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
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Answered by
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
Answered by
1
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
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