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Help me with this fluid Problem Please

  1. Apr 4, 2006 #1
    Help me with this fluid Problem ASAP Please :)

    A long horizontal hose of diameter 3.2 cm is connected to a faucet. At the other end, there is a nozzle of diameter 1.8 cm. Water squirts from the nozzle at velocity 16 m/sec. Assume that the water has no viscosity or other form of energy dissipation.

    Okay so How long will it take to fill a tub of volume 110 liters with the hose ?

    I got the flow rate which is 16.277 liter, 110/162.77 = 6.75 why isnt this the asnwer
  2. jcsd
  3. Apr 4, 2006 #2
    Can you show some work?

    [tex] \dot {m} = \rho \dot{ V_{avg}} A_c [/tex]

    This is a steady state nozzle.
  4. Apr 4, 2006 #3
    o ya what did was

    ((1.8)/ (100))^2 * 3.14 * 16 = .016277 M^3/S = Flow rate

    I then converted it to Liters by multiplying 1000
  5. Apr 4, 2006 #4
    Area of a cross section is [tex] \pi \frac{d^2}{4} [/tex]

    Why did the problem give you the initial diameter? Are you sure this is how the problem is worded?
  6. Apr 4, 2006 #5
    yep exactly, i copied and pasted from online HW, o ya intial diameter was for another part i needed to find Velcotit which i did
  7. Apr 4, 2006 #6
    What? You found the inital diameter in another part? Can you post the origional problem. I have no clue what is what anymore :confused:.
  8. Apr 4, 2006 #7
    sorry one more problem,

    A rectangular block of ice 14 m on each side and 0.5 m thick floats in seawater. The density of the seawater is 1025 kg/m3. The density of ice is 917 kg/m3.

    a) How high does the top of the ice block float above the water level?

    Okay so i know Fb=pgVdis

    pwatergVdis = picegVdis

    (1025)(9.8)(14*1.1*14)+ (x)(m*g) = 917*9.8*X

    What Am i missing, thanks so much
    Last edited: Apr 4, 2006
  9. Apr 4, 2006 #8
    i got the problem, thanks alot i guess i didnt know what the area of a cross section is since all it says on formula sheet is the area of circle.
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