A fire hose exerts a force on the person holding it. This is because the water accelerates as it goes from the hose through the nozzle. How much force is required to hold a 7.5 cm diameter hose delivering 450 L/min through a 0.80cm diameter nozzle?

The velocity will be increased as the inverse ratio of areas:

New velocity/oldvelocity=(7.5/.8)^2

The old velocity can be solved by

450L/min*1dm^3/L * 1min/60sec is flow in dm^3/sec, so divide that by area and you get linear flow.

acceleration=changevelocity/time so on a per second basis,
acceleartion=newvelocity-oldvelocity
=oldvelocity((7.5/.8)^2-1)

and then, Force=mass/second*acceleration
= density*450L/60sec*acceleratin

check my thinking.

To determine the force required to hold the hose, we can use Bernoulli's principle and the equation of continuity.

First, let's find the velocity of the water as it exits the nozzle. We can use the equation of continuity, which states that the product of the area and velocity of a fluid remains constant.

We have the following information:
- Diameter of the hose (d1): 7.5 cm
- Diameter of the nozzle (d2): 0.80 cm
- Flow rate (Q): 450 L/min

Converting the flow rate to m^3/s:
450 L/min = 450/60 L/s = 7.5 L/s = 0.0075 m^3/s

Using the equation of continuity:
A1v1 = A2v2

Where:
A1 = π(d1/2)^2 (Cross-sectional area of the hose)
v1 = velocity of water in the hose
A2 = π(d2/2)^2 (Cross-sectional area of the nozzle)
v2 = velocity of water at the nozzle

Rearranging the equation:
v2 = (A1/A2) * v1

Substituting the values:
v2 = (π(7.5 cm/2)^2) / (π(0.80 cm/2)^2) * v1
v2 = (28.27 cm^2) / (0.502 cm^2) * v1
v2 ≈ 56.29 * v1

Now we can calculate the velocity at the nozzle in terms of v1.

Next, we need to apply Bernoulli's principle, which states that the sum of pressure energy, kinetic energy, and potential energy in a fluid system remains constant.

Assuming the pressure outside the nozzle is atmospheric pressure, we can equate the pressure at the nozzle (P2) to atmospheric pressure (P0) and solve for the pressure difference (ΔP).

P2 + (1/2)ρv2^2 = P0 + (1/2)ρv0^2

Where:
P2 = pressure at the nozzle
ρ = density of water
v2 = velocity at the nozzle
P0 = atmospheric pressure
v0 = velocity of water outside the nozzle (assumed to be negligible)

The term (1/2)ρv0^2 can be ignored since the velocity outside the nozzle is considered negligible.

Simplifying the equation:
P2 = P0 + (1/2)ρv2^2

The force exerted on the person holding the hose is equal to the pressure difference (ΔP) multiplied by the cross-sectional area of the nozzle (A2):

Force = ΔP * A2

Now we can calculate the force required to hold the hose.

Note: In this explanation, we made some assumptions and approximations. In practice, the actual force may vary due to factors such as turbulence, friction, and other effects.

To determine the force required to hold the fire hose, we need to consider the change in momentum of the water as it flows from the hose through the nozzle.

First, we need to calculate the velocity of the water at the hose and the nozzle.

Given:
- Diameter of the hose (D1) = 7.5 cm
- Diameter of the nozzle (D2) = 0.80 cm
- Flow rate of water (Q) = 450 L/min

We can convert the flow rate from liters per minute to cubic meters per second:

Flow rate (Q) = 450 L/min = (450/60) L/s = 7.5 L/s = 0.0075 m^3/s

Next, we can calculate the velocities at the hose (v1) and the nozzle (v2) using the equation for flow rate:

Q = A × v

where A is the cross-sectional area of the hose/nozzle and v is the velocity.

The cross-sectional area of a circle is calculated using the formula:

A = π × (D/2)^2

where π is approximately 3.14 and D is the diameter.

For the hose:

A1 = π × (D1/2)^2 = 3.14 × (7.5 cm/2)^2 = 3.14 × (0.075 m)^2 = 0.01767 m^2

For the nozzle:

A2 = π × (D2/2)^2 = 3.14 × (0.8 cm/2)^2 = 3.14 × (0.008 m)^2 = 0.0000201 m^2

Now, we can calculate the velocities:

Q = A × v

For the hose:

0.0075 m^3/s = 0.01767 m^2 × v1
v1 = 0.0075 m^3/s / 0.01767 m^2 = 0.423 m/s

For the nozzle:

0.0075 m^3/s = 0.0000201 m^2 × v2
v2 = 0.0075 m^3/s / 0.0000201 m^2 = 373.134 m/s

Now that we have the velocities, we can calculate the change in momentum (Δp) of the water:

Δp = m × Δv

where m is the mass of the water and Δv is the change in velocity.

The mass (m) of the water flowing through the hose can be calculated using the density (ρ) of water and the volume flow rate (Q):

m = ρ × Q

The density of water (ρ) is approximately 1,000 kg/m^3.

m = 1,000 kg/m^3 × 0.0075 m^3/s = 7.5 kg/s

Now, we can calculate the change in momentum:

Δp = m × Δv = m × (v2 - v1)

Δp = 7.5 kg/s × (373.134 m/s - 0.423 m/s) = 7.5 kg/s × 372.711 m/s = 2805.833 kg·m/s

Finally, the force (F) required to hold the hose is given by Newton's second law of motion:

F = Δp / Δt

where Δt is the time it takes for the water to flow through the nozzle.

To calculate Δt, we need to know the distance from the hose to the nozzle. Assuming they are close together, we can estimate Δt to be very small.

Therefore, the force required to hold the fire hose is:

F = 2805.833 kg·m/s / Δt

Note that without precise information about Δt, we cannot determine the exact force required to hold the hose.