Fluid Physics Help: Find Water Distance at 10mm Diam.

In summary, the conversation discusses the problem of determining the distance below a faucet at which the water stream has narrowed to 10mm diameter. The equations A1v1=A2v2 and Q=vA are mentioned and the attempt at a solution involves calculating the velocity of the water and using the equation v² = Vo² + 2 g*x to determine the distance. However, the attempt at a solution is incorrect.
  • #1
ch2kb0x
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Homework Statement



Water flowing out of a 16-mm diameter faucet fills a 2L bottle in 10s. At what distance below the faucet has the water stream narrowed to 10mm diameter?

Homework Equations



A1v1=A2v2 Q=vA

The Attempt at a Solution


edit: this is what I TRIED to do:(*note: i ignored some unit conversions at this time)

Q=2L/10s=2x10^-3 m^3/s.

v= Q/A = 2x10^-3 m^3/s / pi(5mm)^2 = 2.54x10^-5

x=vt = 2.54x10^-5 * 10s = 2.54*10^-4.
lol this is so wrong, it doesn't make any sense.
 
Last edited:
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  • #2


The presumption is that as the water falls it will narrow because of it's increase in speed?

Calculate the velocity of the water coming from the opening - 2 liters flowing through the area of the opening in 10 seconds - and then figure what velocity will be required to flow through an area that is reduced to 10 mm. That velocity should occur from gravity at v² = Vo² + 2 g*x where x is the distance below shouldn't it?
 
  • #3


I would approach this problem by first defining the variables and units involved. The diameter of the faucet is given as 16 mm and the volume of water that flows out in 10 seconds is 2 L. I would first convert the volume to cubic meters and the diameter to meters for consistency.

2 L = 0.002 m^3
16 mm = 0.016 m

Next, I would use the equation Q=Av to find the velocity of the water. The cross-sectional area (A) can be calculated using the formula for the area of a circle, A= πr^2.

Q= Av
0.002 m^3/s = (π(0.008m)^2)v
v = 0.318 m/s

Now, we can use the equation v=at to find the distance (x) traveled in 10 seconds.

v=at
0.318 m/s = a(10s)
a = 0.0318 m/s^2

Finally, we can use the equation x=at^2/2 to find the distance below the faucet where the water stream has narrowed to 10 mm diameter.

x=at^2/2
x= (0.0318m/s^2)(10s)^2/2
x= 1.59 m

Therefore, at a distance of 1.59 m below the faucet, the water stream has narrowed to 10 mm diameter. It is important to note that this calculation assumes a constant flow rate and neglects any factors that may affect the flow of water, such as air resistance or turbulence.
 

1. How do you calculate the distance water travels at a 10mm diameter?

The distance water travels at a 10mm diameter can be calculated using the equation d = 4Q/πrv, where d is the distance, Q is the flow rate, r is the radius (5mm), and v is the velocity.

2. What is the flow rate of water at a 10mm diameter?

The flow rate of water at a 10mm diameter can be calculated using the equation Q = Av, where Q is the flow rate, A is the cross-sectional area (πr²), and v is the velocity.

3. How does the velocity of water affect its distance at a 10mm diameter?

The velocity of water directly affects its distance at a 10mm diameter. The faster the water is flowing, the farther it will travel in a given amount of time.

4. How does the diameter of the water affect its distance?

The diameter of the water does not directly affect its distance. However, it does affect the flow rate, which in turn affects the distance the water travels.

5. What is the significance of finding the distance of water at a 10mm diameter?

Understanding the distance water travels at a specific diameter is important in various fields such as engineering, fluid dynamics, and hydrology. It can also be used in practical applications, such as designing water systems or predicting the flow of water in natural systems.

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