Solve Hard Fluid Problem: Calculate Minimum Work & Power of Pump

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The discussion focuses on calculating the minimum work and power required for a pump to move water from a well 35 meters deep to a house 50 meters above the well. The calculations reveal that the minimum work required is 291,550 Joules, and the minimum power rating of the pump is 40.5 Watts, based on a daily water usage of 0.35 m³ over 2 hours. Additionally, the flow velocity at the faucet is determined to be 2.88 m/s, utilizing the principle of conservation of mass and Bernoulli's equation for pressure calculations.

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  • Understanding of fluid mechanics principles, particularly Bernoulli's equation.
  • Knowledge of basic physics concepts such as work, power, and density.
  • Familiarity with unit conversions, especially between metric units (m, kg, s).
  • Ability to apply the continuity equation for fluid flow (A1V1 = A2V2).
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  • Learn about the continuity equation and its applications in fluid dynamics.
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  • Investigate pump efficiency and factors affecting performance in hydraulic systems.
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Engineers, physics students, and professionals involved in fluid dynamics, pump design, and hydraulic systems will benefit from this discussion.

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A pump, submerged at the bottom of a well that is 35 m deep, is used to pump water uphill to a house that is 50 m above the top of the well. The density of water is 1000 kg/cm^3. All pressures are gauge pressures. Neglect the effect of friction, viscosity.

A) REsidents of the house use .35 m^3 of water per day. The day's pumping is completed in 2 hours during the day.
1) calculate the minimum work required to pump the water used per day
2) Calculate the minimum power rating of the pump

B) The average pressure the pump actually produces is 9.2X10^5 N/m^2. Within the well the water flows at .50 m/s and the pipe has a diameter of 3cm. At the house the pipe diameter is 1.25cm
1) Calculate the flow velocity when a faucet in the house is open
2) Explain how you could calculate the minimum pressure at the faucet

http://www.collegeboard.com/prod_downloads/ap/students/physics/b_physics_b_frq_03.pdf
picture here, problem #6

attemp solution
A)
1- M=DV
M=1000(.35)= 350kg = 350000g
W=mgh
W=350000(9.8)(30+50) 30from bottom to ground, 50 from ground to the house
W=274400000J
2- P=W/t
P= 274400000J/7200s 2hour=7200s
P=38111.1 Watt

B)
1- AV=AV
(.03/2)^2*pi*.05= (.0125/2)^2*pi*V
V=2.88 m/s
2- P=F/A
I know the area, but i don't know how to get the F,

I need a confirm, or any suggestion
Appreciate
 
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I'm doing this problem as well!
I just wish to comment on something.
Why did you convert 0.35m^3 to 350000g?
Everything else is in kg... and
on part A1 doesn't h=35+50=85?

Now I think the solution to P=F/A

F=ma
and from Rho(density)=m/V
and a=g
P=*rho*(V)(g)/A


I hope this makes sense.
 
W=Fd
W=(mg)h
W=(rho)Vgh
Note: (rho) = density - in this case of water
W=1000*.35*9.8*85
W=291550 Joules

P=W/t
P=291550/7200
Note 2hrs = 7200s
P=40.5 watts

Next one you have done correctly..
A1v1=A2v2
V2 = 2.88 m/s

Last one.. You must use bernoullis principle
P+(.5(rho)v^2)+((rho)gh)
P is given, rho = density of water(1000kg/m**3)
v = 2.88 m/s, g is 9.8, h is 85m

What you were doing wrong was not converting units properly
Always stick to m, kg, s
 

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