336 THE WATER PROBLEM OF A LARGE CITY 



the walls required. The greater the elevation of the houses, 

 the stronger must be the pumps and the engines which run 

 them. 



In order to understand how these calculations are made, we 

 must learn about the pressure of water. 



One cubic foot of it weighs about 62.5 pounds ; this is equiva- 

 lent to saying that water i foot deep presses on the bottom of 



the containing vessel with a force of 

 62.5 pounds to the square foot. If 

 the water is 2 feet deep, the load 

 supported by the vessel is doubled, 

 and the pressure on each square foot 

 of the bottom of the vessel will be 

 125 pounds, and if the water is 10 

 feet deep, the load borne by each 

 square foot will be 625 pounds. The 

 deeper the water, the greater is the 

 weight sustained by the confining 

 vessel and the greater the pressure 

 exerted by the water. 



Since the pressure borne by I 

 square foot of surface is 62.5 pounds, 

 the pressure supported by I square 

 of 62.5 pounds, or .43 pound, nearly ^ 



FIG. 205. Water i foot deep 

 exerts a pressure of 62.5 pounds 

 a square foot. 



inch of surface is 

 pound. Suppose a vessel held water to the depth of 10 feet, 

 then upon every square inch of the bottom of that vessel there 

 would be a pressure of 4.34 pounds. If a one-inch tap were 

 inserted in the bottom of the vessel so that the water flowed 

 out, it would gush forth with a force of 4.34 pounds. If the 

 water were 20 feet deep, the force of the outflowing water 

 would be twice as strong, because the pressure would be 

 doubled. But the flow would not remain constant. As the 

 water leaves the outlet, less and less of it remains in the 



