To find the work required to pump water from a half-full upright cylindrical tank, we need to integrate the work needed to move each differential volume of water to the specified height.

Parameters:

• Radius of the tank $r = 4$ meters
• Height of the tank $h = 12$ meters
• Mass density of water $\rho = 1000 \, \text{kg/m}^3$
• The tank is half full, so the water's height is $6$ meters (half of 12 meters).
• The water needs to be pumped 2 meters above the top of the tank, so the total height to pump the water is $12 + 2 = 14$ meters.

Steps:

1. Volume of a thin slice of water: The work will involve considering a thin horizontal slice of water at height $y$ from the bottom of the tank. This slice has a thickness $dy$.

   The volume of this slice is: $\text{Volume} = \pi r^2 \, dy = \pi (4)^2 \, dy = 16\pi \, dy \, \text{m}^3$

2. Weight of the slice: The mass of this slice is given by: $\text{Mass} = \rho \times \text{Volume} = 1000 \times 16\pi \, dy = 16000\pi \, dy \, \text{kg}$ The weight of the slice is: $W = \text{Mass} \times g = 16000\pi \, dy \times 9.8 = 156800\pi \, dy \, \text{N}$

3. Height to pump the water: The distance this slice needs to be moved is $14 - y$, since it’s at height $y$ and needs to be pumped to 14 meters.

4. Work done for the slice: The work to move this slice is given by: $dW = W \times \text{Distance} = 156800\pi \, dy \times (14 - y)$

5. Total work: To find the total work, we integrate the work for all slices of water from $y = 0$ to $y = 6$ (since the tank is half full): $W = \int_0^6 156800\pi (14 - y) \, dy$ Let's compute this integral:

   $W = 156800\pi \int_0^6 (14 - y) \, dy$

   First, calculate the integral:

   = \left( 14(6) - \frac{6^2}{2} \right) - (0) = (84 - 18) = 66$$ Now, substitute this back into the equation for work: $$W = 156800\pi \times 66$$ Finally, calculate the total work: $$W = 10348800\pi \, \text{Joules}$$ Using $$\pi \approx 3.1416$$, we get: $$W \approx 32421252.4 \, \text{Joules}$$

Thus, the work required to pump the water from the half-full cylindrical tank is approximately 32.42 million joules.

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Would you like further details on this solution or have any questions?

Here are 5 related questions:

1. How would the work change if the tank were only a quarter full?
2. How do you calculate the work for a different shape of tank (e.g., conical)?
3. What are the effects of changing the density of the fluid (e.g., for oil or other liquids)?
4. How would the formula change if the water were pumped to a height different than 2 meters above the tank?
5. How do you compute the total volume of water in the tank?

Tip: The work to pump water is proportional to both the density of the liquid and the height it needs to be lifted.