Fluid Mech. Chapter14

Fluid Mechanics Chapter 14

Steady Incompressible Flow in Pipe systems

FOSTEMINTI International University

Flow in Pipeline SystemsFlow in Pipeline Systems

Objectives:o To analyze the energy losses occurred in pipelineso To understand total energy and hydraulic grade lineso To develop understanding of the water flow through a single pipe, pipes

in series, in parallel, and in branching

Losses of energy in pipelines

Pipe flow systemsPipe flow systems

o Single pipe flowo Pipes in serieso Pipes in parallelo Pipes in branching

Study the given information and data for the pipeline system including losses

Sketch the pipelines system Apply Bernoullis equation between two locations of interest Apply Continuity equation if necessary Solve for the unknowns

Single pipe flow Single pipe flow [[free discharge into atmosphere at B]free discharge into atmosphere at B]

To determine the mean velocity v and discharge Q: Applying the Bernoullis equation between A and B:

Total energy/wt at A = Total energy/wt at B + Losses

pA and pB = patm = 0 vA = 0 (large reservoir) vB = v

v = ? Q = Av = ?

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Single pipe flowSingle pipe flow[[flow from reservoir A to reservoir B]flow from reservoir A to reservoir B]

To determine the mean velocity v and discharge Q: Applying the Bernoullis equation between A and B:

Total energy/wt at A = Total energy/wt at B + Losses

pA and pB = patm = 0 vA = vB = 0 (large reservoirs)

v = ? Q = Av = ?

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Example 14.1: Single pipe flowExample 14.1: Single pipe flow

Fig. 14.2: Flow through a Siphon

Example 14.1: Single pipe flow

(a) To determine the mean velocity v: Applying the Bernoullis equation between A and C:

Total energy/wt at A = Total energy/wt at C + LossesLosses include entry loss + friction loss in pipe AC

pA and pc = Patm vA = 0 (large reservoir) vc = v

By substituting the values of zA–zc = 4 m, f = 0.08, d = 100 mm = 0.1 m, l1+l2 = l = 15 m

v = 1.26 m/s

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Example 14.1: Single pipe flow

(b) To find the gauge pressure pB :

Applying the Bernoullis equation between A and B:

pA = 0, vA = 0, (zA- zB) = -1.5 m, v = 1.26 m/s, f = 0.08, l1 = 5 m, d=0.1 m

pB/g = -1.5 – 1.416 = - 2.916 m

pB = - 28.60x103Nm-2

= 28.60 kN/m2 below atmospheric pressure = 72.70 kN/m2 (absolute)

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Example 14.2: Pipes in seriesExample 14.2: Pipes in series

Fig. 14.3: Pipes in series

Example 14.2: Pipes in series

(a) Losses of head in the pipeline system are:

1) Loss of head at entry, h1= 0.5 v12/2g

2) Loss of head in friction in AC,

3) Loss of head at sudden enlargement, h2 = (v1 – v2)2/2g

4) Loss of head in friction in CB,

5) Loss of head at exit, h3 = v22/2g.

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Example 14.2: Pipes in series By applying Bernoulli’s equation at A and B of the surface of two reservoirs

Since pA = pB = patm = 0 and, if the reservoirs are large, VA and VB will be negligible,

ZA – ZB = Losses = entry loss + friction loss in pipe AC + sudden enlargement + friction loss in pipe BC + exit loss

By applying Continuity equation,Q = A1v1 = A2v2 = (/4)d1

2v1 = (/4)d22v2

Substituting d1 = 0.2 m and d2 = 0.25 m,

v1 = 1.5625 v2 (or) v2 = 0.64 v1

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Example 14.2: Pipes in series

Putting ZA – ZB = h = 9 m, f = 0.01, l1 = 15 m, l2 = 45 m,

v1 = 5.03 m/s

Q = (/4)d12v1 = (/4) x 0.22 x 5.03

Q= 0.158 m3/s

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Hydraulic Gradient (Hydraulic Grade Line)

Total Energy = Pressure Energy + Kinetic Energy + Potential Energy Total Energy Line = Pressure Head + Velocity Head + Elevation Head

TEL = p/g + v2/2g + z

Hydraulic Grade Line = Total Energy Line – Velocity HeadHGL = TEL - v2/2g

Example 14.3: Pipes in parallelExample 14.3: Pipes in parallel

Fig. 14.4: Pipes in parallel

Example 14.3: Pipes in parallel (a) For flow by way of pipe 1,

Since PA = PB = Patm = 0, and, if the reservoirs are large, VA and VB will be negligible,

Putting ZA – ZB = h = 10 m, f = 0.008, l = 100 m, d1 = 50 mm = 0.05 m,

v12 = 2g x 10/(1.5+64)

v1 = 1.731 m/s

Volume rate of flow through pipe 1, Q1 = (/4)d12v1 = (/4) x 0.052 x 1.731

Q1 = 0.0034 m3/s

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Example 14.3: Pipes in parallel For flow by way of pipe 2,

Since PA = PB and both VA and VB can be assumed negligible,

Putting ZA – ZB = h = 10 m, f = 0.008, l = 100 m, d2 = 100 mm = 0.1 m,

v22 = 2g x 10/(1.5+32)

v2 = 2.42 m/s

Volume rate of flow through pipe 2, Q2 = (/4)d22v2 = (/4) x 0.12 x 2.42

Q2 = 0.0190 m3/s

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Example 14.3: Pipes in parallel

(b) Replacing the two pipes by the equivalent single pipe which will convey the same total flow,

o Volume rate of flow through single pipe,Q = Q1 + Q2 = 0.0034 + 0.0190 = 0.0224 m3/s v

If v is the velocity in the single pipe, Q = (/4)D2v Dia: D length = 100 m

Applying the steady flow energy equation between A and B,

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Example 14.3: Pipes in parallel

• Putting ZA – ZB = h = 10 m, f = 0.008, l = 100 m, V = 0.02852/D2,

10 = (1.5D + 3.2)(0.02852)2/2gD5

241 212D5 – 1.5D – 3.2 = 0 (or) f(D) = 0

This equation can be solved graphically or by successive approximations.

An approximate answer can be obtained by omitting the second term; then,241 212D5 = 3.2 and D = 0.1058 m

• If D = 0.1058 m, then f(D) = 3.198 – 0.159 – 3.2 = -0.161• If D = 0.107 m, then f(D) = 3.383 – 0.161 – 3.2 = +0.022• This result is sufficiently accurate for practical purposes. Diameter of equivalent single pipe = 0.107 m = 107 mm.

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Example 14.4: Pipes in branchingExample 14.4: Pipes in branching

Fig. 14.5: The three reservoir problem

Example 14.4: Pipes in branching

For flow from A to B,

Since PA = PB = atmospheric pressure and, if the reservoirs are large, VA and VB will be negligible,

Putting ZA – ZB = h = 16 m, f = 0.01, l1 = 120 m, d1 = 0.12 m, l2 = 60 m, d2 = 0.075 m,

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For flow from A to C,

Putting ZA – ZC = 24 m, f = 0.01, l1 = 120 m, d1 = 0.12 m, l3 = 40 m, d3 = 0.06 m,

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For continuity of flow at D,Flow through AD = Flow through DB + Flow through DC, Q1 = Q2 + Q3

A1v1 = A2v2 + A3v3

(/4)d12v1 = (/4)d2

2v2 + (/4)d32v3,

v1 = (d2/d1)2 v2 + (d3/d1)2 v3

Substituting numerical values,v1 = (0.075/0.12)2v2 + (0.06/0.12)2v3

v1 – 0.3906v2 – 0.2500v3 = 0 (3)

Values of v1, v2, and v3 are found by solution of the simultaneous equations (1), (2) and (3). From (1), v2 = (9.81 – 1.25v1

2) (4)

From (2) v3 = (17.657 – 1.5v12) (5)


Substituting in equation (3), v1 – 0.3906 (9.81 – 1.25v1

2) – 0.25 (17.657 – 1.5v12) = 0 (6)

Equation (6) can be solved graphically or by successive approximations . .

If the square roots are to be real, the value of V1 cannot exceed the lowest value that make one of the terms under the square root signs equal to zero. This will be given by (9.81 - 1.25v1

2) = 0; i.e., v12= 9.81/1.25 = 7.848, so that

v1 must be less than (7.848) = 2.80 m/s.

If v1 = 2.8 m/s, f(v1) = 2.8 – 0.0391 – 0.6071 = + 2.1538;

v1 = 1.9 m/s, f(v1) = 1.9 – 0.8990 – 0.8747 = + 0.1263;

v1 = 1.8 m/s, f(v1) = 1.8 – 0.9374 – 0.8943 = - 0.0317;

v1 = 1.82 m/s, f(v1) = 1.82 – 0.9300 – 0.8905 = -0 .0005


Taking v1 = 1.82 m/s as a sufficiently accurate result,

Volume rate of flow in AD, Q1 = (/4)d1

2v1 = (/4) x 0.122 x 1.82 = 0.0206 m3/s

From eqn. (4), v2 = (9.81 – 1.25x1.822) = 2.381 m/s,

Volume rate of flow in DB, Q2 = (/4)d2

2v2 = (/4) x 0.0752 x 2.381 = 0.0105 m3/s

From eqn. (5), v3 = (17.657 – 1.5x1.822) = 3.562 m/s,

Volume rate of flow in DC, Q3 = (/4)d3

2v3 = (/4) x 0.062 x 3.562 = 0.0101 m3/s

Checking for continuity at D,Q2 + Q3 = 0.0105 + 0.0101 = 0.0206 = Q1 O.K.

The End

Fluid Mech. Chapter14

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