Question

Fire hoses used in major structure fires have inside diameters of 6.0 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of 2.2×106 N/m2. The hose goes 10.0 m up a ladder to a nozzle having an inside diameter of 2.8 cm.
a.   What is the pressure in the nozzle?
b.   What is the speed of water coming out of the nozzle?

Answer 1

a)

V = Volume flow rate = 40 L/s = 40 x 10^-3 m³/s

d_n = diameter of nozzle = 2.8 cm = 0.028 m

A_n = area of cross-section of the nozzle = (0.25) $\pi$ d_n² = (0.25) (3.14) (0.028)² = 0.00061544 m²

v_n = speed of water coming out of nozzle = ?

Volume flow rate is given as

V = A_n v_n

(40 x 10^-3) = (0.00061544) v_n

v_n = 65 m/s

d_h = diameter of hose = 6 cm = 0.06 m

A_h = area of cross-section of the hose = (0.25) $\pi$ d_h² = (0.25) (3.14) (0.06)² = 0.002826 m²

v_h = speed of water coming out of hose = ?

Volume flow rate is given as

V = A_h v_h

(40 x 10^-3) = (0.002826) v_h

v_h = 14.2 m/s

P_h = gauge pressure at hose = 2.2 x 10⁶ Pa

P_n = gauge pressure at nozzle

h₁ = 0

h₂ = 10 m

Using Bernoulli's theorem

P_h + $\rho$ g h₁ = P_n + $\rho$ g h₂

2.2 x 10⁶ + (1000) (9.8) (0) = P_n + (1000) (9.8) (10)

P_n = 2.102 x 10⁶ Pa