parallel that begin at the same junction and terminate at the same junction One of the most common cases is to replace a segment of pipe with two smaller pipes This may be a stopgap measure that just seems to remain in place Pipes in Parallel 2 Monday, October 22, 2012
Transcript
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Civil Engineering Hydraulics
Pipes in Parallel
Calvin: You know, Hobbes, some days even your lucky rocket ship underpants don’t help.
Parallel Pipes
¢ There are often cases where we find pipes laid in parallel that begin at the same junction and terminate at the same junction
¢ One of the most common cases is to replace a segment of pipe with two smaller pipes
¢ This may be a stopgap measure that just seems to remain in place
Pipes in Parallel 2 Monday, October 22, 2012
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Parallel Pipes
¢ If you have had circuits in physics or an electrical engineering class, you may recognize some of the same methods for solution that are used for solving parallel resistances
Pipes in Parallel 3 Monday, October 22, 2012
Parallel Pipes
¢ The fundamentals are also used in solving for water distribution systems through closed loop piping networks
Pipes in Parallel 4 Monday, October 22, 2012
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Parallel Pipes
¢ We will start with the simplest form of a parallel network containing two pipes
Pipes in Parallel 5 Monday, October 22, 2012
Parallel Pipes
¢ The volumetric flow rate approaching point A is known and due to conservation of mass that is the same as the flow exiting at point B
Pipes in Parallel 6 Monday, October 22, 2012
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Parallel Pipes
¢ We usually know the length and cross section area of both of the pipes from A to B and we want to know how much flow is going through each pipe.
Pipes in Parallel 7 Monday, October 22, 2012
Parallel Pipes
¢ Now the view we have of this system is shown from above so all the pipes shown are on the same elevation
Pipes in Parallel 8 Monday, October 22, 2012
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Parallel Pipes
¢ If we neglect minor losses, all the pressure drops along the two pipes from A to B are due to friction losses
Pipes in Parallel 9 Monday, October 22, 2012
Parallel Pipes
¢ The pressure at A is the same for both pipes and the pressure at B is also the same for both pipes
Pipes in Parallel 10 Monday, October 22, 2012
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Parallel Pipes
¢ This means that the pressure drop along both pipes must be the same
Pipes in Parallel 11
Δp3 = Δp2
Monday, October 22, 2012
Parallel Pipes
¢ Since both of these pressure drops are due to friction we can replace both of them by the pressure loss due to friction loss terms
Pipes in Parallel 12
Δp3 = Δp2f3L3D3
v32
2g= f2L2D2
v22
2g
Monday, October 22, 2012
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Parallel Pipes
¢ The diameter and length of each of the pipes is known but we have unknown friction factors and velocities on each side of the expression
Pipes in Parallel 13
Δp3 = Δp2f3L3D3
v32
2g= f2L2D2
v22
2gMonday, October 22, 2012
Parallel Pipes
¢ Since the friction factor is dependent on the Reynolds number and the Re is a function of the velocity, we have an expression that cannot be solved directly
Pipes in Parallel 14
Δp3 = Δp2f3L3D3
v32
2g= f2L2D2
v22
2gMonday, October 22, 2012
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Parallel Pipes
¢ Most of the solution methods assume a flow in one of the pipes (which sets the flow in the other) and then iterates the solution until the difference in pressure drops is within tolerance.
Pipes in Parallel 15
Δp3 = Δp2f3L3D3
v32
2g= f2L2D2
v22
2gMonday, October 22, 2012
Parallel Pipes ¢ One method that is used it to manipulate the expression so that we get an equivalent to an electrical circuit
Pipes in Parallel 16
f3L3
d3
v32
2g=
f2L2
d2
v22
2g
f3L3
d3
Q32
A32
2g=
f2L2
d2
Q22
A22
2g
f3L3
d3
Q32
πd32
4⎛
⎝⎜⎞
⎠⎟
2
2g=
f2L2
d2
Q22
πd22
4⎛
⎝⎜⎞
⎠⎟
2
2gMonday, October 22, 2012
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Parallel Pipes ¢ One method that is used it to manipulate the expression so that we get an equivalent to an electrical circuit
Pipes in Parallel 17
f3L3
d3
Q32
πd32
4⎛
⎝⎜⎞
⎠⎟
2
2g=
f2L2
d2
Q22
πd22
4⎛
⎝⎜⎞
⎠⎟
2
2g
f3L3
d3
Q32
2gπd3
2
4⎛
⎝⎜⎞
⎠⎟
2 =f2L2
d2
Q22
2gπd2
2
4⎛
⎝⎜⎞
⎠⎟
2
Q32 =
f2L2
2g π4
⎛⎝⎜
⎞⎠⎟
2
d25
⎛
⎝
⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟
f3L3
2g π4
⎛⎝⎜
⎞⎠⎟
2
d35
⎛
⎝
⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟
Q22
Monday, October 22, 2012
Parallel Pipes ¢ This is similar to the division of current between two parallel resistances.
Pipes in Parallel 18
Q32 =
f2L2
d25
⎛
⎝⎜⎞
⎠⎟
f3L3
d35
⎛
⎝⎜⎞
⎠⎟
Q22
Q3 =
f2L2
d25
⎛
⎝⎜⎞
⎠⎟
0.5
f3L3
d35
⎛
⎝⎜⎞
⎠⎟
0.5 Q2
Monday, October 22, 2012
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Parallel Pipes ¢ The difference is that f is a function of Q so this is just an estimate.
Pipes in Parallel 19
Q32 =
f2L2
d25
⎛
⎝⎜⎞
⎠⎟
f3L3
d35
⎛
⎝⎜⎞
⎠⎟
Q22
Q3 =
f2L2
d25
⎛
⎝⎜⎞
⎠⎟
0.5
f3L3
d35
⎛
⎝⎜⎞
⎠⎟
0.5 Q2
Monday, October 22, 2012
Parallel Pipes ¢ We can start by assuming a split between the two pipes, calculate the fractions and use that as the first iteration.
Pipes in Parallel 20
Q32 =
f2L2
d25
⎛
⎝⎜⎞
⎠⎟
f3L3
d35
⎛
⎝⎜⎞
⎠⎟
Q22
Q3 =
f2L2
d25
⎛
⎝⎜⎞
⎠⎟
0.5
f3L3
d35
⎛
⎝⎜⎞
⎠⎟
0.5 Q2
Monday, October 22, 2012
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Parallel Pipes
Pipes in Parallel 21
Q32 =
f2L2
d25
⎛
⎝⎜⎞
⎠⎟
f3L3
d35
⎛
⎝⎜⎞
⎠⎟
Q22
Q3 =
f2L2
d25
⎛
⎝⎜⎞
⎠⎟
0.5
f3L3
d35
⎛
⎝⎜⎞
⎠⎟
0.5 Q2
Example Problem 5.12 Equivalent Resistance ρ 1.94 slug1>-‐3 μ 1.90E-‐05 lbf1s1>-‐2 Q 3.00E-‐01 >3s-‐1
FracNon of Q in pipe 2 0.671330633 0.328669367 Q 0.2014 0.0986 >3s-‐1 v 0.367785226 0.283825015 >1s-‐1 Re 3.14E+04 2.42E+04
f 2.37E-‐02 2.52E-‐02
Δp 0.408859847 0.408859704 lbf1>-‐2
Δp2-‐Δp3 1.42917E-‐07 lbf1>-‐2
Monday, October 22, 2012
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Parallel Pipes ¢ This is where EXCEL and goal seek are
very useful.
Pipes in Parallel 31
Goal seek has found a flow balance that has a very small difference between the pressure loss across the two pipes.
Monday, October 22, 2012
Parallel Piping
¢ While that these methods work well with two pipes in parallel, when you have three or more pipes in parallel, since there is more than one flow partition to make, the methods we have considered will not work.
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Parallel Piping
¢ There are mathematical approximations and methods that can be used to generate an approximate solution but there is not method that I know of that can be broken down into a simplification like the two pipe problem.
Pipes in Parallel 33 Monday, October 22, 2012
Parallel Piping
¢ In Hydrology, you will be introduced to a method of looking at pipe networks that involved a mathematical method known as a relaxation method.
¢ The particular method that you will see is known as the Hardy-Cross method.
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Parallel Piping
¢ We will look as a method in EXCEL that can generate a solution using an EXCEL tool that you may not have seen before.
Pipes in Parallel 35 Monday, October 22, 2012
Parallel Piping
¢ If the problem required that we add yet another pipe to the system in parallel but this time we added a 4000 ft section of 6 nominal pipe in parallel to the other two sections.
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Parallel Piping
¢ We could just add another column to the table with the characteristics of the new pipe.
Pipes in Parallel 37 Monday, October 22, 2012
Parallel Piping
¢ To set the flows we now need two fractions. ¢ The first will partition the flows between the
original (pipe 2) and the remaining pipes. ¢ The second will partition the flow not going
into pipe 2 between pipes 3 and 4.
Pipes in Parallel 38
( )2 1
3 2 2
4 2 3
Q fraction QQ fraction Q QQ Q Q Q
== −= − −
Monday, October 22, 2012
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Flow Partitions in 3 Pipe Problem
Pipes in Parallel 39
FracNons 0.5 0.5 Q 0.1500 0.0750 0.0750 >3s-‐1
Monday, October 22, 2012
Parallel Piping Problem
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Parallel Piping Problem
¢ Now we cannot subtract one pressure drop from another but we can develop an expression to makes sure all the pressure drops are equal.
¢ We do this by looking at the difference between all the pairs of pipes and making sure that they are all about equal.
Pipes in Parallel 41 Monday, October 22, 2012
Parallel Piping Problem
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Parallel Piping Problem
¢ Rather than using Goal Seek, we will use a feature of EXCEL called Solver which allows for the modification of more than one variable to achieve a goal.
Pipes in Parallel 43 Monday, October 22, 2012
Pipes in Parallel 44
Solver is part of the Data menu. You may have to add it into your EXCEL.
Monday, October 22, 2012
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Parallel Piping Problem
Pipes in Parallel 45 Monday, October 22, 2012
Parallel Piping Problem
Pipes in Parallel 46 Monday, October 22, 2012
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Parallel Piping Problem
Pipes in Parallel 47 Monday, October 22, 2012
Parallel Piping Problem
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Homework 21-1
¢ Using the iteration method we developed, solve for the flow in each pipe is the system if one pipe is 3000 ft long and is Schedule 40 nominal 3 wrought iron and the second pipe is 2800 ft long and is Schedule 40 nominal 2 ½ wrought iron. The total flow in the system is 1800 gpm. The fluid is water.
Pipes in Parallel 49 Monday, October 22, 2012
Homework 21-2
¢ Using the EXCEL and Goal Seek to solve for the flow in each pipe is the system if one pipe is 1200 m long and is Schedule 80 nominal 4 wrought iron and the second pipe is 1500 m long and is Schedule 80 nominal 2 wrought iron. The total flow in the system is 6.0 m3/min. The fluid is water.
Pipes in Parallel 50 Monday, October 22, 2012
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Homework 21-3
¢ Use EXCEL and Solver to solve for the flow in each pipe in a system of three pipes. All the pipes are 2500 ft long and Schedule 40 steel. The first pipe is 1 Nominal, the second is 2 Nominal, and the third is 3 Nominal. Total flow in the system is 2500 gpm.
