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

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In summary, the conversation discusses the use of a pump to transport water from a well to a house. The minimum work required to pump the daily water usage and the minimum power rating of the pump are calculated. Additionally, the flow velocity when a faucet in the house is open and the minimum pressure at the faucet are calculated using Bernoulli's principle. The conversation also addresses proper unit conversions and the use of density and gravity in calculations.
  • #1
logglypop
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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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  • #2
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.
 
  • #3
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
 

1. What is a hard fluid problem?

A hard fluid problem is a complex and challenging issue that involves the flow of a fluid through a system. It often requires advanced mathematical and scientific techniques to solve and can be found in various industries, such as engineering, physics, and chemistry.

2. Why is it important to calculate the minimum work and power of a pump?

Calculating the minimum work and power of a pump is crucial for understanding and optimizing the performance of a system. It can help determine the most efficient and cost-effective way to move fluids through a system, leading to energy savings and reduced costs.

3. How do you approach solving a hard fluid problem?

Solving a hard fluid problem requires a systematic approach, starting with defining the problem and identifying the relevant variables. Then, mathematical equations and principles are applied to the problem to develop a solution. Finally, the solution is tested and refined to ensure accuracy and relevance.

4. What factors affect the minimum work and power of a pump?

The minimum work and power of a pump are influenced by various factors, including the type of fluid, flow rate, pressure, and the design and efficiency of the pump itself. Other external factors, such as temperature and viscosity, can also impact these calculations.

5. Can computer software be used to solve hard fluid problems?

Yes, computer software can be a powerful tool for solving hard fluid problems. Many programs are specifically designed to handle complex fluid dynamics calculations, making the process more efficient and accurate. However, it is essential to have a thorough understanding of the problem and the software to ensure the validity of the results.

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