• Support PF! Buy your school textbooks, materials and every day products Here!

Fluid Mechanics: Water leak in a water tower

  • Thread starter mattst88
  • Start date
29
0
Problem: A water tank develops a leak 20 meters below the surface of the water. The tank is 20 meters above the ground and the hole leaking water is 0.5cm in diameter. How far from the tank will the water hit the ground?

I used P1 + (1/2)(ρ)(v1^2) + ρgy = P2 + (1/2)(ρ)(v2^2) + ρgy

I then removed (1/2)(ρ)(v1^2) since the water inside the tank has no velocity. ρgy on the right side was also removed since y = 0.

I assumed the tank was open topped and therefore P1 = P2.

We're left with:
ρgy = (1/2)(ρ)(v2^2)

so I canceled the ρ, leaving:

(9.8 m/s^2)(20 m) = (1/2)(v2^2)

so v = sqrt(98) at an angle of 0 degrees, right?

-------

yf = yi + (vyi)(t) + (1/2)(ay)(t^2)
0 = 20 + (0)(t) + (1/2)(-9.8)(t^2)
-20 = (1/2)(-9.8)(t^2)
-40/-9.8 = t^2
t = sqrt(40/9.8) ~ 2.02 seconds for the water to hit the ground

if there is no y component to the initial velocity, then it must be all x. therefore sqrt(98) m/s * 2.02 s = 20 meters away from the tank.

Can someone verify my answer or find any errors in my logic or calculations? Thanks
 

Answers and Replies

Galileo
Science Advisor
Homework Helper
1,989
6
It looks ok.
 

Related Threads for: Fluid Mechanics: Water leak in a water tower

  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
7
Views
5K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
3K
Replies
4
Views
4K
  • Last Post
Replies
13
Views
619
Replies
1
Views
1K
  • Last Post
Replies
7
Views
3K
Top