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Apply this equation to the fluid element in Figure 2.3 leads to:
P
2
x
D
2
f u a
14
(2.3)
u u
Since the change in the velocity may be expressed as du dx dt which will
x t
du u dx u u u
give an acceleration of a u . Assuming incompressible
dt x dt t x t
fluid simplifies the velocity and acceleration to be independent to position along the
u conduit. Thus the partial differentiate 0 and acceleration becomes:
x
Thus Eq. (2.3) can be written as:
du
a (2.4)
dt
dP du
dx D dt
2
2 f u (2.5)
dP is the pressure drop across the controlled volume at any time. Integration of Eq. (2.5)
along the conduit length gives the pressure drop on the whole conduit:
2 du 2 du
P (2 f u ) dx (2 f u ) L
(2.6)
D dt D dt
L
Where P is the instantaneous fluctuating pressure drop over the conduit of length L as
a result of velocity fluctuation.
For the model to be able to cope with positive and negative velocity i.e. reverse flow,
term
2
u in Eq. (2.6) may be rewritten asu u , so u alone can be used to determine the
sign of this item and u u together for value. Rearranging Eq. (2.6) gives:
du P
2 f
u u
dt L D
(2.7)
This equation is to calculate the internal flow velocity fluctuation along the changing
pressure drop.