Problem: A sump pump (used to drain water from the basement of houses built below the water table) is draining flooded basement at a rate of 0.750 L/s, with an output pressure of 3.00x105 N/m2.(a) The water enters a hose with a 3.00-cm inside diameter and rises 2.50 m above thew pump. What is the pressure at this point?(b) the hose goes over the foundation wall, losing 0.500 m in height, and widens to 4.00 cm in diameter. what is the pressure now? You may neglect frictional losses in both parts of the problem.

FREE Expert Solution

From Bernoulli's principle:

$\boxed{{\mathbf{P}}_{\mathbf{1}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\rho }}_{\mathbf{1}}{{\mathbf{v}}_{\mathbf{1}}}^{\mathbf{2}}\mathbf{+}{\mathbf{\rho }}_{\mathbf{1}}{\mathbf{h}}_{\mathbf{1}}\mathbf{g}\mathbf{=}{\mathbf{P}}_{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\rho }}_{\mathbf{2}}{{\mathbf{v}}_{\mathbf{2}}}^{\mathbf{2}}\mathbf{+}{\mathbf{\rho }}_{\mathbf{2}}{\mathbf{h}}_{\mathbf{2}}\mathbf{g}\mathbf{=}\mathbf{c}\mathbf{o}\mathbf{n}\mathbf{s}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{t}}$

(a)

Since it has the same velocity, the equation is reduced to:

${\mathbf{P}}_{\mathbf{1}}\mathbf{+}{\mathbf{\rho }}_{\mathbf{1}}{\mathbf{h}}_{\mathbf{1}}\mathbf{g}\mathbf{=}{\mathbf{P}}_{\mathbf{2}}\mathbf{+}{\mathbf{\rho }}_{\mathbf{2}}{\mathbf{h}}_{\mathbf{2}}\mathbf{g}$

P₂ is obtained as:

$\begin{array}{rcl}{\mathbf{P}}_{\mathbf{2}}& \mathbf{=}& {\mathbf{P}}_{\mathbf{1}}\mathbf{+}{\mathbf{\rho }}_{\mathbf{1}}{\mathbf{h}}_{\mathbf{1}}\mathbf{g}\mathbf{-}{\mathbf{\rho }}_{\mathbf{2}}{\mathbf{h}}_{\mathbf{2}}\mathbf{g}\\ & \mathbf{=}& {\mathbf{P}}_{\mathbf{1}}\mathbf{+}\mathbf{\rho g}\mathbf{(}\mathbf{∆}\mathbf{h}\mathbf{)}\\ & \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{)}\mathbf{(}\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{)}\end{array}$

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