Some Engineering Applications
D.J. Jeffrey
January 11, 2006
These notes supplement those in your textbook.
1
Flow of water in a pipe network
Consider the network of pipes shown in figure 1.
?
21
?
12
6
6
10
8


15
16
13
25
i
1
i
2
i
3
i
4
A
B
C
D
Figure 1: The layout of a simple fourpipe flow problem.
The arrows show the direction of flow. The numbers show us rates of flow (litres/minute) and we have to
find the unknown rates
i
1
, i
2
, i
3
, i
4
. We begin by giving directions to each of the unknown flows. This
requires a
sign convention
. We guess the direction in which the flow is going, and if we are right, then we
get a positive answer. If we are wrong, the only effect is that we get a negative answer. For example, in
figure 2 we see two ways of representing water flowing to the left.

i
1
= +3
i
2
=

3
Figure 2: Sign convention. Both
i
1
and
i
2
represent water flowing to the left.
1
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Draw arrows going left on AC and BD; down on AB and CD.
Now we apply conservation of mass at junction A. Draw a circle around point A. What goes in must come
out. Therefore we have
At A:
+ 21

15

i
1
+
i
3
= 0
.
Repeat this at junctions B,C,D.
At B:
+
i
1
+ 13

12 +
i
4
= 0
,
At C:

8

i
3

i
2
+ 16 = 0
,
At D:
+
i
2

i
4
+ 10

25 = 0
.
That gives us 4 equations in 4 unknowns. You may have heard the rule “For
n
unknowns, you need
n
equations.” So it seems we are finished. However, let us try to solve these equations.
We start with the equation for junction B. It becomes
i
1
=

1

i
4
.
(1)
Now we skip to junction D and get
i
2
= 15 +
i
4
.
(2)
Now junction C says that
i
3
= 8

i
2
, and substituting for
i
2
from equation (2) we obtain
i
3
=

7

i
4
.
(3)
We still need an equation for
i
4
. All we have left is the equation at A. This says
i
3
=

6 +
i
1
. Well, we
have an expression for
i
1
, so let us substitute.
i
3
=

6 +
i
1
=

6

1

i
4
=

7

i
4
.
This is (3) again. It seems we are going around in circles. Maybe we have the wrong 4 equations. Can we
get more?
In addition to conservation at each junction, we can apply conservation over a pipe. Apply it now to the
pipe AB. Draw a circle enclosing both A and B.
For AB
+21

15 + 13

12 +
i
4
+
i
3
= 0
.
This is just (3) yet again.
In fact, 4 equations are not enough for these 4 unknowns, and neither are 5 equations. These equations
have multiple solutions. The way we can prove this will be taught later. These observations tell you
engineering information. They tell you that this problem of pipe flow is underdetermined. If you study
fluid mechanics later in your engineering courses, you will realize that we have left out the pressure. To see
why this is so, look back at figure 1. Without the effect of pressure, there is nothing to stop water making
a circuit around ABCD. The speed of this closed loop current is an arbitrary parameter in the solution.
This example teaches several important lessons for the study of linear algebra. First, the old rule “For
n
unknowns you need
n
equations” is not always correct. Second, we need a systematic method for deciding
when equations have solutions, and we cannot just rely on special methods that may only work for one
problem. Answering these questions is part of linear algebra.
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