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Homework Help: Pressure and mass

  1. May 23, 2007 #1
    1. The problem statement, all variables and given/known data
    A water tower is a familiar sight in many towns. The purpose of such a tower is to provide storage capacity and to provide sufficient pressure in the pipes that deliver the water to customers. A spherical reservoir contains 4.75 x 10^5 kg of water when full, and is 15m above the ground. The reservoir is vented to the atmosphere at the top. There are two houses connected to the tower, the first at ground level, and the second 7.2 m higher. For a full reservoir, find the gauge pressure that the water has at the faucet in each house. Ignore the diameter of the delivery pipes.
    Here is an image:
    http://img405.imageshack.us/img405/6920/p1187yc5.gif [Broken]
    2. Relevant equations
    Pressure - atmospheric pressure = (water density * 9.8 * h)

    3. The attempt at a solution
    I've gotten 1000 * 9.8 * 15m for the pressure in the tower.
    I've tried pgh + atmospheric pressure, and it doesn't work. I don't know what to do with the mass of the water in the tower.

    I am really stumped on this one.

    Thanks a million for any help.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. May 23, 2007 #2


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    Staff Emeritus
    Science Advisor

    Given 475000 kg of water, use density of water (1000 kg/m3) to determine the volume of the sphere, from which one will obtain the R or D. The height of the sphere is D or 2R.

    With the vent at the top of the sphere, the gage pressure is 0 psig, since it is at atmospheric pressure. At the bottom of the sphere, the gage pressure is [itex]\rho[/itex]gD, then going down further to the bottom of the standpipe, add the pressure due to 15 m.

    I think you can handle the rest.
  4. Feb 7, 2012 #3
    Question 1.
    Bill and Bob have been testing a new aluminised PET (Polyethylene Terephthalate) film and they now use the material to make helium filled balloons. Bill’s balloon is less inflated than Bob’s, and the latter jokes about it. However, Bill insists he has to be careful because if he were to take the balloon on hills and mountains, the atmospheric pressure would decrease and the balloon would expand and burst. Bill then makes a bet that he can walk, drive or climb somewhere high enough for his balloon to burst...

    The balloons are partially filled with helium and they unfold as helium expands into them. The balloons burst once they reach a volume of 22.0 dm3; the pressure and temperature of the helium inside the balloons are always equal to the surrounding atmospheric pressure and temperature. Bill and Bob added respectively 8600 cm3 and 18000 cm3 of helium in their balloons, by the sea side where the altitude is 0 meters, the temperature is 25.0oC and the atmospheric pressure is 760mm Hg.

    The conditions on Mount Everest during climbing season are typically: P = 0.330atm and
    T = -20.0oC.

    (a) Calculate the volume that would be occupied by the helium in Bill’s balloon on Mount Everest (3 sig. fig.). Can Bill ever hope to win his bet? [3 marks]

    (b) The dependence of the atmospheric pressure with altitude can be approximated by the following formula:
    Patm = -7.57x10-2 h + 1
    where h is the altitude in km, Patm is the atmospheric pressure in atm.

    Calculate the altitude at which Bob’s balloon would burst, assuming that the temperature remains constant and equal to 250C (3 sig. fig).
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