Water is flowing through a horizontal tube. The pressure of the liquid...
We know that Pv remains constant for any fluid and for non compressible fluids like water we get that Pv is always constant.
Also as 76cm of Hg = 105 Pa
We get 2m Hg = 200/76 x 105 Pa
Thus from conservation of Pv, we get that
2 x 200/76 x 105 = 4 x P
Thus we get P = 100/76 x 105
Thus we get P = 1.31 x 105
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Water is flowing through a horizontal tube. The pressure of the liquid...
To solve this problem, we can use Bernoulli's equation, which states that the total pressure of a fluid is constant along a streamline. The equation is given by:
P + 1/2 * ρ * v^2 + ρ * g * h = constant
Where:
P is the pressure of the fluid,
ρ is the density of the fluid,
v is the velocity of the fluid,
g is the acceleration due to gravity, and
h is the height of the fluid above a reference point.
In this case, the pressure in the portion where the velocity is 2 m/s is given as 2 m of Hg. We can assume that the height of the fluid above a reference point is zero for both portions of the tube. Therefore, the equation becomes:
P1 + 1/2 * ρ * v1^2 = P2 + 1/2 * ρ * v2^2
We can substitute the values given:
2 + 1/2 * ρ * (2)^2 = P2 + 1/2 * ρ * (4)^2
Simplifying the equation:
2 + 2ρ = P2 + 8ρ
Combining like terms:
P2 = 2 - 2ρ + 8ρ
P2 = 2 + 6ρ
We do not have the value for the density of the fluid, so we cannot determine the exact pressure in the portion where the velocity is 4 m/s.