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  • #46
    Re: Here is one for you math nuts

    Don't confuse PSI with the total force acting on a surface, which is PSI TIMES square inches.

    We had a central fan room on submarines, which was pressurized to something like 2" WC with the fan on high. The door was about 24" by 60". You could not open the door with the fan on high. Too much force on the door. You could shift the fan to low speed, and then open the door.

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    • #47
      Re: Here is one for you math nuts

      Originally posted by lovetheUSA View Post
      Don't confuse PSI with the total force acting on a surface, which is PSI TIMES square inches.

      We had a central fan room on submarines, which was pressurized to something like 2" WC with the fan on high. The door was about 24" by 60". You could not open the door with the fan on high. Too much force on the door. You could shift the fan to low speed, and then open the door.

      I was on a sub in a previous lifetime. I heard a story while we were in refit in the Holy Loch once... Imagine if you will, someone smoking a joint in the fan room. His theory was, reportedly, that the fans would dissipate the smell so much all over the boat that nobody would notice. His hypothesis didn't survive the experiment, and he was arrested.
      Time flies like an arrow.

      Fruit flies like a banana.

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      • #48
        Re: Here is one for you math nuts

        Originally posted by lovetheUSA View Post
        Don't confuse PSI with the total force acting on a surface, which is PSI TIMES square inches.

        .
        Yes, You and tool are correct, this is what I was looking for.
        The clog (paper towels and wipes) was packed like I've never seen before, they minds well have been flushing wool blankets.

        Also a section of very cheap pipe looked as though it blew from the inside out so I was trying to get an idea of the forces involved to see if this was even possible.

        I jumped the gun a bit on the angel. One very short section was close to a 45 but for the most part it was closer to 22 1/2.
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        • #49
          Re: Here is one for you math nuts

          Originally posted by G3sprinklers View Post
          Are you sure??
          not saying I am correct so check my figures and thinking

          Agree with what you have coming at you, that is volume.
          And once it gets to moving velocity pressure will come into play and this can be measured with a pitot tube

          Lets just say you have a piece of tubing that is 100 feet tall and filled with water, ID of 0.25 inch (you can cover that with you thumb).
          surface area of plug (or cross sectional area of the pipe) = (pi)(r squared)
          = (3.14159)(0.125" squared)
          = 0.049 sq. inches

          volume = (area)(height)
          = (0.049 sq. in.)(1200 in.)
          =58.8 cubic inches

          weight of the water = 0.0361 pounds / cu. in.
          = (58.8 cu. in.)(0.0361 lb / cu.in.)
          = 2.12 pounds

          pressure head = (height of water column in feet)(0.434 psi / foot)
          = (100 feet)(0.434 psi / foot)
          = 43.4 pounds per sq. inch

          Now same set up with ID of 3 inch (you should be able to cover that with the palm of you hand)
          area of plug = 7.069 sq. in.
          volume = 8,482.8 cu. in.
          weight of water = 306.22 pounds
          pressure head = 43.4 psi

          The pressure head is the same, right?
          right

          Which is easier to hold back? and why? (we are just talking static pressure not velocity pressure).
          the 0.25" tube, takes less pressure to overcome the weight of the water

          Is there something different going on with the suface area of the plug?
          head pressure is the same, so why is it harder to hold back the 3 inch tube

          pressure on the face of the plug
          area of 0.25" tube = 0.049 sq. in.
          head pressure = 43.4 pounds / sq. in.
          = (0.049 sq. in.)(43.4 lb. / sq. in.)
          = 2.12 pounds

          area of 3" tube = 7.069
          head pressure = 43.4
          = (7.069)(43.4)
          = 306.7 pounds

          What about the weight difference in the water, in volume?
          difference in weight, 2 compared to 306

          Would there be any difference in the force required to hold it back if there was a 0.25 inch orifice at the bottom of the 3 inch pipe (pressure head would be the same I know)?
          the pressure required to hold it back would be the same for the 0.25" plug reguardless of the size of the pipe diameter above.

          G3
          Originally posted by lovetheUSA View Post
          Don't confuse PSI with the total force acting on a surface, which is PSI TIMES square inches.
          If you will look at the mat above you will see that is exactly what was done...........

          you are saying .........(pounds / sq. in.)(sq. inches)...where do we get the figures to plug into the variables??

          the figures you are talking about are as follows....
          psi = pressure head = 43.4 psi
          sq. in. = the surface area of the plug = 7.069 and 0.049 sq. in.

          (pounds / sq. in.)(sq. inches)

          [when you multiply this out the sq. in. cancell each other out]

          Plug those in, work it out and what you end up with is pounds (see figures above in red), which is a force (static energy, it is at rest, not moving). The force (or weight) of the water is acting directly on the surface area of the plug.

          If you had the 3" tube reduced down to a 0.25" outlet with a plug in it the force acting on the face of the plug will be smaller since the plug is smaller, even though the pressure head is the same. The shoulders of the reducer would be holding the rest of the force or weight.

          Clear as mud, right??

          G3

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