Basics of Fluid Mechanics, 2014a

7.5. EXAMPLES OF INTEGRAL ENERGY CONSERVATION 219
the pipe, vanish again because the value of z is constant. Also, as can be noticed from
equation (7.I.a), the velocity is identical (in and out). Hence the second term becomes
∫
A
⎛ ⎛ ⎞
⎝h + ⎝
U
✟ ✟✟ ✟✯ constant ⎞
2
∫
✟ 2 + gz ⎠⎠ ρU rn dA =
A
h
{(
}} {
E u + P )
ρU rn dA
ρ
(7.I.d)
Equation (7.I.d) can be further simplified (since the area and averaged velocity are
constant, additionally notice that U = U rn )as
∫ (
E u + P )
∫
ρU rn dA =ΔPUA+ ρE u U rn dA (7.I.e)
A ρ
A
The third term vanishes because the boundaries velocities are zero and therefore
∫
PU bn dA =0
(7.I.f)
Combining all the terms results in
˙Q = ρU dU
dt
equation (7.I.g) can be rearranged as
A
Lπr
{ }} {
2
V pipe +ρ d ∫
∫
E u dV +ΔPUdA+ ρE u UdA
dt V pipe A
−K U2
2
{ ∫ }} ∫ {
d (E u )
˙Q − ρ
dV −
V pipe
dt
A
ρE u UdA= ρLπr 2 U dU
dt +(P in − P out ) U
(7.I.g)
(7.I.h)
The terms on the LHS (left hand side) can be combined. It common to assume (to
view) that these terms are representing the energy loss and are a strong function of
velocity square 19 . Thus, equation (7.I.h) can be written as
−K U 2
2 = ρLπr2 U dU
dt +(P in − P out ) U
(7.I.i)
Dividing equation (7.I.i) byKU/2 transforms equation (7.I.i) to
U + 2 ρLπr2
K
Equation (7.I.j) is a first order differential equation.
described in the appendix and which is
0
U = e tK
1 ⎛
0
− @
2 πr 2 A
@
ρL
⎜
⎝ 2(P in − P out )e
tK
2 πr 2 ρL
K
dU
dt = 2(P in − P out )
(7.I.j)
K
The solution this equation is
1
A
⎞ 0
@ ⎟
+ c⎠ e
2 πr2 ρtL
K
1
A
(7.I.k)
19 The shear work inside the liquid refers to molecular work (one molecule work on the other molecule).
This shear work can be viewed also as one control volume work on the adjoined control volume.