Answer:
(a) 1) When V = 1000 L, the height of the liquid in the tank is approximately 62 cm
2) When V = 5,000 L, the height of the liquid in the tank is approximately 106 cm
3) When V = 10,000 L, the height of the liquid in the tank is approximately 202 cm
b) The drawing of the spherical tank showing the height of the liquid level for each volume created with Microsoft Excel is attached
Step-by-step explanation:
(a) The height of the spherical tank = 3 m
The diameter of the spherical tank, d = 3 m
The capacity of the spherical tank = 14,000 L
The volume of a sphere, V = (4/3) × π × r³ =
Where, r = d/2
∴ r = 3 m/2 = 1.5 m
V = (4/3) × π × (1.5)³ ≈ 14.1371669412 m³
Using calculus, we have;
[tex]V = 2 \times \pi \times \int\limits^r_{0} {(r^2 - y^2)} \, dx = 2 \times \pi \times \left[r^2 \cdot y - \dfrac{y^3}{3} \right]^y_0[/tex]
For the two halves, we have;
[tex]V = \pi \times \left[r^2 \cdot y - \dfrac{y^3}{3} \right]^{\dfrac{y}{2}} _0 + \pi \times \left[r^2 \cdot y - \dfrac{y^3}{3} \right]^y_{\dfrac{y}{2}}[/tex]
[tex]V = 2 \times \pi \times \left[y^3- \dfrac{y^3}{3} \right]^y_0 = \dfrac{4 \cdot \pi \cdot y^3}{3}[/tex]
1) For V = 1000 L, we have;
[tex]1000 \ L = \dfrac{4 \cdot \pi \cdot y^3}{3}[/tex]
1000 L =
When V = 1000 L, we have;
[tex]y = \sqrt[3]{\dfrac{3 \times 1 \, m^3}{4 \times \pi } } \approx 0.62 \ m[/tex]
The liquid level, y ≈ 0.62 m = 62 cm
2) When V = 5,000 L, we have;
[tex]y = \sqrt[3]{\dfrac{3 \times 5 \, m^3}{4 \times \pi } } \approx 1.06 \ m[/tex]
The liquid level, y ≈ 1.06 m = 106 cm
3) When V = 10,000 L, we use the height of the unfilled volume at the top which is the volume of the sphere less the 10,000 L as follows;
For the volume of the sphere of 14,000 L, we have;
Volume at the top = 14,000 L - 10,000 L = 4,000 L = 4 m³
Therefore
[tex]y_{top} = \sqrt[3]{\dfrac{3 \times 4 \, m^3}{4 \times \pi } } \approx 0.985 \ m[/tex]
The height of the liquid in the tank, y = d - [tex]y_{top}[/tex]
∴ y = 3 m - 0.985 m = 2.015 m
The liquid level, y ≈ 2.015 m = 201.5 cm ≈ 202 cm
b) Please find attached the drawing of the spherical tank created with Microsoft Excel
a) (i) The level line for 1000 liters is set at a height of 0.486 meters.
(ii) The level line for 5000 liters is set at a height of 1.203 meters.
(iii) The level line for 10000 liters is set at a height of 1.926 meters.
b) The sphere is represented by the equation [tex]x^2+(y-1.5)^{2} = 2.25[/tex] and the level lines are represented by the linear functions [tex]y = 0.486[/tex], [tex]y = 1.203[/tex] and [tex]y = 1.926[/tex]. Please see the second image attached.
a) First, we derive an expression for the volume of a spherical section ([tex]V[/tex]), in cubic meters, in terms of tank radius ([tex]R[/tex]), in meters, and liquid height ([tex]h[/tex]), in meters. The volume can be found by using the slicing method integral equation:
[tex]V = \int\limits^{H}_{0} {A(h)} \, dh[/tex] (1)
Where:
If we know that [tex]A = \pi\cdot [R^{2}-(R-h)^{2}][/tex], then the volume formula is:
[tex]V = \pi\int\limits^H_0 {h\cdot (2\cdot R-h)} \, dh[/tex]
[tex]V = 2\pi\cdot R\int\limits^{H}_{0} {h} \, dh - \pi \int\limits^{H}_{0} {h^{2}} \, dh[/tex]
[tex]V = \pi\cdot R\cdot H^{2}-\frac{\pi\cdot H^{3}}{3}[/tex]
Then we obtain the following third order polynomial:
[tex]\frac{\pi\cdot H^{3}}{3}-\pi\cdot R\cdot H^{2}+V = 0[/tex] (2)
Now we proceed to solve numerically for each case:
[tex]H = 0.486\,m[/tex]
The level line for 1000 liters is set at a height of 0.486 meters. [tex]\blacksquare[/tex]
[tex]H = 1.203\,m[/tex]
The level line for 5000 liters is set at a height of 1.203 meters. [tex]\blacksquare[/tex]
[tex]H = 1.926\,m[/tex]
The level line for 10000 liters is set at a height of 1.926 meters. [tex]\blacksquare[/tex]
b) The sphere is represented by the equation [tex]x^2+(y-1.5)^{2} = 2.25[/tex]level lines are represented by the linear functions [tex]y = 0.486[/tex], [tex]y = 1.203[/tex] and [tex]y = 1.926[/tex]. By a graphing tool we have the following result. [tex]\blacksquare[/tex]
To learn more on volumes, we kindly invite to check this verified question: https://brainly.com/question/1578538
