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Answer :
The total amount of turpentine spilled is approximately 761.06 cubic feet, the maximum spill rate is approximately 0.00116 cubic feet per second, and the total time the leak was active is approximately 364.3 hours.
To determine the total amount of turpentine spilled, the maximum spill rate, and the total time the leak was active, we can follow these steps:
(a) Total Amount of Turpentine Spilled
First, calculate the volume of turpentine lost due to the leak.
1. Initial Volume [tex](V_initial):[/tex]
[tex]\[ V_{\text{initial}} = \pi \left(\frac{D}{2}\right)^2 h_{\text{initial}} \][/tex]
Where [tex]\(D = 15 \, \text{ft}\) and \(h_{\text{initial}} = 17.3 \, \text{ft}\):[/tex]
[tex]\[ V_{\text{initial}} = \pi \left(\frac{15}{2}\right)^2 \times 17.3 \][/tex]
[tex]\[ V_{\text{initial}} = \pi \times 7.5^2 \times 17.3 \][/tex]
[tex]\[ V_{\text{initial}} = \pi \times 56.25 \times 17.3 \][/tex]
[tex]\[ V_{\text{initial}} \approx 3057.45 \, \text{ft}^3 \][/tex]
2. Final Volume[tex](V_final):[/tex]
[tex]\[ V_{\text{final}} = \pi \left(\frac{D}{2}\right)^2 h_{\text{final}} \][/tex]
Where [tex]\(h_{\text{final}} = 13.0 \, \text{ft}\):[/tex]
[tex]\[ V_{\text{final}} = \pi \left(\frac{15}{2}\right)^2 \times 13.0 \][/tex]
[tex]\[ V_{\text{final}} = \pi \times 7.5^2 \times 13.0 \][/tex]
[tex]\[ V_{\text{final}} = \pi \times 56.25 \times 13.0 \][/tex]
[tex]\[ V_{\text{final}} \approx 2296.39 \, \text{ft}^3 \][/tex]
3. Volume of turpentine spilled [tex](V_spilled):[/tex]
[tex]\[ V_{\text{spilled}} = V_{\text{initial}} - V_{\text{final}} \][/tex]
[tex]\[ V_{\text{spilled}} = 3057.45 - 2296.39 \][/tex]
[tex]\[ V_{\text{spilled}} \approx 761.06 \, \text{ft}^3 \][/tex]
(b) Maximum Spill Rate
To find the maximum spill rate, use Torricelli’s law for the speed of efflux of a fluid under gravity:
[tex]\[v = \sqrt{2gh}\][/tex]
[tex]\[v = \sqrt{2 \times 32.2 \times 7}\][/tex]
[tex]\[v \approx \sqrt{450.8}\][/tex]
[tex]\[v \approx 21.22 \, \text{ft/s}\][/tex]
The area of the leak hole (A) is given by:
[tex]\[A = \pi \left(\frac{d}{2}\right)^2\][/tex]
Where [tex]\(d = 0.1 \, \text{in} = \frac{0.1}{12} \, \text{ft} = 0.00833 \, \text{ft}\):[/tex]
[tex]\[A = \pi \left(\frac{0.00833}{2}\right)^2\][/tex]
[tex]\[A \approx \pi \times (0.004165)^2\][/tex]
[tex]\[A \approx 5.45 \times 10^{-5} \, \text{ft}^2\][/tex]
The maximum spill rate [tex](\(Q_{\text{max}}\)):[/tex]
[tex]\[Q_{\text{max}} = A \times v\][/tex]
[tex]\[Q_{\text{max}} \approx 5.45 \times 10^{-5} \times 21.22\][/tex]
[tex]\[Q_{\text{max}} \approx 0.00116 \, \text{ft}^3/\text{s}\][/tex]
(c) Total Time the Leak was Active
To determine the total time (t), use the total volume spilled and the average spill rate. Assuming the spill rate is the average of the initial and final rates:
[tex]\[Q_{\text{avg}} = \frac{Q_{\text{max}}}{2}\][/tex]
[tex]\[Q_{\text{avg}} \approx \frac{0.00116}{2}\][/tex]
[tex]\[Q_{\text{avg}} \approx 0.00058 \, \text{ft}^3/\text{s}\][/tex]
Total volume spilled [tex](\(V_{\text{spilled}}\)):[/tex]
[tex]\[t = \frac{V_{\text{spilled}}}{Q_{\text{avg}}}\][/tex]
[tex]\[t \approx \frac{761.06}{0.00058}\][/tex]
[tex]\[t \approx 1,311,483 \, \text{s}\][/tex]
Convert time to hours:
[tex]\[t \approx \frac{1,311,483}{3600}\][/tex]
[tex]\[t \approx 364.3 \, \text{hours}\][/tex]
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Answer:
a) [tex]V=759.8727\ ft^3[/tex]
b) [tex]\dot V=1.403\times 10^{-3}\ ft^3.s^{-1}[/tex]
c) [tex]t\approx29.541\ s[/tex]
Explanation:
Given:
- diameter of hole in the tank, [tex]d'=0.1\ in=\frac{1}{120}\ ft[/tex]
- position of the hole form the tank bottom, [tex]h' =7\ ft[/tex]
- initial level of turpentine in the tank before the leakage, [tex]h_i=17.3\ ft[/tex]
- level of turpentine in the tank after the repair of leakage, [tex]h_f=13\ ft[/tex]
- diameter of the tank, [tex]d=15\ ft[/tex]
- density of turpentine oil, [tex]\rho=55\ lbm.ft^3[/tex]
a)
Now, volume of turpentine spilled:
[tex]V=A.(h_i-h_f)[/tex]
where:
[tex]A=[/tex] area of the cross section of the tank's volume
[tex]V=\pi .\frac{d^2}{4} \times(h_i-h_f)[/tex]
[tex]V=\pi\times\frac{15^2}{4} \times(17.3-13)[/tex]
[tex]V=759.8727\ ft^3[/tex]
b)
When the tank was full the liquid level was highest:
so velocity form the height of the hole will be given as:
[tex]v=\sqrt{2g.(h_i-h')}[/tex]
[tex]v=\sqrt{2\times 32.12\times (17.3-7) }[/tex]
[tex]v=25.722\ ft.s^{-1}[/tex]
Now we have the flow rate of the spillage given by:
[tex]\dot V=(\pi.\frac{d'^2}{4}) \times v[/tex]
[tex]\dot V=\pi\times \frac{(\frac{1}{120})^2 }{4} \times 25.722[/tex]
[tex]\dot V=1.403\times 10^{-3}\ ft^3.s^{-1}[/tex]
c)
Total time the leak was active can be calculated as:
[tex]t=\frac{V}{\dot V}[/tex]
[tex]t=\frac{759.8727}{25.722}[/tex]
[tex]t\approx29.541\ s[/tex]
