The hemispherical tank shown is full of water. Given that water weighs 62.5 lb/ft3, find the work required to pump the water out of the tank.
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What is shown is just the tank (a hemisphere) with a radius of 5 ft.
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First I calculated the Volume of the hemisphere, V = (2/3)*pi*r^3
V = (2/3)*pi*125 = (250/3)*pi
Then I took the integral of: Volume*5y*dy from 0 to 5.
Which equals: ((250/3)*pi)*(5/2)y^2 evaluated at 5 and 0.
I came up with 16362.5 ft*lb.
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Am I using the wrong method?
The work required depends upon where the water is extracted. I assume you are pumping out the top.
I don't see why you claim that the energy is
Volume*5y*dy from 0 to 5
The work required is the weight of each differential slab of height dy, multiplied by the distance it must be lifted, 5 - y, integrated from 0 to 5. The area of each slab is different. It depends upon y.
The area of the slab depends upon y; the area at height y from the base is given by the product of the square of the radius of the slice and pi, i.e., A(y) = pi*r^2. However, we need to determine r as a function of y so that the area can be expressed as a function of y. To do this, we can use similar triangles.
To determine r(y), consider a cross-sectional view of the hemisphere. This cross-section is a semicircle of radius 5. Vertical lines at heights y and 0 are similar. The radius of the slice at height y can be called r. Therefore, we have the proportion:
(5 - y) / r = 5 / 5
which simplifies to r = 5 - y.
Now we can write the area A(y) as:
A(y) = pi*(5-y)^2.
Next, we find the volume of the differential slab of height dy as:
dV = A(y)*dy.
Now, to find the work required to pump the slab of water out of the tank, we need to multiply its weight by the distance that it needs to be lifted. The weight of the slab is given by:
dW = dV * 62.5
and the distance it needs to be lifted is (5 - y).
Thus, the total work required to pump out that slab is:
dU = dW * (5 - y) = 62.5 * A(y) * dy * (5 - y).
To find the total work required to pump out all the water, we integrate dU over the interval [0, 5]:
U = integral(62.5 * pi * (5 - y)^2 * dy * (5 - y)) over [0, 5].
U = 62.5 * pi * integral((5 - y)^3 * dy) over [0, 5].
Now, we integrate and find the value of the integral:
U = 62.5 * pi * [(-1/4)*(5 - y)^4] evaluated from 0 to 5.
U = 62.5 * pi * ([-(1/4)*(0)^4] - [(-1/4)*(5)^4])
U = 62.5 * pi * (-(0) + 625/4)
U = 62.5 * 625 * (pi / 2)
U = 19531.25 * pi ft*lb.
Therefore, the work required to pump the water out of the tank is 19531.25 * pi ft*lb.
To find the work required to pump the water out of the tank, we can use the concept of differential work.
First, let's determine the weight of a differential slab of water. The weight of the water can be calculated as the product of its volume and the density of water. The volume of a differential slab can be approximated as the area of the cross-section multiplied by its thickness.
Since we are dealing with a hemispherical tank with a radius of 5 ft, the cross-section of the tank at a certain height y can be represented by a circle with radius R, where R is given by the equation R = √(5^2 - y^2).
The area of the cross-section can be calculated as A = πR^2.
The volume of the differential slab can be approximated as dV = A * dy.
Now, we can calculate the weight of the water in the differential slab as W = dV * density = A * dy * density.
To find the work required, we need to multiply the weight of the water by the distance it needs to be lifted, which is given by the difference between the height of the slab and the top of the tank, 5 - y.
The work done by pumping the water out can be expressed as dW = (W * (5 - y)).
Finally, to find the total work required, we need to integrate the expression for dW from y = 0 to y = 5. This integration will sum up the work done by each differential slab over the entire height of the tank.
∫(0 to 5) [(A * dy * density) * (5 - y)]
Now we substitute the values of A and dy:
∫(0 to 5) [(π * (√(5^2 - y^2))^2 * dy * density) * (5 - y)]
Simplifying further:
∫(0 to 5) [(π * (25 - y^2) * dy * density) * (5 - y)]
Now we can perform the integration and substitute the value of the density, which is given as 62.5 lb/ft³:
∫(0 to 5) [(π * (25 - y^2) * dy * 62.5) * (5 - y)]
Evaluating this integral will give us the work required to pump the water out of the tank.
To find the work required to pump the water out of the tank, you need to calculate the weight of each differential slab of water and the distance it needs to be lifted.
The weight of each differential slab of water can be found by multiplying the density of water, which is 62.5 lb/ft^3, by the volume of each slab.
The volume of each slab can be calculated by finding the surface area of each differential ring and multiplying it by the thickness of the slab (dy).
The surface area of each differential ring can be found by multiplying the circumference of the hemisphere at a certain height (2πr) by the thickness of the slab (dy).
Using this information, the work required can be expressed as the integral of the weight of each slab (62.5 lb/ft^3 * volume of each slab) multiplied by the distance it needs to be lifted (5 - y) from 0 to 5:
∫(62.5 lb/ft^3 * volume of each slab * (5 - y)) dy from 0 to 5.
To calculate the volume of each slab, you will need to calculate the surface area of each differential ring and multiply it by the thickness of the slab (dy):
Surface area of each differential ring = 2πr * dy.
Therefore, the volume of each slab = (2πr * dy) * dy.
Substituting this into the integral:
∫(62.5 lb/ft^3 * (2πr * dy) * dy * (5 - y)) dy from 0 to 5.
This integral will give you the work required to pump the water out of the tank.