Microscopic Momentum Transport in Laminar Flow

Transport Phenomena

Rahma Muthia, S.T., M.Sc., Ph.D.

Department of Chemical Engineering, Faculty of Engineering

Universitas Indonesia 2023

Microscopic Momentum Transport part 2 Laminar Flow

This PPT is for internal lectures at DTK-UI

What examples we will look at

• Steady-state momentum balance without generation

➢ Plane Couette flow

• Steady-state momentum balance with generation

➢ Flow between parallel plates

➢ Flow in a vertical circular tube

➢ Flow in a horizontal circular tube

➢ Falling film on a vertical plate

Steady-state microscopic

momentum balance with generation

Inventory rate of equation

Inventory rate of equation

gravitational and pressure forces

Flow between parallel plates

Consider the flow of a

Newtonian fluid between two parallel plates separated by a distance B under steady-state conditions. The pressure is imposed in the z-direction while both plates are held stationary.

v_z = v_z(x) and v_x = v_y = 0 The components of the total momentum flux are:

Equation of motion:

The equation of motion is:

Flow between parallel plates

ቚ (𝜋_𝑧𝑧

𝑧 𝑊 ∆𝑥 + 𝜋_𝑥𝑧ቚ

𝑥 𝑊 ∆𝑧) −(𝜋_𝑧𝑧ቚ

𝑧+∆𝑧 𝑊 ∆𝑥 + 𝜋_𝑥𝑧ቚ

𝑥+∆𝑥 𝑊 ∆𝑧) + (𝑃ቚ

𝑧 − ቚ𝑃

𝑧+∆𝑧)𝑊 ∆𝑥 = 0

(𝜋_𝑧𝑧ቚ

𝑧 𝑊 ∆𝑥 − 𝜋_𝑧𝑧ቚ

𝑧+∆𝑧 𝑊 ∆𝑥) + (𝜋_𝑥𝑧ቚ

𝑥 𝑊 ∆𝑧 −𝜋_𝑥𝑧ቚ

𝑥+∆𝑥 𝑊 ∆𝑧) + (𝑃ቚ

𝑧 − ቚ𝑃

𝑧+∆𝑧)𝑊 ∆𝑥 = 0

Dividing the above equation by W Δx Δz and taking the limit as Δx → 0 and Δz → 0 give

The equation of motion is:

Flow between parallel plates

Assumption: the flow is fully developed, which means that velocity variations only take place in the direction perpendicular to the flow.

Therefore, dv_z/d_z = 0.

Flow between parallel plates

The substitution of equations gives:

Integration of the equation twice gives:

Identify boundary conditions:

Apply the boundary conditions to the solution obtained giving

Flow between parallel plates

Flow between parallel plates

The volumetric flow rate is the integration of the velocity distribution over the cross-sectional area:

The substitution of to the above equation gives:

Dividing the volumetric flow rate by the flow area gives the average velocity as:

Flow in a vertical circular tube

Consider the flow of a Newtonian fluid in a vertical circular pipe under steady-state conditions.

The pressure gradient is imposed in the z-direction.

v_z = v_z(r) and v_r = v_θ = 0

The components of the total momentum flux are:

Flow in a vertical circular tube

As the pressure in the pipe depends on z, it is important to consider the pressure gradient only to the z-direction.

The equation of motion is:

ቚ (𝜋_𝑧𝑧

𝑧 2𝜋𝑟 ∆𝑟 + 𝜋_𝑟𝑧ቚ

𝑟 2𝜋𝑟 ∆𝑧) − (𝜋_𝑧𝑧ቚ

𝑧+∆𝑧2𝜋𝑟 ∆𝑟 + 𝜋_𝑟𝑧ቚ

𝑟+∆𝑟 2𝜋(𝑟 + ∆𝑟) ∆𝑧) + (𝑃ቚ

𝑧 − ቚ𝑃

𝑧+∆𝑧)2𝜋𝑟 ∆𝑟 + 𝜌𝑔2𝜋𝑟 ∆𝑟∆𝑧 = 0 Note that:

Flow in a vertical circular tube

The equation of motion is:

ቚ (𝜋_𝑧𝑧

𝑧 2𝜋𝑟 ∆𝑟 + 𝜋_𝑟𝑧ቚ

𝑟 2𝜋𝑟 ∆𝑧) − (𝜋_𝑧𝑧ቚ

𝑧+∆𝑧2𝜋𝑟 ∆𝑟 + 𝜋_𝑟𝑧ቚ

𝑟+∆𝑟 2𝜋(𝑟 + ∆𝑟) ∆𝑧) + (𝑃ቚ

𝑧 − ቚ𝑃

𝑧+∆𝑧)2𝜋𝑟 ∆𝑟 + 𝜌𝑔2𝜋𝑟 ∆𝑟∆𝑧 = 0

ቚ (𝜋_𝑧𝑧

𝑧 2𝜋𝑟 ∆𝑟 − 𝜋_𝑧𝑧ቚ

𝑧+∆𝑧2𝜋𝑟 ∆𝑟) + (𝜋_𝑟𝑧ቚ

𝑟 2𝜋𝑟 ∆𝑧 − 𝜋_𝑟𝑧ቚ

𝑟+∆𝑟 2𝜋(𝑟 + ∆𝑟) ∆𝑧) + (𝑃ቚ

𝑧 − ቚ𝑃

𝑧+∆𝑧)2𝜋𝑟 ∆𝑟 + 𝜌𝑔2𝜋𝑟 ∆𝑟∆𝑧 = 0

Dividing the above equation by 2πr Δr Δz and taking the limit as Δr → 0 and Δz → 0 give

Flow in a vertical circular tube

Or,

Assumption: the flow is fully developed, which means that velocity variations only take place in the direction perpendicular to the flow.

Therefore, dv_z/d_z = 0.

giving: at r = 0, dv_z/dr = 0

r = R, v_z = 0

Flow in a vertical circular tube

The modified pressure is defined by:

Substitution to gives

Let’s assume that this side is a constant

Flow in a vertical circular tube

Substitution to gives

Integration twice gives:

Where C₁ and C₂ are constants.

Flow in a vertical circular tube

Identify boundary conditions for

at r = 0, dv_z/dr = 0 r = R, v_z = 0

giving:

The maximum velocity at the centre of the tube:

Flow in a vertical circular tube

The volumetric flow rate is the integration of the velocity distribution over the cross-sectional area:

The substitution of to the above equation gives:

Dividing the volumetric flow rate by the flow area gives the average velocity as:

𝑤 = 𝜋(Ƥ_𝑜 − Ƥ_𝐿)𝑅⁴𝜌

8𝜇𝐿 Hagen-Poiseuille equation

Flow in a horizontal circular tube

v_z = v_z(r) and v_r = v_θ = 0

The components of the total

momentum flux are: z

r

The equation of motion is:

ቚ (𝜋_𝑧𝑧

𝑧 2𝜋𝑟 ∆𝑟 +𝜋_𝑟𝑧ቚ

𝑟 2𝜋𝑟 ∆𝑧)

− (𝜋_𝑧𝑧ቚ

𝑧+∆𝑧 2𝜋𝑟 ∆𝑟 + 𝜋_𝑟𝑧ቚ

𝑟+∆𝑟 2𝜋(𝑟 + ∆𝑟) ∆𝑧) + (𝑃ቚ

𝑧 − ቚ𝑃

𝑧+∆𝑧)2𝜋𝑟 ∆𝑟 = 0

Note that:

Flow in a horizontal circular tube

ቚ (𝜋_𝑧𝑧

𝑧 2𝜋𝑟 ∆𝑟 − 𝜋_𝑧𝑧ቚ

𝑧+∆𝑧 2𝜋𝑟 ∆𝑟) + (𝜋_𝑟𝑧ቚ

𝑟 2𝜋𝑟 ∆𝑧 − 𝜋_𝑟𝑧ቚ

𝑟+∆𝑟 2𝜋(𝑟 + ∆𝑟) ∆𝑧) + (𝑃ቚ

𝑧 − ቚ𝑃

𝑧+∆𝑧)2𝜋𝑟 ∆𝑟 = 0

Dividing the above equation by 2πr Δr Δz and taking the limit as Δr → 0 and Δz → 0 give

Or,

Assumption: the flow is fully developed. Therefore, dv_z/d_z = 0.

Flow in a horizontal circular tube

Let’s assume that this side is a constant

Integration twice gives:

Where C₁ and C₂ are constants.

at r = 0, dv_z/dr = 0 r = R, v_z = 0

Flow in a horizontal circular tube

Identify boundary conditions for

at r = 0, dv_z/dr = 0 r = R, v_z = 0

giving:

The maximum velocity at the centre of the tube:

Flow in a horizontal circular tube

The volumetric flow rate is the integration of the velocity distribution over the cross-sectional area:

The substitution of to the above equation gives:

Dividing the volumetric flow rate by the flow area gives the average velocity as:

Falling film on a vertical plate

Consider a film of liquid falling down a vertical plate under the action of gravity. Since the liquid is in contact with air, it is necessary to consider both phases.

For the liquid phase:

The superscript of ^Lrefers to the liquid phase

Falling film on a vertical plate

For a rectangular control volume:

Dividing each term by W ΔxΔz and taking the limit as Δx → 0 and Δz → 0:

or:

Falling film on a vertical plate

Assumption: the flow is fully developed, which means that velocity variations only take place in the direction perpendicular to the flow.

Therefore, dv_z^L/d_z = 0.

Falling film on a vertical plate

For the stagnant air, over a control volume of thickness Δx, Δz, and W:

The superscript of ^A refers to the stagnant air

Dividing each term by W Δx Δz and taking the limit as Δz → 0:

or:

Falling film on a vertical plate

At the liquid-air interface,

As both P^L and P^A depend only on z:

One can conclude that:

giving:

Falling film on a vertical plate

As ρ^L >> ρ^Athen ρ^L – ρ^A ≈ ρ^A

The above equation can be simplified as:

The integration of the above equation gives:

with the boundary conditions that can be applied:

Falling film on a vertical plate

The application of those boundary conditions give:

The maximum velocity (at which x=0) can be expressed as:

Falling film on a vertical plate

The volumetric flow rate is the integration of the velocity distribution over the cross-sectional area:

The substitution of to the above equation gives:

Dividing the volumetric flow rate by the flow area gives the average velocity as:

Flow in a horizontal circular tube

z r

Water flows inside a horizontal circular pipe at 25^oC. Determine the average velocity and plot the velocity distribution inside the pipe. The length and inside the radius of the pipe are 15 and 0.009 m, respectively. The viscosity of water is 8.937 x 10^-4 kg/(m.s),

and the pressure drop along the pipe is 520 Pa.