# Help me with this fluid Problem Please

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 isnt this the asnwer

## Answers and Replies

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Can you show some work?

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

This is a steady state nozzle.

o ya 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 ya 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 .

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 .
i got the problem, thanks alot 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.