Transport Laboratory

Quantities and units

The coloured block represents a one-dimensional slab or fluid gap, not a three-dimensional solution. All conservation totals are per unit cross-sectional area. A positive flux points in the direction of increasing x (or increasing y for momentum).

Set the problem

Choose the material property, initial condition and boundary conditions. The calculation is one-dimensional.

SOLUTION TYPE∂ψ/∂t ≠ 0

The field changes with time from the specified initial condition.

Conductivity, k15 W m⁻¹ K⁻¹

INITIAL CONDITION T(x,0) = 300 K

The value at every position at time t = 0.

Initial temperature300 K

INITIAL PROFILE

The same initial value is assigned to every cell.

Flow velocity, u0.00 m s⁻¹

BOUNDARY CONDITIONSfor t > 0

Left-face temperature650 K

RIGHT FACE

Right face: zero diffusive flux. Advective outflow, if present, may leave.

Volumetric heat source0 W m⁻³

Layered materialLower property in the centre

Calculated field

SIMULATION TIME0.00 s

650 K

low k

OPEN

300 K650 K

Temperature field, T(x)Change a control to calculate a new profile

Transient fieldA straight temperature profile is not generally expected before steady state.

Plot axes and labels

Horizontal-axis label Vertical-axis label Vertical minimum Vertical maximum

Automatically fit vertical limits to the displayed field The horizontal limits remain the physical domain, 0 to L. Changing them would change the problem, not merely the graph.

Check the conservation balance

Compare the total stored amount with the amount entering, leaving and generated.

CONTINUITY & LOCAL CONSERVATION

∂ρ/∂t + d(ρv_x)/dx = 0

Mass continuity is always enforced. For constant-density, one-dimensional flow in a constant-area channel, dv_x/dx = 0.

ρc_p ∂T/∂t = −dq_x/dx + S_T

Energy accumulation equals net heat-flux inflow plus generation.

FOURIER'S LAW

q_x = −k dT/dx

Conductivity connects the temperature gradient to heat flux.

Conservation balance BALANCED

Initially stored amount—

Entered through faces—

Left through faces—

Generated internally—

Currently stored amount—

initial + entered − left + generated= current

Numerical balance error0.00%

INFORMATION FROM THE SPATIAL PROFILE Maximum gradient: 0 K m⁻¹

The total energy can remain unchanged while internal temperature gradients drive heat transfer.

Fourier number0.00

Péclet number0.00

WORKED EXAMPLES

Worked-out examples

Follow the reduction from the conservation statement to the one-dimensional transport equation, then load the case into the solver.

HEAT · TRANSIENT01

Heating a cold slab from one face

A plane wall of thickness L = 20 mm is initially cold. At t = 0 its left face is suddenly brought to 650 K while the right face is insulated.

Domain: 0 ≤ x ≤ L; one-dimensional plane wall
Initial condition: T(x,0) = 300 K
Boundary conditions: T(0,t) = 650 K; ∂T/∂x(L,t) = 0
Properties: k = 9.57 W m⁻¹ K⁻¹; ρc_p = 3.0 MJ m⁻³ K⁻¹
Assumptions: Constant properties; no advection; no internal heat generation

Conservationaccumulation = net conduction in
Local balanceρcₚ ∂T/∂t = −dqₓ/dx
Fourier lawqₓ = −k dT/dx
Governing equation∂T/∂t = α ∂²T/∂x²
Diffusivity definitionα = k/(ρcₚ)

Prediction and checkHeat penetrates from the left over a scale √(αt). The temperature remains between 300 K and 650 K, and the rise in stored energy must equal the integrated heat entering at x = 0.

Quantities and units for this example

HEAT · COMPOSITE02

Two materials carrying the same heat flux

A 20 mm composite wall contains a 4 mm low-conductivity central layer. Its two faces are maintained at different temperatures until steady state is reached.

Domain: 0 ≤ x ≤ 20 mm; interfaces at x = 8 and 12 mm
Initial condition: Not needed for the final steady solution; use T(x,0) = 300 K to watch the transient approach
Boundary conditions: T(0) = 675 K; T(L) = 300 K
Properties: Outer layers: k; central layer: 0.12k
Interface conditions: T is continuous and q_x is continuous

Steady conservationdqₓ/dx = 0
Flux continuityqₓ,1 = qₓ,2
Fourier lawdT/dx = −qₓ/k
Consequencelower k ⇒ larger |dT/dx|

Key distinctionAt steady state, heat flux is the same through every layer. The temperature gradient is largest in the layer with the lowest conductivity.

Quantities and units for this example

MASS · DIFFUSION03

Solute diffusing into a layer

A solute-free layer of thickness L = 2 mm is exposed at its left face to a maintained solute concentration; the right face is impermeable.

Domain: 0 ≤ x ≤ 2 mm
Initial condition: c_A(x,0) = 0 mol m⁻³
Boundary conditions: c_A(0,t) = 850 mol m⁻³; ∂c_A/∂x(L,t) = 0
Properties: D_A = 10⁻¹⁰ m² s⁻¹
Assumptions: Constant D_A; no bulk flow; no homogeneous reaction

Species balance∂cₐ/∂t = −dJₐ,x/dx
Fick lawJₐ,x = −Dₐ dcₐ/dx
Constant Dₐ∂cₐ/∂t = Dₐ ∂²cₐ/∂x²
Time scaletᴅ ~ L²/Dₐ

Prediction and checkThe penetration distance grows as √(D_At), while the diffusion time scales as L²/D_A. Concentration must remain between 0 and 850 mol m⁻³.

Quantities and units for this example

MOMENTUM · COUETTE04

Momentum diffusion from a moving wall

An incompressible Newtonian fluid occupies a 10 mm gap. At t = 0 the lower wall begins moving in the x direction while the upper wall remains fixed.

Domain: 0 ≤ y ≤ H = 10 mm; velocity v_x(y,t)
Initial condition: v_x(y,0) = 0
Boundary conditions: v_x(0,t) = V_w = 1.4 m s⁻¹; v_x(H,t) = 0
Properties: ρ = 1000 kg m⁻³; μ = 0.0316 Pa s
Assumptions: Incompressible; no pressure gradient; no slip; v_y = 0

Continuity∂v_x/∂x + ∂v_y/∂y = 0 → 0 = 0
Momentum balanceρ ∂v_x/∂t = −dJ_yx^(p)/dy
Momentum-flux lawJ_yx^(p) = −μ dv_x/dy
Index meaningy direction; x-momentum
Governing equation∂vₓ/∂t = ν ∂²vₓ/∂y²
Steady solutionvₓ(y) = Vw(1 − y/H)

Sign convention and checkJ_yx^(p) is the flux of x-momentum in the y direction and points down the velocity gradient. Momentum penetrates over √(νt); the final profile is linear.

Quantities and units for this example