The diameter of a capillary is 4 mm. If Reynolds's number of the tube is 1000 and the co-efficient of viscosity for the water is 0.02 poise, determine the maximum speed of water for the stream line flow in the tube?

Question

Question

This question was previously asked in

AAI ATC Junior Executive 25 March 2021 Official Paper (Shift 3)

Answer 1

Given:

The diameter of the capillary (d) = 4 mm or (0.004 m)

Reynolds's number of the tube (R) = 1000

Co-efficient of viscosity for the water (η) = 0.02 poise

Concept:

We get to know about the laminar and turbulent nature of flow through the Reynolds number.

Critical velocity is the maximum velocity above which the nature of flow no longer remains Laminar.

Formula:

The Reynolds number is given by : \(R = \frac{ρ v d}{η}\)

Where v is the average velocity of the flow.

1 Poise = 0.1 Pa-s

Density of water (ρ) = 1000 kg/m³

Calculation:

• 0.02 poise = 0.02 × 0.1 Pa-s

= 0.002 Pa-s

∵ R = ρvd/η

⇒ v = Rη/ρd

⇒ v = (1000 × 0.002)/(1000 × 0.004)

= 0.5 m/s or 50 cm/s

So, v_max = 2v = 2 × 50

⇒ vmax = 100 cm/s