1
Civil Engineering Hydraulics Mechanics of Fluids
Simple Piping Problems
Pipe Geometries
In the text, problems are worked out for pipes of noncircular cross sections.
While the derivations are fine and work well, you can also use an equivalent hydraulic diameter to work on these problems.
Simple Piping Problems 2
2
Pipe Geometries
An annulus is developed by putting one pipe inside of another so that the center of the pipes are aligned. This is a very common setup in heat exchangers.
Simple Piping Problems 3
The blue area is the flow area we are concerned with.
Pipe Geometries
Outer pipe is a 4-nominal, schedule 40 pipe. Inner pipe is a 2-nominal, schedule 40 pipe. we can get the inner and outer diameters for the pipes from Table C.1.
Simple Piping Problems 4
3
Pipe Geometries
To calculate a hydraulic diameter, we need the area and the wetted perimeter.
Simple Piping Problems 5
Pipe Geometries
So the equivalent Dh is
Simple Piping Problems 6
4
Pipe Geometries
Calculate the average flow velocity using our normal method
Simple Piping Problems 7
Pipe Geometries
Now we can calculate the Re for the flow
Simple Piping Problems 8
5
Pipe Geometries
For a cast iron pipe the ε is 0.00085 from Table 5.2 so we can calculate the ε/Dh
Simple Piping Problems 9
Pipe Geometries
We can use the Moody Diagram to calculate the friction factor, f
Simple Piping Problems 10
6
Moody Diagram
Simple Piping Problems 11
f is approximately 0.032
Pipe Geometries
Now we can use this friction factor to calculate the head loss in this section of pipe.
Simple Piping Problems 12
In this case, all the pressure loss was due to friction.
7
Pipe Geometries
This is the most common way to work this type of problem. If the pipes made of different materials you could develop an average relative roughness
Simple Piping Problems 13
!
Daverage="Douter
!Douter
+"Dinner!Dinner
"Douter +"Dinner
Pipe Geometries
Example problem 5.5 gives a different type of problem where we know the pressure drop over a length of pipe and we want to find the volumetric flow rate.
Simple Piping Problems 14
8
Pipe Geometries
Benzene flows through a 12-nominal schedule 80 wrought iron pipe. The pressure drop measured at points 350 m apart is 34 kPa. Determine the flow rate through the pipe.
Simple Piping Problems 15
Pipe Geometries
Using something like Solver in EXCEL, this isn’t a difficult problem. However, if you need to do this problem by an iterative method, it can be solved.
Simple Piping Problems 16
9
Pipe Geometries
Since the pipe has a constant cross section and is horizontal, all the of the pressure loss is due to friction so we have the expression
Simple Piping Problems 17
p1 ! p2 =!v 2
2fLDh
Pipe Geometries
The problem we have is the f is a function of v through the Re and we don’t have a direct solution
Simple Piping Problems 18
p1 ! p2 =!v 2
2fLDh
10
Pipe Geometries
From what is given in the problem and from what we can look up in tables we have
Simple Piping Problems 19
Pipe Geometries
Rearranging the Bernoulli expression
Simple Piping Problems 20
!p = !v 2
2fLDh
2Dh!p!L
= v 2 f
v =2Dh!pf !L
11
Pipe Geometries
Now we can start by assuming a friction factor, using the roughness ratio, and getting a Re from there.
Simple Piping Problems 21
Moody Diagram
Simple Piping Problems 22
Assumed f is 0.015
Your assumed f must be such that the point on the relative roughness curve is within the diagram.
12
Pipe Geometries
From the assumed f, calculate the v and the Re
Simple Piping Problems 23
Pipe Geometries
Now, using this Re, you can go back to the Moody diagram and calculate a new estimate for f
Simple Piping Problems 24
13
Moody Diagram
Simple Piping Problems 25
New estimate for f is 0.0142
Pipe Geometries
And you can iterate the process based on the new estimate for f
Simple Piping Problems 26
14
Moody Diagram
Simple Piping Problems 27
New estimate for f is 0.0142
Using this Moody diagram, you really can’t get much closer an estimate than we already have. Since the new Re is so close to the Re we just used, this would not be a bad termination point.
Pipe Geometries
So out volumetric flow rate is
Simple Piping Problems 28
15
Pipe Geometries
There are expressions for approximating the friction factor that do not rely on the Moody diagram. These are all empirical equations but they do have the ability to be solved directly. The forms are given on page 246 and 247 of the 4th edition of the text and page 267 of the 3rd edition (only two are given in the 3rd edition).
Simple Piping Problems 29
Pipe Geometries
One of the ones not given in the 3rd edition is the Swamee-Jain equation
Simple Piping Problems 30
f = 0.250
log !3.7D
+ 5.74Re0.9
!"#
$%&
'
()
*
+,
2
16
Pipe Geometries
One of the ones not given in the 3rd edition is the Swamee-Jain equation
Simple Piping Problems 31
f = 0.250
log !3.7D
+ 5.74Re0.9
!"#
$%&
'
()
*
+,
2
Re = 5.74
e0.250f - !
3.7D
!
"
####
$
%
&&&&
10.9
Pipe Geometries
Using the equation for f rather than the Moody diagram for iteration
Simple Piping Problems 32
17
Pipe Geometries
Second Iteration
Simple Piping Problems 33
Pipe Geometries
Third Iteration
Simple Piping Problems 34
18
Pipe Geometries
A third type of problem is where we are given the flow and the head loss and asked to choose a pipe size. Example 5.7
Simple Piping Problems 35
Pipe Geometries
Simple Piping Problems 36
19
Pipe Geometries
Simple Piping Problems 37
Pipe Geometries
Simple Piping Problems 38
20
Pipe Geometries
Simple Piping Problems 39
p1 ! p2 =!v 2
2fLDh
Collect all the known information on the left side2"p!L
= v 2 fDh
Since we are dealing with a pipe with a circular cross section, Dh is the pipe diameter
v = QA
= Q!D 2
4
2"p!L
= Q!D 2
4
#
$
%%%
&
'
(((
2
fDh
) 2"p!L
= Q 2
!4
#$%
&'(
2
fDh
5 )2"p!L
!4
#$%
&'(
2
Q 2 = fDh
5
Pipe Geometries
Simple Piping Problems 40
This just sets up all the calculations as functions to make things a bit easier.
21
Pipe Geometries
Simple Piping Problems 41
This just sets up all the calculations as functions to make things a bit easier.
Pipe Geometries
Simple Piping Problems 42
This just sets up all the calculations as functions to make things a bit easier.
22
Pipe Geometries
Simple Piping Problems 43
This just sets up all the calculations as functions to make things a bit easier.
Pipe Geometries
Simple Piping Problems 44
This just sets up all the calculations as functions to make things a bit easier.
23
Pipe Geometries
So to carry the flow what this pressure drop or lower, we would need a pipe with an inside diameter of 0.515 ft as a minimum. This is a pipe with an ID of 6.176 in.
To get an internal diameter at least this large, you would specify a 8 nominal pipe
Simple Piping Problems 45
Homework
Simple Piping Problems 46
24
Homework
Simple Piping Problems 47