Water is flowing in a pipe with a circular cross-section area, and at all points
the water completely fills the pipe. At point 1, the cross-section area of the
pipe is 0.080-m2 and the fluid velocity is 4.50 m/s.
(a) What is the fluid speed at a point in the pipe where the cross-sectional area
is 0.101 m2?
(b) Calculate the volume of the water discharge from the open end of the pine
in 500 hours.
Q = flow rate at every cross section (water is incompressible for our purposes) = v A = 4.50 m/s * 0.080 m^2 = 0.36 m^3/s
so
answer part b first
0.36 m^3/s * 500 hr * 3600 s/hr = 648,000 m^3 in 500 hours (heavens to Betsy, what a waste ! )
now part a
4.50 * ( .08/.101) = 3.56 m/s
a mirror with a parabolic cross section is used to collect sunlight on a pipe located at the focus of the mirror the pipe is located 8 inches from the vertex of the mirror write an equation of the parabola that models the cross section of the mirror assume that the parabola opens upward
Let's assume that the parabola has its vertex at the origin, with the axis of symmetry along the y-axis. Then, the equation of the parabola is of the form y = ax^2, where a is a constant that determines the shape of the parabola.
To find the value of a, we need to use the information given about the mirror and the pipe. We know that the pipe is located at the focus of the mirror, which is a point on the axis of symmetry. We also know that the pipe is 8 inches from the vertex of the mirror, which means it is located at a distance of 8 inches along the y-axis.
The distance between the vertex and the focus of a parabola is given by the constant a/4. In this case, we have a distance of 8 inches, so we can write:
a/4 = 8
Solving for a, we find:
a = 32
Therefore, the equation of the parabola that models the cross section of the mirror is:
y = 32x^2
a mirror with a parabolic cross section is used to collect sunlight on a pipe located at the focus of the mirror the pipe is located 8 inches from the vertex of the mirror write an equation of the parabola that models the cross section of the mirror assume that the parabola opens upward
A. y=1/48x^2
B. y=1/64x^2
C. y=1/16x^2
D. y=1/32x^2
The correct answer is D. y = 1/32x^2.
We can start by using the formula for a parabolic mirror, which is given by:
y = a*x^2 / (4*f)
where y is the vertical axis, x is the horizontal axis, a is the curvature of the mirror, and f is the focal length.
In this problem, we are given that the pipe is located at the focus of the mirror, which means that f = 8 inches. We are also told that the mirror has a parabolic cross section that opens upward, so a > 0.
To find the value of a, we can use the fact that the distance from the vertex of the parabola (the point where the slope is zero) to the focus is equal to f/2. In this problem, the vertex is located at the origin, so we can write:
a = 4*f / d^2
where d is the distance from the vertex to the pipe. In this case, d = 8 inches, so we have:
a = 4*8 / 8^2 = 1/2
Substituting this value of a and f = 8 into the equation for a parabolic mirror, we get:
y = (1/2)*x^2 / (4*8) = 1/64 * x^2
Therefore, the equation of the parabola that models the cross section of the mirror is:
y = 1/64x^2
So the answer is B, which is different from the answer I gave earlier. I apologize for the mistake in my previous response.
To answer part (a) of the question, we can use the principle of continuity, which states that the volume flow rate of an incompressible fluid is constant along a pipe, assuming no leaks. The equation for continuity is:
A1v1 = A2v2
Where A1 and A2 are the cross-sectional areas at points 1 and 2, and v1 and v2 are the fluid velocities at those points.
Given:
A1 = 0.080 m^2 (cross-sectional area at point 1)
v1 = 4.50 m/s (fluid velocity at point 1)
A2 = 0.101 m^2 (cross-sectional area at the point where we want to find the fluid speed)
We can rearrange the equation to solve for v2:
v2 = (A1v1) / A2
Substituting the values:
v2 = (0.080 m^2 * 4.50 m/s) / 0.101 m^2
v2 ≈ 3.56 m/s
Therefore, the fluid speed at the point where the cross-sectional area is 0.101 m^2 is approximately 3.56 m/s.
Moving on to part (b) of the question, we need to calculate the volume of water discharge from the open end of the pipe in 500 hours. To do this, we need to know the flow rate of the water.
The volume flow rate (Q) is given by:
Q = Av
Where A is the cross-sectional area of the pipe and v is the fluid velocity.
Given:
A (cross-sectional area at any point) = 0.080 m^2 (from part (a))
v = 4.50 m/s (from point 1)
We can now calculate the flow rate:
Q = 0.080 m^2 * 4.50 m/s
Q = 0.36 m^3/s
Since the flow rate is constant, we can calculate the volume of water discharge in 500 hours by multiplying the flow rate by the time:
Volume = Q * time
Given:
time = 500 hours = 500 * 3600 seconds (since 1 hour = 3600 seconds)
Volume = 0.36 m^3/s * (500 * 3600 seconds)
Volume = 0.36 m^3/s * 1,800,000 seconds
Volume ≈ 648,000 cubic meters
Therefore, the volume of water discharged from the open end of the pipe in 500 hours is approximately 648,000 cubic meters.