Page 32 - Fundamentals of The Finite Element Method for Heat and Fluid Flow

SOME BASIC DISCRETE SYSTEMS
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Table 2.1
Details of pipe network
Component Number Diameter, cm Length, m
1 2.50 30.00
2 2.00 20.00
3 2.00 25.00
4 1.25 20.00
Example 2.2.1 In a pipe network as shown in Figure 2.2, water enters the network at a
-3
3
2
rate of 0.1 m /s with a viscosity of 0.96 × 10 Ns/m . The component details are given in
Table 2.1. Determine the pressure values at all nodes.
5
On substitution of the various values, we get the following resistances in Ns/m from
Equation 2.15
R 1 = 0.3 × 10 7
R 2 = 0.5 × 10 7
R 3 = 0.6 × 10 7
R 4 = 3.2 × 10 7
Now Equation 2.23 can be formulated as

     
5.33 −2.00 −3.33 p 1  0.1
10 −7  −2.00 3.98 −1.98  p 2 = 0.0 (2.24)
−3.33 −1.98 5.31  p 3   −1.0 
The solution of the above simultaneous system of equations with p 3 = 0.0(assumedas
reference pressure) gives
6
p 1 = 0.231 × 10 N/m 2
6
p 2 = 0.116 × 10 N/m 2
From Equations 2.18, 2.19, 2.20 and 2.21, we can calculate the flow quantities as
p 1 − p 3 3
q 1 = = 0.0769 m /s
R 1
p 1 − p 2 3
q 2 = = 0.0231 m /s
R 2
p 2 − p 3 3
q 3 = = 0.0193 m /s
R 3
p 2 − p 3 3
q 4 = = 0.0036 m /s (2.25)
R 4
It is possible to take into account the entrance loss, exit loss, bend loss, and so on, in
the calculation of nodal pressures and flows in each circuit. If the fluid flow in the network
is turbulent, it is still possible to define an element, but the element equations are no longer
linear as can be seen from an empirical relation governing fully developed turbulent pipe
flow (Darcy-Weisbach formula (Shames 1982))
2
8fLQ ρ
p 1 − p 2 = (2.26)
πD 5