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Example Problem: Pipe Flow and Energy Loss | AME 30331, Study notes of Fluid Mechanics

Material Type: Notes; Professor: Sucosky; Class: Fluid Mechanics; Subject: Aerospace and Mechanical Engr.; University: Notre Dame; Term: Fall 2009;

Typology: Study notes

2009/2010

Uploaded on 02/24/2010

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AME 30331 Fall 09
EXAMPLE PROBLEM: PIPE FLOW AND ENERGY LOSS
A pump delivers water (
3
9.80 10

N/m3,
6
1.12 10

m2/s) at a gage pressure
1550P
kPa
to a pair of inclined pipes made of commercial steel (roughness
0.045
mm). The pipes make
an angle
30
with respect to the horizontal. The characteristics of the pipes are:
12
3LL
m;
12.5D
cm;
25.0D
cm.
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
m/s for the speed in the smaller-diameter pipe and perform two additional
iterations. Does the process appear to be converging?
pf3
pf4
pf5

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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

PPP (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.