state and prove bervonlli therom
Answers
Answer:
Hope this helps you mate xD ☺️
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
❤❤Answer ❤❤
◾◽Bernoulli's principle formulated by Daniel Bernoulli states that as the speed of a moving fluid increases (liquid or gas), the pressure within the fluid decreases. Although Bernoulli deduced the law, it was Leonhard Euler who derived Bernoulli's equation in its usual form in the year 1752.