The inside diameters of the larger portions of the horizontal pipe depicted in Figure P9.47 are 2.80 cm. Water flows to the right at a rate of 1.80 10-4 m3/s. Determine the inside diameter of the constriction.
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To determine the inside diameter of the constriction, we can use the principle of continuity which states that the volume flow rate is constant in an incompressible fluid flowing through a pipe of varying cross-sectional area.
The equation to calculate the volume flow rate (Q) is:
Q = A1 * v1 = A2 * v2
where A1 and A2 are the cross-sectional areas of the pipe at two different points, and v1 and v2 are the velocities of the water at those points.
In this case, we are given the values for A1, A2, and v1, and we need to find v2.
First, we need to convert the given flow rate from cm^3/s to m^3/s:
Q = (1.80 * 10^(-4)) m^3/s
Next, we need to find the velocity v2. Since the pipe narrows at the constriction, the water velocity increases at that point.
Using the equation:
Q = A1 * v1 = A2 * v2
we can rearrange it to solve for v2:
v2 = (A1 * v1) / A2
The cross-sectional area A1 is given as 2.80 cm, which we can convert to m:
A1 = (2.80 cm)^2 * (1 m/100 cm)^2 = 7.84 * 10^(-5) m^2
The velocity v1 is given as 1.80 * 10^(-4) m^3/s.
Now, we can substitute the known values into the equation to solve for v2:
v2 = (7.84 * 10^(-5) m^2 * 1.80 * 10^(-4) m^3/s) / A2
Finally, rearranging the equation again, we can solve for A2:
A2 = (7.84 * 10^(-5) m^2 * 1.80 * 10^(-4) m^3/s) / v2
Since we know v2, we can calculate A2. The resulting value will be the cross-sectional area at the constriction.
To find the diameter of the constriction, we can use the formula for the area of a circle:
A = π * r^2
where A is the area and r is the radius.
Since we have the area (A2), we can rearrange the equation to solve for the radius (r2):
r2 = √(A2 / π)
Finally, we can double the radius to obtain the inside diameter of the constriction.
Note: Please note that the specific values and calculations might change depending on the actual values given in Figure P9.47.