Question

# 28 A sump pump (used to drain water from the basement of houses built below the...

28

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 0.610 L/s, with an output pressure of 2.80×105 N/m2. The water enters a hose with a 2.90–cm inside diameter and rises 2.70 m above the pump. What is its pressure at this point?

 3.06*10^5 N/m^2
 Incorrect. Tries 2/40 Intentos Anteriores

The hose goes over the foundation wall, losing 0.550 m in height, and widens to 3.70 cm in diameter. What is the pressure now? You may neglect frictional losses in both parts of the problem.

 Tries 0/40

(a)

as there is no change in cross section area, the velocity will be same

v1 = v2

so,

we have (as per Bernoulli's equation)

P1 + gh1 = P2 + gh2

so,

P2 = P1 - g ( h2 - h1)

P2 = 2.80e5 - 1000 * 9.8 * (2.70 - 0)

P2 = 2.535e5 Pa

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(b)

Now, area is changing, so velocity will also change

Q = 0.61 L/s = 0.00061 m3 /s

so,

v1 = 0.00061 / pi * 0.01452 = 0.9235 m/s

v2 = 0.00061 / pi * 0.01852 = 0.5676 m/s

so,

P2 = 2.80e5 + 1/2 * 1000 *  (0.92352 - 0.56762 ) + 1000 * 9.8 * (0 - 2.15)

P2 = 2.591e5 Pa