# 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 x 10^-3 kg s^-1, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 x 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].

## Question

Glycerine flows steadily through a horizontal tube of length $1.5 \mathrm{~m}$ and radius 1.0 $\mathrm{cm}$. If the amount of glycerine collected per second at one end is $4.0 \times 10^{-3} \mathrm{~kg} \mathrm{~s}^{-1}$, what is the pressure difference between the two ends of the tube? (Density of glycerine $=1.3 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ 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].

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## Answer

### Step 1: Calculate the volume flow rate ( Q )

Given:

$ \dot{m} = 4.0 \times 10^{-3} , \text{kg/s} $

$ \rho = 1.3 \times 10^{3} , \text{kg/m}^3 $

[ Q = \frac{\dot{m}}{\rho} = \frac{4.0 \times 10^{-3} , \text{kg/s}}{1.3 \times 10^{3} , \text{kg/m}^3} = 3.077 \times 10^{-6} , \text{m}^3/\text{s} ]

### Step 2: Calculate the pressure difference (\Delta P) using Poiseuille's Law

Given:

$\eta = 0.83 , \text{Pa s}$

$L = 1.5 , \text{m}$

$ r = 1.0 \times 10^{-2} , \text{m} $

$ Q = 3.077 \times 10^{-6} , \text{m}^3/\text{s} $

[ \Delta P = \frac{8 \eta L Q}{\pi r^4} ] [ \Delta P = \frac{8 \times 0.83 \times 1.5 \times 3.077 \times 10^{-6}}{\pi \times (1.0 \times 10^{-2})^4} ] [ \Delta P = \frac{30.623 \times 10^{-6}}{\pi \times 10^{-8}} ] [ \Delta P \approx 9.75 \times 10^2 , \text{Pa} ]

Thus, **the pressure difference between the two ends of the tube is approximately ( 975 , \text{Pa} )**.

### Check for Laminar Flow

To check if the flow is laminar, we need to calculate the Reynolds number ( Re ): [ Re = \frac{\rho v d}{\eta} ]

where:

$\rho = 1.3 \times 10^{3} , \text{kg/m}^3 $

$ d = 2r = 2 \times 1.0 \times 10^{-2} , \text{m} $

$ \eta = 0.83 , \text{Pa s}$

$v = \frac{Q}{A} $

$ A = \pi r^2 $

Let's first determine the velocity ( v ):

[ A = \pi (1.0 \times 10^{-2})^2 = \pi \times 10^{-4} , \text{m}^2 ] [ v = \frac{Q}{A} = \frac{3.077 \times 10^{-6}}{\pi \times 10^{-4}} = \frac{3.077 \times 10^{-6}}{3.1416 \times 10^{-4}} \approx 9.80 \times 10^{-3} , \text{m/s} ]

Now calculate ( Re ):

[ Re = \frac{1.3 \times 10^{3} \times 9.80 \times 10^{-3} \times 2 \times 1.0 \times 10^{-2}}{0.83} ] [ Re = \frac{2.5482 \times 10^{-1}}{0.83} \approx 3.07 ]

Since the Reynolds number is less than 2000, the flow is **laminar**.

So, the assumption of laminar flow in the tube is correct.

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