2 problems of fluids help

[timon00]

16. The normal atmosphercic pressure is 1.013 x 10^5. A storm causes that the height of the barometer of mercury reduce 20.0 mm form the normal height. What is the atmosferic preasurre ? ( the density of mercury is 13.59 g/cm3


50. Out of a fire extingisher comes out water under air presurre. What manometric pressure is required in the tank for the water burst to have a speed of 30 m/s when the water level is 0.5 m below the water pump.


as hard as i try i really cant solve is problems can anyone help me???

[timon00]

i got the first one , can anyone help me with the other one?

[Andrew Mason]

50. Out of a fire extingisher comes out water under air presurre. What manometric pressure is required in the tank for the water burst to have a speed of 30 m/s when the water level is 0.5 m below the water pump.
It is not entirely clear on the question but it seems that the air pressure has to lift the water .5 m and send it out at 30 m/sec.

Think of pressure as an energy density (energy/unit volume):
P = \frac{\delta E}{\delta V}

In order to lift an element \delta m = \rho \delta v of water h = .5 m. and accelerate it to v = 30 m/sec energy of:

\delta E = \delta mgh + \frac{1}{2}\delta mv^2

is required. So an energy density of:

\frac{\delta E}{\delta V} = \rho gh + \frac{1}{2}\rho v^2 = P

P = \rho (gh + \frac{1}{2}v^2)

P = 10^3 (9.8 \times .5 + .5 \times 900)

P = 4.55 \times 10^2 KPa or about 4.55 atm. (4.55 bars)

AM

[timon00]

thanks andrew but i have another question

so i did P = 10^3(9.8*0.5 +900*0.5) = and it gives me another answer diffrent that yours it gives me 4.55 x 10^5 half of what you got , why did u multiply it by two ??

[Andrew Mason]

thanks andrew but i have another question

so i did P = 10^3(9.8*0.5 +900*0.5) = and it gives me another answer diffrent that yours it gives me 4.55 x 10^5 half of what you got , why did u multiply it by two ??
Just forgot to multiply by the .5. See edited reply above

AM