a u shaped tube is partially filled with water. one end of the tube is then capped so that the air in that end is trapped. the tube has a bore diameter of 10 mm. the initial height of the entrapped air gap is 200mm. a guage pressure of 1599mm hg is then applied to the other end of the tube. CALCULATE THE DIFFDRENCE IN HEIGHT OF THE WATER COLUMN ONEACH SIDE OF THE TUBE AFTER THE PRESSURE IS APPLIED

Question

a u shaped  tube is partially filled with water. one end of the tube is then capped so that the air in that end is trapped. the tube has a bore diameter of 10 mm. the initial height of the entrapped air gap is 200mm. a guage pressure of 1599mm hg is then applied to the other end of the tube. CALCULATE THE DIFFDRENCE IN HEIGHT OF THE WATER COLUMN ONEACH SIDE OF THE TUBE AFTER THE PRESSURE IS APPLIED

mstoldegon · Answer

I will try to give some help.
   Assume that the U shaped tube is sufficiently tall enough to pressurize the open end without the water being pushed down to the U and letting the air leak past.
   That said, I think the 10 mm diameter is not important as the U shape would cancel out any effect the "weight" of water would have on the diameter of the water at the bottom of the U - thats my thinking.
   Atmospheric pressure is 760 mm. That is the pressure that is now in the closed portion as it was the pressure on both sides before one side was capped. If the water is pushed up 100 mm, the pressure will double to to 1520 mm hg but only 760 mm hg of this would push back against the 1599 mm hg pressure applied. As gauge pressure is o when under 760 mm hg of atmospheric pressure.
   A quick, but I feel slightly inaccurate calculation, would be:
   D1 = 200 mm, D2 is the new height of the air in the closed column above the water.
   D2 = (200)*((760)/(1599+760)) = 64.43 mm
   D3 = The difference in the columns (2 * D2) = 2*(200-64.43) = 271.14 mm
That looks like it should solve the problem but I think one also has to take into consideration the weight of the water in the height difference between the columns. If so that makes the equation far more complex - I THINK??
   What do you think?

mstoldegon · Answer

BTW, was this the question you wanted to ask in the Physics section?

mstoldegon · Answer

I must sign out now but I will work on this tomorrow. It seems like there might be a simple equation to combine the force of the compressed air and the weight of the water.

anonymous · Answer

a 3m x 3m x 4m room contains air with a moisture content of 11.6 g h20/dry air at 1 atm (absolute) pressure and 25 degree Celsius. the dry air component can be assumed to have a composition of 21 mol% oxygen and 79 mol% nitrogen.
1) calculate the mass fraction of water vapour in the room.2
2) mole fraction of water vapour in the room?
3) ppm of N2?
4) average molecular wt of air?
5) molar concentration of water vapour in the room(lbmol/ft3)?