Question

A biogas digester is to be built for producing biogas from animal dung at a farm. Determine how much cement mixture will be required in m3 to construct a 10ft deep well with 3.5ft inner and 4ft outer diameter, including the top and bottom lid with 4inch thickness? How many liters of paint will be required to paint its outer surfaces if paint coverage is 12m2/liter? Neglect any reinforcement and piping used in the structure. (Draw geometry first and use theorem of Pappus for solution)

Answer 1

Here $r_i = 3.5/2 = 1.75ft$ is a inner and $r_o = 4/2 = 2ft$ is a outer radii. $h = 10 ft$ is the height of the well.

1) According to the volume theorem of Pappus, the volume of a solid of revolution generated by rotating a plane figure F about an external axis is equal to the product of the area A of F and the distance d traveled by the geometric centroid of F:

$$V = Ad$$

The centroid of the rectangle F is located in the middle between two cylinders. Thus:

$$R = r_i + \dfrac{r_o - r_i}{2} = 1.75 + \dfrac{2 - 1.75}{2} = 1.875 ft$$

Then the distance traveled by the geometric centroid is:

$$d = 2\pi R = 3.75\pi\space ft$$

The area of the rectangle F is:

$$A = h\times (r_o - r_i) = 10\times (2-1.75) = 2.5 ft^2$$

Thus, obtain:

$$V = Ad = 3.75\pi\cdot 2.5 = 9.375\pi\space ft^3$$

The volume of lid is the volume of the ordinary cylindet wiht height $h_{lid} = 4\space inch \approx 0.33ft$ and radius $r_o = 2\space ft$. Thus:

$$V_{lid} = \pi r_o^2h_{lid} = 1.33\pi \space ft^3$$

Thus, the total volume is:

$$V_{well} = V + 2V_{lid} = 9.375\pi\space ft^3 +2\cdot 1.33\pi \space ft^3\approx 37.83\space ft^3 = 1.07m^3$$

2) According to the area theorem of Pappus, the surface area A of a surface of revolution generated by rotating a plane curve C about an axis external to C and on the same plane is equal to the product of the arc length s of C and the distance d traveled by the geometric centroid of C:

$$A = sd$$

In our case, the curves C for the both cylinders are the sides of the rectangle F (see figure). Their lenghts are $s = h = 10ft$. Both centroids are located in the middle of these sides, thus, for a inner cylinder $d_i = 2\pi r_i$ and for the outer one $d_o =2\pi r_o$. Thus, the total surface area will be:

$$A = 2\pi h r_i + 2\pi h r_o = 2\pi h(r_i + r_o)\\ A = 2\pi \cdot 10\cdot (1.75 +2) \approx 235.62 \space ft^2$$

The area of the lid is the area of surface of cylinder (side surface plus one outer cap):

$$A_{lid} = 2\pi r_o h_{lid} + \pi r_0^2 \approx 5.33\pi\space ft^2$$

The total area is:

$$A_{well} = A + 2A_{lid} \approx 240.95 \space ft^2$$

The paint coverage is $12 \space m^2/liter = 129.167\space ft^2/liter$ . Then the amount of paint required to paint well's outer surfaces will be:

$$Q = \dfrac{A_{well}}{129.167\space ft^2/liter} \approx 1.87 \space liter$$

Answer. 1.07 m^3, 1.87 liter.