Chapter 9, Problem 47P

Old Faithful geyser in Yellowstone Park erupts at approximately 1-hour intervals, and the height of the fountain reaches 40.0 m (Fig. P9.47). (a) Consider the rising stream as a series of separate drops. Analyze the free-fall motion of one of the drops to determine the speed at which the water leaves the ground. (b) Treat 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) What is the pressure (above atmospheric pressure) in the heated underground chamber 175 m below the vent? You may assume the chamber is large compared with the geyser vent.

To determine

The speed at which the water leaves the ground.

Answer to Problem 47P

The speed at which the water leaves the ground is $28.0\text{m/s}$.

Explanation of Solution

When one of the drop reaches the maximum height, its speed would be $v_y^2=0=v_{0y}^2+2a_y\left(\Delta y\right)\Rightarrow v_{0y}=\sqrt{-2a_y\left(\Delta y\right)}$.

Given info: The deceleration of the water drop is $-9.80{\text{m/s}}^{\text{2}}$ and the height of the fountain is $40.0\text{m}$.

The formula for the speed of the water drop is,

$\begin{array}{c}{v}_{0y}=\sqrt{-2\left(-9.80{\text{m/s}}^{\text{2}}\right)\left(40.0\text{m}\right)}\\ =28.0\text{m/s}\end{array}$

Thus, the speed at which the water leaves the ground is $28.0\text{m/s}$.

Conclusion:

Therefore, the speed at which the water leaves the ground is $28.0\text{m/s}$.

To determine

The speed of the speed of the water as it leaves ground level.

Answer to Problem 47P

The speed of the speed of the water as it leaves ground level is $28.0\text{m/s}$.

Explanation of Solution

By treating this as an ideal fluid in stream line flow using Bernoulli’s equation, the speed of the water is $\left(1/2\right)\rho v_{\text{geyser}}^2=0+\rho g\left(y_{\text{top}}-y_{\text{vent}}\right)$ and $\Rightarrow v_{\text{geyser}}=\sqrt{2g\left(y_{\text{top}}-y_{\text{vent}}\right)}$.

Given info: Acceleration due to gravity is $9.80{\text{m/s}}^{\text{2}}$ and the height of the fountain is $40.0\text{m}$.

The formula for the speed of geyser is,

$v_{\text{geyser}}=\sqrt{2g\left(y_{\text{top}}-y_{\text{vent}}\right)}$

• $g$ is acceleration due to gravity.
• $\left(y_{\text{top}}-y_{\text{vent}}\right)$ is height of the fountain.

Substitute $9.80{\text{m/s}}^{\text{2}}$ for $g$ and $40.0\text{m}$ for $\left(y_{\text{top}}-y_{\text{vent}}\right)$ to find $v_{\text{geyser}}$.

$\begin{array}{c}{v}_{\text{geyser}}=\sqrt{2\left(9.80{\text{m/s}}^{\text{2}}\right)\left(40.0\text{m}\right)}\\ =28.0{\text{m/s}}^{\text{2}}\end{array}$

Thus, the speed of the speed of the water as it leaves ground level is $28.0\text{m/s}$

Conclusion:

Therefore, the speed of the speed of the water as it leaves ground level is $28.0\text{m/s}$

To determine

The pressure in the heated underground chamber 175 m below the vent.

Answer to Problem 47P

The pressure in the heated underground chamber 175 m below the vent is $2.1\text{1MPa}$.

Explanation of Solution

Bernoulli’s equation between the chamber and the geyser vent by considering $v_{\text{chamber}}\simeq 0$ gives $P+0+\rho g{y}_{\text{chamber}}=P_{\text{atm}}+\left(1/2\right)\rho v_{\text{vent}}^2+\rho g{y}_{\text{vent}}$ and the pressure difference is $P-P_{\text{atm}}=\rho \left[\left(1/2\right)v_{\text{vent}}^2+g\left(y_{\text{vent}}-y_{\text{chamber}}\right)\right]$. This measurement represents the gauge pressure.

Given info: Density of water is $1.00{\text{kg/m}}^{\text{3}}$, speed of vent is $28.0\text{m/s}$, acceleration due to gravity is $9.80{\text{m/s}}^{\text{2}}$, and the distance of the underground chamber is $175\text{m}$.

The formula for the pressure in the heated underground chamber is,

$P-P_{\text{atm}}=\rho \left[\left(1/2\right)v_{\text{vent}}^2+g\left(y_{\text{vent}}-y_{\text{chamber}}\right)\right]$

• $\rho$ is density of geyser.
• $v_{\text{vent}}$ is speed of vent.
• $g$ is acceleration due to gravity.
• $y_{\text{vent}}-y_{\text{chamber}}$ is the distance of underground chamber.

Substitute $1.00{\text{kg/m}}^{\text{3}}$ for $\rho$, $28.0\text{m/s}$ for $v_{\text{vent}}$, $9.80{\text{m/s}}^{\text{2}}$ for $g$, and $175\text{m}$ for $y_{\text{vent}}-y_{\text{chamber}}$ to find $P-P_{\text{atm}}$.

$\begin{array}{c}P-P_{\text{atm}}=\left(1.00{\text{kg/m}}^{\text{3}}\right)\left[\left(1/2\right){\left(28.0\text{m/s}\right)}^2+\left(9.80{\text{m/s}}^{\text{2}}\right)\left(175\text{m}\right)\right]\\ =2.11\text{MPa}\end{array}$

Thus, the pressure in the heated underground chamber 175 m below the vent is $2.1\text{1MPa}$.

Conclusion:

Therefore, the pressure in the heated underground chamber 175 m below the vent is $2.1\text{1MPa}$.