In an industrial cooling process, water is circula...

In an industrial cooling process, water is circulated through a system. If the water is pumped with a speed of 0.45 m/s under a pressure of 400 torr from the first floor through a 6.0-cm diameter pipe, what will be the pressure on the next floor 4.0 m above in a pipe with a diameter of 2.0 cm?
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    Answers ( 6 )

    1. margaux vidal
      margaux vidal

      Hey sorry for taking so long to get back to you, I tried workingit out yesterday and I couldn't get the bernoulli equation to comeout right because I used the wrong relation between area andvelocity. They have the following realtion:

      Then you can solve for V2, or the velocity of the pip with a2.0 cm diameter


      Now use the bernoulli equation..
      You will want
      And now you are looking for the Pressure of the second fluidso your equation is
      Solving for the pressure Px gives you:
    2. enni salonen
      enni salonen
      Put all units ----in SI system
      In an industrial cooling process, water is circulated through asystem.
      If the water is pumped with a speed of( v1 = 0.45 m/s)
      under a pressure of ( P1 = 400 torr) from the first floor througha( D1 = 6.0cm) diameter pipe,
      Let ' P2 ' be the pressure on the next floor ( h = 4.0 m ) above in a pipe with adiameter of ( D2 = 2.0 cm)
      Density of water ( ? ) = 1000  kg / m3
      From the equation of continuity wehave
      The Flow rateis          A1 v1 =A2v2

                         0r,                v2 = A1 v1/ A 2

                         Or,              v2= ( D1 / D2 ) 2 v1

      Apply ; Bernoulli's Theorem
             P 1 + ( 1/2 )*?
      * v 1 2 + 0   =    P2+ ( 1/2 )* ? * v22 + ? * h * g
      Put the above givenvalues in the above expression , to get
      ' P2 '

    3. isabella gutierrez
      isabella gutierrez
      45.6
    4. hannah scherer
      hannah scherer
      use the bernoulli equation from your book or get it from the webpage
      http://en.wikipedia.org/wiki/Bernoulli%27s_principle#Incompressible_flow_equation

      0.5(v squared) + gh + P/r = 0.5(v squared) + gh + p/r

      the left side is for the first floor where
      (v squared) is (o.45 squared) and g is 9.8 but h is zero
      and P is given as 400torr which must be converted to pascals
      and r, the density of water is 1g per ml, you have to convert it tokg per cubic meter or either do some algebra and divide it out ofthe equation.

      right side
      calculate velocity. area x vel on the little pipe equals area xvel
      on the big pipe. area is pi x r x r. pi is 3.14 and r isdiameter/2. convert the diameter from cm to meters and find areathen find velocity for the right side.

      centimeters x 1meter/(100 cm)= meters
      6cm=o.06 meters
      radius on left is 0.03meters
      area on left is 3.14 x 0.03 x 0.03
      area x vel on left is
      3.14(0.03 squared)(0.45)=3.14(0.01 squared)(veloc upstair)
      calculate your velocity upstairs and put it in the right side ofthe bernoulli equation. h is 4 meters for the right side of thebernouli

      http://www.onlineconversion.com/density_common.htm
      this is how you convert grams per ml into kg per cubic meters
      just mult by 1000. the density, r, of water is 1 gram per ml

      Source:http://au.answers.yahoo.com/question/index?qid=20080203121146AAHwdwX
    5. giulia rodrigues
      giulia rodrigues
      Whenever you have time could you please finish the answer youstarted on the question,

      "In an industrial cooling process, water is circulated through asystem. If the water is pumped with a speed of 0.45 m/s under apressure of 400 torr from the first floor through a 6.0-cm diameterpipe, what will be the pressure on the next floor 4.0 m above in apipe with a diameter of 2.0 cm?"
      thanks
    6. mustafa kavakl?o?lu
      mustafa kavakl?o?lu
      You will need bernoullies equation for this one.
      As well as the area/velocity relation


      You will need the velocity of the fluid in the pipe for the 6 cmcase.
      A2 =0.00282 m^2
      A1 = .000314m^2
      V 1 = 1.8m/s


      Now use the fluid/energy equation..

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