Page:Elementary Text-book of Physics (Anthony, 1897).djvu/153

§ 120] '''120. Diminution of Pressure.'''—The Sprengel air-pump, an important piece of apparatus to be described hereafter, depends for its operation on the diminution of pressure at points along the line of a flowing column of liquid. Let us consider a large reservoir filled with liquid, which runs from it by a vertical tube entering the bottom of the reservoir. From equation (53) the value of $$p$$, the pressure at any point in the tube, is $$p = p_{1} + (V_{1} - V)d - \tfrac{1}{2}dv^2 \left(1 - \frac{A^2}{A_{1}^2}\right)\cdot$$ The ratio $$\frac{A^2}{A_{1}^2}$$ may be set equal to zero. If $$h$$ (Fig. 40) represent the height of the upper surface above the point in the tube at which we desire to find the pressure, then $$(V_{1} - V) = gh$$. We then have $$p = p_{a} + dgh - \tfrac{1}{2}dv^2$$. If the tube be always filled with the liquid, $$Av = A_{0}v_{0}$$, where $$A$$ and $$A_{0}$$ represent the areas of the cross-sections of the tube at the point we are considering and at the bottom of the tube, and $$v$$ and $$v_{0}$$ represent the corresponding velocities. Further, $$v_{0}^2 = 2gh_{0}$$ if $$h_{0}$$ represent the distance from the upper surface to the bottom of the tube. We obtain, by substitution, If $$h$$ equal $$\frac{A_{0}^2}{A^2}h_{0}$$, we have $$p = p_{a};$$ and if an opening be made in the wall of the tube, the moving liquid and the air will be in equilibrium. If $$h$$ be less than $$\frac{A_{0}^2}{A^2}h_{0}$$, the pressure $$p$$ will be less than $$p_{a}$$, and air will flow into the tube. Since this inequality exists when $$A_{0} = A$$, it follows that, if a liquid flow from a reservoir down a cylindrical tube, the pressure at any point in the wall of the tube is less than the atmospheric pressure by an amount equal to the pressure of a column of the liquid, the height of which is equal to the distance between the point considered and the bottom of the tube.