Old Faithful Geyser in Yellowstone National Park erupts at...

Old Faithful Geyser in Yellowstone National Park erupts at approximately one-hour intervals, and the height of the water $\begin{array}{llll}\text { column } & \text { reaches } & 40.0 & \mathrm{~m}\end{array}$ (Fig. $\mathrm{P} 14.25)$ (a) Model the rising stream as a series of separate droplets. Analyze the free-fall motion of one of the droplets to determine the speed at which the water leaves the ground. (b) What If? Model the rising stream as an ideal fluid in streamline flow. Use Bernoulli's equation to determine the speed of the water as it leaves ground level. (c) How does the answer from part (a) compare with the answer from part (b)? (d) What is the pressure (above atmospheric) in the heated underground chamber if its depth is $175 \mathrm{~m}$ ? Assume the chamber is large compared with the geyser's vent.

Old Faithful Geyser in Yellowstone National
Park erupts at approximately one-hour intervals, and the height of the water $\begin{array}{llll}\text { column } & \text { reaches } & 40.0 & \mathrm{~m}\end{array}$
(Fig. $\mathrm{P} 14.25)$
(a) Model the rising stream as a series of separate droplets. Analyze the free-fall motion of one
of the droplets to determine the speed at which the water leaves the ground. (b) What If? Model the rising stream as an ideal fluid in streamline flow. Use Bernoulli's equation to determine the speed of the water as it leaves ground level.
(c) How does the answer from part (a) compare with the answer from part
(b)? (d) What is the pressure (above atmospheric) in the heated underground chamber if its depth is $175 \mathrm{~m}$ ? Assume the chamber is large compared with the geyser's vent.

Key Concepts

Projectile Motion in Free Fall

This concept involves analyzing the motion of an object moving under the influence of gravity alone. By considering the droplet as a point mass and applying kinematic equations, particularly the relationship between gravitational acceleration, initial velocity, and maximum height (v² = 2gh), one can determine the speed with which an object must be projected to reach a given height when air resistance is neglected.

Bernoulli's Equation

Bernoulli's principle relates the pressure, velocity, and height in a flowing ideal fluid along a streamline, stating that the sum of static pressure, dynamic pressure, and hydrostatic pressure is constant. It is used to determine fluid speeds and pressures in scenarios where viscous effects are negligible, and it provides a basis for comparing the behavior of an ideal continuous flow to that of discrete droplets in free-fall.

Comparison of Fluid Models

This concept involves understanding how different modeling approaches—treating a fluid as a collection of individual droplets versus an ideal continuous fluid—can yield different results due to underlying assumptions. The analysis highlights the importance of fluid dynamics assumptions, such as the absence of interactions between droplets in one model and the presence of streamline flow and energy conservation in the other, impacting the predicted speeds and pressures.

Hydrostatic Pressure

Hydrostatic pressure arises from the weight of a fluid column and is calculated as the product of the fluid density, gravitational acceleration, and the height or depth of the fluid. This concept is crucial when determining the pressure at a specific depth in any large body of fluid, where the pressure increase relative to a reference level (e.g., atmospheric pressure) can be significant.

Transcript

00:01 Hi there, so for this problem, we are given that height of this water column is given, and that is the change in the height that is equal to 40 meters.

00:13 So for part of this problem, we are asked to model the rising stream as a series of separate door pearls, analyze the free fall of motion of one of the droplets to determine the speed at which water lifts the ground.

00:27 So we need to obtain that initial speed.

00:32 I'm treating this as a droplet.

00:35 So for the upward flight of a water drop projectile from a geyser bend to a fountain top, we can use kinematics that states that the final speed square is equal to the initial speed square.

00:53 This plus, well, minus two times the acceleration to gravity times the change in the altitude.

01:00 Now we know that at the matter cement point the final speed is equal to zero then solving for the initial speed that initial speed is equal to the square root of two times acceleration to the gravity times the altitude of this water column so that will be the square root of two times 9 .8 meters per second square times the altitude of this which is 40 meters so from this will obtain a value of of 28 meters per second.

01:39 So that's a solution for part a of this problem.

01:43 Now for part b, we are asked about, what if model the rising stream as an ideal fluid in a stream -life flow? we need to use the bernoulez equation.

01:57 Ok, so in this case, we need to use the bernoulez equation to determine the initial speed.

02:02 So we know that between the geyser band and in a fountain top, we have the following situation from the evernullis equation that is.

02:13 The pressure at the point 1, that is the pressure at the geyser, plus 1 divided by 2 times the density of the liquid, times the speed at that point to the square.

02:25 This plus the density times the acceleration due to gravity, times the height at the gaiser.

02:36 And this is equal for, well, we have the same, but for the fountain top, speed 2 to 2 square, plus the density, acceleration, gravity, and the high 2.

02:54 Now, in this case, we said that, well, air is so low in density that very nearly the pressure 1 is equal to the pressure 2, that is just one atmosphere.

03:12 So we can cancel this one with this one.

03:16 And then we are left with the following.

03:20 So we can also cancel the density because all of the densities are the same.

03:27 And this last term in here, the final velocity, is equal to 0.

03:32 So we are left, which is 1 divided by 2 times the initial speed at the point 1 to the square.

03:40 Then we can pass this turn to the other side, so that will be the acceleration due to gravity times the difference between the height 2 minus the height 1.

03:50 And this corresponds to the change in the height that is given, the 40 meters given.

03:56 So now finally solving for the speed at 1, as you can see, we obtained the same expression as before, 2 times the acceleration due to the gravity times the change in the height, the square root of that.

04:08 So that will give us the same value as before.

04:11 28 meters per second.

04:13 So that's the solution for part b of this problem...

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