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 the pipe is greatest where the cross-sectional area is greatest using the Bernoulli equation, we can start by writing down the equation for steady, incompressible flow along a streamline:
P + 1/2ρv^2 + ρgh = constant
Where:
- P is the pressure of the fluid
- ρ is the density of the fluid
- v is the fluid velocity
- g is the acceleration due to gravity
- h is the height above some reference point
Assuming the pipe is horizontal (h = 0), and neglecting any changes in elevation, we are left with:
P + 1/2ρv^2 = constant
Let's consider two points in the pipe, A and B, where A represents a location with a smaller cross-sectional area and B represents a location with a greater cross-sectional area.
At point A, the pressure is P_A and the fluid velocity is v_A.
At point B, the pressure is P_B and the fluid velocity is v_B.
Since the flow is steady and along a streamline, the constant term is the same at both points A and B.
So we can write:
P_A + 1/2ρv_A^2 = P_B + 1/2ρv_B^2
Now, we need to relate the fluid velocities at points A and B with the cross-sectional areas.
According to the principle of continuity, the mass flow rate of fluid should be constant along the pipe, which means that the product of the cross-sectional area (A) and velocity (v) should be constant:
A_A * v_A = A_B * v_B
Rearranging this equation, we get:
v_A/v_B = A_B/A_A
Now, substituting this relationship back into the Bernoulli equation, we have:
P_A + 1/2ρv_A^2 = P_B + 1/2ρv_A^2 * (A_B/A_A)^2
Since the fluid density (ρ) is constant, we can simplify the equation to:
P_A - P_B = 1/2ρv_A^2 * [(A_B/A_A)^2 - 1]
From this equation, we can see that (A_B/A_A)^2 - 1 will always be positive since the area ratio squared is always greater than 1.
So this implies that P_A - P_B is also positive, which means that the pressure at point A (where the cross-sectional area is smaller) is less than the pressure at point B (where the cross-sectional area is greater).
Hence, we can conclude that the pressure in the pipe is greatest where the cross-sectional area is greatest, as shown using the Bernoulli equation.
To show that the pressure in a pipe is greatest where the cross-sectional area is greatest, we will use the Bernoulli's equation and consider a streamline flow of fluid through a horizontal pipe with varying cross-sections.
The Bernoulli's equation relates the pressure, velocity, and height of a fluid at any point along a streamline. It states that the sum of the pressure energy, kinetic energy, and gravitational potential energy per unit volume is constant along the streamline.
The Bernoulli's equation is expressed as:
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,
h is the height of the fluid above an arbitrary reference point.
Let's consider two different points along the pipe, point 1 and point 2, with corresponding cross-sectional areas A1 and A2. Let the velocity and pressure at these points be v1, v2, P1 and P2, respectively.
According to Bernoulli's equation, we have:
P1 + 1/2 * ρ * v1^2 + ρ * g * h1 = P2 + 1/2 * ρ * v2^2 + ρ * g * h2
Since the pipe is horizontal, the heights h1 and h2 can be considered equal, so they cancel out from the equation.
P1 + 1/2 * ρ * v1^2 = P2 + 1/2 * ρ * v2^2
Now, let's focus on the pressure terms. Assuming the fluid is incompressible (constant density), we can rewrite the equation as:
P1 + 1/2 * v1^2 = P2 + 1/2 * v2^2
Since we are comparing the pressure at two different points in the pipe, let's assume that the velocity is higher at point 2 (v2 > v1). Consequently, the term 1/2 * v2^2 will be larger than 1/2 * v1^2.
Therefore, to satisfy the equation, P2 must be greater than P1 to compensate for the larger velocity term. In other words, the pressure in the pipe is greatest where the velocity is greatest.
Now, as per the continuity equation, which states that the mass flow rate of a fluid remains constant along a streamline:
A1 * v1 = A2 * v2
Since v2 > v1, we can conclude that A2 must be smaller than A1 to maintain a constant mass flow rate. Thus, the cross-sectional area is smallest where the velocity is greatest.
Combining this with the conclusion from Bernoulli's equation, we can infer that the pressure in the pipe is greatest where the cross-sectional area is greatest.
In summary, by applying Bernoulli's equation and considering the continuity equation, we have shown that the pressure in a pipe is greatest where the cross-sectional area is greatest.