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

Answer:

the question is too long to understand.

Answer 2

Given :-

Length of the horizontal tube = 2 m

Radius of the tube = 1 cm = 0.01 m

Diameter of the tube = 0.02 m

To Find :-

The pressure difference between the two ends of the tube.

Solution :-

We know that,

• l = Length
• r = Radius
• d = Diameter
• v = Volume
• m = Mass

Given that,

Length (l) = 2 m

Radius (r) = 0.01 m

Diameter (d) = 2r = 0.02 m

According to the question,

Glycerin is flowing at the rate of $\sf 2.0 \times 10^{-3} \ kg/s$

$\sf M = 2.0 \times 10^{-3} \ kg/s$

Density of glycerin, $\sf \rho = 1.3 \times 10^{3} \ kg \ m^{-3}$

Viscosity of glycerin, $\sf \eta = 0.83 \ Pa \ s$

We know, volume of glycerin flowing per sec

$\sf V=\dfrac{M}{Density}$

Substituting their values,

$\sf V=\dfrac{2 \times 10^{-3}}{1.3 \times 10^{3}}$

$\sf V=1.54 \times 10^{-6} \ m^{3} /s$

Using Poiseville’s formula,

$\sf V=\dfrac{\pi P' r^{4}}{8 \eta l}p'=\dfrac{V8 \eta l}{\pi r^{4}}$

Where, p’ = Pressure difference between the two ends of the pipe

$\sf p'=\dfrac{1.54 \times 10^{-6} \times 8 \times 0.83 \times 2}{\pi \times 0.01^{4}}$

$\sf p'=6.51 \times 10^{2} \ Pa$

And, we know,

$\sf R=\dfrac{4 \rho P}{\pi d \eta}$

Where, R = Reynolds’s number

$\sf R=\dfrac{4 \times 1.3 \times 10^{3} \times 1.53 \times 10^{-6}}{\pi \times 0.83 \times 0.02}$

$\sf R=0.153$

Since the Reynolds’s number is 0.153 which is way smaller than 2000, the flow of glycerin in the pipe is laminar.