# Work needed to pump water out of spherical tank

• Seraph404
In summary, the problem involves finding the work required to pump water out of a spherical tank with a radius of 3m and a height of 1.5m. Using the weight density of water and the volume of each slice of the sphere, the work can be calculated by integrating over the appropriate interval and adding the work done on each layer of water.
Seraph404

## Homework Statement

A spherical tank is full of water. If radius r = 3m and height h = 1.5m, find the work W required to pump the water out of the spout. (Use 9.8 for g and 3.14 $$\pi$$)

## Homework Equations

Work is the integral from a to b of f(x)dx.

## The Attempt at a Solution

Volume= area * thickness, so I got V= pi*r2dx for any given slice of that sphere. I could be wrong, but I think the density of water is 1000kg/m3, so I multiplied that by the volume to get mass of the water, and then multiplied the mass by 9.8 to get the weight of water. Then I set up my integral from 0 to 1.5 of the answer I got from multiplying all those numbers out * xdx.
But I got the wrong answer. So how do I work a problem such as this?

Seraph404 said:

## Homework Statement

A spherical tank is full of water. If radius r = 3m and height h = 1.5m, find the work W required to pump the water out of the spout. (Use 9.8 for g and 3.14 $$\pi$$)

## Homework Equations

Work is the integral from a to b of f(x)dx.

## The Attempt at a Solution

Volume= area * thickness, so I got V= pi*r2dx for any given slice of that sphere. I could be wrong, but I think the density of water is 1000kg/m3, so I multiplied that by the volume to get mass of the water, and then multiplied the mass by 9.8 to get the weight of water. Then I set up my integral from 0 to 1.5 of the answer I got from multiplying all those numbers out * xdx.
But I got the wrong answer. So how do I work a problem such as this?
First of all I'm not sure what you mean by the 'height' of a spherical tank.

Is the situation that water is essentially being lifted vertically to the top ofa spherical tank?

Assuming that is the case, the work done is the integral (sum) of the weight of each slice of the sphere.

Draw a diagram: a circle centred on (0,0) with radius 3. Let x be the y-ordinate of a slice. Then the radius of a slice is given by sqrt(3^2 - x^2).

You can then find the volume, and hence the mass by multiplying the density, of each slice. I'll let you sort out the units.

Integrate that over the appropriate interval that x ranges over.

Show us your work if you get stuck.

Your mistake is that you have the weight of all of the water in the sphere being lifted from every height in the sphere: in other words, the same weight is being multiple times.

Analyse the problem using things that are constant- that's one reason for emphasizing "Riemann sums" as the definition of integral: that's often how we set up integrals in applications.

As Unco suggeted, draw a circle representing the sphere. Put it in "standard position" in a coordinate system so the center is at (0,0) and the radius is 3: $x^2+ y^2= 9$. Draw a horizontal line at any level in the circle, at y, representing one "layer" of water, with thickness "dy". All the water in that "layer" is at the same height, y, and so must be lifted the same distance, 3- y (notice that y may be from -3 to 3 so this distance is from 3-(-3)= 6 at the bottom of the tank to 3-3= 0 at the top).

Now, in 3 dimensions, that "layer" of water is a disk whith radius equal to the x coordinate where the line at y crosses the circle: $$\displaystyle x^2+ y^2= 9$$ so "radius squared" is $]x^2= 9- y^2$ and the area of the disk is $\pi(9- y^2)$. The volume of the thin layer is $\pi(9-y^2)$ and since weight density of water is 1000(9.81)= 9810 Newtons/l3, the weight of that layer is $9810\pi(9- y^2)dy$ and, finally, the work done lifting that weight a distance 3- y is $9810\pi(9- y^2)(3- y)dy$.

You get work required to lift all of the water to the top of the sphere by "adding" the work done on every "layer" of water in the sphere. That is, you integrate that from y= -3 to y= 3.

HallsofIvy said:
Your mistake is that you have the weight of all of the water in the sphere being lifted from every height in the sphere: in other words, the same weight is being multiple times.

Analyse the problem using things that are constant- that's one reason for emphasizing "Riemann sums" as the definition of integral: that's often how we set up integrals in applications.

As Unco suggeted, draw a circle representing the sphere. Put it in "standard position" in a coordinate system so the center is at (0,0) and the radius is 3: $x^2+ y^2= 9$. Draw a horizontal line at any level in the circle, at y, representing one "layer" of water, with thickness "dy". All the water in that "layer" is at the same height, y, and so must be lifted the same distance, 3- y (notice that y may be from -3 to 3 so this distance is from 3-(-3)= 6 at the bottom of the tank to 3-3= 0 at the top).

Now, in 3 dimensions, that "layer" of water is a disk whith radius equal to the x coordinate where the line at y crosses the circle: $$\displaystyle x^2+ y^2= 9$$ so "radius squared" is $]x^2= 9- y^2$ and the area of the disk is $\pi(9- y^2)$. The volume of the thin layer is $\pi(9-y^2)$ and since weight density of water is 1000(9.81)= 9810 Newtons/l3, the weight of that layer is $9810\pi(9- y^2)dy$ and, finally, the work done lifting that weight a distance 3- y is $9810\pi(9- y^2)(3- y)dy$.

You get work required to lift all of the water to the top of the sphere by "adding" the work done on every "layer" of water in the sphere. That is, you integrate that from y= -3 to y= 3.

But, what about the work done to get the water out of the spigget at the top? Isn't (3-y) just the distance to the top of the sphere, but before the spigget?

Last edited:

## 1. What is the formula for calculating the work needed to pump water out of a spherical tank?

The formula for calculating the work needed to pump water out of a spherical tank is W = (πh/2)(3R - h)ρg, where W is the work in joules, h is the height of the water in the tank, R is the radius of the tank, ρ is the density of the water, and g is the acceleration due to gravity.

## 2. How does the height of the water in the tank affect the amount of work needed to pump it out?

The height of the water in the tank directly affects the amount of work needed to pump it out. As the height increases, the work needed also increases. This is because the higher the water level, the more potential energy it has, thus requiring more work to overcome gravity and pump it out.

## 3. Can the work needed to pump water out of a spherical tank be reduced?

Yes, the work needed to pump water out of a spherical tank can be reduced by decreasing the height of the water in the tank or by using a pump with a higher efficiency. Additionally, using a larger tank with a smaller height can also reduce the amount of work needed.

## 4. How does the density of water affect the work needed to pump it out of a spherical tank?

The density of water does not have a significant effect on the work needed to pump it out of a spherical tank. This is because the formula for calculating work takes into account the density of water as ρ, which is a constant value. However, the mass of the water will affect the work needed, as a larger mass will require more work to pump out.

## 5. Are there any other factors that can affect the work needed to pump water out of a spherical tank?

Other factors that can affect the work needed to pump water out of a spherical tank include the efficiency of the pump being used, the height and diameter of the pipe through which the water is being pumped, and any frictional losses in the system. Additionally, temperature and pressure variations can also have a minor impact on the work needed.

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