fluid_mechanics

8.5 Pipe Flow Examples 439
A
B
D 1 , 1
D 2 , 2
D 3 , 3
(1)
(2)
C
(3)
F I G U R E 8.37
A three-reservoir system.
For some pipe systems,
the direction
of flow is not
known a priori.
for fluid that travels through pipes 112 and 132. These can be combined to give h L2
h L3
. This is a
statement of the fact that fluid particles that travel through pipe 122 and particles that travel through
pipe 132 all originate from common conditions at the junction 1or node, N2 of the pipes and all end
up at the same final conditions.
The flow in a relatively simple looking multiple pipe system may be more complex than
it appears initially. The branching system termed the three-reservoir problem shown in Fig. 8.37
is such a system. Three reservoirs at known elevations are connected together with three pipes
of known properties 1lengths, diameters, and roughnesses2. The problem is to determine the
flowrates into or out of the reservoirs. If valve 112 were closed, the fluid would flow from
reservoir B to C, and the flowrate could be easily calculated. Similar calculations could be
carried out if valves 122 or 132 were closed with the others open.
With all valves open, however, it is not necessarily obvious which direction the fluid flows.
For the conditions indicated in Fig. 8.37, it is clear that fluid flows from reservoir A because the
other two reservoir levels are lower. Whether the fluid flows into or out of reservoir B depends
on the elevation of reservoirs B and C and the properties 1length, diameter, roughness2 of the
three pipes. In general, the flow direction is not obvious, and the solution process must include
the determination of this direction. This is illustrated in Example 8.14.
E XAMPLE 8.14
Three-Reservoir, Multiple-Pipe System
GIVEN Three reservoirs are connected by three pipes as are
shown in Fig. E8.14. For simplicity we assume that the diameter
of each pipe is 1 ft, the friction factor for each is 0.02, and
because of the large length-to-diameter ratio, minor losses are
negligible.
A
Elevation = 100 ft
D 1 = 1 ft
(1) 1 = 1000 ft
FIND
SOLUTION
Determine the flowrate into or out of each reservoir.
It is not obvious which direction the fluid flows in pipe 122.
However, we assume that it flows out of reservoir B, write the
governing equations for this case, and check our assumption.
The continuity equation requires that
which,
since the diameters are the same for each pipe, becomes simply
V 1 V 2 V 3
Q 1 Q 2 Q 3 ,
The energy equation for the fluid that flows from A to C in pipes
112 and 132 can be written as
p A
g V 2 A
2g z A p C
g V 2 C
2g z C f / 1 V 2 1
1
D 1 2g f / 3 V 2 3
3
D 3 2g
(1)
F I G U R E E8.14
By using the fact that p A p C V A V C z C 0, this becomes
For the given conditions of this problem we obtain
100 ft
D 3 = 1 ft
3 = 400 ft
z A f / 1 V 2 1
1
D 1 2g f / 3 V 2 3
3
D 3 2g
0.02
2132.2 fts 2 2
D 2 = 1 ft
2 = 500 ft
(3)
(2)
Elevation =
20 ft
Elevation =
0 ft
1
11 ft2 311000 ft2V 2 1 1400 ft2V 2 34
C
B