Question

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].

Answer 1

9.8 × 10² Pa

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 a rate of 4.0 × 10^–3 kg s^–1.

M = 4.0 × 10^–3 kg s^–1

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

Viscosity of glycerine, η = 0.83 Pa s

Volume of glycerine flowing per sec:

`V = M/rho`

`=(4.0 xx 10^(-3))/(1.3 xx 10^3)`

= 3.08 × 10^–6 m³ s^–1

According to Poiseville’s formula, we have the relation for the rate of flow:

`V = (pipr^4)/8etal`

Where, p is the pressure difference between the two ends of the tube

`:.  p = (V8etal)/pir^4`

`= (3.08 xx 10^(-6) xx 8 xx 0.83 xx 1.5)/(pi xx (0.01)^4)`

= 9.8 × 10² Pa

Reynolds’ number is given by the relation:

`R =(4rhoV)/pideta`

`= (4xx1.3xx10^3xx3.08 xx 10^(-6))/(pixx(0.02)xx0.83) = 0.3`

Reynolds’ number is about 0.3. Hence, the flow is laminar.

Answer 2

l = 1.5 m, `r = 1 xx 10^(-2)m`

`"Volume/s", V = "Mass/s"/"Density" = (4xx10^(-3))/(1.3 xx 10^3) m^3 s^(-1)`

`= 4/1.3 xx 10^(-6) m^3 s^(-1)`

`eta = 0.83`  Pa s

Now, `V = (pipr^4)/(8etak)`, where p is the pressure difference across the capillary.

or `p = (8Vetal)/(pir^4)`

Substituting values,

`p = 8 xx 4/1.3 xx 10^(-6) xx 0.83 xx 1.5 xx 7/22 xx 1/10^(-8) Pa`

`= 9.75 xx 10^2 Pa`

The Reynolds number is 0.3. So the flow is laminar