Water is forced out of a fire extinguisher by air pressure. How much gauge air pressure in the tank (above atmospheric) is required for the water jet to have a speed of 34.6 m/s when the water level is d = 0.520 m below the nozzle?

To determine the gauge air pressure required for the water jet to have a speed of 34.6 m/s, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid. The equation is as follows:

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

Here, P₁ is the initial pressure in the fire extinguisher tank, v₁ is the initial velocity of the water jet, h₁ is the initial height of the water level below the nozzle, P₂ is the atmospheric pressure, v₂ is the final velocity of the water jet (34.6 m/s), h₂ is the final height of the water level below the nozzle, ρ is the density of water, and g is the acceleration due to gravity.

Since we are looking for the gauge air pressure (above atmospheric), we can rearrange the equation as follows:

P₁ - P₂ = ½ρv₂² + ρgh₂ - ½ρv₁² - ρgh₁

To solve for the gauge air pressure, we need to substitute the given values into the equation and solve for P₁ - P₂.

Given:
v₁ = 0 m/s (initial velocity at rest)
v₂ = 34.6 m/s
h₁ = 0.520 m (initial height below the nozzle)
h₂ = 0 m (final height, at the nozzle)
ρ = 1000 kg/m³ (density of water)
g = 9.8 m/s² (acceleration due to gravity)

Substituting these values into the equation, we have:

P₁ - P₂ = ½ * 1000 * (34.6)² + 1000 * 9.8 * 0 - ½ * 1000 * (0)² - 1000 * 9.8 * 0.520

Simplifying the equation gives:

P₁ - P₂ = ½ * 1000 * (1197.16) - 1000 * 9.8 * 0.520

P₁ - P₂ = 598580 - 5096

P₁ - P₂ ≈ 593484 Pa

Therefore, the gauge air pressure required in the tank (above atmospheric) for the water jet to have a speed of 34.6 m/s is approximately 593,484 Pa.

To determine the gauge air pressure required for the water jet to have a speed of 34.6 m/s when the water level is 0.520 m below the nozzle, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid.

Bernoulli's equation can be written as:

P + (1/2)ρv^2 + ρgh = constant

Where:
P is the pressure of the fluid (in this case, air pressure)
ρ is the density of the fluid (in this case, water)
v is the velocity of the fluid (in this case, the water jet velocity)
g is the acceleration due to gravity
h is the height of the fluid (in this case, the distance between the nozzle and the water level)

At the level of the nozzle, there is no height difference (h = 0). Thus, the equation simplifies to:

P + (1/2)ρv^2 = constant

Since we are interested in the air pressure (P), we can rearrange the equation as follows:

P = constant - (1/2)ρv^2

To find the value of the constant, we need to know the air pressure at some point (let's say atmospheric pressure, which we'll denote as Patm). Thus, we can rewrite the equation as:

P = Patm - (1/2)ρv^2

Substituting the given values:
v = 34.6 m/s (velocity of the water jet)
ρ = 1000 kg/m^3 (density of water)
h = 0.520 m (distance between the nozzle and water level)
g = 9.8 m/s^2 (acceleration due to gravity)
Patm = atmospheric pressure (varies, typically around 101325 Pa)

We can now calculate the gauge air pressure (above atmospheric) required:

P = Patm - (1/2)ρv^2

P = Patm - (1/2)ρv^2
P = Patm - (1/2)(1000 kg/m^3)(34.6 m/s)^2
P = Patm - 600156 Pa

The gauge air pressure required for the water jet to have a speed of 34.6 m/s when the water level is 0.520 m below the nozzle is Patm - 600156 Pa.