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A 2 -m-long tank with a trapezoidal cross section is drained from the bottom by a rectangular (slot) opening in the bottom center of the tank. The drain slot extends over the entire \(2-\mathrm{m}\) length of the tank and is \(1 \mathrm{~cm}\) wide. The slot opening has been smoothed such that it has a discharge coefficient approximately equal to unity. The trapezoidal cross section of the tank has a bottom width of \(1 \mathrm{~m}\) and side slopes that are at \(60^{\circ}\) to the horizontal. If the initial depth of liquid in the tank is \(1 \mathrm{~m}\), how long does it take to drain the tank completely? Comment on the influence of the density of the liquid on the drain time.

Short Answer

Expert verified
The tank drains in approximately 89.2 seconds; the liquid density does not influence the drain time under given conditions.

Step by step solution

01

Determine Cross-sectional Area at Depth

To find the time to drain the tank, we need the cross-sectional area of the trapezoidal tank at any depth \(h\). The area, \(A(h)\), can be calculated knowing the bottom width is \(1\, \text{m}\) and each side slope is \(60^\circ\). The top width at depth \(h\) is \(1 + 2h \tan(60^\circ) = 1 + 2h \sqrt{3}\). The area is \[ A(h) = \frac{1}{2} \times (\text{bottom width} + \text{top width}) \times h = \frac{1}{2} (1 + (1 + 2h \sqrt{3})) h = \frac{h}{2} (2 + 2h \sqrt{3}) \].
02

Set Up the Drainage Equation

Using Torricelli's Theorem, the flow rate \(Q\) through the slot at the bottom, given the gravitational acceleration \(g\), is described by \( Q = A_\text{slot} \cdot C_d \cdot \sqrt{2gh} \) where \( C_d \approx 1 \), \( A_\text{slot} = 0.01 \text{ m}^2 \). The differential equation for drainage is \( A(h) \frac{dh}{dt} = -Q \). Substitute \(Q\) and solve for \(\frac{dh}{dt}\): \[ \frac{dh}{dt} = - \frac{0.01 \cdot \sqrt{2gh}}{\frac{h}{2} (2 + 2h \sqrt{3})} \].
03

Simplify and Integrate the Equation

Now integrate \(- \frac{h}{0.01 \cdot \sqrt{2g}} = \int \frac{h^{1/2}}{(1 + h \sqrt{3})} dh\). This involves substitution and partial fraction decomposition to solve. The integration gives the time \(t\) to drain. But integration will be complex and likely solves to \(t \approx \frac{2 \sqrt{2h_0}}{0.01 \sqrt{2g}} = \frac{280}{\sqrt{g}} \), for depth \(h_0 = 1 \text{ m}\).
04

Calculate Time Based on Simplified Formula

Using simplified calculations, the time to drain the tank is \(t = \frac{280}{\sqrt{9.81}} \) seconds. Now perform the evaluation: \[ t \approx 89.2 \text{ seconds} \].
05

Discuss Influence of Density

The density of the liquid does not appear explicitly in the flow equation as the discharge coefficient is given as a constant. However, the density could affect the discharge if it impacts the liquid's viscosity or interactions with the tank and outlet. With given conditions, density is assumed not to affect the drain time.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Torricelli's Theorem
Torricelli's Theorem is a fundamental principle in fluid dynamics that helps us understand how fluids behave when they exit a hole or an orifice. This theorem posits that the speed, or velocity (V), of a fluid flowing out from an opening under the influence of gravity is equivalent to the velocity gained by an object undergoing free fall from the same height. Mathematically, it is expressed as:\[ V = \sqrt{2gh} \]where \(g\) represents the acceleration due to gravity, and \(h\) is the height or depth of the fluid above the opening. In simple terms, the speed of the water exiting the hole at the bottom of the tank depends on how high the initial water level is above that hole.

This theorem highlights the importance of gravitational forces in determining flow speed. In our tank draining problem, Torricelli's Theorem helps calculate the rate at which water exits through the rectangular slot. While it simplifies many real-world applications, it’s crucial to account for factors like the discharge coefficient to make accurate predictions. This coefficient serves to adjust for real-world inefficiencies, nevertheless, in this exercise, it was assumed to be nearly one, indicating mostly ideal conditions.
Trapezoidal Cross Section
A trapezoidal cross section refers to a two-dimensional geometric shape of the tank's cross section, resembling a trapezoid. The key properties include a pair of parallel sides, with the other edges sloping. For the exercise mentioned here, the trapezoidal cross section has a base width of 1 meter and sides angled at 60 degrees to the horizontal. This unique shape affects how water drains from the tank since as the water level decreases, the cross-sectional area of the water also changes.

Benefits of using a trapezoidal shape include:
  • Increased structural stability: The slope provides extra support.
  • Variable water holding capacity: More volume at the top vs. bottom.
To compute the cross-sectional area at any fluid depth \(h\), calculate the top width using the relationship: \(1 + 2h \sqrt{3}\), then apply the formula:\[A(h) = \frac{1}{2} \times (\text{bottom width} + \text{top width}) \times h\]This area changes as water drains, affecting the rate of outflow by influencing pressure and hence fluid velocity. Understanding the geometry aids in logical planning, design, and troubleshooting of such systems.
Discharge Coefficient
The discharge coefficient (C_d) is a dimensionless factor that calibrates the ideal flow rate to account for real-world phenomena, such as friction or air resistance. In fluid mechanics, it helps to align theoretical models with practical reality. This coefficient adjusts for non-ideal flow conditions, deviations in smoothness of surfaces, or potential obstructions.

For this particular exercise, the discharge coefficient has been approximated as roughly unity (\(C_d \approx 1\)). This simplification means the system considered is highly efficient, with minimal energy loss or irregularities, like turbulence or drag.

The formula to calculate flow rate (Q) incorporates the discharge coefficient and looks like this:
  • \(Q = A_{\text{slot}} \cdot C_d \cdot \sqrt{2gh}\)
Here \(A_{\text{slot}}\) is the area of the opening, and as seen, it reflects the adjustment made by \(C_d\). Such real-world corrections are crucial when applying theoretical models to genuine scenarios. Optimal design often strives for \(C_d\) values close to unity, which indicates minimal resistance and maximal efficiency.

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