Answer to Question #184391 in Mechanics | Relativity for Konoha

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 3.00×10⁵ N/m².

1. The water enters a hose with a 3.00-cm inside diameter and rises 2.50 m above the pump. What is its pressure at this point? Blank 1
2. 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? Blank 2

Round your answer to the nearest 0 decimal place

Answer 1

Answer :-

from Bernoulli's Equation

$p+\frac{1}{2}\rho v^2 +\rho gh = cons.$

$P \ is \ pressure \ of \ the \ fluid \\ \rho \ is \ the \ densty \ of \ the \ fluid \\ g \ is \ the \ acceleration \ due \ to \ gravity \\ v \ is \ the \ velocity \ of \ the \ fluid \\ \Delta h \ is \ the \ change \ into \ the \ height .$

(1)

$P_1 +\frac{1}{2}\rho v^2 +\rho gh_1 = P_2+\frac{1}{2}\rho v^2 +\rho gh_2$ (velocity is same)

$P_2= P_1 + \rho gh_1 - \rho gh_2$

$= P_1 + \rho g(h_1-h_2)$

$=P_1 + \rho g(\Delta h)$

$= 3 \times 10^5 + 1000 \times 9.8 \times 2.5$

$\boxed{P = 3.24 \times 10 ^5 N/m^2}$

(2)

Fron the rate law

$Q = Av$

by given information

we know that

$Q = A_1v_1 \\ v_1 = \frac Q A_1 \\ \\$

= $\frac Q {\pi r_1^2}$

$= \frac{0.750\times 10^{-3}}{\pi \times 0.015^2}$

$= 1.06$

$\boxed{V_1 = 1.06 m/s}$

now

$Q = A_2v_2 \\ v_2 = \frac Q A_2 \\ \\$

= $\frac Q {\pi r_2^2}$

$= \frac{0.750\times 10^{-3}}{\pi \times 0.02^2}$

$= 0.597$

$\boxed{V_2 =0.597m/s}$

now

using

$P_2 = P_1+\frac{1}{2}\rho (v_1-v_2) + \rho g (h_1-h_2)\\$

$= 3\times 10^5 - \frac{1}{2}\times 1000(1.06^2-0.597^2) +1000\times 9.8 \times 2.5$

$= 2.75\times 10^5$

$\boxed{P_2 = 2.75 \times 10^5 N/m^2}$