a fluid flows through a horizontal pipe of varying cross-section. Assuming the flow is streamline and applying the bernoulli equation, show that pressure in the pipe is greatest where cross-sectional area is greatest.
To show that the pressure in a horizontal pipe is greatest where the cross-sectional area is greatest, we can apply the Bernoulli equation, which relates the pressure, velocity, and height of a fluid within a streamline flow.
The Bernoulli equation states:
P + 1/2 * ρ * V^2 + ρ * g * h = constant
Where:
P is the pressure of the fluid,
ρ (rho) 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 from an arbitrary reference point.
Now let's consider two different points in the horizontal pipe: a point where the cross-sectional area is smaller (point 1) and a point where the cross-sectional area is larger (point 2).
At point 1:
Let's assume the pressure at point 1 is P1, the velocity is V1, and the height is h1.
Using the Bernoulli equation, we have:
P1 + 1/2 * ρ * V1^2 + ρ * g * h1 = constant
At point 2:
Let's assume the pressure at point 2 is P2, the velocity is V2, and the height is h2.
Again, using the Bernoulli equation, we have:
P2 + 1/2 * ρ * V2^2 + ρ * g * h2 = constant
Since the two points are in the same streamline flow, the constant in both equations remains the same.
Now, let's analyze the terms in the equation that involve velocity (1/2 * ρ * V^2):
Since the flow is streamline, the velocity at point 1 (V1) is less than the velocity at point 2 (V2). So V2^2 > V1^2.
Since the density (ρ) and the acceleration due to gravity (g) are constant, we can ignore their contribution for this analysis.
So at point 2, we have:
P2 + (larger term due to velocity) = constant
And at point 1, we have:
P1 + (smaller term due to velocity) = constant
Since the terms involving velocity are the only difference between the equations, we can conclude that point 2 (where the cross-sectional area is larger) will have a smaller term involving velocity, thus a larger pressure value (P2) compared to point 1 (where the cross-sectional area is smaller) with a larger term involving velocity (P1).
Therefore, we can conclude that the pressure in the pipe is greatest where the cross-sectional area is greatest.