Solving for Water Flow Distance Under Faucet

Click For Summary
SUMMARY

The discussion focuses on calculating the distance below a faucet where water stream diameter narrows from 13.0 mm to 10.0 mm, with an initial velocity of 2.30 m/s. The user correctly applied the continuity equation, resulting in a secondary velocity (v2) of 0.915 m/s. However, the application of Bernoulli's equation contained errors leading to an incorrect height calculation of 0.222 m. The final resolution emphasized the importance of accounting for gravitational acceleration in the calculations.

PREREQUISITES
  • Understanding of fluid dynamics principles, specifically Bernoulli's equation.
  • Knowledge of the continuity equation in fluid flow.
  • Familiarity with basic physics concepts such as velocity and gravitational acceleration.
  • Ability to perform unit conversions and area calculations for circular cross-sections.
NEXT STEPS
  • Review the derivation and applications of Bernoulli's equation in fluid mechanics.
  • Learn about the continuity equation and its implications in varying cross-sectional areas.
  • Study the effects of gravitational acceleration on fluid flow in vertical systems.
  • Explore practical examples of fluid dynamics in real-world applications, such as plumbing systems.
USEFUL FOR

Students and professionals in physics, engineering, and fluid mechanics, particularly those interested in practical applications of fluid flow principles in systems like faucets and plumbing.

momogiri
Messages
52
Reaction score
0
Question:
Water flows out of a 13.0mm-diameter sink faucet at 2.30m/s.
At what distance below the faucet has the water stream narrowed to 10 mm diameter?

What I've done is that...
I found A_1 (faucet) to be 0.0001327m^2
and A_2 (bottom) to be 0.000314159m^2

So I did the whole A_1*v_1 = A_2*v_2 thing
and found v_2 to be 0.915m/s

Then I plugged it in, into Bernoulli's equation
1/2*rho*v_1^2 + rho*g*y_1 = 1/2*rho*v_2^2 + rho*g*y_2 (cancelling out the P_1 and P_2 because they.. cancel..)

In which it is..
0.5*1000*2.3^2 + 1000*9.8*y_1 = 0.5*1000*0,9715^2 + 1000*9.8*y_2

2645 + 9800*y_1 = 471.906 + 9800*y_2
9800*(y_1 - y_2) = -2174.094

So.. h = 0.222m

Or so I thought... It's wrong, so what am I doing wrong? Is it my A_1 and A_2's, etc.?
 
Physics news on Phys.org
and A_2 (bottom) to be 0.000314159m^2
Check this number. What diameter was used?

If the initial velocity is 2.3 m/s, how does one obtain 0.95 m/s for V2? The water accelerates with gravity.
 
Astronuc said:
Check this number. What diameter was used?

If the initial velocity is 2.3 m/s, how does one obtain 0.95 m/s for V2? The water accelerates with gravity.

:D Thanks for the reminder that water accelerates with gravity
That helped, and I got it :D
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 24 ·
Replies
24
Views
7K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
3K
  • · Replies 15 ·
Replies
15
Views
11K
Replies
1
Views
2K
Replies
2
Views
2K
Replies
4
Views
6K
Replies
3
Views
3K