State and prove Bernoulli’s theorem.
Answers
Answer:
Consider a fluid of negligible viscosity moving with laminar flow, as shown in Figure 1. This is Bernoulli's theorem You can see that if there is a increase in velocity there must be a decrease of pressure and vice versa.
Explanation:
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Explanation:
Bernoulli's Theorem:
According to Bernoulli's theorem, the sum of the energies possessed by a flowing ideal liquid at a point is constant provided that the liquid is incompressible and non-viseous and flow in streamline.
Potential energy + Kinetic energy + Pressure energy = Constant
P+
2
1
pv
2
+pgh=Constant
gh+
2
1
v
2
+
p
P
=C ............(11.11)
Where C is a constant.
This relation is called Bernoulli's theorem.
Dividing eqn. (11.11) by g, we get
h+
pg
P
+
2
1
g
v
2
=C
′
............(11.12)
Where C is another constant.
For horizontal flow, h remains same throughout.
So,
pg
P
+
2g
v
2
=Constant
or; P+
2
1
pv
2
=Constant
P is static pressure of the liquid and
2
1
pv
2
is its dynamic and velocity pressure.
Thus, for horizontal motion, the sum of static and dynamic pressure is constant. If p
1
v
1
and p
2
v
2
represent pressure and velocities at two points. Then
P
1
+
2
1
pv
1
2
=P
2
+
2
1
pv
2
2
Concepts : 1. In Bernoulli's theorem P+pgh+
2
1
ev
2
= constant. The term (p+pgh) is called static pressure and the term
2
1
pv
2
is the dynamic pressure of the fluid.
2. Bernoulli theorem is fundamental principle of the energy.
3. The equation
pg
P
+
2
1
g
v
2
+h=constant
the term
pg
P
= pressure head
the term
2g
v
2
= velocity head
h = gravitational head.
Derivation of Bernoulli's Theorem :
The energies possessed by a flowing liquid are mutually convertible. When one type of energy increases, the other type of energy decreases and vice-versa. Now, we will derive the Bernoulli's theorem using the work-energy theorem.
Consider the flow of liquid. Let at any time, the liquid lies between two areas of flowing liquid A
1
and A
2
. In time interval Δt, the liquid displaces from A
1
by Δx
1
=v
1
Δt and displaces from A
2
by Δx
2
=v
2
Δt. Here v
1
and v
2
are the velocities of the liquid at A
1
and A
2
.
The work done on the liquid is P
1
A
1
Δx
1
by the force and P
2
A
2
Δx
2
against the force respectively.
Net work done,
W=P
1
A
1
Δx
1
−P
2
A
2
Δx
2
⇒W=P
1
A
1
v
1
Δt−P
2
A
2
v
2
Δt
=(P
1
−P
2
)ΔV .........(11.13)
Here, ΔV→ the volume of liquid that flows through a cross-section is same (from equation of continuity).
But, the work done is equal to net change in energy (K.E. + P.E.) of the liquid, and
ΔK=
2
1
pΔV(v
1
2
−v
2
2
) ........ (11.14)
and ΔU=pΔV(h
2
−h
1
) ........ (11.15)
∴(P
1
−P
2
)ΔV=
2
1
pΔV(v
1
2
−v
2
2
)+pgΔV(h
2
−h
1
)
P
1
+
2
1
v
1
2
+pgh
1
=P
2
+
2
1
v
2
2
+pgh
2
or P+
2
1
pv
2
+pgh=constant .......... (11.16)
This is the required relation for Bernoulli's theorem.
∴A
1
v
1
=A
2
v
2
So, more the cross-sectional area, lesser is the velocity and vice-versa.
So, Bernoulli's theorem,
P
1
+
2
1
p
1
v
1
2
=P
2
+
2
1
p
2
v
2
2