The diagram depicts a surge tank with clear sides for illustration purposes, showing how it is used to manage variable flows. During high flow periods, excess water is directed to the surge tank and flows out more slowly over a weir.### Diagram Details:- **Dimensions of the Tank**: - Height: 7.5 m - Width: 2.0 m - Length: 2.0 m - **Weir Details**: - Width of the weir: 1.0 m - Height from base to the weir opening: 6.0 m - Water flows out through an opening above the base by height \( h \).### Volumetric Flow Over the Weir:The volumetric flow (\( \dot{V} \)) over the weir is described by the equation:\[\dot{V} = 0.011 \times b \times g^{1/2} \times h^{3/2}\]Where:- \( b \) = weir width (m)- \( g \) = gravitational acceleration (m/s\(^2\))- \( h \) = height of water above the weir edge (m)### Problem Statement:Assuming no excess flow is currently being diverted to the surge tank, determine the time required for the water level in the tank to decrease from an initial height of 7.5 m to 6.25 m.

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The surge tank pictured (shown with clear sides for illustration purposes) is used to even out variable flows. During periods of high flow, excess water is diverted to the surge tank where it flows out more slowly over the weir. The volumetric flow over the weir is V = 0.011 * b*g¹/2h³/2 where b = weir width (m) 2 g = gravitational acceleration (m/s) h=height of water above the weir edge (m) Assuming no excess flow is currently being diverted to the surge tank, determine the time required for the water level in the tank to become 6.25 m if the initial height is 7.5 m

The surge tank pictured (shown with clear sides for illustration purposes) is used to even out variable flows. During periods of high flow, excess water is diverted to the surge tank where it flows out more slowly over the weir. The volumetric flow over the weir is V = 0.011 * b*g¹/2h³/2 where b = weir width (m) 2 g = gravitational acceleration (m/s) h=height of water above the weir edge (m) Assuming no excess flow is currently being diverted to the surge tank, determine the time required for the water level in the tank to become 6.25 m if the initial height is 7.5 m

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

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Concept explainers

Fluid Dynamics

The study of the flow of gases and liquids, usually in and around solid surfaces, is known as fluid dynamics in physics and engineering. Fluid dynamics, for example, can be used to evaluate the movement of air through an airplane wing or the surface of a vehicle. It can also be used in ship design to increase the speed at which ships pass through water.

Question

The diagram depicts a surge tank with clear sides for illustration purposes, showing how it is used to manage variable flows. During high flow periods, excess water is directed to the surge tank and flows out more slowly over a weir.

### Diagram Details:
- **Dimensions of the Tank**:
- Height: 7.5 m
- Width: 2.0 m
- Length: 2.0 m
- **Weir Details**:
- Width of the weir: 1.0 m
- Height from base to the weir opening: 6.0 m
- Water flows out through an opening above the base by height \( h \).

### Volumetric Flow Over the Weir:

The volumetric flow (\( \dot{V} \)) over the weir is described by the equation:

\[
\dot{V} = 0.011 \times b \times g^{1/2} \times h^{3/2}
\]

Where:
- \( b \) = weir width (m)
- \( g \) = gravitational acceleration (m/s\(^2\))
- \( h \) = height of water above the weir edge (m)

### Problem Statement:

Assuming no excess flow is currently being diverted to the surge tank, determine the time required for the water level in the tank to decrease from an initial height of 7.5 m to 6.25 m.

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