Derive bernoulli equation from steady flow energy equation
Answers
Answer:
Consider the motion of a fluid particle in a flow field in steady flow.
Applying Newton’s second law (which is referred to as the linear momen-
tum equation in fluid mechanics) in the s-direction on a particle moving
along a streamline gives
(12–3)
In regions of flow where net frictional forces are negligible, there is no
pump or turbine, and there is no heat transfer along the streamline, the sig-
nificant forces acting in the s-direction are the pressure (acting on both
sides) and the component of the weight of the particle in the s-direction
(Fig. 12–3). Therefore, Eq. 12–3 becomes
(12–4)
where u is the angle between the normal of the streamline and the vertical z-
axis at that point, m -
rV -
r dA ds is the mass, W -
mg -
rg dA ds is
the weight of the fluid particle, and sin u -
dz/ds. Substituting,
(12–5)
Canceling dA from each term and simplifying,
(12–6)
Noting that V dV -
d(V2) and dividing each term by r gives
(12–7)
dP
r
1
2 d 1V2 2 g dz -
0
1
2
dP rg dz -
rV dV
dP dA rg dA ds dz
ds -
r dA ds V
dV
ds
P dA 1P dP2 dA W sin u -
mV
dV
ds
a Fs -
mas
as -
dV
dt -
0V
0s
ds
dt -
0V
0s V -
V
dV
ds
dV -
0V
0s ds
0V
0t
dt and dV
dt -
0V
0s
ds
dt
0V
0t
FIGURE 12–2
During steady flow, a fluid may not
accelerate in time at a fixed point, but
it may accelerate in space.
z
x
W
n s
P dA
(P + dP)dA
Steady flow along a streamline
dx
dz
ds
u
u
ds
g
FIGURE 12–3
The forces acting on a fluid
particle along a streamli