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
All elements of flow converging at 2 WILL have the same head
loss.
The flow will adjust automatically so that the head loss in each
branch pipe WILL BE THE SAME!
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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 (2
unknowns – 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 2
equations available.
Solutions for NETWORKS – HARDY CROSS method.
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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).
You will have to ASSUME a friction factor.
4. Equate the head losses for the pipes.
ha = hb
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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.
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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 heat
exchanger 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
between points 1 and 2.
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Assignment # 8
• 12.1M
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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.
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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
2. Assume values of Q for each pipe.
3. Divide the network into closed circuits.
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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 –
h
∑
ΔQ =
∑ 2kQ
9. for each pipe calculate –
Q’ = Q – ΔQ
10. Repeat until ΔQ becomes very small.
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