Parallel Flow

8/8/2019 Parallel Flow

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PARALLEL PIPELINE SYSTEMS

Where flow occurs through along more than one pathway.

The principle of continuity still applies -

Q1 = Q2 = Qa + Qb + Qc

Again the GE equation can be applied between points 1 and 2

p1/γ + z1 + v12/2g + – hL = p2/γ + z2 + v22/2g



hL = ha = hb = hc

So essentially we have 2 equations to solve parallel pipe systems

- discharge equation

- head loss equation

So when we have a parallel system with two branches (2unknowns – Q) we can easily compute the values.

However, for system with more than 2 branches the

computations become complex!

Systems with 3 or more branches – called NETWORKS

NETWORKS are INDETERMINATE!

Why – because the number of unknowns are more than the 2equations available.

Solutions for NETWORKS – HARDY CROSS method.



System with 2 branches

Procedure for computations –

1. Set up the conservation of mass equation

Q1 = Q2 = Qa + Qb

2. Express Q in terms of area and velocity of pipe

Qa = Aa * Va; Qb = Ab * Vb

3 Determine the hL for each pipe in terms of the velocity (V)



Using this equation get Va in terms of Vb or vice-versa.

5. Now introduce this info for equation in step 1 and compute

one of the velocities.

6. Using equation from step 4, compute the other velocity.

7. Compute Q values.

8. Check your friction factors; iterate if necessary.



EXAMPLE PROBLEM 12.1

100 gal/min of water at 60F is flowing in a 2-inch sch 40 pipe at

section 1. The heat exchanger in branch “a” has a loss

coefficient K = 7.5 based on the velocity head in the pipe. All

three valves are wide open. Branch “b” is a 1 ¼ inch sch 40

pipe. Elbows are standard. The length of pipe between points 1

and 2 for branch b is 20 ft. Because of the size of the heatexchanger the length of the pipe in branch “a” is short and

friction loss can be neglected. For this arrangement, determine –

(a) volume flow rate Q in each branch; (b) pressure drop





Assignment # 8

• 12.1M



HARDY CROSS METHOD for 3 branches or more

(NETWORKS) –

Procedure set up to solve Networks.

You have to express the head loss h in terms of Q

h = kQn

We already express h in terms of v; and v = Q/A

We have estimate the volume Q in the pipes. Estimation is

based on the principle of conservation of volume and the likely

losses that may occur in the pipe.

You have to divide the network into closed circuits and assign

+ve and –ve values to head loss in the pipes.



Rule –

If flow in a given pipe of a circuit is clockwise, Q and h are

positive. If flow is counterclockwise, Q and h are negative.

Stepwise procedure for Hardy Cross Method –

1. Express energy loss in each pipe by h = kQ2



4. For each pipe calculate head loss using h = kQ2

5. For each circuit, algebraically sum the values of h using the

direction convention/rule.

Find ∑h.

6. For each pipe calculate 2kQ

7. Sum all values of 2kQ for each circuit, assuming all are

positive = ∑2kQ

8. For each circuit calculate –

∑∑

=ΔkQ

hQ

2

9. for each pipe calculate –

Q’ = Q – ΔQ

10. Repeat until ΔQ becomes very small.





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Parallel Flow