Question

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 2.80×10^{5}
N/m^{2}. 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 |

Answer #1

**(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

_______________________________________________

**(b)**

Now, area is changing, so velocity will also change

Q = 0.61 L/s = 0.00061 m^{3} /s

so,

v1 = 0.00061 / pi * 0.0145^{2} = 0.9235 m/s

v2 = 0.00061 / pi * 0.0185^{2} = 0.5676 m/s

so,

P2 = 2.80e5 + 1/2 * 1000 * (0.9235^{2} -
0.5676^{2} ) + 1000 * 9.8 * (0 - 2.15)

P2 = 2.591e5 Pa

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