Chapter 7

Steady Flow in Pipes

7.1 Single pipe problems

7.2 Multiple-Pipe Systems

7.1 Hydraulic Computation of Simple Pipes

Free Discharge

the acting head :

It is the sum of the head from the pipe outlet centroid to the water surface in the upstream reservoir and the approach velocity head in the upstream reservoir.

the coefficient of discharge in the piping system (管系流量系数)

Submerged Discharge

the coefficient of the discharge in the piping system

Question

Two identical long pipes with different setting height are shown in the figure, The relationship between the two discharges in the pipes is:

A. Q1<Q2； B. Q1>Q2；C. Q1=Q2； D. uncertain

Question

A L-length short pipe is connected to an external wall of the water tank. The free discharge with a Q1-discharge is shown in Fig. A. Submerged discharge with a Q2-discharge is shown in figure B, the relationship between Q1 and Q2 is:

A.Q1=Q2； B. Q1>Q2；C.Q1<Q2； D. uncertain.

Example

The reinforced concrete inverted siphon through the road base is shown in figure . The pipe length lAB=60m, lBC=80m, lCD=60m,α=20°. Frictional factor

λ= 0.0167

Determine: (1) When the water surface elevation difference 27.4m-19.4m=8m，the pipe diameter d=2m, check the discharge Q;

(2) If the discharge Q'=25.14m3/s , the pipe diameter and downstream water surface elevation are constant, what is the upstream water surface elevation?

Solution:

(1) the energy equation

Minor loss coefficient: K in=0.5, Kout=1.0, Kbend=0.046

the velocity

The discharge

(2)

Because the length, diameter, material of the pipe elements and the piping system don't change, the loss coefficient is not changed, so

Thus H'>H, the upstream water surface elevation increases to 30.06m.

Consider the example of a reservoir feeding a pipe, as shown in figure .

The pipe diameter is 100mm and has length 15m and feeds directlyinto the atmosphere at point C 4m below the surface of the reservoir (i.e. za – zc = 4.0m). The highest point on the pipe is a B which is 1.5m above the surface of the reservoir (i.e. zb – za = 1.5m) and 5 m along the pipe measured from the reservoir. Assume the entrance and exit to the pipe to be sharp and the value of friction factor λto be 0.32. Calculate a) velocity of water leaving the pipe at point C, b) pressure in the pipe at point B.

Example

a) for entry loss kL = 0.5 , Pressure at A and C are both atmospheric, uAis very small so can be set to zero, giving

⎟⎠⎞

⎜⎝⎛ ++=−

+++=

dL

guzz

gu

gu

dlz

guz

cA

CA

λ

λ

5.012

25.0

222

222

Substitute in the numbers from the question

smu

u

/26.11.0

1532.05.181.92

42

=

⎟⎠⎞

⎜⎝⎛ ×

+×

=

b)

23

2

12

221

2

1058.28

)1.0

0.52.35.1(81.92

26.181.91000

5.1

5.012

25.0

22

mNp

pdl

gu

gpzz

gu

gdulz

gu

gpz

B

B

BBA

BB

A

×−=

×+

×+

×=−

⎟⎠⎞

⎜⎝⎛ +++=−

++++=

λρ

λρ

That is 28.58 kN/m2 below atmospheric.

7.2 Multiple-Pipe Systems

Basic concepts of pipe system analysis apply also to multiple pipe systems. However, the solution procedure is more involved and can be iterative.

Consider the following:

1. pipes in series

2. pipes in parallel

1 Pipes in Series

•Volume flow rate is constant•Head loss is the summation of parts

the piping that is connected by several pipe elements of different diameters in order.

Basic discharge formula for Pipes in Series

（1）Energy equation

Where n is the numbers of pipe elements ; m is the numbers of minor resistance.

（2）Continuity equation

Without discharge output

With discharge output

The pipe elements of the pipes in series have:

A. the same head loss；B. the same total head loss；C. the same hydraulic slope；D. the same discharge through them

Consider the two reservoirs shown in figure, connected by a single pipe that changes diameter over its length. The surfaces of the two reservoirs have a difference in level of 9m. The pipe has a diameter of 200mm for the first 15m (from A to C) then a diameter of 250mm for the remaining 45m (from C to B).

Example

For the entrance use kL = 0.5 and the exit kL = 1.0. The join at C is sudden. For both pipes use λ = 0.04.

Total head loss for the system H = height difference of reservoirs

and solve for Q, to give Q = 0.158 m3/s

2 pipes in parallel

Parallel Piping is the piping that the fluid can flow from one location to another location by two or more pipe elements.

Volume flow rate is the sum of the components

Pressure loss across all branches is the same

The analysis can be carried out by simply treating each pipe individually and summing flow rates at the end.

• For parallel pipes, perform CV analysis between points A and B

• Since Δp is the same for all branches, head loss in all branches is the same

Question

From the figure, we can see that the relationship among the head losses from A to B in the pipes in parallel 1, 2, 3 is:

A. hfAB＝hfl+hf2+hf3；

B. hfAB＝hfl+hf2；

C. hfAB＝hf2+hf3；

D. hfAB＝hfl=hf2=hf3.

ExampleTwo pipes connect two reservoirs (A and B) which have a height difference of 10m. Pipe 1 has diameter 50mm and length 100m. Pipe 2 has diameter 100mm and length 100m. Both have entry loss kL = 0.5 and exit loss kL=1.0 and Darcy f of 0.008.

Calculate: a) rate of flow for each pipe b) the diameter D of a pipe 100m long that could replace the two pipes

and provide the same flow.