Final answer:
To determine the gauge pressure required for water to emerge from the small end of a tapered pipe with a speed of 12 m/s, Bernoulli's equation can be used, taking into account the elevation difference. The gauge pressure, which is the difference between the pressure at the large end and the pressure at the small end, can be calculated using the given dimensions and the equation (1/2)(ρ)(v₂)² - (1/2)(ρ)(v₁)².
Step-by-step explanation:
The gauge pressure is the difference between the pressure at the large end and the pressure at the small end of the pipe. To determine the gauge pressure required for water to emerge from the small end with a speed of 12 m/s, we need to consider Bernoulli's equation and the given elevation difference.
Bernoulli's equation states that the sum of the pressure, the kinetic energy per unit volume, and the potential energy per unit volume is constant along a streamline. Since the pipe is tapered, the velocities at the large and small ends are different. We can equate the pressure at the large end, p₁, to the pressure at the small end, p₂, plus the difference in kinetic energy and potential energy.
Using the given information, we have that the velocity at the small end, v₂, is 12 m/s and the elevation difference, h, is 8 m. The dimensions of the pipe are also given: the outer diameter is 2a and the inner diameter is a. We can use these dimensions to find the cross-sectional areas at the large end, A₁, and the small end, A₂.
Using the relationship A = πr², where r is the radius, we find that the radius at the large end is 2a/2 = a and the radius at the small end is a/2. Therefore, the cross-sectional area at the large end is A₁ = π(a)² and the cross-sectional area at the small end is A₂ = π(a/2)².
Now we can set up the equation for Bernoulli's equation. At the large end, we have p₁ + (1/2)ρv₁² + ρgh = p₂ + (1/2)ρv₂² + ρgh. Since the pipe is open to the atmosphere, the pressure at both ends is equal to atmospheric pressure, which we can assume is 0. Therefore, the equation simplifies to (1/2)ρv₁² + ρgh = (1/2)ρv₂².
Substituting the given values, we have (1/2)(ρ)(v₁)² + (ρ)(g)(h) = (1/2)(ρ)(v₂)². Rearranging the equation, we can solve for the gauge pressure, which is the difference between the pressure at the large end and the pressure at the small end. Therefore, the gauge pressure is given by p₂ - p₁ = (1/2)(ρ)(v₂)² - (1/2)(ρ)(v₁)².