Calculate the Reynolds numbers for the flow of water through a nozzle with a radius of 0.250 cm and a garden hose with a radius of 0.900 cm, when the nozzle is attached to the hose. The flow rate through hose and nozzle is 0.500 L/s. Can the flow in either possibly be laminar

In both cases, reynolds number is greater than 2000,so the flows can't be laminar.

Explanation:

A) For flow in a tube of uniform diameter, the reynolds number is defined as;

Re = 2ρvr/η

where;

ρ is the fluid density,

v its speed,

η is viscosity

r is the tube radius.

In this question,

We are given;

r = 0.25cm = 0.25 × 10^(-2) m

η of water has a standard value of 1.005 × 10^(-3)

ρ of water has a standard value of 1000 kg/m³

In the reynolds equation, we don't know the velocity. So let's calculate it from;

Q' = vA

Where; Q' is flow rate = 0.5 L/s = 0.0005 m³/s

Area = πr² = π × (0.25 × 10^(-2))²

Area = 1.963 × 10^(-5) m²

So, v = Q/A = 0.0005/(1.963 × 10^(-5)) = 25.5 m/s

So, Re = 2ρvr/η = (2*1000*25.5*0.25 × 10^(-2))/(1.005 × 10^(-3))

Re = 126865.67

Re > 2000 and so the flow is not laminar

B) Now,

radius = 0.9cm = 0.9 × 10^(-2) m

So, A = πr² = π × (0.9 × 10^(-2))²

A = 2.5447 × 10^(-4) m²

v = Q/A = 0.0005/(2.5447 × 10^(-4))

v = 1.965 m/s

Re = 2ρvr/η = (2*1000*1.965*0.9 × 10^(-2))/(1.005 × 10^(-3))

Re = 35194.03

Re > 2000. So flow is not laminar.

twist boundary

explanation:

answer:

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