3. Water enters a horizontal pipe of non-
uniform cross section with a velocity of
0.6m/s and leaves the other end with a
velocity of 0.4m/s. At the first end, pressure
of water is 1200N/m'. Calculate the pressure
of water at the other end Density of water is
1000kg/ml
(Ans: 100N/m')
Answers
Answer:
The answer is 1300 N/m^2.
Explanation:
Bernoulli's equation for a horizontal pipe can be used to find the pressure at the second point of the pipe, which is
1/2ρ V1^2 + P1 = 1/2ρ V2^2 + P2
Given data:
where, ρ is the density of water = 1000kg/m3
v1 is the velocity at first point = 0.6m/s
P1 is the pressure at first point = 1200N/m2
v2 is the velocity at second point = 0.4m/s
P2 is the pressure at second point = ?
Putting these values in the above equation:
1/2 (1000)(0.6)^2 + 1200 = 1/2 (1000)(0.4)^2 + P2
P2 = 500 [ (0.6)^2 - (0.4)^2 ] + 1200
P2 = 500 (0.36 - 0.16) + 1200
P2 = (500 X 0.2) + 1200
P2 = 100 + 1200
P2 = 1300 N/m^2
Answer:
The answer will be 1300 N/m^2
Explanation:
According to the problem the velocity of the water in the front end of the pipe is given let be v1, and the other end be v2
The pressure at both side is also given
Now by applying Bernoulli's equation for horizontal pipe,
1/2ρv1^2+ b1 = 1/2ρv2^2 +b2 [ b1 and b2 are the pressure and the ρ is the density of the water]
=> 1/2 x 1000x 0.6 x 0.6 + 1200 = 1/2 x 1000 x 0.4 x 0.4 + b2
=> b2 = 1300
Therefore the pressure at the other end s 1300 N/m^2