Usually fluid 1 will be a liquid and fluid 2 a gas so the density requirement

Usually fluid 1 will be a liquid and fluid 2 a gas so

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incapable of mixing, in order for the manometer to function. Usually, fluid 1 will be a liquid and fluid 2 a gas; so the density requirement is easily satisfied. It is also typically the case that in static situations as we are treating here, mixing of gases with a liquid is negligible unless pressures are extremely high. We will assume that pressures are low to moderate in the present example. Our task is to calculate the pressure p 2 using the measured values of height h i , i = 1 , 2 , 3, given in the figure under the assumption that the densities are known. We will further assume that the measurement is being done at sea level, and at a known temperature, so that both p atm and the gravitational acceleration g are specified. To determine p 2 we will first write the equation of fluid statics for each of the separate fluids. We note that this should be expected because the height of the first fluid in the manometer will be determined mainly by pressure in the second fluid. Thus, we write ∂p 1 ∂z = ρ 1 g , ∂p 2 ∂z = ρ 2 g . In order to solve these we must specify boundary conditions. It is clear from the figure and the above discussion that the boundary condition for the first equation should be p 1 ( h 3 ) = p atm . Furthermore, we must always require that pressure be continuous across an interface between two static fluids, for if it were not continuous there would be unbalanced forces (neglecting surface tension) at the interface tending to move it until the forces balanced. Continuity of pressure provides the boundary condition needed to solve the second equation, and it leads to coupling of the equations that is necessary for the height of fluid 1 to yield information on the pressure of fluid 2. The interface between the two fluids is at z = h 1 , and continuity of pressure at this location implies p 2 ( h 1 ) = p 1 ( h 1 ) . We are now prepared to solve the two equations. Integration of each of these yields p 1 ( z ) = ρ 1 gz + C 1 , and p 2 ( z ) = ρ 2 gz + C 2 . We can find the integration constant C 1 by imposing the first boundary condition; we have: p atm = p 1 ( h 3 ) = ρ 1 gh 3 + C 1 C 1 = p atm + ρ 1 gh 3 . Thus, the complete ( particular ) solution to the first equation is p 1 ( z ) = p atm + ρ 1 g ( h 3 z ) . At this point we recognize that evaluation of this at z = h 1 will provide the boundary condition needed for the second equation via continuity of pressure across the interface between the two fluids; we obtain p 1 ( h 1 ) = p atm + ρ 1 g ( h 3 h 1 ) = p 2 ( h 1 ) ,
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4.1. FLUID STATICS 107 with the last equality coming from the second boundary condition. We can now determine C 2 by evaluating the equation for p 2 ( z ) above at h 1 and using the result just obtained. Thus, we have p 2 ( h 1 ) = ρ 2 gh 1 + C 2 = p 1 ( h 1 ) = p atm + ρ 1 g ( h 3 h 1 ) . The only unknown in this sequence of equalities is C 2 , so we can solve for it by equating the second and last of the above expressions: ρ 2 gh 1 + C 2 = p atm + ρ 1 g ( h 3 h 1 ) , which results in C 2 = p atm + ρ 1 g ( h 3 h 1 ) + ρ 2 gh 1 .
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