Questions: Problem 2. An inverted conical tank is 3 m tall and 1 m in diameter at its widest point. The water is being pumped out of a spout 2 m above the top of the tank. Recall that the metric density of water is ρ=1000 kg / m^3. (a) Find the work needed to empty the tank if it is full. Include units. (b) Find the work to empty half the height of the tank (assuming it is full to begin with). Make a prediction before you calculate!

Problem 2. An inverted conical tank is 3 m tall and 1 m in diameter at its widest point. The water is being pumped out of a spout 2 m above the top of the tank. Recall that the metric density of water is ρ=1000 kg / m^3.
(a) Find the work needed to empty the tank if it is full. Include units.
(b) Find the work to empty half the height of the tank (assuming it is full to begin with). Make a prediction before you calculate!

Transcript text: Problem 2. An inverted conical tank is 3 m tall and 1 m in diameter at its widest point. The water is being pumped out of a spout 2 m above the top of the tank. Recall that the metric density of water is $\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}$. (a) Find the work needed to empty the tank if it is full. Include units. (b) Find the work to empty half the height of the tank (assuming it is full to begin with). Make a prediction before you calculate!

Solution

Solution Steps

Step 1: Understanding the Problem

We need to find the work required to pump water out of an inverted conical tank. The tank is 3 meters tall and 1 meter in diameter at its widest point. The water is being pumped out of a spout 2 meters above the top of the tank. The density of water is given as $ \rho = 1000 \, \text{kg/m}^3 $.

Step 2: Setting Up the Integral for Work

The work $ W $ required to pump the water out can be calculated using the integral: \[ W = \int_{0}^{3} \rho g A(y) (h + 2 - y) \, dy \] where:

$ \rho = 1000 \, \text{kg/m}^3 $ (density of water)
$ g = 9.8 \, \text{m/s}^2 $ (acceleration due to gravity)
$ A(y) $ is the cross-sectional area of the water at height $ y $
$ h = 3 \, \text{m} $ (height of the tank)
$ 2 \, \text{m} $ is the additional height above the tank

Step 3: Finding the Cross-Sectional Area $ A(y) $

The radius of the tank at height $ y $ is proportional to $ y $. Since the tank is 3 meters tall and 1 meter in diameter at the top, the radius $ r(y) $ at height $ y $ is: \[ r(y) = \frac{y}{3} \] The cross-sectional area $ A(y) $ is then: \[ A(y) = \pi \left( \frac{y}{3} \right)^2 = \frac{\pi y^2}{9} \]

Step 4: Setting Up the Integral with the Cross-Sectional Area

Substitute $ A(y) $ into the integral: \[ W = \int_{0}^{3} 1000 \cdot 9.8 \cdot \frac{\pi y^2}{9} \cdot (5 - y) \, dy \] Simplify the constants: \[ W = \frac{1000 \cdot 9.8 \cdot \pi}{9} \int_{0}^{3} y^2 (5 - y) \, dy \] \[ W = \frac{9800 \pi}{9} \int_{0}^{3} (5y^2 - y^3) \, dy \]

Step 5: Evaluating the Integral

Evaluate the integral: \[ \int_{0}^{3} (5y^2 - y^3) \, dy = \left[ \frac{5y^3}{3} - \frac{y^4}{4} \right]_{0}^{3} \] \[ = \left( \frac{5 \cdot 27}{3} - \frac{81}{4} \right) - \left( 0 - 0 \right) \] \[ = \left( 45 - 20.25 \right) \] \[ = 24.75 \]

Step 6: Calculating the Work

Substitute back into the expression for $ W $: \[ W = \frac{9800 \pi}{9} \cdot 24.75 \] \[ W = \frac{9800 \pi \cdot 24.75}{9} \] \[ W = 26850 \pi \, \text{Joules} \]