Old Faithful geyser in Yellowstone Park erupts at approximately 1 hour intervals

Question

Old Faithful geyser in Yellowstone Park erupts at approximately 1 hour intervals, and the height of the fountain reaches 50.0 m

(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.
m/s
(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.
m/s
(c) What is the pressure (above atmospheric pressure) in the heated underground chamber 173 m below the vent? You may assume that the chamber is large compared with the geyser's vent.
MPa

Explanation / Answer

For the upward fl ight of a water-drop projectile from geyser vent to fountain-top

vy^2 = voy^2 + 2ay ( del y)

since vy = 0

voy = sqrt 2 ay del y max

voy = sqrt -2 ( -9.8) ( 50 m)

=31.3 m/s

Because of the low density of air and the small change in altitude, atmospheric pressure at
the fountain top will be considered equal to that at the geyser vent. Bernoulli’s equation

v top = 0

1/2 rho v _vent^2 = 0 + rho g( y top - y vent)

v_vent = sqrt 2 g ( y top- y vent)

= sqrt 2 ( 9.8) ( 50 m)

= 31.3 m/s

(c)

Apply Bernoulis equation

( P+ 0 + rho gy) chamber = P atm + 1/2 rho v vent^2 + rho g y vent

p- patm = rho ( 1/2 v_vent ^2 + g ( y vent- y chamber)

= 1000 (( 31.3)^2/2+ 9.8 ( 173)

=2.18^ 10^6 Pa

= 2.18 MPa

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