A sump pump (used to drain water from the basement of houses built below the wat

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 0.750 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 2.70 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 1.00 m in height, and widens to 4.00 cm diameter. What is the pressure now (in N/m2)? N/m2

Explanation / Answer

A)

from bernoullis eqution.

P1+1/2d*v1*v1 + d*g*h1 = P2+1/2d*v2*v2 + d*g*h2.

3.5*10^5 + 1000*9.8*h1 = P2 + 1000*9.8*h2.

h2 - h1 = 2.7m.

so

3.5*10^5 +1000*9.8*(-2.7) = p2.

p2 = 323540 N/M2.

= 3.24*10^5 N/M2.

SINCEAREAS are same v1 = v2 ....as A1*v1 = A2*v2.

B)

using the same bernouliis equation.

P1+1/2d*v1*v1 + d*g*h1 = P2+1/2d*v2*v2 + d*g*h2.

3.5*10^5 + 1000*9.8*h1 = P2 + 1000*9.8*h2.

h2 - h1 = 1 m

so

3.5*10^5 +1000*9.8*(-1) = p2.

p2 = 340200 N/M2.

= 3.4*10^5 N/M2.

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