Suppose that the tank is hemispherical, of radius R, initially full of water, and has an outlet of 5 cm² cross-sectional area at the bottom. Set up the model for outflow. Find the... Suppose that the tank is hemispherical, of radius R, initially full of water, and has an outlet of 5 cm² cross-sectional area at the bottom. Set up the model for outflow. Find the time t to empty the tank (a) for any R, (b) for R = 1 m. Plot t as a function of R. Find the time when h = R/2 (a) for any R, (b) for R = 1 m.
Understand the Problem
The question is focused on applying Torricelli's Law to a hemispherical tank, asking to find the time it takes to empty the tank under certain conditions. The user needs to set up a model for outflow, utilize previous work, and analyze two specific scenarios for the radius.
Answer
The time to empty the tank, $t$, depends on the integral derived from Torricelli's Law and the volume equation. For \( R = 1 \text{ m} \), numerical methods provide specific values.
Answer for screen readers
The time to empty the tank for radius ( R ) is derived from: $$ t = \int_{H}^{0} -\frac{0.0005 \sqrt{2gh}}{\frac{dV}{dh}} dh $$
For ( R = 1 \text{ m} ), calculate using numerical or analytical methods.
Steps to Solve
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Understanding Torricelli's Law
Torricelli's Law states that the speed of fluid flowing out of an orifice under the force of gravity is given by: $$ v = \sqrt{2gh} $$ where ( g ) is the acceleration due to gravity and ( h ) is the height of the fluid above the outlet.
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Set Up the Equation for Outflow
The flow rate through the outlet can be written as: $$ Q = A v $$ where ( A ) is the cross-sectional area of the outlet. Given ( A = 5 , \text{cm}^2 = 0.0005 , \text{m}^2 ), we have: $$ Q = 0.0005 \sqrt{2gh} $$
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Relate Flow Rate to Change in Volume
The volume ( V ) of water in a hemispherical tank can be represented as: $$ V = \frac{1}{2} \pi h^2 (3R - h) $$ The change in volume is related to the change in height by: $$ \frac{dV}{dt} = -Q $$
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Set Up Volume Differential Equation
By substituting ( Q ) into the volume equation, we have: $$ -0.0005 \sqrt{2gh} = \frac{dV}{dt} $$ and since ( dV = \frac{dV}{dh} dh ): $$ -0.0005 \sqrt{2gh} = \frac{dV}{dh} \frac{dh}{dt} $$
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Determine ( \frac{dV}{dh} ) and Rearrange
Compute ( \frac{dV}{dh} ): $$ \frac{dV}{dh} = \frac{d}{dh} \left( \frac{1}{2} \pi h^2 (3R - h) \right) $$ This gives: $$ \frac{dV}{dh} = \pi h(3R - h) - \frac{1}{2} \pi h^2 $$
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Solve the Differential Equation
Set up the differential equation with variables separated: $$ dt = -\frac{0.0005 \sqrt{2gh}}{\frac{dV}{dh}} dh $$
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Integrate to Find Time t
Integrate to solve for ( t ): $$ t = \int_{h=H}^{h=0} -\frac{0.0005 \sqrt{2gh}}{\frac{dV}{dh}} dh $$
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Calculate Specific Cases
For ( R = 1,m ), input the specific values into the integrated equation to find the time ( t ).
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Plot ( t ) as a Function of ( R )
Use computational tools or graphing to visualize ( t ) against various values of ( R ).
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Find Time When ( h = \frac{R}{2} )
Substitute ( h = \frac{R}{2} ) into the equation derived to solve for specific time values.
The time to empty the tank for radius ( R ) is derived from: $$ t = \int_{H}^{0} -\frac{0.0005 \sqrt{2gh}}{\frac{dV}{dh}} dh $$
For ( R = 1 \text{ m} ), calculate using numerical or analytical methods.
More Information
The resulting time to empty the tank depends on the tank's dimensions and the specific cross-sectional area of the outlet. The integration results can be computed using numerical methods for varying values of ( R ).
Tips
- Neglecting to account for changes in the shape of the tank as it empties can lead to inaccuracies.
- Miscalculating the area of the outlet; ensure correct unit conversions.
- Forgetting to apply limits correctly during integration.
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