### Q.13:- 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 × 10^{3} 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:-**Length of the horizontal tube, *l* = 1.5 m

Radius of the tube, *r* = 1 cm = 0.01 m

Diameter of the tube, *d* = 2*r* = 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 × 10^{3} kg m^{–3}

Viscosity of glycerine, *η* = 0.83 Pa s

Volume of glycerine flowing per sec:

*V* = *M* / *ρ*

= 4 × 10^{-3} / (1.3 × 10^{3}) = 3.08 × 10^{-6} m^{3}s^{-1}

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

*V* = πpr^{4} / 8*η*l

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

∴ *p* = *V*8*ηl* / πr^{4}

= 3.08 × 10^{-6} × 8 × 0.83 × 1.5 / [ π × (0.01)^{4} ]

= 9.8 × 10^{2} Pa

Reynolds’ number is given by the relation:

*R* = 4*pV* / πd*η*

= 4 × 1.3 × 10^{3} × 3.08 × 10^{-6} / ( π × 0.02 × 0.83)

= 0.3

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