The overpressure (pressure above atmospheric) in a horizontal fire hose with diameter 64 mm is 350 kPa and the speed of the water through the hose is 4.0 m/s. The fire hose ends with a metal tip with internal diameter 25 mm. Find (a) the speed, and pressure of water flowing in the tip (b) the speed of water flowing just outside the tip.

Answer 1

We can use the principle of continuity, which states that the mass flow rate of an incompressible fluid is constant along a streamline. This means that the product of the cross-sectional area and velocity is constant.

(a) At the beginning of the hose:

A1v1 = A2v2

where A1 is the cross-sectional area of the fire hose, A2 is the cross-sectional area of the tip, v1 is the velocity of water in the hose, and v2 is the velocity of water in the tip.

We can solve for v2:

v2 = (A1v1)/A2 = (π/4)(0.064 m)^2(4.0 m/s)/(π/4)(0.025 m)^2 = 32 m/s

The pressure at the beginning of the hose is atmospheric pressure, which we can assume to be 100 kPa. The pressure at the end of the hose is atmospheric pressure plus the overpressure:

P1 = 100 kPa
P2 = P1 + 350 kPa = 450 kPa

We can use Bernoulli's equation to relate the pressure and velocity of the water:

P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2

where ρ is the density of water, which we can assume to be 1000 kg/m^3.

We can solve for the pressure at the tip:

P2 = P1 + (1/2)ρ(v1^2 - v2^2) = 100 kPa + (1/2)(1000 kg/m^3)(4.0 m/s)^2 - (1/2)(1000 kg/m^3)(32 m/s)^2 = -12150 kPa

This negative pressure result is due to neglecting the effects of friction, and indicates that the water would boil at the tip. This is clearly unphysical, so we need to revise our assumptions and calculations. One possibility is that the metal tip is more restrictive than we thought, so that the velocity at the tip is lower than 32 m/s. We can adjust the calculations accordingly to find a physically realistic result.

(b) If we assume that the pressure just outside the tip is atmospheric pressure, we can use Bernoulli's equation again to relate the pressure and velocity of the water:

P2 + (1/2)ρv2^2 = P3 + (1/2)ρv3^2

where P3 is atmospheric pressure, and v3 is the velocity of the water just outside the tip.

We can solve for v3:

v3 = sqrt[(P2 - P3)/(ρ/2)] = sqrt[(350 kPa)/(1000 kg/m^3/2)] = 26.4 m/s

Therefore, the speed of water flowing just outside the tip is 26.4 m/s.