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

Given Data

Parameter	Value
Length (\(L\))	1.5 m
Radius (\(r\))	1.0 cm = 0.01 m
Mass flow rate (\(\dot{m}\))	4.0 × 10-3 kg/s
Density (\(\rho\))	1.3 × 103 kg/m³
Viscosity (\(\eta\))	0.83 Pa·s

Solution: Pressure Drop

Steady laminar flow through a horizontal tube uses Poiseuille's law:

$$\Delta P = \frac{8\eta L Q}{\pi r^4}$$

First, calculate volume flow rate \(Q\):

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

Now compute \(\Delta P\):

$$\Delta P = \frac{8 \times 0.83 \times 1.5 \times (3.077 \times 10^{-6})}{\pi \times (0.01)^4}$$

Step-by-step:

• Numerator: \(8 \times 0.83 \times 1.5 \times 3.077 \times 10^{-6} = 3.048 \times 10^{-5}\)
• Denominator: \(\pi \times 10^{-8} = 3.142 \times 10^{-8}\)
• \(\Delta P = \frac{3.048 \times 10^{-5}}{3.142 \times 10^{-8}} = 970 \, \text{Pa}\)

Pressure Difference

\(\Delta P = 970 \, \text{Pa}\)

Laminar Flow Check

Calculate Reynolds number \(Re = \frac{\rho v D}{\eta}\), where \(D = 2r = 0.02\) m.

Average velocity \(v = \frac{Q}{A} = \frac{Q}{\pi r^2}\):

$$v = \frac{3.077 \times 10^{-6}}{\pi \times (0.01)^2} = 0.0982 \, \text{m/s}$$

$$Re = \frac{1.3 \times 10^3 \times 0.0982 \times 0.02}{0.83} = 307$$

Flow Verification

\(Re \approx 307 \ll 2000\) → Laminar flow assumption is correct.