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

  • Thread starter Thread starter Alt+F4
  • Start date Start date
  • Tags Tags
    Fluid
AI Thread Summary
The discussion revolves around calculating the time required to fill a 110-liter tub using a hose with a 3.2 cm diameter and a nozzle of 1.8 cm diameter, where water exits at 16 m/sec. The flow rate was calculated to be approximately 16.277 liters per second, leading to an initial estimate of 6.75 seconds to fill the tub. However, confusion arose regarding the use of the initial diameter and the correct application of formulas, particularly the area of a cross-section. Participants clarified the need to understand the area calculation for the nozzle and its role in determining flow rate. The conversation also briefly shifted to a separate problem involving the buoyancy of a block of ice, indicating a broader context of fluid mechanics challenges.
Alt+F4
Messages
305
Reaction score
0
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
 
Physics news on Phys.org
Can you show some work?

\dot {m} = \rho \dot{ V_{avg}} A_c

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 \pi \frac{d^2}{4}

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 origional 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 origional 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.
 
Back
Top