What is the speed of the water exiting the nozzle?

[mparsons06]

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.

 

[rock.freak667]

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.

 

[mparsons06]

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.

 

[AtticusFinch]

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?

 

[mparsons06]

It would be m/s correct? But are my numbers correct? Or are my formulas wrong?

 

[AtticusFinch]

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?

 

[mparsons06]

It states that pressure is constant. But I'm still confused?

 

[AtticusFinch]

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)

 

[mparsons06]

A*v describes the volume rate of low (m^3/s)... Right?

 

[AtticusFinch]

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.

 

[mparsons06]

So is v1 = v2 = 0.019 m^3/s?

 

[AtticusFinch]

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?

 

[mparsons06]

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?

 

[AtticusFinch]

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.

 

[mparsons06]

It is wrong. I have no idea why. Ughhhh.

 

[AtticusFinch]

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.

 

[mparsons06]

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.

 

[AtticusFinch]

Ah I see what is wrong. The question gives you diameter not radius. Divide by 2 before finding the area.

 

[mparsons06]

But do you really need radius if A = pi * d² / 4?

 

[AtticusFinch]

Well no, but you didn't divide by 4 when you used the diameter.

 

[mparsons06]

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?

 

[AtticusFinch]

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.

 

[mparsons06]

It is correct. Once again, thank you for spending the time to help me with this. I appreciate it tremendously.

 

[AtticusFinch]

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.