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The tank shown drains through a triangular weir whose flowrate is described by the equation given. Determine the time it will take for the tank to drain from completely full (h= 5 ft) to 5.25 feet deep (h = 0.25 ft) assuming there is no flow into the tank. Volumetric Flowrate (²³) = S 5 ft ft3 h=5 ft = 10 ft 2.48 h2.5 (for h in feet) -20 ft- -20 ft-

The tank shown drains through a triangular weir whose flowrate is described by the equation given. Determine the time it will take for the tank to drain from completely full (h= 5 ft) to 5.25 feet deep (h = 0.25 ft) assuming there is no flow into the tank. Volumetric Flowrate (²³) = S 5 ft ft3 h=5 ft = 10 ft 2.48 h2.5 (for h in feet) -20 ft- -20 ft-

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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The tank shown drains through a triangular weir whose flowrate is described by the equation given. Determine the time it will take for the tank to drain from completely full (h = 5 ft) to 5.25 feet deep (h = 0.25 ft) assuming there is no flow into the tank. **Volumetric Flowrate** \(\left(\frac{\text{ft}^3}{\text{s}}\right)\) = \(2.48 \, h^{2.5}\) (for h in feet) The diagram consists of two illustrations: 1. **View of the Weir**: It shows a cross-section of the triangular weir with an angle \(\theta\) at the top, indicating the opening where the water exits the tank. The initial depth of the water (h) is marked as 5 feet. 2. **3D Representation of the Tank**: This is an isometric view showing the tank's dimensions. The tank is a rectangular prism, 20 feet in length, 20 feet in width, and 10 feet in height. Water exits through the triangular weir on one side of the tank.
Transcribed Image Text:The tank shown drains through a triangular weir whose flowrate is described by the equation given. Determine the time it will take for the tank to drain from completely full (h = 5 ft) to 5.25 feet deep (h = 0.25 ft) assuming there is no flow into the tank. **Volumetric Flowrate** \(\left(\frac{\text{ft}^3}{\text{s}}\right)\) = \(2.48 \, h^{2.5}\) (for h in feet) The diagram consists of two illustrations: 1. **View of the Weir**: It shows a cross-section of the triangular weir with an angle \(\theta\) at the top, indicating the opening where the water exits the tank. The initial depth of the water (h) is marked as 5 feet. 2. **3D Representation of the Tank**: This is an isometric view showing the tank's dimensions. The tank is a rectangular prism, 20 feet in length, 20 feet in width, and 10 feet in height. Water exits through the triangular weir on one side of the tank.
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