How Long Does It Take to Fill a Tub Using a Hose with Different Diameter Ends?

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Homework Help Overview

The discussion revolves around a fluid dynamics problem involving a hose with varying diameters and the time required to fill a tub with water. The original poster presents a scenario with specific measurements and velocities, seeking assistance in calculating the time to fill a tub of 110 liters.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss flow rates and the application of formulas related to fluid dynamics, including cross-sectional area calculations. There are inquiries about the relevance of the initial diameter and the clarity of the problem statement.

Discussion Status

Some participants are attempting to clarify their understanding of the problem and the calculations involved. There is a mix of interpretations regarding the initial diameter's role, and one participant expresses confusion about the problem's wording. Guidance has been offered in the form of formula references and calculations, but no consensus has been reached.

Contextual Notes

There is mention of a potential misunderstanding regarding the problem's requirements and the need for clarity on the area of cross-section. Additionally, another unrelated problem about a block of ice is introduced, which may contribute to the confusion in the discussion.

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Help me with this fluid Problem ASAP Please :)

A long horizontal hose of diameter 3.2 cm is connected to a faucet. At the other end, there is a nozzle of diameter 1.8 cm. Water squirts from the nozzle at velocity 16 m/sec. Assume that the water has no viscosity or other form of energy dissipation.


Okay so How long will it take to fill a tub of volume 110 liters with the hose ?

I got the flow rate which is 16.277 liter, 110/162.77 = 6.75 why isn't this the asnwer
 
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Can you show some work?

[tex]\dot {m} = \rho \dot{ V_{avg}} A_c[/tex]

This is a steady state nozzle.
 
o you what did was

((1.8)/ (100))^2 * 3.14 * 16 = .016277 M^3/S = Flow rate

I then converted it to Liters by multiplying 1000
 
Area of a cross section is [tex]\pi \frac{d^2}{4}[/tex]

Why did the problem give you the initial diameter? Are you sure this is how the problem is worded?
 
yep exactly, i copied and pasted from online HW, o you intial diameter was for another part i needed to find Velcotit which i did
 
What? You found the inital diameter in another part? Can you post the original problem. I have no clue what is what anymore :confused:.
 
sorry one more problem,

A rectangular block of ice 14 m on each side and 0.5 m thick floats in seawater. The density of the seawater is 1025 kg/m3. The density of ice is 917 kg/m3.

a) How high does the top of the ice block float above the water level?

Okay so i know Fb=pgVdis

pwatergVdis = picegVdis

(1025)(9.8)(14*1.1*14)+ (x)(m*g) = 917*9.8*X



What Am i missing, thanks so much
 
Last edited:
cyrusabdollahi said:
What? You found the inital diameter in another part? Can you post the original problem. I have no clue what is what anymore :confused:.
i got the problem, thanks a lot i guess i didnt know what the area of a cross section is since all it says on formula sheet is the area of circle.
 

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