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BASIC PRINCIPLES
OF PIPE FLOW
Dr/ Ahmed safwat
Amir Ashraf sayed
Branching Pipes
A simple branching-pipe system
Adding a parallel pipe to the original pipe
A δZ
B
Example
A pipe joins two reservoirs whose head difference is 10m. The pipe is 0.2 m
diameter, 1000m in length and has a f value of 0.008. a) What is the flow in the
pipeline? b) It is required to increase the flow to the downstream reservoir by
30%. This is to be done adding a second pipe of the same diameter that
connects at some point along the old pipe and runs down to the lower reservoir.
Assuming the diameter and the friction factor are the same as the old pipe, how
long should the new pipe be?
Solution
A
δZ
B
Case 1:
Case 2:
Solution
2
Solution
Problem Description:
Given:
Reservoir elevations, sizes of pipes ,friction factor,
minor loss coefficients and fluid properties
Required: Flow through each pipe and head losses
Expect the flow direction in the flowing figure?
Flow Between Reservoirs
Step analysis:
 Note the sign convention: QJA is the flow from J to
A; it will be negative if the flow actually goes from
A to J.
 The direction of flow in any pipe is always from
high head to low head.
 The problem and its solution method can be
generalized to any number of reservoirs.
Solution Procedure:
1. Establish the head loss vs discharge equations
for each pipe.
2. Guess an initial head at the junction, HJ .
3. Calculate flow rates in all pipes (from the
head differences)
4. Calculate net flow out of J.
5. If necessary, adjust HJ to reduce any flow
imbalance and repeat from (3)
If the direction of flow in a pipe, say JB, is
not obvious then a good initial guess is to
set HJ = HB so that there is initially no flow
in this pipe. The first flow-rate calculation
will then establish whether H J should be
lowered or raised and hence the direction
of flow in this pipe.
Important Note
Example 1
Reservoirs A, B and C have constant water levels of 150, 120
and 90 m respectively above datum and are connected by
pipes to a single junction J at unknown elevation. The
length (L), diameter (D), friction factor (f ) and minor-loss
coefficient (K) of each pipe are given below. Calculate:
(a) Calculate the flow in each pipe.
(b) Calculate the pressure at the junction J.
Solution
Prepare head loss vs discharge relations for each pipe
The value of HJ is varied until the net flow out of J is 0·
•If there is net flow into the junction then HJ needs to be
raised.
•If there is net flow out of the junction then HJ needs to be
lowered.
After the first two guesses at HJ, subsequent iterations are
guided by interpolation.
The quantity of flow in each pipe is given in the bottom
row of the table, with the direction implied by the sign.
Examples 2
Solution
Solution
pipe lines lec additional.pptx

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pipe lines lec additional.pptx

  • 1. BASIC PRINCIPLES OF PIPE FLOW Dr/ Ahmed safwat Amir Ashraf sayed
  • 2. Branching Pipes A simple branching-pipe system Adding a parallel pipe to the original pipe A δZ B
  • 3. Example A pipe joins two reservoirs whose head difference is 10m. The pipe is 0.2 m diameter, 1000m in length and has a f value of 0.008. a) What is the flow in the pipeline? b) It is required to increase the flow to the downstream reservoir by 30%. This is to be done adding a second pipe of the same diameter that connects at some point along the old pipe and runs down to the lower reservoir. Assuming the diameter and the friction factor are the same as the old pipe, how long should the new pipe be?
  • 7. Problem Description: Given: Reservoir elevations, sizes of pipes ,friction factor, minor loss coefficients and fluid properties Required: Flow through each pipe and head losses Expect the flow direction in the flowing figure? Flow Between Reservoirs
  • 9.  Note the sign convention: QJA is the flow from J to A; it will be negative if the flow actually goes from A to J.  The direction of flow in any pipe is always from high head to low head.  The problem and its solution method can be generalized to any number of reservoirs.
  • 10. Solution Procedure: 1. Establish the head loss vs discharge equations for each pipe. 2. Guess an initial head at the junction, HJ . 3. Calculate flow rates in all pipes (from the head differences) 4. Calculate net flow out of J. 5. If necessary, adjust HJ to reduce any flow imbalance and repeat from (3)
  • 11. If the direction of flow in a pipe, say JB, is not obvious then a good initial guess is to set HJ = HB so that there is initially no flow in this pipe. The first flow-rate calculation will then establish whether H J should be lowered or raised and hence the direction of flow in this pipe. Important Note
  • 12. Example 1 Reservoirs A, B and C have constant water levels of 150, 120 and 90 m respectively above datum and are connected by pipes to a single junction J at unknown elevation. The length (L), diameter (D), friction factor (f ) and minor-loss coefficient (K) of each pipe are given below. Calculate: (a) Calculate the flow in each pipe. (b) Calculate the pressure at the junction J.
  • 13. Solution Prepare head loss vs discharge relations for each pipe
  • 14. The value of HJ is varied until the net flow out of J is 0· •If there is net flow into the junction then HJ needs to be raised. •If there is net flow out of the junction then HJ needs to be lowered. After the first two guesses at HJ, subsequent iterations are guided by interpolation.
  • 15. The quantity of flow in each pipe is given in the bottom row of the table, with the direction implied by the sign.