1. GUIDED BY:- ANERI CHAvan.
BRANCH:- CIVIL ENGINEERING.
BATCH:- C YEAR: 2nd
GUIDED BY:- ANERI CHAvan.

2. subject:- fluid mechanics
subject:- fluid mechanics. Topic of presentation:-bernoulli’s equation & SOLVE EXAMPLE.
NAME ENROLLENMENT NO
YADAV DIVYA M (D2D=64)
RANA HETAL R (D2D=65)
RANA SAMIR L (D2D=66)
PATEL NIRMAL S (D2D=67)
PATEL ARPAN J (D2D=68)

3. BERNOULLI’S EQUATION
“ In a steady, ideal flow of an incompressible fluids, the total energy at any point of the fluid is constant.”

4. The total energy consists of potential energy, kinetic energy and pressure energy. Mathematically ,
z = constant
where,
z = potential energy
= kinematic energy

5. = pressure energy Proof :-
- consider a perfect incompressible liquid flowing through a non-uniform pipe as shown in figure.
- Let us consider two sections AA and BB of the pipe.
- Now let us assume that the pipe is running full and there is a continuity of flow between the two sections.

6. z1 = Height of AA above the datum
P1 = pressure at AA
V1 = velocity of liquid at AA
A1 = C/S area of the pipe at AA
Z2,P2,V2 and A2 are corresponding values at BB.
Let the liquid between the two section AA and BB move to A’A’ and B’B’ through very small lengths dl1 and dl2 as shown in figure (6.3).
Let W = weight of liquid between AA and A’A’
We Know that,
density (w) = …………v = volume

7. W = w × V
W = w × a1 × dl1
a1 × dl1 = ………………….……..(1)
Similarly, W = w × a2 ×dl2
a2 × dl2 = ………………………(2)
a1 × dl1 = a2 × dl2 =
- we know that work done by pressure at AA, in moving the liquid to A’A’
= Force × Displacement
= p1 × a1 × dl1

8. - Similarly , work done pressure at BB in moving the liquid to B’B’
= -p2 × a2 × dl2
Total work done by pressure,
= p1 × a1 × dl1 – p2 × a2 × dl2
= p1 × a1 × dl1 – p2 × a1 × dl1
= a1 × dl1 × (p1 – p2)
= ( p1 – p2) ….. Pressure energy
Loss of potential energy,
= WZ1 – WZ2
= W ( Z1 – Z2)

9. Gain in kinetic energy,
= W
  =
- We know that,
- loss of potential energy + work done by pressure.
= Gain in kinetic energy

10. W (Z1-Z2) + (P1 –P2) = ( V2² - V1²)
…...….Bernoulli’s equation

11. EXAMPLE
The water is flowing through a pipe having diameter 30 cm and 12cm at section 1 and 2 respective. The rate of flow through pipe is 40 lit/sec. the section 1 is 8m above datum and section 2 is 6 m above datum. If the pressure at section 1 is 42 N/CM², find the intensity of pressure at section 2.
solution:-
d1= 30 cm = 0.3 m
A1= × ( 0.30)²=0.0706m²

12. d2 = cm = 0.12
A2 = × ( 0.12 )² = m²

13. … Z1 = 8 m
… Z2 = 6 m
… Q = 40 lit / sec = 0.4 m³/s\
… P1 = 42 N / cm ²
= 420 KN / m² =
… Q = A1 × V1
0.04 = × v1
V1 = m/s
… Q = A2 × V2
0.04 = × v2
V2 = 3.54 m/s

14. As per Bernoulli’s equation,
Z = Z =
( 8 × 9.81 ) = ( 6 × 9.81 ) + p
= p
∴ P2 = KN / M²

15. 2). The top and bottom diameter of a 2 m long vertical pipe are 10 cm and 5 cm respectively. Water flows down the pipe at 30 lit/ sec. Find pressure difference between the two ends of the pipe.
solution:
Q = 30 lit / sec = 0.03 m³/s
A1= × ( 0.10)² = 7.85 × 10­³ m²
A2 = × ( 0.05 )² =1.963 × 10-³ m²

17. ∴ Q = A1 × V1
0.03 = 7.85× 10¯³ × v1
∴ V1 = 3.82 m/sec
∴ Q = A2 × V2
0.03 = × 10¯³ × v2
∴ V2 = m/sec

18. As per Bernoulli’s equation,
Z = Z
Z =
∴ ( 2.0 × 9.80 ) + P = 0 + P
∴ P1 - P2 = – 7.29 – 19.6
∴P1 - P2 = KN/m²

19. d1 = 80 mm = 0.08m A1 = × (0.08)² = 5.03 × 10-³ d2 = 240 mm = 0.24 m
3) A pipe of 5.0 m length is inclined at an angle of 15°with horizontal surface. Smaller section of pipe has diameter of 80 mm and is at lower end. The large section of pipe has diameter 240mm and is at higher end.Determine the difference of pressure between two sections of pipe if velocity at smaller section is 1 m/sec.
Solution:-
d1 = 80 mm = 0.08m
A1 = × (0.08)² = 5.03 × 10-³
d2 = 240 mm = 0.24 m
A2 = × ( 0.24 )² = 1.29 m
Z1 = 0
Z2 = = m

21. v1 = 1 m/s
we know that,
A1V1 = A2V2
5.03 × 10-³ × 1.0 = × v2
V2 = m/s
As per Bernoulli’s equation,
Z = Z =
V2 = m/s

22. P = ( 1.29 × 9.81 ) + P
P = P
P1 – P2 = KN/M²
P1 – P2 = KN/M²

23. Thank you