What is the speed of the water exiting the nozzle?

In summary, the question asks for the speed at which water exits a fire hose with a diameter of 7.15 cm and a volume flow rate of 0.019 m3/s. Using the principle of continuity, the volume flow rate is found to be constant at any point, and by manipulating the equation A*v = 0.019 m3/s, the velocity of the water exiting the nozzle with an inner diameter of 2.17 cm is calculated to be 51.4 m/s. This calculation is based on the assumption that the diameter given in the question is actually the radius, and not the diameter.
  • #1
mparsons06
61
0
1. Homework Statement :

Water flows through a fire hose of diameter 7.15 cm at a rate of 0.019 m3/s. The fire hose ends in a nozzle of inner diameter 2.17 cm. What is the speed with which the water exits the nozzle?2. The attempt at a solution:

So, I tried using Principle of Continuity:

A1 * v1 = A2 * v2

A = pi * d^2 / 4

A1 = 0.00402 m
A2 = 0.00037 m

A1 * v1 = A2 * v2
(0.00402 m) * (0.019 m^3/s) = (0.00037 m) * v2
0.0000764 m^4/s = (0.00037 m) * v2
v2 = (0.0000764 m^4/s) / (0.00037 m)
v2 = 0.206 m^3/s

But my answer is incorrect. Do I need to apply another equation? Or did I mess up somewhere? Please help.
 
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  • #2
mparsons06 said:
1. Homework Statement :

Water flows through a fire hose of diameter 7.15 cm at a rate of 0.019 m3/s. The fire hose ends in a nozzle of inner diameter 2.17 cm. What is the speed with which the water exits the nozzle?

But my answer is incorrect. Do I need to apply another equation? Or did I mess up somewhere? Please help.

Assuming you typed the question properly, then 0.019 m3/s i not the velocity of the water entering the nozzle.
 
  • #3
rock.freak667 said:
Assuming you typed the question properly, then 0.019 m3/s is not the velocity of the water entering the nozzle.

So are you saying 0.019 m3/s is not my v1, but my v2? I'm confused.
 
  • #4
mparsons06 said:
So are you saying 0.019 m3/s is not my v1, but my v2? I'm confused.

Think about it. Are those the units you would expect a velocity term to have?
 
  • #5
It would be m/s correct? But are my numbers correct? Or are my formulas wrong?
 
  • #6
mparsons06 said:
It would be m/s correct? But are my numbers correct? Or are my formulas wrong?

Your equations are correct. What does the principal of continuity state should be continuous?
 
  • #7
It states that pressure is constant. But I'm still confused?
 
  • #8
mparsons06 said:
It states that pressure is constant. But I'm still confused?

That's not what the principle states. What physical quantity does A*v describe? (Look at the units)
 
  • #9
A*v describes the volume rate of low (m^3/s)... Right?
 
  • #10
mparsons06 said:
A*v describes the volume rate of low (m^3/s)... Right?

Exactly, so the principle states that the volume flow rate is constant at any point.

Use that fact to answer the question.
 
  • #11
So is v1 = v2 = 0.019 m^3/s?
 
  • #12
mparsons06 said:
So is v1 = v2 = 0.019 m^3/s?

Nope. You just realized that A*v (not v) is 0.019 m^3/s. Since this quantity must be conserved how would you find v2?
 
  • #13
A*v = 0.019 m^3/s
v = (0.019 m^3/s) / A = (0.019 m^3/s) / (3.14*(0.0217m)^2) = 12.9 m/s

Is that correct?
 
  • #14
mparsons06 said:
A*v = 0.019 m^3/s
v = (0.019 m^3/s) / A = (0.019 m^3/s) / (3.14*(0.0217m)^2) = 12.9 m/s

Is that correct?

Yes.
 
  • #15
It is wrong. I have no idea why. Ughhhh.
 
  • #16
mparsons06 said:
It is wrong. I have no idea why. Ughhhh.

Hmm that's odd. Are you sure you wrote the problem down correctly? Also, though I'm not sure whether this matters, but I'm getting 12.8 m/s rather than 12.9.
 
  • #17
Yes, I wrote it down right. I will 12.8, but I'm not sure if it'll make a difference. But anyways, thank you for all of your help. =) I really appreciate it.
 
  • #18
Ah I see what is wrong. The question gives you diameter not radius. Divide by 2 before finding the area.
 
  • #19
But do you really need radius if A = pi * d2 / 4?
 
  • #20
Well no, but you didn't divide by 4 when you used the diameter.
 
  • #21
So if I fix the equation and recalculate:

v = (0.019 m^3/s) / A = (0.019 m^3/s) / (3.14*(0.0217m)^2 / 4) = 51.4

Is the units m/s or m3/s?
 
  • #22
mparsons06 said:
So if I fix the equation and recalculate:

v = (0.019 m^3/s) / A = (0.019 m^3/s) / (3.14*(0.0217m)^2 / 4) = 51.4

Is the units m/s or m3/s?

Still m/s, we just threw in a number that won't affect anything.
 
  • #23
It is correct. Once again, thank you for spending the time to help me with this. I appreciate it tremendously.
 
  • #24
mparsons06 said:
It is correct. Once again, thank you for spending the time to help me with this. I appreciate it tremendously.

Yeah no problem, good job.
 

What is the speed of water coming out of a nozzle?

The speed of water coming out of a nozzle can vary depending on several factors such as the pressure of the water, the shape and size of the nozzle, and the viscosity of the water. Generally, it can range from a few meters per second to over 100 meters per second.

How can the speed of water coming out of a nozzle be measured?

The speed of water coming out of a nozzle can be measured using various techniques such as using a flow meter, a pitot tube, or by using the Bernoulli's equation. Each method has its own advantages and limitations, and the choice of method depends on the specific situation.

Does the speed of water coming out of a nozzle depend on the angle of the nozzle?

Yes, the speed of water coming out of a nozzle can be affected by the angle of the nozzle. Changing the angle of the nozzle can change the shape of the water stream and the direction of the flow, which can impact the speed of the water coming out.

What factors can affect the speed of water coming out of a nozzle?

Several factors can impact the speed of water coming out of a nozzle. These include the pressure of the water, the shape and size of the nozzle, the angle of the nozzle, and the viscosity of the water. Other external factors such as wind can also affect the speed of the water coming out.

Why is the speed of water coming out of a nozzle important?

The speed of water coming out of a nozzle is important in various fields such as fluid dynamics, engineering, and industrial processes. It is a crucial parameter in designing efficient water systems, and it can also impact the performance of devices that use water, such as pumps and sprinkler systems.

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