# 2 problems of fluids help

1. Dec 3, 2004

### 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???

2. Dec 3, 2004

### timon00

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

3. Dec 4, 2004

### Andrew Mason

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

Last edited: Dec 4, 2004
4. Dec 4, 2004

### 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 ??

5. Dec 4, 2004

### Andrew Mason

Just forgot to multiply by the .5. See edited reply above

AM