Problem 72 - University Physics with Modern Physics

Question

A closed and elevated vertical cylindrical tank with diameter 2.00 m contains water to a depth of 0.800 m. A worker accidently pokes a circular hole with diameter 0.0200 m in the bottom of the tank. As the water drains from the tank, compressed air above the water in the tank maintains a gauge pressure of \(5.00 \times 10^{3}\) Pa at the surface of the water. Ignore any effects of viscosity. (a) Just after the hole is made, what is the speed of the water as it emerges from the hole? What is the ratio of this speed to the efflux speed if the top of the tank is open to the air? (b) How much time does it take for all the water to drain from the tank? What is the ratio of this time to the time it takes for the tank to drain if the top of the tank is open to the air?

Step-by-Step Solution

Answer

(a) 5.069 m/s, speed ratio is 1.28. (b) 1578 s, time ratio is 0.78.

Step 1: Identifying Given Values

First, gather the given values: the tank diameter is 2.00 m which implies a radius of 1.00 m, the hole's diameter is 0.0200 m implying a radius of 0.0100 m, the depth of water is 0.800 m, and the gauge pressure is \(5.00 \times 10^{3}\, \text{Pa}\). Note that the density of water \( \rho \) is \(1000\, \text{kg/m}^3\) and the gravitational acceleration \( g \) is approximately \(9.81\, \text{m/s}^2\).

Step 2: Applying Torricelli’s Theorem with Gauge Pressure

The speed \( v \) of the water emerging just after the hole is made is determined by applying Torricelli's theorem, modified for additional pressure: \( v = \sqrt{2g h + \frac{2P}{\rho}} \), where \( h = 0.800\, \text{m} \) and \( P = 5.00 \times 10^{3} \, \text{Pa}\). Substituting these values, we have \( v = \sqrt{2(9.81)(0.800) + \frac{(2)(5.00 \times 10^{3})}{1000}} \).

Step 3: Calculating Efflux Speed

Substituting into the formula, \( v = \sqrt{2 \times 9.81 \times 0.800 + \frac{2 \times 5.00 \times 10^{3}}{1000}} \) which simplifies to \( v = \sqrt{15.696 + 10} = \sqrt{25.696} \approx 5.069 \text{ m/s}\).

Step 4: Determining Efflux Speed with Open Tank

If the top of the tank is open to the air, the speed of efflux is given by \( v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.800} \). Substituting, we get \( v = \sqrt{15.696} = 3.96 \text{ m/s}\).

Step 5: Calculating Speed Ratio

The ratio of the speed with the gauge pressure to that with an open tank is \( \frac{5.069}{3.96} \approx 1.28\).

Step 6: Finding Drainage Time with Pressure

Use the flow rate \( Q = A_{hole}v \), where \( A_{hole} = \pi \times (0.01)^2 = 3.14 \times 10^{-4} \text{ m}^2\) is the area of the hole and \( v = 5.069 \text{ m/s}\). \( Q = 3.14 \times 10^{-4} \times 5.069 = 1.592 \times 10^{-3} \text{ m}^3/\text{s}\).

Step 7: Calculating Volume and Time Required with Pressure

The initial water volume \( V_0 = \pi R_{tank}^2 \times h = \pi \times (1.00)^2 \times 0.800 = 2.513 \text{ m}^3\). The time \( t = \frac{V_0}{Q} = \frac{2.513}{1.592 \times 10^{-3}} \approx 1578\, \text{s} \).

Step 8: Determining Time to Drain with Open Tank

Similarly, when the tank is open, the flow rate \( Q = 3.14 \times 10^{-4} \times 3.96 \approx 1.245 \times 10^{-3}\text{ m}^3/\text{s}\). Hence, time required \( t_{open} = \frac{2.513}{1.245 \times 10^{-3}} \approx 2018\, \text{s} \).

Step 9: Calculating Time Ratio

The time ratio is \( \frac{1578}{2018} \approx 0.78\).

Key Concepts

Torricelli’s theorem

Torricelli's theorem is a principle from fluid dynamics that is used to calculate the speed of a fluid flowing out of an orifice or hole in a tank. The theorem is derived from Bernoulli's principle. It relates the speed of efflux (outflow) to the height from which the fluid falls, assuming no external pressure and frictionless flow.
However, when gauge pressure is present, adjustments need to be made. The generalized formula becomes: \[ v = \sqrt{2g h + \frac{2P}{\rho}} \]- **Without additional pressure:** The speed is computed by \( v = \sqrt{2gh} \), where \( g \) is the acceleration due to gravity \( 9.81\, \text{m/s}^2 \) and \( h \) is the height of the water column.- **With additional pressure:** Gauge pressure also contributes to the speed, and is factored in through the term \( \frac{2P}{\rho} \), where \( P \) is the gauge pressure, and \( \rho \) is the fluid density.
This adjustment means that in conditions with elevated gauge pressure, the speed of efflux will be higher compared to an open-to-atmosphere scenario. This is crucial for calculating dynamic conditions of fluid systems.

Gauge pressure

Gauge pressure is the pressure relative to atmospheric pressure. It is different from absolute pressure, which is what you would measure with a vacuum as the reference point. In practical terms, gauge pressure is the pressure "above" atmospheric pressure.
In this exercise, gauge pressure compensates for the additional force exerted on the fluid by a compressed gas over the water surface. Some important points to note are:

• **Role in fluid flow:** Higher gauge pressure increases the efflux speed of a fluid through an opening, as it adds extra force beyond buoyancy and gravity alone.
• **Measurement:** It is often measured in Pascals (Pa), a unit of pressure in the International System of Units (SI).
• **Equation application:** It contributes an extra term in Torricelli’s theorem to calculate the speed of efflux.

Gauge pressure is crucial for understanding how fluids behave in systems with added pressures, such as tanks with compressed air, which this exercise perfectly illustrates.

Efflux speed

Efflux speed refers to the speed at which fluid exits through an opening. It is determined by the forces acting on the fluid, including gravity, external pressure (if present), and the height difference in fluid levels.
In the context of the exercise, efflux speed was calculated initially under the influence of gauge pressure and then compared to what it would be without such pressure:

• **With gauge pressure:** The efflux speed was calculated using Torricelli’s modified theorem: \( v = \sqrt{2gh + \frac{2P}{\rho}} \).
• **Without gauge pressure:** The calculation simplifies to \( v = \sqrt{2gh} \), as only height influences the speed when the tank is open and exposed to the atmosphere.
• **Impact of pressure:** The result shows that efflux speed is significantly increased by the presence of gauge pressure further driving out water.

Understanding efflux speed is critical for designing systems that control the release of fluids, which involve the engineering and safety considerations of maintaining certain flow rates.