Chapter 15, Problem 40P

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. P15.40). (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 P15.40

To determine

The speed at which the water leaves the ground by modelling the stream as a series of separate droplets.

Answer to Problem 40P

The speed at which the water leaves the ground by modelling the stream as a series of separate droplets is $28.0\text{ m/s}$.

Explanation of Solution

Consider the upward flight of a water-drop projectile from geyser vent to fountain top.

Take the upward direction to be $+y$ direction.

Write the equation of motion in the vertical direction.

  $v_{yf}^2=v_{yi}^2+2a_y\Delta y$        (I)

Here, $v_{yf}$ is the $y$ component of the final velocity, $v_{yi}$ is the $y$ component of the initial velocity, $a_y$ is the $y$ component of the acceleration and $\Delta y$ is the vertical displacement.

At the maximum height the velocity of the water drop will be zero so that the value of $v_{yf}$ will be $0\text{ m/s}$. The value of $a_y$ will be the negative of the acceleration due to gravity since the upward direction is taken to be $+y$ direction.

Substitute $0\text{ m/s}$ for $v_{yf}$ and $-g$ for $a_y$ in equation (I) and rewrite it for $v_{yi}$.

  $\begin{array}{c}0\text{ m/s}=v_{yi}^2+2\left(-g\right)\Delta y\\ v_{yi}^2=2g\Delta y\\ v_{yi}=\sqrt{2g\Delta y}\end{array}$        (II)

Conclusion:

The value of $g$ is $9.80{\text{ m/s}}^2$ .

Substitute $9.80{\text{ m/s}}^2$ for $g$ and $40.0\text{ m}$ for $\Delta y$ in equation (II) to find $v_{yi}$ .

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

Therefore, the speed at which the water leaves the ground by modelling the stream as a series of separate droplets is $28.0\text{ m/s}$.

To determine

The speed of the water as it leaves the ground by modelling the stream as an ideal fluid in streamline flow.

Answer to Problem 40P

The speed of the water as it leaves the ground by modelling the stream as an ideal fluid in streamline flow is $28.0\text{ m/s}$.

Explanation of Solution

Write the Bernoulli’s equation.

  $P_1+\frac{1}{2}\rho v_1^2+\rho g{y}_1=P_2+\frac{1}{2}\rho v_2^2+\rho g{y}_2$        (III)

Here, $P_1$ is the pressure at the point 1, $\rho$ is the density of the liquid, $v_1$ is the speed of the liquid at point 1, $g$ is the acceleration due to gravity, $y_1$ is the height of the point 1 above the reference position, $P_2$ is the pressure at the point 2, $v_2$ is the speed of the liquid at point 2 and $y_2$ is the height of the point 2 above the reference position.

Air has very low density so that the pressure at both geyser vent and fountain-top will be atmospheric pressure.

Substitute $P_a$ for $P_1$, $P_2$, $0\text{ m}$ for $y_1$ and $0\text{ m/s}$ for $v_2$ in equation (III) and rewrite it for $v_1$.

  $\begin{array}{c}{P}_a+\frac{1}{2}\rho v_1^2+\rho g\left(0\text{ m}\right)=P_a+\frac{1}{2}\rho \left(0\text{ m/s}\right)+\rho g{y}_2\\ \frac{1}{2}{v}_1^2=g{y}_2\\ v_1^2=2g{y}_2\\ v_1=\sqrt{2g{y}_2}\end{array}$        (IV)

Conclusion:

Substitute $9.80{\text{ m/s}}^2$ for $g$ and $40.0\text{ m}$ for $y_2$ in equation (IV) to find $v_1$.

  $\begin{array}{c}{v}_1=\sqrt{2\left(9.80{\text{ m/s}}^2\right)\left(40.0\text{ m}\right)}\\ =28.0\text{ m/s}\end{array}$

Therefore, the speed of the water as it leaves the ground by modelling the stream as an ideal fluid in streamline flow is $28.0\text{ m/s}$.

To determine

The comparison of answers of part (a) and part (b).

Answer to Problem 40P

The answers of part (a) and part (b) agree precisely.

Explanation of Solution

It is asked to find the speed with which the water leaves the ground. In part (a), the rising stream was modelled as free-fall motion of one of the droplets. The value of the speed found using this model is $28.0\text{ m/s}$ .

In part (b), the rising stream was modelled as an ideal fluid in streamline flow. The value of the speed found using this model is also $28.0\text{ m/s}$. The two values agree precisely. This implies that the two models are consistent with each other.

Conclusion:

Thus, the answers of part (a) and part (b) agree precisely.

To determine

The pressure in the heated underground chamber.

Answer to Problem 40P

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

Explanation of Solution

Take point 1 in the Bernoulli’s equation to be the chamber and the point 2 to be fountain-top.

The velocity of the water drop at both the points are $0\text{ m/s}$. The value of $P_2$ will be atmospheric pressure.

Substitute $0\text{ m/s}$ for $v_1$ and $v_2$ in equation (III).

  $\begin{array}{c}{P}_1+\frac{1}{2}\rho \left(0\text{ m/s}\right)+\rho g{y}_1=P_2+\frac{1}{2}\rho \left(0\text{ m/s}\right)+\rho g{y}_2\\ P_1+\rho g{y}_1=P_2+\rho g{y}_2\\ P_1-P_2=\rho g\left(y_2-y_1\right)\end{array}$        (V)

Conclusion:

The density of water is $1000{\text{ kg/m}}^3$ . Since the chamber is in depth underground, the value of $y_1$ will be negative.

Substitute $1000{\text{ kg/m}}^3$ for $\rho$, $9.80{\text{ m/s}}^2$ for $g$ , $40.0\text{ m}$ for $y_2$ and $-175\text{ m}$ for $y_1$ in equation (V) to find the pressure above atmospheric pressure.

  $\begin{array}{c}{P}_1-P_2=\left(1000{\text{ kg/m}}^3\right)\left(9.80{\text{ m/s}}^2\right)\left(40.0\text{ m}-\left(-175\text{ m}\right)\right)\\ =\left(1000{\text{ kg/m}}^3\right)\left(9.80{\text{ m/s}}^2\right)\left(215\text{ m}\right)\\ =2.11\times {10}^6\text{ Pa}\frac{1\text{ MPa}}{{10}^6\text{ Pa}}\\ =2.11\text{ MPa}\end{array}$

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