Maximum height for water from a fire hose

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Homework Statement


Fire hose has diameter of 4.0 cm and flow rate of 10 L/s. There is pressure of 2.2 bar inside the hose. How high the water can go at best? Water density is 1.00E3 kg/m^3 and air pressure outside the hose is 1.0 bar.

Homework Equations


Flow rate
$$ Q = Av $$
Newtons equations?

The Attempt at a Solution



I tired to find speed for water particle when it exits the hose. I can get speed directly from flow rate.

$$ v = \frac{Q}{A} = \frac{Q}{(\frac{d}{2})^2 \pi} $$
$$ v \approx 7.957 \text{ m/s} $$

If we would aim the hose at 45 angle it would follow height as function of time as:

$$ h(t) = vt-gt^2 $$

i solved the maxima with first derivative $$ t = \frac{v}{2g} \approx 0.4055 \text{ s} $$

then you would compute the height from the function $$ h(0.41\text{ s}) \approx 1.6 \text{ m} $$

However the correct answer should be 26m. What I'am doing wrong?
 
on Phys.org
hoses have nozzles to convert pressure to speed
kuru was faster to point you in the right direction
note they ask for maximum height, so I d aim a little higher :wink:
 
BvU said:
hoses have nozzles to convert pressure to speed
kuru was faster to point you in the right direction
note they ask for maximum height, so I d aim a little higher :wink:
kuruman said:
Try Bernoulli's equation and don't forget the necessary unit conversions.

I tried Bernoulli's equation and ended up with this.
$$ p_1 + \frac{1}{2}pv^2_1 = p_2 + \frac{1}{2}pv_2^2 $$
Speed can be expressed as
$$ v = \frac{Q}{\pi (\frac{d}{2})^2} $$
$$ \implies v_2 = \sqrt{\frac{p_1-p_2}{\rho}+(\frac{Q}{(\frac{d}{2})}^2 \pi)^2\cdot 2} $$
when i compute with numbers i get 19m/s for speed.

Computing maximum height with mgh = 1/2 mv² we get that h = v²/2g, which gives approx 18.3 m.
Correct answer was 26 m. Any suggestion on what I'am doing wrong?
 
Since the problem is asking for the speed, use Bernoulli's equation between a point just inside the hose and the point at maximum height. You don't really need the speed. Also, I get 26 m if I assume that the pressure outside is zero. It looks like there is an error in the quoted answer that needs to be checked.
 
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