Bernoulli's Equation and fire hose

[ChemIsHard]

A fire hose must be able to shoot water to the top of a building 35.0 m tall when aimed straight up. Water enters this hose at a steady rate of 0.500 m3/s and shoots out of a round nozzle.

i) What is the maximum diameter this nozzle can have?

ii) If the only nozzle available has a diameter twice as great, what is the highest point the water can reach?

I know that Flow rate=0.500 m3/s=A*V. I know the pressure needed to obtain this height from P=35*9.8*1000=3.43X10^5 Pa.

I know I should be applying Bernoulli's equation but I'm stuck. Any pointers are appreciated.

 

[ChemIsHard]

I just had an idea.

P+1/2PV^2+Pgh=pgh+1/2pv^2+P

We can cross out both p's since they're both exposed to atmosphere. The first equation has no potential while the second has no kinetic.

1/2pv^2=pgh

v=root(2g(35m))
v=26.2 m/s

0.5=26.2*Pi(r^2)

Calculate for r then times by two for maximum diameter. Then just recalculate using the previous equation with 2r.

Looks good?

 

[tomcue]

Hello! You are on the right track by applying Bernoulli's equation in this situation. I would like to provide some guidance to help you solve this problem.

First, let's review Bernoulli's equation, which states that the total energy of a fluid flowing through a pipe is constant. This can be expressed as:

P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2

Where P is the pressure, ρ is the density of the fluid, v is the velocity, and h is the height.

Now, let's apply this equation to the problem at hand. We know that the water is entering the hose at a steady rate of 0.500 m3/s and shooting out of a round nozzle. This means that the area of the nozzle (A) is the same as the area of the hose where the water is entering. We can then rewrite the equation as:

P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2

Since the water is shooting straight up, the velocity at the entrance of the nozzle (v1) is zero. This means that the first term on the left side of the equation (P1 + 1/2ρv1^2) becomes simply P1. We can also assume that the pressure outside of the hose (P2) is atmospheric pressure, which we can approximate as 101,325 Pa. With these simplifications, our equation becomes:

P1 + ρgh1 = 101,325 + 1/2ρv2^2 + ρgh2

We know that the pressure needed to reach a height of 35.0 m is 3.43x10^5 Pa, so we can substitute this value for P1 and solve for v2. This gives us a velocity of 24.5 m/s at the exit of the nozzle.

Now, let's move on to the first part of the question. We are looking for the maximum diameter that the nozzle can have in order to reach a height of 35.0 m. To do this, we can use the equation for the flow rate (Q = Av) and substitute in the velocity we just calculated (v2 =