A large tank of water has a hose connected to it. The tank is sealed at the top

Question

A large tank of water has a hose connected to it. The tank is sealed at the top and has compressed air between the water surface and the top. When the water height h has the value 3.50m , the absolute pressure p of the compressed air is 4.20 x 10^5 {{ Pa}}. Assume that the air above the water expands at constant temperature, and take the atmospheric pressure to be 1.00 imes 10^5 {{ Pa}}. Part A What is the speed with which water flows out of the hose when h = 3.50m ? Part B As water flows out of the tank, h decreases. Calculate the speed of flow for h = 3.10m . Part C Calculate the speed of flow for h =2.10m .

Explanation / Answer

I AM GOING TO ASSUME THAT THE HEIGHT OF THE TANK IS 4 METERS. I THINK THERE IS A PICTURE THAT IS SUPPOSED TO GO WITH THIS PROBLEM, BUT YOU DIDNT INCLUDE IT. IF THIS PROBLEM IS FROM YOUNG & FREEDMAN, I THINK I HAVE THE RIGHT ONE... BUT I HAVE TO GUESS. PLEASE PROVIDE ALL RELEVANT INFO SO WE DONT HAVE TO GUESS TO ANSWER YOUR QUESTIONS.

the pressure of the air changes as the height changes. Calc the height of the air in each case

Part A: air height = 4 - 3.5 = 0.5

Part B: 4-3.1 = 0.9

Part C: 4-2.1 = 1.9

Air pressure in each case then is

Part A: 420000 given

Part B: 420000*0.5/0.9 = 233333

Part C: 420000*0.5/1.9 = 110526

Now use bernoullis... in each case:

pressure of air + density*g*height = 100000 + (1/2) density * v^2 + density*g*1.00

or

pressure + 1000*9.8*height = 100000 + 500v^2 + 1000*9.8*1.00

p + 9800 h = 100000 + 500v^2 + 9800

or

v^2 = ( p + 9800h - 91200)/500

So now...

Part A: v^2 = (420000 + 9800*3.5 - 91200 ) / 500 = 726.2

v = 26.95 m/s

Part B: v^2 = (233333+ 9800*3.1 - 91200 ) / 500 = 726.2

v = 18.57 m/s

Part C: v^2 = (110526 + 9800*2.1 - 91200 ) / 500 = 726.2

v = 8.934 m/s

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