Question

A fire hose has an inside diameter of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of 1.62×106 N/m2 . The hose goes 10.0 m up a ladder to a nozzle having an inside diameter of 3.00 cm. Calculate the Reynolds numbers for flow in the fire hose and nozzle to show that the flow in each must be turbulent

Answer 1

Answer:

for the hose

$R_e=1.269\times 10^{5}$

hence, greater than 2000. thus the flow is turbulent

for the nozzle

$R_e=1.694\times 10^{6}$

hence, greater than 2000. thus the flow is turbulent

Explanation:

Given:

Diameter of hose, d₁ = 6.40 cm = 0.064 m

Flow, Q = 40.0 L/s = 40.0 × 10⁻³ m³/s

Gauge pressure, P = 1.62 × 10⁶ N/m²

Diameter of the nozzle, d₂ = 3.0 cm = 0.03 m

Now,

Q = A₁V₁

A₁ is the area at the hose

V₁ is velocity at the hose

Now,

V₁ = Q/A₁

on substituting the values, we get

V₁ =$\frac{40\times 10^{-3}}{\frac{\pi d_1^2}{4}}$

or

V₁ =$\frac{40\times 10^{-3}}{\frac{\pi 0.064^2}{4}}$

or

V₁ = 12.43 m/s

Also

Q = A₂V₂

A₂ is the area at the nozzle

V₂ is velocity at the nozzle

Now,

V₂ = Q/A₂

on substituting the values, we get

V₂ =$\frac{40\times 10^{-3}}{\frac{\pi d_1^2}{4}}$

or

V₂ =$\frac{40\times 10^{-3}}{\frac{\pi 0.03^2}{4}}$

or

V₂ = 56.6 m/s

Now,

The Reynold's number (Re) is given as:

$R_e=\frac{2\rho v r}{\eta}$

where,

ρ = density of water = 1000 kg/m³

η = viscosity coefficient = 1.002 × 10⁻³

thus,

for the hose

$R_e=\frac{2\times1000\times12.43\times (0.064/2)}{1.002\times 10^{-3}}$

or

$R_e=1.269\times 10^{5}$

hence, greater than 2000. thus the flow is turbulent

for the nozzle

$R_e=\frac{2\times1000\times56.6\times (0.03/2)}{1.002\times 10^{-3}}$

or

$R_e=1.694\times 10^{6}$

hence, greater than 2000. thus the flow is turbulent