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AME 30331 – Fall 09
EXAMPLE PROBLEM: PIPE FLOW AND ENERGY LOSS
A pump delivers water (
3
9.80 10 N/m
3 ,
6
m
2 /s) at a gage pressure P 1 (^) 550 kPa
to a pair of inclined pipes made of commercial steel (roughness 0.045mm). The pipes make
an angle 30 with respect to the horizontal. The characteristics of the pipes are:
L 1 (^) L 2 3 m; D 1 (^) 2.5cm; D 2 (^) 5.0cm.
Determine the velocity of water being ejected to the atmosphere at the upper end, ignoring
minor losses.
Hint: since the velocity is to be determined, this problem requires you to use iterations. Use a
first guess of V 1 (^) 9 m/s for the speed in the smaller-diameter pipe and perform two additional
iterations. Does the process appear to be converging?
Step 1: Deriving an expression for the velocity in the pipe
Start with the energy equation between the entrance of the first pipe and the exit of the second
pipe:
2 2 1 1 2 2 1 2 2 2
L
P V P V
z z h
g g g g
To write this equation, we have assumed turbulent flow in each pipe (i.e., 1 2 1 ), and
nearly uniform velocity profile in each pipe (i.e., V 1 (^) V 1 and V 2 (^) V 2 ).
The head loss term must be decomposed into a head loss in pipe 1 and a head loss in pipe 2:
1 2
2 2 1 1 2 2 1 2 1 2 22
L L L
L V L V
h h h f f D g D g
Therefore, the energy equation can be rewritten:
2 2 1 2 2 2 1 1 2 2 1 2 1 2 1 2 1 2
P P L V L V
V V z z f f
g g D g D g
Since P 2 Patm , then 1 2 g
pump
P P P (i.e., gage pressure delivered by the pump)
Also, given the geometry of the pipe, z 1^ ^ z 2^ ^ L 1^ L 2 sin^
Therefore, the previous equation can be simplified as:
^ ^
2 2 (^2 2 1 1 2 ) 1 2 1 2 1 2 1 2
sin 2 2 2
P g (^) pump L V L V V V L L f f g g D g D g
Instead of expressing this relation in terms of two different velocities V 1 and V 2 , we can use
conservation of mass to keep only one unknown velocity.
Continuity: Q 1 (^) Q 2
AV 1 1 (^) A V 2 2
2 1 2 1 2
D
V V
D
Substituting back into the energy equation:
Therefore:
2 1 2 2
f
f
^
Using the Colebrook formula, we obtain a refined value for the friction factor:
2.0log 3.7 (^) Re
D
f (^) f
^
Therefore:
2 1 2 2
f
f
^
Now, we can substitute these frictions factors into the expression that was derived for V 1
( Equation 1 ).
1 2
1 1 2
V
f f
^
m/s
Using continuity:
2 1 2 1 2
D
V V
D
m/s
This is the end of the “guess” step.
Step 3: First iteration
We start with the new values of V 1 and V 2 , and we calculate the corresponding Reynolds
numbers:
1 1 1
2 2 2
Re 510714
Re 255357
V D
V D
Using the approximate Colebrook formula:
2 1 2 2
f
f
^
Using the Colebrook formula:
2 1 2 2
f
f
^
Once again, substituting these values into the expression for V 1 ( equation 1 ) yields:
1 2
1 1 2
V
f f
^
m/s
Using continuity:
2 1 2 1 2
D
V V
D
m/s
This is the end of the 1
st iteration.
Step 4: Second iteration
Based on the new velocity values:
1 1 1
2 2 2
Re 519643
Re 259821
V D
V D
Note that:
1
2
Re 519643 2300
Re 259821 2300
^ ^
^
. Therefore, the initial assumption that the flow is turbulent in
each pipe is satisfied.
Using those updated Reynolds numbers in the approximate Colebrook formula:
2 1 2 2
f
f
^
Using the Colebrook formula:
2 1 2 2
f
f
^
Substitute back into the expression for V 1 ( equation 1 ):
1 2
1 1 2
V
f f
^
m/s
Using continuity:
2 1 2 1 2
D
V V
D
m/s
This is the end of the 2
nd iteration.
