2HW: Fluid Dynamics Begin Date: 11/8/2017 11:59:00 PMDue Date: IISVT7 TT59:00 PM

Question

2HW: Fluid Dynamics Begin Date: 11/8/2017 11:59:00 PMDue Date: IISVT7 TT59:00 PM End Date: TIEUTY TTSU0 PM (10%) Problem 5: 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 0.75 L/s, with an output pressure of 2.8 x 10 N/m2 50% Part (a) The water enters a hose with a 2.6 cm inside diameter and rises 2 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 0.5s m in height, and widens to 43 crn in diameter what is the pressure now in Nim2? You may neglect frictional losses Grade Summary

Explanation / Answer

Given,

volume flow rate = 0.75 L/s = ; P = 2.8 x 10^5 N/m^2

a)Pressure due to the height:

P = rho g H

P' = 1000 x 9.81 x 2.1 = 20601 Pa

So the required pressure at that point will be:

Pr = P - P' = 2.8 x 10^5 - 20601 = 2.594 x 10^5 Pa

Hence, Pr = 2.594 x 10^5 Pa = 2.6 x 10^4 Pa

b)H = 2.1 - 0.55 = 1.55

P = rho g H

P' = 1000 x 9.81 x 1.55 = 15205.5 Pa

So the required pressure at that point will be:

Pr = P - P' = 2.8 x 10^5 - 15205.5 = 2.65 x 10^5 Pa

Hence, Pr = 2.65 x 10^5 Pa

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