Water flows from a fire truck through a hose that is 11.7 cm in diameter and has a nozzle that is 2.0 cm in diameter. The firemen stand on a hill 5.0 m above the level of the truck. When the water leaves the nozzle, it has a speed of 20.0 m/s.
1)
Determine the minimum gauge pressure in the truck’s water tank.
h=5m v=20 m/s hose@r=.0585m and nozzle@radius=.01m
g=9.8 p=density of water=1000
P-Po=pgh
P-Po=1000*9.8*5
P-Po=49000 Pa
Minimum gauge pressure in the truck’s water tank=49000 Pa
Well, it seems like the firefighters have really elevated the situation! And I don't just mean because they're standing on a hill. Let's calculate the minimum gauge pressure in the truck's water tank.
To start with, we'll use the equation P - Po = pgh, where P is the pressure at the nozzle, Po is the pressure at the water tank, p is the density of water, g is the acceleration due to gravity, and h is the height difference between the nozzle and the water tank.
Given:
h = 5.0 m (height difference)
v = 20.0 m/s (velocity at the nozzle)
r_hose = 0.117 m (radius of the hose)
r_nozzle = 0.02 m (radius of the nozzle)
p = 1000 kg/m³ (density of water)
g = 9.8 m/s² (acceleration due to gravity)
Now, let's determine the velocity of the water at the hose using the principle of conservation of mass. Since the water is incompressible, the flow rate is constant throughout.
A_hose * v_hose = A_nozzle * v_nozzle
Where A_hose and A_nozzle are the respective cross-sectional areas of the hose and the nozzle.
A_hose = π * r_hose²
A_nozzle = π * r_nozzle²
Substituting the given values:
π * (0.117 m)² * v_hose = π * (0.02 m)² * 20.0 m/s
Simplifying the equation:
v_hose = (0.02 m)² * 20.0 m/s / (0.117 m)²
v_hose ≈ 2.92 m/s
We now have the velocity of the water at the hose, and we can proceed to calculate the minimum gauge pressure using the equation mentioned earlier.
P - Po = pgh
P - Po = p * g * h
Substituting the given values:
P - Po = 1000 kg/m³ * 9.8 m/s² * 5.0 m
Simplifying:
P - Po = 49000 Pa
So, the minimum gauge pressure in the truck's water tank is 49000 Pa. That's quite a lot of pressure! The firefighters better make sure they have a good grip on those hoses. Safety first!
First, we need to find the height difference (h) between the firemen and the nozzle. Given that the firemen stand on a hill 5.0 m above the level of the truck, this is our starting point.
Next, we need to calculate the change in pressure (P - Po) using the formula P - Po = pgh, where p is the density of water (1000 kg/m³), g is the acceleration due to gravity (9.8 m/s²), and h is the height difference.
Substituting the given values into the equation, we have (P - Po) = (1000 kg/m³)(9.8 m/s²)(5.0 m).
Calculating this expression, we find (P - Po) = 49,000 kg m²/s².
Therefore, the minimum gauge pressure in the truck's water tank is 49,000 Pa (or 49 kPa).
To determine the minimum gauge pressure in the truck's water tank, we can use the equation P - P₀ = ρgh, where P is the pressure at the bottom of the hose, P₀ is the pressure at the nozzle, ρ is the density of water, g is the acceleration due to gravity, and h is the height difference between the nozzle and the bottom of the hose.
Given:
- h = 5.0 m (height difference)
- v = 20 m/s (speed of water leaving the nozzle)
- hose diameter = 11.7 cm = 0.117 m (radius = 0.0585 m)
- nozzle diameter = 2.0 cm = 0.02 m (radius = 0.01 m)
- g = 9.8 m/s² (acceleration due to gravity)
- ρ = 1000 kg/m³ (density of water)
First, we need to find the pressure at the nozzle (P₀) using Bernoulli's equation:
P₀ + 1/2ρv₀² + ρgh₀ = P + 1/2ρv² + ρgh,
where P is the pressure at the bottom of the hose, v is the velocity of water at the bottom of the hose (which is zero), h₀ is the height at the bottom of the hose, and v₀ is the velocity at the nozzle.
Since v₀ = 20.0 m/s, and the height at the bottom of the hose (h₀) is zero (as it is at the same level as the bottom of the hose), the equation simplifies to:
P₀ + 1/2ρv₀² = P.
Now we can substitute the given values and solve for P₀:
P₀ = P - 1/2ρv₀² = P - 1/2 * 1000 * (20.0)² = P - 10^6 P.
Next, we can calculate the pressure at the bottom of the hose (P) using the equation P - P₀ = ρgh:
P = P₀ + ρgh = (P - 10^6 P) + 1000 * 9.8 * 5.0.
Simplifying this equation, we find:
P = 10^6 P + 49000.
Now, we can isolate P:
10^6 P - P = 49000,
(10^6 - 1)P = 49000,
P = 49000 / (10^6 - 1).
Calculating this, we find:
P ≈ 0.049073 atm.
Therefore, the minimum gauge pressure in the truck's water tank is approximately 0.049073 atm.