Q.Glycerine flows steadily through a horizontal tube of length $1.5\ \text{m}$ and radius $1.0\ \text{cm}$. If the amount of glycerine collected per second at one end is $4.0 \times 10^{-3}\ \text{kg s}^{-1}$, what is the pressure difference between the two ends of the tube? (Density of glycerine $= 1.3 \times 10^{3}\ \text{kg m}^{-3}$ and viscosity of glycerine $= 0.83\ \text{Pa s}$). [You may also like to check if the assumption of laminar flow in the tube is correct].

The key idea is Poiseuille’s law for viscous flow through a horizontal tube, where the pressure difference drives the flow against viscous resistance.

Step 1: Find the volume flow rate.

Mass flow rate $\dot{m} = 4.0 \times 10^{-3}\ \text{kg s}^{-1}$, density $\rho = 1.3 \times 10^{3}\ \text{kg m}^{-3}$.

Volume flow rate $Q = \frac{\dot{m}}{\rho} = \frac{4.0 \times 10^{-3}}{1.3 \times 10^{3}} = 3.077 \times 10^{-6}\ \text{m}^3\text{s}^{-1}$.

Step 2: Apply Poiseuille’s law.

For a tube of radius $r = 1.0\ \text{cm} = 0.01\ \text{m}$, length $L = 1.5\ \text{m}$, viscosity $\eta = 0.83\ \text{Pa s}$:

Rearrange for $\Delta P$:

Step 3: Substitute and calculate.

$= \frac{3.064 \times 10^{-5}}{3.1416 \times 10^{-8}} \approx 975\ \text{Pa}$ …

The key here is that the flow is steady and driven entirely by a pressure difference — gravity plays no role because the tube is horizontal. The viscous force in the fluid opposes the motion, and when the flow is steady, the driving pressure force exactly balances the viscous drag. That balance is what Poiseuille’s equation captures.

We are given the mass flow rate, not the volume flow rate. That’s a small but important twist — we must convert using the density. Also, the problem asks us to check whether the flow is laminar, which means we need to compute the Reynolds number after finding the average speed.

Let’s go step by step.

1. Convert mass flow rate to volume flow rate

Mass flow rate $\dot{m} = 4.0 \times 10^{-3}\ \text{kg s}^{-1}$.

Density $\rho = 1.3 \times 10^{3}\ \text{kg m}^{-3}$.

Volume flow rate $Q$ is:

We’ll keep it as $Q \approx 3.08 \times 10^{-6}\ \text{m}^3\text{s}^{-1}$.

2. Apply Poiseuille’s equation

For a horizontal tube of radius $r$, length $L$, with fluid of viscosity $\eta$, the volume flow rate for laminar flow is:

Here $r = 1.0\ \text{cm} = 0.01\ \text{m}$, $L = 1.5\ \text{m}$, $\eta = 0.83\ \text{Pa s}$.

Rearrange for $\Delta P$:

3. Plug in the numbers

First compute $r^4 = (0.01)^4 = 1.0 \times 10^{-8}\ \text{m}^4$.

Now:

Calculate numerator stepwise:

$8 \times 0.83 = 6.64$

$6.64 \times 1.5 = 9.96$

$9.96 \times 3.0769 \times 10^{-6} = 3.0646 \times 10^{-5}$

So:

Using $\pi \approx 3.1416$:

Rounding to two significant figures (matching the given data): $\Delta P \approx 9.8 \times 10^2\ \text{Pa}$. …

Q=mdot/rho=(4.0e-3)/(1.3e3)=3.08e-6 m^3/s. Poiseuille: deltaP=8etaLQ/(pir^4)=9.8e2 Pa. Check Re: v=Q/(pi*r^2)~=9.8e-3 m/s, Re= …