A conical tank has a diameter of 9ft and is 12 ft deep. If the tank is filled with water of density 62.4 ft/cubic ft, how much work is required to pump the water over the top?
Which end of the cone is on top?
Multiply the mass of water in the cone by the depth of the center of mass, and by g.
To find out how much work is required to pump the water over the top of the conical tank, we will follow these steps:
Step 1: Calculate the volume of the conical tank.
Step 2: Determine the mass of the water in the tank.
Step 3: Convert the mass of the water to weight.
Step 4: Calculate the work required to lift the water over the top.
Now, let's calculate the work step by step:
Step 1: Calculate the volume of the conical tank.
The volume of a cone can be calculated using the formula: V = (1/3) * π * r^2 * h
Given that the diameter of the tank is 9 ft, the radius (r) can be calculated as r = diameter / 2 = 9 ft / 2 = 4.5 ft.
The depth of the tank (h) is given as 12 ft.
Substituting the values in the formula, the volume can be calculated as:
V = (1/3) * π * (4.5 ft)^2 * 12 ft
V ≈ 254.47 cubic ft
Step 2: Determine the mass of the water in the tank.
The density of water is given as 62.4 lb/cubic ft.
Since the density is given in lb/cubic ft, we will convert the volume from cubic ft to cubic inch (1 cubic ft = 1728 cubic inches).
Mass = Density * Volume
Mass = 62.4 lb/cubic ft * 254.47 cubic ft * 1728 cubic inches/cubic ft
Mass ≈ 2,066,631.95 lb
Step 3: Convert the mass of the water to weight.
The weight of an object is calculated using the formula: W = m * g
where W is the weight, m is the mass, and g is the acceleration due to gravity.
The acceleration due to gravity is approximately 32.2 ft/s^2.
W = 2,066,631.95 lb * 32.2 ft/s^2
W ≈ 66,562,035.19 ft.lb
Step 4: Calculate the work required to lift the water over the top.
The work required to lift the water over the top is equivalent to the weight of the water multiplied by the height it needs to be lifted.
The height it needs to be lifted is the depth of the tank, which is given as 12 ft.
Work = W * h
Work = 66,562,035.19 ft.lb * 12 ft
Work ≈ 798,744,422.3 ft.lb
Therefore, approximately 798,744,422.3 ft.lb of work is required to pump the water over the top of the conical tank.
To calculate the work required to pump the water over the top of the tank, we need to determine the volume of water in the tank.
The volume of a conical tank can be calculated using the formula V = (1/3)πr^2h, where r is the radius of the tank's base and h is the height of the tank.
Given that the diameter of the tank is 9 ft, the radius (r) would be half of the diameter, which is 4.5 ft. And the height (h) of the tank is given as 12 ft.
Now we can calculate the volume of water in the tank:
V = (1/3)π(4.5^2)(12)
V ≈ 254.47 cubic ft
Next, we need to calculate the weight of the water in the tank. The weight (W) of an object can be calculated using the formula W = m * g, where m is the mass of the object and g is the acceleration due to gravity.
In this case, mass (m) can be calculated by multiplying the volume (V) of water by the density (ρ):
m = V * ρ
m = 254.47 * 62.4
m ≈ 15,872.03 lb.
Lastly, we need to calculate the work (W) required to pump the water over the top of the tank. The work done in pumping liquid against gravity is given by the formula W = m * g * h, where h is the height through which the water is lifted.
W = m * g * h
W = 15,872.03 * 32.17 * 12
W ≈ 6,129,568.79 ft-lbs
Therefore, approximately 6,129,569 ft-lbs of work is required to pump the water over the top of the conical tank.