Question

Water is flowing with pressure 4*10 ^4 at a point where cross section area is 0.02 and velocity is 2 find pressure at cross sectional area 0.01

Answer 1

Answer:

As we know A1V1 = A2V2 Where A1 = 0.02 m2 V1 = 2 m/s A2 = 0.01 m2 0.02 × 2 = 0.01 × V2 V2 = 4 m/s Now using Bernoulli's theorem P1ρ + 12V12 = P2ρ

Answer 2

Water is flowing steadily through a horizontal tube of non-uniform cross-section.If the pressure of water is 4×10

4

N/mm

2

at a point when cross section is 0.02m

2

velocity of flow is 2m/s what is pressure at a point where cross section reduces to 0.01m

As we know

A

1

V

1

=A

2

V

2

Where A

1

=0.02 m

2

V

1

=2 m/s

A

2

=0.01 m

2

⟹0.02×2=0.01×V

2

⟹V

2

=4 m/s

Now using Bernoulli's theorem

ρ

P

1

+

2

1

V

1

2

=

ρ

P

2

+

2

1

v

2

2

⟹P

2

=P

1

+

2

ρ

1

[V

1

2

−V

1

2

]

⟹P

2

=4×10

10

+

2

1000

[2

2

−4

2

]−4×10

10

=3.4×10

4

N/m

2