To calculate the work required to pump the liquid to the top of the tank, we need to follow these steps:

Step 1: Understanding the Problem

The tank is a vertical right circular cylinder with:

• Height ($h$) = 32 ft
• Diameter ($d$) = 18 ft, so the radius ($r$) = 9 ft
• The liquid's weight density ($\rho$) = 50.6 lb/ft$^3$

The liquid must be pumped from various heights $y$, where the top of the tank is $y = 32$ ft and the bottom is $y = 0$ ft.

Step 2: Volume of a Differential Slice

Consider a thin horizontal slice of the liquid at a height $y$ and thickness $dy$.

• The volume of this slice is: $dV = \pi r^2 dy = \pi (9)^2 dy = 81\pi dy$
• The weight of the slice is: $dW = \rho dV = 50.6 \times 81\pi dy$

Step 3: Work to Pump the Slice

The work required to pump this slice of liquid to the top of the tank is the weight of the slice multiplied by the distance it needs to be raised, which is $32 - y$. Thus, the work for the slice is: $d\text{Work} = 50.6 \times 81\pi (32 - y) dy$

Step 4: Total Work

To find the total work, integrate from $y = 0$ to $y = 32$: $\text{Work} = \int_0^{32} 50.6 \times 81\pi (32 - y) \, dy$ This integral can now be solved.

Step 5: Solving the Integral

We need to compute: $\text{Work} = 50.6 \times 81\pi \int_0^{32} (32 - y) \, dy$ The integral is: $\int_0^{32} (32 - y) \, dy = \left[ 32y - \frac{y^2}{2} \right]_0^{32}$ Substituting the limits: $32(32) - \frac{32^2}{2} = 1024 - 512 = 512$

Thus, the total work is: $\text{Work} = 50.6 \times 81\pi \times 512$

Step 6: Calculation

Let's compute the total work using these values.The total work required to pump the liquid to the top of the tank is approximately 6,592,579 ft-lb.

Would you like more details or have any questions about the process?

Here are 5 related questions to expand on this:

1. How would the work change if the tank's height were doubled?
2. How would the work change if the liquid had a different density?
3. What if we only wanted to pump half the liquid to the top?
4. How does the radius of the tank affect the total work?
5. What happens if we need to pump the liquid to a level above the tank's top?

Tip: The larger the height difference, the more work is needed since work is proportional to the distance each liquid element must be lifted.