Question

A sump pump (used to drain water from the basement of houses built below the water table) is draining a flooded basement at the rate of 0.600 L/s, with an output pressure of 3.50 ✕ 10⁵ N/m². You may neglect frictional losses in both parts of this problem.

(a) The water enters a hose with a 3.00 cm inside diameter and rises 3.00 m above the pump. What is its pressure (in N/m²) at this point?

_________ N/m²

(b) The hose goes over the foundation wall, losing 1.30 m in height, and widens to 4.00 cm diameter. What is the pressure now (in N/m²)?

__________ N/m²

Answer 1

a. Initially y = 0, Finally y = 3 m

velocity will remain constant since there is no change in cross section

therefore, we can calculate P' as

3.5 x 105 Pa = P' + (3*9.8 * 1000)

P' = 329600 Pa.

b.

we can calculate velocity of water

Cross Sectional Area of Pipe = pi * 1.5 * 1.5 = 7.07cm2

Flow Rate = 600 cm3/s

therefore velocity = 600/7.07 = 84.86 cm/s = 0.8486 m/s
final velocity = 0.9194 * (9/16) = 0.4773 m/s
so we get eqn as
329600 + (1/2 * 1000 * (0.84)^2) = P'' - (9.8 * 1000 * 1.3) + (1/2 * 1000 * (0.47)^2)
So we get P'' = 342582 Pa