Chapter 14, Problem 14.50P

Review. Old Faithful Geyser in Yellowstone National Park erupts at approximately one-hour intervals, and the height of the water column reaches 40.0 m (Fig. P14.25). (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 m? Assume the chamber is large compared with the geyser’s vent.

Figure P14.25

To determine

The speed at which the water leaves the ground when it is modeled as separate droplets.

Answer to Problem 14.50P

The speed at which the water leaves the ground when it is modeled as separate droplets is $28\text{m}/\text{s}$.

Explanation of Solution

The height of the water column is $40.0\text{m}$.

For the free fall, the equation of motion for upward flight is,

    $v_{\text{f}}^{\text{2}}=v_{\text{i}}^{\text{2}}-2gh$

Here, $v_{\text{f}}$ is the speed of water at the top of the column, $v_{\text{i}}$ is the speed of water leaving the ground, $g$ is the acceleration due to gravity and $h$ is the height of the water column.

Substitute $0\text{m}/\text{s}$ for $v_{\text{f}}$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $40.0\text{m}$ for $h$ to find $v_{\text{i}}$.

    $\begin{array}{c}0=v_{\text{i}}^{\text{2}}-2\times 9.8\text{m}/{\text{s}}^{\text{2}}\times 40.0\text{m}\\ v_{\text{i}}=\sqrt{784{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ =28\text{m}/\text{s}\end{array}$    

Conclusion:

Therefore, the speed at which the water leaves the ground when it is modeled as separate droplets is $28\text{m}/\text{s}$.

To determine

The speed of water as it leaves the ground using Bernoulli’s equation.

Answer to Problem 14.50P

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

Explanation of Solution

Apply Bernoulli’s equation at the ground level and at the top of the column.

    $\begin{array}{c}{P}_0+\frac{1}{2}\rho v_1^2+\rho g{y}_1=P_0+\frac{1}{2}\rho v_2^2+\rho g{y}_2\\ \frac{1}{2}\rho v_1^2+\rho g{y}_1=\frac{1}{2}\rho v_2^2+\rho g{y}_2\\ \frac{1}{2}\left(v_1^2-v_2^2\right)=g\left(y_2-y_1\right)\\ v_1^2-v_2^2=2g\left(y_2-y_1\right)\end{array}$

Here, $P_0$ is the atmospheric pressure, $\rho$ is the density of water, $v_1$ is the speed of water leaving the ground, $v_2$ is the speed of water at the top of the column, $h_1$ is the height of the bottom of the column from the ground and $h_2$ is the height of the top of the column from the ground.

Substitute $0\text{m}/\text{s}$ for $v_2$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $0\text{m}$ for $y_1$ and $40.0\text{m}$ for $y_2$ to find $v_1$.

    $\begin{array}{c}{v}_1^2-0=2\times 9.8\text{m}/{\text{s}}^{\text{2}}\left(40.0\text{m}-0\text{m}\right)\\ v_1=\sqrt{784{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ =28.0\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed of water as it leaves the ground using Bernoulli’s equation is $28.0\text{m}/\text{s}$.

To determine

The answer obtained in part (a) with the answer obtained in part (b).

Answer to Problem 14.50P

Both the models are consistence with each other as the speed of water leaving the ground is same in both models.

Explanation of Solution

Along a streamline for an ideal fluid, the sum of the pressure, kinetic energy per unit volume and the potential energy per unit volume is same at all the points.

The speed of the water leaving the ground obtained in both part (a) as well as in part (b) is $28\text{m}/\text{s}$.

This shows that the both the models are consistence with each other.

Conclusion:

Therefore, both the models are consistence with each other as the speed of water leaving the ground is same in both models.

To determine

The pressure in the heated underground chamber.

Answer to Problem 14.50P

The pressure in the heated underground chamber is $2.11\text{MPa}$.

Explanation of Solution

Apply the Bernoulli’s equation between the chamber and the geyser vent.

    $P_0+\frac{1}{2}\rho v_1^2+\rho g{y}_1=P+\frac{1}{2}\rho v_3^2+\rho g{y}_3$

Here, $P$ is the pressure in the heated underground chamber, $v_3$ is the speed of water in the heated underground chamber and $y_3$ is the depth of the heated underground chamber.

Substitute $0\text{m}$ for $y_1$ and $0\text{m}/\text{s}$ for $v_3$ in the above equation.

    $\begin{array}{c}{P}_0+\frac{1}{2}\rho v_1^2+\rho g\left(0\right)=P+\frac{1}{2}\rho \left(0\right)+\rho g{y}_3\\ P-P_0=\frac{1}{2}\rho \left(v_1^2-2g{y}_3\right)\end{array}$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$, $28\text{m}/\text{s}$ for $v_1$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $175\text{m}$ for $y_3$ in the above equation.

    $\begin{array}{c}P-P_0=\frac{1}{2}\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left\{{\left(28\text{m}/\text{s}\right)}^2-2\times 9.8\text{m}/{\text{s}}^{\text{2}}\times \left(-175\text{m}\right)\right\}\\ =\frac{1}{2}\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\times 4214{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\\ =2.11\times {10}^6\text{Pa}\times \frac{{10}^{-6}\text{MPa}}{1\text{Pa}}\\ =2.11\text{MPa}\end{array}$

Conclusion:

Therefore, the pressure in the heated underground chamber is $2.11\text{MPa}$.