1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A problem about fluid mechanics

  1. Nov 27, 2016 #1
    1. The problem statement, all variables and given/known d
    At what speed, the velocity head of water is equal to pressure head of 40 cm of Hg?

    (1) 10.32 m/s
    (2) 2.8 m/s
    (3) 5.6 m/s
    2. Relevant equations
    v^2/2g = p/rho*g (velocity head= pressure head).

    3. The attempt at a solution
    since pressure head is given as 40 cm/s, i did it like v^2 = 40*2*g which i got after taking square root and all as 28.8 cm/s. i couldn't find my answer matching with any of the given options how to do this?
     
  2. jcsd
  3. Nov 27, 2016 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Where did you bring in the fact that the pressure is given in terms of Hg?
    Also, please show all your working.
     
  4. Nov 27, 2016 #3
    it was given in the ques that the pressure is in Hg. and i have already shown my working in "my attempt at the solution" .
     
  5. Nov 27, 2016 #4
    i have already shown my working in "my attempt at the solution" section. thats all i did..i equated both the equations and got an answer which is irrelevant to the options!
     
  6. Nov 27, 2016 #5

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    How do you get that from the relevant equation you quoted? What happened to p and rho?
     
  7. Nov 27, 2016 #6
    so as per the question, isn't the pressure head = 40 cm of Hg which means p/ rho*2 =40 and we know v^2/2g= presssure head = 40 cm of Hg. this is how i proceeded..is that right?
     
  8. Nov 28, 2016 #7

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I assume you meant p/ (rho*g) =40, but there are two things you need to be careful about there.
    1. The 40 is in cm - always keep track of units;
    2. The rho is the density of Hg, whereas in this equation:
    the rho is the density of the flowing fluid.
     
  9. Nov 28, 2016 #8
    so if rho is the density of flowing fluid (water) is 1 g/cm^3 right? but i m still confused whether the given value of 40 cm of Hg is the value of pressure or that of pressure head. 'cause as you told if i have to take care of rho as the density of fluid in flow (1 gm/cm^3) and p=40 cm of Hg which means 40/(1*980) =0.041 and then equating it with v^2/2g, i am not getting the result :/
     
  10. Nov 28, 2016 #9

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    "40cm" is neither here. If the problem were about flowing Hg it could be the pressure head.
    "40cm of Hg" is the pressure.
    You have p/ (rhoHg*g) =40cm. Look up what rhoHg is and calculate the pressure p.
     
  11. Nov 28, 2016 #10
    why should i calculate pressure there since i have been asked to find velocity only.can't i simply equate v^2/2g=40 cm of Hg ?
     
  12. Nov 28, 2016 #11

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Sure, but you need the pressure head expressed in terms of such-and-such a height of the flowing fluid, i.e. water. The pressure head is NOT 40cm.
     
  13. Nov 28, 2016 #12
    so what is the "40 cm of Hg" given in the question?
     
  14. Nov 28, 2016 #13

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    It is the gauge pressure at a depth of 40cm in a column of Hg. What depth of water would produce the same pressure?
     
  15. Nov 28, 2016 #14
    *40 cm of Hg column! so P=13.56(rho)*40*980(g) right?
     
  16. Nov 28, 2016 #15

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Assuming consistent units for P, yes.
     
  17. Nov 28, 2016 #16
    so i m getting 531552 in the cgs units. should i equate this with v^2/2g ? the answer is exceeding the relevant options beyond limits :/
     
  18. Nov 28, 2016 #17

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That's as a pressure in cgs units, right?
    No, that formula needs a distance, i.e. the pressure head of water that results in the above pressure.
     
  19. Nov 28, 2016 #18
    how to find that? i am baffled :/ so i have found the pressure value in cgs unit and now how to convert it as such distance value?
     
  20. Nov 28, 2016 #19

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You have found the value of the pressure, p. To find what that is as pressure head of water, you can use p/(ρg) again, but now using the density of water.
     
  21. Nov 28, 2016 #20
    ok now finding the pressure head of water 531552(pressure as found in cgs unit) /[1 (rho of water) *980] =542 and equating this with v^2/2g i m getting 1032 .the answer is not this as per the options right?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: A problem about fluid mechanics
Loading...