Piping System 1

CIVL 2310 Fluid Mechanics Lecture 21: Piping Systems (1) 11.1 to 11.3 of “Mechanics of Fluids,” M. C. Potter and D. C. Wiggert, Brooks/Cole, 2001. Topics: 1. Losses 2. Simple pipe systems 1. Losses It is convenient to express friction losses in pipes as hL = rQ 2 (1) where according to lecture 19 r has the form (Darcy Weisbach formula) r=

fL (2) 2gDA2

with A being the pipe cross sectional area. Using the expression of Swamee and Jain for f f =

1.325

[ln(e / 3.7 D ) + 5.74 / R )] 0.9

2

(3) for 0.01>e/D>10-8 and 108>R>5000

gives r=

L 1.07 (4) 5 gD ln( e / 3.7 D ) + 5.74 / R 0.9 ) 2

[

]

where we have replaced A=πD2/4. Note that for fully rough regime (large R), f (and thus r) becomes only a function of e/D. Using the equivalent length LE, the minor losses can also be incorporated into r. r=

f ( L + LE ) 2gDA2

2. Simple pipe systems Pipe systems are comprised of several pipe lines. Depending on the line arrangement three cases will be analyzed: series, parallel and branch piping. a) Series piping The beginning of each line coincides with the end of the previous line. The simple pipe problems solved in lectures 19 and 20 belong to this case.

Mass conservation states that the discharge is constant for all lines, i.e., Q1 = Q2 = ..... = Q N = Q

(5)

N= number of pipe lines

The energy equation applied between the two ends A and B of the system reads  pA  p  + z A  −  B + z B  = r1Q12 + r2Q22 + ..... + rN Q N2 (6)   γ   γ  which can be written, using (5), as N  pA  p  + z A  −  B + z B  = Q 2 ∑ ri  i =1  γ   γ 

(7)

Example

b) Parallel piping All pipe line beginnings and ends coincide.

Mass conservation at A and B states N

Q = Q1 + Q2 + ..... + Q N = ∑ Qi

(8)

i =1

One energy equation can be written for each line, i.e.,

 pA  p  + z A  −  B + z B  = ri Qi2 (9)   γ   γ  Combining (8) and (9) gives p  p  Q2 W =  A + z A  −  B + z B  =  γ   γ   N 1 ∑  i =1 r i 

   

2

(10)

Example

c) Branch piping Systems are connected but without forming a closed loop

Flow directions need to be assumed in this case. The unknowns are the p  piezometric head at B HB=  B + z B  and the discharges Q1, Q2 and Q3. Mass  γ  conservation at the junction B reads

Q1 − Q2 − Q3 = 0 (11)

for the elements connected at the junction, the energy equation gives:

 pA  p  + z A  −  B + z B  = r1Q12   γ   γ   pB  p  + z B  −  C + zC  = r2Q22   γ   γ 

(12, 13, 14)

 pB  p  + z B  −  D + z D  = r3Q32   γ   γ  we have four equations with four unknowns. The solution can be obtained by successive approximations as follows: a) assume Q1 and get HB or assume HB and get Q1 using (12) b) get Q2 and Q3 from (13) and (14), respectively c) use (11) to check for continuity. Instead of 0, (11) will give ∆Q. d) adjust Q1 or HB accordingly and repeat a) to c) until ∆Q ≈ 0 within a specified error. If a pump exists, (12) is modified

 pA  p  + z A  −  B + z B  + H P = r1Q12 (15)   γ   γ  bringing to the problem an additional unknown Hp. The additional equation needed is given by the head-discharge curve for the pump (lecture 20). Example