Find the amount of work (in Joules) required to empty the trough by pumping the water over the top.

Question

A trough is 2 meters long, 3 meters wide, and 3 meters deep. The vertical cross-section of the trough parallel to an end is shaped like an isosceles triangle (with height 3 meters, and base, on top, of length 3 meters). The trough is full of water (density 1000kg/m³ ). Find the amount of work (in Joules) required to empty the trough by pumping the water over the top. (Note: Use g = 9.8m/s^2 as the acceleration due to gravity.)

______Joules

Answer 1

Let us first label the basic directions in the problem :

Let us call the longitudinal direction (along which lies the 'spine' of the tank) y.

We call the vertical direction z, choosing the positive direction to be downwards (so that z = 0 describes the top surface of the water, z > 0 denotes some depth z below the surface of the water),

the x-direction is the direction along which the base of the triangular cross section lies. We choose the base to lie in the interval x ∈ [- 3/2 , 3/2] , so that the deepest section of the trough has x-coordinate = 0

Now, let us find the equation of the surfaces bounding the water from below :

The top edges of these surfaces coincide with the edges of the base, at z = 0 , x = ± 3/2 ;

The bottom edge of these surfaces coincide with each other at z = 3 , x = 0.

Since the surfaces are flat, their intersection with the x-z plane is given by the lines z + 2|x| = 3

Now, we are in a position to think about the problem.

The question asks us the work done in emptying the trough by pumping from the top.

If we imagine an infinitesimal element of volume dV = dx dy dz located at depth z, the work done in bringing it to the top surface (using a pump) = dW = dm z g = (ρ dx dy dz) z g ;

where : dm is the mass of the element , ρ is the density , and g is the acceleration due to gravity.

So, in order to find the total work done, we need to find the volume integral of dW above :

W = ∫∫∫dV ρ z g.

In order to perform the integral, we want to perform the x, y, z integrations with appropriate limits.

• First, note that due to left - right symmetry (left : x < 0 and right : x > 0), the work done in lifting the left half of the water = the work done in lifting the right half. So, we only need to perforn the integration for x > 0 and double the result.

• For the y-direction, nothing interesting happens. The limits are just y = 0 to y = 2.

• For a given x, we see that the maximum depth to which elements of water are allowed to exist =

z_max = 3 - 2x (there is no water below the lower bounding surface).

• Finally, let us note that since the upper limit of the z integral depends on x, we need to perform the z-integral before the x-integral. No restrictions on the y-integral, since its limits are independent of the other directions.

Finally, putting everything together, we get :

W = 2 ρ g ∫₀² dy ∫₀^3/2 dx ∫₀^(3-2x)dz = 4ρg ∫₀^3/2 dx (3 -2 x)

Hope this helps.

Answer 2

There are several ways to do this. One way is to find the centroid of the isosceles triangle in the y coordinate. The work will be equal to mg(distance from the centroid to the top of the trough. You obtain m by multiplying density time volume of the trough.

Another technique is to lift differential horizontal slices up a height and integrate:

Integral of ρg(3-y)dV

x(y) is .5y for one side of the triangle (origin is at bottom of the trough)

dV is 2zx(y)dy = 2*2*(.5y)dy = 2ydy

The work is ρg * Integral from 0 to 3 of (3-y)(2y)dy

I'll leave the rest to you. Please consider a tutor. take care.