Class Management Help 12HW: Fluid Dynamics Begin Date: 11/8/2017 11:59:00 PM-Due
ID: 1782964 • Letter: C
Question
Class Management Help 12HW: Fluid Dynamics Begin Date: 11/8/2017 11:59:00 PM-Due Date: 11/15/2017 11:59:00 PM End Date: 11/172017 11:59:00 PM (10%) Problem 5: A sump pump (used to drain water from the basement ofhouses built below the water table) is draining a flooded basement at the rate of 0.75 Lis, with an output pressure of 2.8 x 105 N/m2 50% Part (a) The water enters a hose with a 2.6 cm inside diameter and rises 21 m above the pump what is its pressure at this point N/m2? You may neglect frictional losses. 50% Part (b) The hose goes over the foundation wall, losing 0.55 m in height, and widens to 4 l cm in diameter, what is the pressure now in N/m2? You may neglect frictional losses. Grade Summary Deductions 0% Potential 100% Submissions coso tan() | |Explanation / Answer
part A:
from bernoullis theorem
P1 + 0.5 rho v1^2 + rho g h1 = P2 + 0.5 rho v2^2 + rho g h2
here A1 = A2 and from eqn of continuity A1 V1 = A2 V2
gives v1 = V2
P1 + rho g h1 = P2 + rho g h2
P1 + 0 = P2 + rho g h2
P2 = (2.8*10^5) -(1000* 9.8 * 2.1)
P2 = 2.49 *10^5 Pa
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let the initial speed of water flow at the hose be v1
volume flow rate dV = v1 * A1
given dV = 0.75 *10^-3
Diamter D = 0.026 m
Radius r = D/2 = 0.026/2 = 0.013 m
v1 * 3.14 * (0.013)^2 = 0.75 *10^-3
v1 = 1.413 m/s
now , for the hose at the foundation let the speed be v2
A2 V2 = V
v2 * 3.14 * (0.043/2)^2 = 0.75 *10^-3
V2 = 0.516 m/s
apply from bernoulis theorem
P2 - P1 = p * g * h + 0.5 * p * (v1^2 - v2^2)
P2 = P1 + p * g * h + 0.5 * p * (v1^2 - v2^2)
P2 = (2.8 *10^5) + (1000 * 9.8 * 0.55) + (0.5 *1000 *(1.413^2 - 0.516^2)
P2 = 2.862 *10^5 Pa is the pressure.