a) the product of flow area and velocity is constant
V' = 0.88*(2.1/1.0)^2 = 3.88 m/s is the velcoity in the smaller-diameter pipe.
b) use the Bernoulli principle with variable height
P + (1/2) (density)*V^2 + (density)* y = constant
y = 7.1 m on the second floor and 0 at the input point
water density = 1000 kg/m^3
169*10^3 Pa + (1/2)(1000)(0.88)^2 + 0
= P' + (1/2)(1000)V'^2 + (1000 kg/m^3)(9.8 m/s^2)*7.1 m
Solve for P'
A water pipe having a 2.1 cm inside diameter carries water into the basement of a house at a speed of 0.88 m/s and a pressure of 169 kPa. The pipe tapers to 1.0 cm and rises to the second floor 7.1 m above the input point.
(a) What is the speed at the second floor?
? m/s
(b) What is the water pressure at the second floor?
? kPa
I have lost my book
1 answer
1 answer