Water flows at 12 m/s in a horizontal pipe with a pressure of 3.0 x 10 4 N/m2 . If the pipe widens to twice its original radius, what is the pressure in the wider section?
To determine the pressure in the wider section of the pipe, we can use the principle of continuity. According to this principle, the mass flow rate of an incompressible fluid remains constant as it flows through different sections of a pipe.
The mass flow rate can be expressed as the product of the density of the fluid (ρ), the cross-sectional area of the pipe (A), and the velocity of the fluid (v):
m = ρ * A * v
Since the mass flow rate remains constant, we can equate the expressions for the different sections of the pipe:
m1 = m2
ρ1 * A1 * v1 = ρ2 * A2 * v2
Here, we know that the velocity (v1) in the original section of the pipe is 12 m/s, and the radius of the wider section is twice the radius of the original section, which means the new cross-sectional area (A2) will be four times the original area (A1).
Now, let's go step by step to find the pressure in the wider section:
Step 1: Given that the velocity (v1) in the original section is 12 m/s, and the radius (r1) is known, we can find the cross-sectional area (A1) using the formula:
A1 = π * r1^2
Step 2: Calculate the cross-sectional area (A2) of the wider section by substituting the new radius (r2 = 2 * r1) into the area formula:
A2 = π * r2^2
Step 3: Since the density of water is approximately constant, we can cancel it out from our equation:
A1 * v1 = A2 * v2
Step 4: Rearrange the equation to solve for the velocity (v2) in the wider section:
v2 = (A1 * v1) / A2
Step 5: Finally, substitute the known values for A1, A2, and v1 into the equation and solve for v2.
Once we have the velocity (v2) in the wider section, we can use Bernoulli's equation to find the pressure difference between the two sections:
P1 + 0.5 * ρ * (v1)^2 = P2 + 0.5 * ρ * (v2)^2
Since the velocity (v1) and pressure (P1) in the original section are known, we can rearrange the equation to isolate the pressure in the wider section (P2).
Once we have the pressure difference between the two sections, we can calculate the pressure in the wider section by adding the pressure difference to the pressure in the original section.
97500 Pa
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The water speed will drop by a factor 1/4 because of the larger pipe area. That follows from the incompressible fluid continuity equation, V * Area = constant. The water speed in the wider section is 3 m/s
Use the change in velocity and the Bernoulli equation to get the change in pressure.
P + (1/2)(density) V^2 = constant
P2 - P1 = (1/2)(water density)(V1^2 - V2^2
Solve for P2.