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Speed of Water Out Nozzle. Please Help?

  1. Apr 11, 2010 #1
    1. The problem statement, all variables and given/known data:

    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.
     
    Last edited: Apr 12, 2010
  2. jcsd
  3. Apr 11, 2010 #2

    rock.freak667

    User Avatar
    Homework Helper

    Assuming you typed the question properly, then 0.019 m3/s i not the velocity of the water entering the nozzle.
     
  4. Apr 11, 2010 #3
    So are you saying 0.019 m3/s is not my v1, but my v2? I'm confused.
     
  5. Apr 11, 2010 #4
    Think about it. Are those the units you would expect a velocity term to have?
     
  6. Apr 11, 2010 #5
    It would be m/s correct? But are my numbers correct? Or are my formulas wrong?
     
  7. Apr 11, 2010 #6
    Your equations are correct. What does the principal of continuity state should be continuous?
     
  8. Apr 11, 2010 #7
    It states that pressure is constant. But I'm still confused?
     
  9. Apr 12, 2010 #8
    That's not what the principle states. What physical quantity does A*v describe? (Look at the units)
     
  10. Apr 12, 2010 #9
    A*v describes the volume rate of low (m^3/s)... Right?
     
  11. Apr 12, 2010 #10
    Exactly, so the principle states that the volume flow rate is constant at any point.

    Use that fact to answer the question.
     
  12. Apr 12, 2010 #11
    So is v1 = v2 = 0.019 m^3/s?
     
  13. Apr 12, 2010 #12
    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?
     
  14. Apr 12, 2010 #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?
     
  15. Apr 12, 2010 #14
    Yes.
     
  16. Apr 12, 2010 #15
    It is wrong. I have no idea why. Ughhhh.
     
  17. Apr 12, 2010 #16
    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.
     
  18. Apr 12, 2010 #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.
     
  19. Apr 12, 2010 #18
    Ah I see what is wrong. The question gives you diameter not radius. Divide by 2 before finding the area.
     
  20. Apr 12, 2010 #19
    But do you really need radius if A = pi * d2 / 4?
     
  21. Apr 12, 2010 #20
    Well no, but you didn't divide by 4 when you used the diameter.
     
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