A level pipeline is required to pass through a hill having a parabolic profile d

Question

A level pipeline is required to pass through a hill having a parabolic profile described by equation 3.1. Distances in this equation are in meters. The origin of the x and y coordinates is fixed at elevation zero near the base of the hill as shown in Figure 3.1 Y =-0.005 x^2 + 0.35 x m Setup: Describe the of the setup of this problem (e.g., Include a diagram of problem, a list of what is known/given and a list of what is being found). Calculation: Write the quadratic equation for a pipeline of elevation y = 3 m in standard form. Solve the quadratic equation from part (a) to determine the positions of the tunnel entry x_A and the exit x_B using: the quadratic formula and completing the square.

Explanation / Answer

(a) The pipeline will take the form of the hill before entering the tunnel.

So, given are - 1. the equation of pipeline profile before entering the hill

2. Pipeline entering point and exit point of the tunnel

3. Zero elevation

To be found are - 1. Pipeline entry point (Xa) from hill starting point (O)

2. Pipeline exit point (Xb) from hill starting point (O)

(b) Hill profile : y = -0.005x2 + 0.35x

As the pipeline also takes the same profile we assume before entering the tunnel, so pipeline equationat an elevation of 3m will be :

3 = -0.005x2 + 0.35x

or, 3000 = -5x2 + 350x

or, 5x2 - 350x + 3000 = 0

This is the standard equation.

Now,   5x2 - 350x + 3000 = 0

Either, x = [350 + sqrt(3502 - 2X5X3000)] / (2X5) or, x = [350 - sqrt(3502 - 2X5X3000)] / (2X5)

= (350 + 304.14) / 10 = (350 - 304.14) / 10

= 65.414 = 4.586

From the above equation we get the entry point of tunnel is Xa = 4.586m and exit point of tunnel is Xb = 65.414m.

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