1. The problem statement, all variables and given/known data An inverted U-tube monometer, as shown in Figure attached , has air at the top of the tube. If the pipes contain oil (s.g. = 0.9), h1= 0.6 m, h2= 1.8 m and h= 0.45 m, determine the difference in pressure between point B and point A. 2. Relevant equations 3. The attempt at a solution for my analysis i dedicded to use the equation defining the pressures from point a to be as : Pa - (900kg/m3*9.81m/s2*0.6m)- (101325Pa) +((0.6m+1.8m)*9.81m/2*900kg/m3)) = Pb = which implied (Pa-Pb) = (900kg/m3*9.81m/s2*0.6m) + 101325Pa - ((0.6m+1.8m)*9.81m/2*900kg/m3)) = 133109.4Pa = 133Kpa. the thing about the solution is i think atmospheric pressure of 101325Pa can be used as 0 gauge. which would define the equation as Pa - (900kg/m3*9.81m/s2*0.6m) +((0.6m+1.8m)*9.81m/2*900kg/m3)) = Pb. also Could an analysis just be made from point A to point D then this pressure just equated to the pressure from ( Pb to point D ) excluding atmospheric pressure term (101325) and the term ((0.45m)*9.81m/2*900kg/m3)) since the fluid at these points are at the same horizontal height so equal pressure. that is Pa - (900kg/m3*9.81m/s2*0.6m)= Pb -(900kg/m3*9.81m/s2*1.8m). verification of the right analysis would be appreciated as im a bit confused which is the most suited approach.