### Problem DescriptionA spherical tank with a radius of 3 meters is half full of a liquid that has a density of 900 kg/m³. The tank has a spout at its top that is 1 meter tall. Determine the amount of work \( W \) required to pump the liquid out of the spout.**Given:**- Radius of the spherical tank \( r = 3 \) m- Density of the liquid \( \rho = 900 \) kg/m³- Spout height \( h = 1 \) m- Gravitational acceleration \( g = 9.8 \) m/s²### Formula to Find Work \( W \)To calculate the work required to pump the liquid out, use the equation for work due to gravity:\[ W = \int_{a}^{b} (Volume \cdot Density \cdot Gravity \cdot Distance) \, dx \]### Diagram DescriptionThe image includes a spherical tank with a radius \( r \) of 3 meters. It is depicted as half full with liquid. There is a vertical spout of height \( h \) equal to 1 meter at the top of the tank. The distance \( y \) from any element of the liquid to the spout will be vital for our calculations. If we set up the coordinate system with the origin at the center of the sphere:- The tank’s bottom is at \( y = -3 \)- The tank's top is at \( y = 3 \)### Final Work CalculationYou need to find the work done to pump the liquid from the current level to a height of 1 meter above the tank, taking into account the weight of the liquid and the height it is being moved to. The results can be organized and computed with integral calculus, where every small element of liquid is considered for its contribution to the total work required:\[ W = \_ \_ \_ \, J \]Fill in the necessary integrals and calculations to complete the problem. This will give you the exact value of work \( W \) required to pump the liquid out of the spout.

R = 3 mh = 1 mρ = 900 Kg/m³g = 9.8 m/s²

A spherical tank with radius 3 m is half full of a liquid that has a density of 900 kg/m³. The tank has a 1 m spout at the top. Find the work W required to pump the liquid out spout. (Use 9.8 m/s² for g.) W = h

A spherical tank with radius 3 m is half full of a liquid that has a density of 900 kg/m³. The tank has a 1 m spout at the top. Find the work W required to pump the liquid out spout. (Use 9.8 m/s² for g.) W = h

Elements Of Electromagnetics

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ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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Concept explainers

Applied Fluid Mechanics

A fluid is a substance that is continuously deformed by the application of shear stress. The rate of deformation of a fluid depends on its viscosity. The fluid tries to flow when it interacts with some other system.

Question

### Problem Description

A spherical tank with a radius of 3 meters is half full of a liquid that has a density of 900 kg/m³. The tank has a spout at its top that is 1 meter tall. Determine the amount of work \( W \) required to pump the liquid out of the spout.

**Given:**
- Radius of the spherical tank \( r = 3 \) m
- Density of the liquid \( \rho = 900 \) kg/m³
- Spout height \( h = 1 \) m
- Gravitational acceleration \( g = 9.8 \) m/s²

### Formula to Find Work \( W \)

To calculate the work required to pump the liquid out, use the equation for work due to gravity:

\[ W = \int_{a}^{b} (Volume \cdot Density \cdot Gravity \cdot Distance) \, dx \]

### Diagram Description

The image includes a spherical tank with a radius \( r \) of 3 meters. It is depicted as half full with liquid. There is a vertical spout of height \( h \) equal to 1 meter at the top of the tank.

The distance \( y \) from any element of the liquid to the spout will be vital for our calculations. If we set up the coordinate system with the origin at the center of the sphere:
- The tank’s bottom is at \( y = -3 \)
- The tank's top is at \( y = 3 \)

### Final Work Calculation

You need to find the work done to pump the liquid from the current level to a height of 1 meter above the tank, taking into account the weight of the liquid and the height it is being moved to. The results can be organized and computed with integral calculus, where every small element of liquid is considered for its contribution to the total work required:

\[ W = \_ \_ \_ \, J \]

Fill in the necessary integrals and calculations to complete the problem. This will give you the exact value of work \( W \) required to pump the liquid out of the spout.

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