# Hard fluid problem

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

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 dont know how to get the F,

I need a confirm, or any suggestion
Appreciate

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 doesnt 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