Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10–3 kg s–1, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 × 103 kg m–3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].

Asked by Pragya Singh | 1 year ago |  62

1 Answer

Solution :-

Given:
Length of the horizontal tube, l = 1.5 m

Radius of the tube, r = 1 cm = 0.01 m

Diameter of the tube, d = 2r = 0.02 m

Glycerine is flowing at the rate of 4.0 × 10-3 kg/s

M = 4.0 × 10-3 kg/s

Density of glycerine, ρ = 1.3 × 103 kg m-3

Viscosity of glycerine, η = 0.83 Pa s

We know, volume of glycerine flowing per sec:

V =\( \dfrac{ M }{ density }= \dfrac{ 4\; \times \;10^{ -3 } }{ 1.3\; \times \;10^{ 3 } }\)

= 3.08 x 10 -6 m3 /s

Using Poiseville’s formula, we get:

V = \( \dfrac{ \pi p^{‘} r^{ 4 }}{ 8\; \eta \;l}p^{‘} = \dfrac{ V \;8\; \eta \;l}{ \pi r^{ 4 }}\)

Where p’ is the pressure difference between the two ends of the pipe.

p’ = \( \dfrac{3.08\; \times \;10^{-6} \;\times \;8\; \times \;0.83\; \times \;2 }{\pi \;\times \;(0.01)^{4} }\)

​ = 9.8 x 10Pa

And, we know:
R = \( \dfrac{ 4 \;\rho \;V }{ \pi \;d\; \eta }\)​, [Where R = Reynolds’s number]

R = \( \dfrac{4 \times 1.3 \times 10^{3} \times 3.08\times 10^{-6} }{\pi \times 0.83 \times 0.02 }= 0.3\)

Since the Reynolds’s number is 0.3   which is way smaller than 2000 , the flow of glycerine in the pipe is laminar.

Answered by Abhisek | 1 year ago

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