Question

Water from Old Faithful Geyser shoots upward to a height of about 50 m once every 90 minutes. What must be the pressure in the hot springs below the ground in order for the water to go that high?

Answer 1

Given:

height of geyser, h = 50 m

speed of water at ground can be given by third eqn of motion with v = 0 m/s:

`u^(2) = v^(2) + 2gh`

putting v = 0 m/s in the above eqn, we get:

u =
`√(2gh)` (1)

Now, pressure is given by:

`\Delta p = (1)/(2)u^(2)\rho \\` (2)

where,

`\Delta p` = density of water = 1000 kg/
`m^(3)`

g = 9.8 m/
`s^(2)`

Using eqn (1) and (2):

`\Delta p = (1)/(2)(√(2gh))^(2)\rho`

⇒
`\Delta p = (1)/(2)\rho gh`

`\Delta p = (1)/(2)*2* 1000* 9.8* 50` =
`\Delta p` = 490000 Pa

`p_(atm)` = 101325 Pa

For absolute pressure, p:

p =
`\Delta p` +
`p_(atm)`

p = 490000 - 101325

p = 388675 Pa

Therefore, pressure in hot springs for the water to attain the height of 50m is 388675 Pa