Question 9.4: The bottle of champagne shown in Figure 9.21 is under pressu...
The bottle of champagne shown in Figure 9.21 is under pressure, as indicated by the mercury-manometer reading. Compute the net vertical force on the 2-in.-radius hemispherical endcap at the bottom of the bottle.
Given: Pressure measurement; bottle geometry.
Find: Net vertical force on hemispherical surface.
Assume: No relative motion of fluid elements (hydrostatics); fluids have constant, uniform density, and it is appropriate to evaluate densities at 68˚F. We look up values for champagne and mercury at this temperature and find that (ρg)_{_C} = 59.9 lbf/ft³ and that (ρg)_{_M} = 847 lbf/ft³.
We have a manometer that gives us the champagne pressure at a height of 6 in. (We denote values at this position by the subscript “*”), if we work through the U-tube as we did in Example 9.1:
p _{ \ast } +(ρ g)_c \left\lgroup\frac{2}{12} ft\right\rgroup – (ρ g)_M \left\lgroup\frac{4}{12} ft\right\rgroup = P_{atm}=0 (gage),
so
p _{ \ast } =(ρ g)_M(0.333 \textrm{ ft})-(ρ g)_c (0.167 \textrm{ ft})= (847 lbf/ft³)(0.333 ft) – (59.9 lbf/ft³)(0.167 ft)
= 272 lbf/ft² = 272 psf,
This pressure p _{ \ast } acts on the circular cross-sectional area of the champagne bottle at a height of 6 in., imparting a resultant force of
p _{ \ast } A=(272 \textrm{ psf})(\frac{\pi}{4} (\frac{4}{12}\textrm{ ft} )^2 ) =23.74 lbfon the champagne below it. In addition, the champagne below this 6-in. height imparts its own force on the hemispherical surface. The vertical component of this force, which we are looking for, can also be interpreted as the weight of this champagne above the surface. The net vertical force on the endcap will thus be the p _{ \ast }A force already calculated, plus the weight of the fluid below * and above the hemispherical surface. Since we know the specific weight of the champagne, we need only to find the volume between * and the endcap (Figure 9.22).
F_V = p_{ \ast }A + weight of champagne
p_{ \ast }A + (ρg)_C [ π(0.167)^2(0.5)-\frac{2π}{3} (0.167)^3]
= 23.74 lbf + [2.61 – 0.58] lbf
= 25.8 lbf.
