A water pipe having a 1.0 inch inside diameter carries water into the basement of a house at a speed of 3.0 ft/s and a pressure of 25 lb/in^2. If the pipe tapers to 0.5 inch and rises to a second floor 25 ft above the input point,
(a) what is the speed at the second floor?
Work:
d1 = pipe diameter in basement = 1.0 inch
p1 = pipe pressure in basement = 25 lb/in^2
v1 = water speed in pipe in basement = 3.0 ft/s
d2 = pipe diameter on second floor = 0.5 inch
h = height that pipe rises = 25 ft
R = AV
A1V1 = A2V2
V2 = A1V1/A2
V2 = (0.08 ft)^2*(3 ft/s)/(0.04 ft)^2
V2 = 12 ft/s
(b) the water pressure at the second floor?
Work:
p2 = p1 + ro*g*h,
where p = pressure, ro = density, g = gravitational constant, h = height
p2 = 25 lb/in^2 + (1,000 kg/m^3)*(9.8 m/s^2)*(7.6 m)
P2 = 25 lb/in^2 - (25 kg/in^3)*(385.2 in/s^2)*(300 in)
p2 = 25 lb/in^2 - 12 lb/in^2
p2 = 13 lb/in^2
Solution
(a) V2 = 12 ft/s
(b) p2 = 13 lb/in^2