A 0.8 m diameter tank is filled with water to a depth of 2.2 m and is open to the atmosphere at the top. The water drains through a 1.0 cm diameter pipe at the bottom; that pipe then joins a 1.5 cm diameter pipe open to the atmosphere, as shown in Fig. 18-53.

|0.8 m | ||
| | ||<------vacuum tube for b
| | ||
| | ||
| | || ---------
| ============---------
Tank^ ^pipe1 ^pipe2
(1cm) (1.5cm)

(not to scale)

(a) Find the flow speed in the narrow section.

(b) What is the water height in the sealed vertical tube shown?

Hmm didn't keep the spaces darn ;\, basically the tank was on the left, 1 cm pipe stretches for a bit out of the bottom right of the tank, midway through that ascending is a similar diameter vacuum tube for part b, and then to the right of the 1 cm pipe, it basically turns into a 1.5 cm pipe as described in the problem.

To solve this problem, we can make use of Bernoulli's equation, which states that the sum of the pressure, kinetic energy per unit volume, and potential energy per unit volume is constant along a streamline in a fluid flow. Let's break down the steps to find the answers to both parts of the question:

(a) Finding the flow speed in the narrow section:

1. We can use Bernoulli's equation between two points: the surface of the water in the tank and the narrow section of the pipe where the diameter is 1.0 cm. Let's call these points 1 and 2, respectively.

2. At the surface of the water (point 1), the pressure is atmospheric pressure, and the speed and height are zero since the water is at rest.

3. At the narrow section of the pipe (point 2), the pressure is still atmospheric pressure, but the speed and height are unknown.

4. Bernoulli's equation can be written as follows:

P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2

where P is the pressure, ρ is the density of water, v is the speed of water, g is the acceleration due to gravity, and h is the height of the water.

5. Since the height and atmospheric pressure are the same at both points, we can simplify the equation to:

1/2ρv1^2 = 1/2ρv2^2

6. Rearranging the equation, we get:

v2 = v1 * √(D1^2 / D2^2)

where v1 is the speed of the water at the surface and D1 and D2 are the diameters of the respective sections.

7. The diameter of the narrow section is 1.0 cm, which corresponds to a radius of 0.5 cm or 0.005 m. The diameter of the tank is 0.8 m.

v2 = v1 * √(0.005^2 / 0.8^2)

Calculate this expression to find the flow speed in the narrow section.

(b) Finding the water height in the sealed vertical tube:

1. At the bottom of the narrow section, the pressure is atmospheric pressure, and the speed is the value we obtained in part (a).

2. At the top of the sealed vertical tube, the speed is zero since the water is not flowing.

3. Using Bernoulli's equation again between these two points:

P2 + 1/2ρv2^2 + ρgh2 = P3 + 1/2ρv3^2 + ρgh3

where P3 is also atmospheric pressure, and v3 and h3 are unknown.

4. As before, since the pressure and speed are the same at points 2 and 3, we can simplify the equation to:

h2 + 1/2(v2^2 / g) = h3

5. Rearranging the equation, we get:

h3 = h2 + 0.5(v2^2 / g)

where h2 is the height of the water at the bottom of the narrow section and g is the acceleration due to gravity.

6. The height h2 can be calculated by subtracting the depth of the water in the tank from the total height.

h2 = total height - depth of water

7. Substitute the values of h2, v2, and g into the equation to find the water height in the sealed vertical tube.

By following these steps, you should be able to find the answers to both parts (a) and (b) of the question.