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SCIENCE · INVESTIGATE AND EXPLAIN

Connect a local conduction law to a global entropy balance

A steady slab has conductivity k=2 W/(m K), area 0.5 m² and length L, with hot boundary Th and cold boundary 300 K. There is no volumetric heating, material motion or time-dependent storage. Under the supplied constant-conductivity/local-equilibrium assumptions, T(x)=Th+(300−Th)x/L.

Learning outcome: Derive the local expression, integrate it independently and reconcile both boundary fluxes. Change length while retaining boundary temperatures, and distinguish no storage from no production.

Starting recap

Review logarithmic integrals, equilibrium states, absolute temperature and the first-law work convention. State the domain and distinguish calibration data from unused validation.

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Try

At Th=600 K, L=2 m, Qdot=150 W. Entropy influx is 0.25 W/K and outflux 0.5 W/K, so generation is 0.25 W/K. Local density is 0.125 W/(m³ K) at the hot end and 0.5 at the cold end; multiplying either endpoint by the whole volume would not integrate the varying density.

Open the experiment

Predict

What does zero steady entropy storage imply here?

Experiment

Derive the local expression, integrate it independently and reconcile both boundary fluxes. Change length while retaining boundary temperatures, and distinguish no storage from no production.

Reconcile local generation and boundary entropy flux

Worked teaching model · synthetic values · Steady one-dimensional Fourier conduction, k=2 W/(m K), area=0.5 m², cold boundary300 K, no volumetric heating or material motion. Local equilibrium and constant k are supplied assumptions. Linear T(x); no physical exposure or apparatus instructions.

Reset keeps your prediction, captures and practice history.

Local generation under steady Fourier conduction

entropy generation density (W/(m³ K))position (m)020.50

Graph axes rescale independently; read units and values, not screen angles. 1: k (dT/dx)²/T² (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Local generation under steady Fourier conduction

k (dT/dx)²/T²

position (m)entropy generation density (W/(m³ K))
00.125
0.0250.1265773
0.050.1281846
0.0750.1298227
0.10.1314924
0.1250.1331946
0.150.13493
0.1750.1366996
0.20.1385042
0.2250.1403447
0.250.1422222
0.2750.1441377
0.30.146092
0.3250.1480864
0.350.150122
0.3750.1521998
0.40.154321
0.4250.1564869
0.450.1586987
0.4750.1609577
0.50.1632653
0.5250.1656229
0.550.1680319
0.5750.1704939
0.60.1730104
0.6250.175583
0.650.1782134
0.6750.1809034
0.70.1836547
0.7250.1864693
0.750.1893491
0.7750.1922961
0.80.1953125
0.8250.1984004
0.850.2015621
0.8750.2048
0.90.2081165
0.9250.2115143
0.950.214996
0.9750.2185643
10.2222222
1.0250.2259727
1.050.229819
1.0750.2337643
1.10.2378121
1.1250.241966
1.150.2462296
1.1750.2506069
1.20.255102
1.2250.2597192
1.250.2644628
1.2750.2693376
1.30.2743484
1.3250.2795004
1.350.2847989
1.3750.2902494
1.40.295858
1.4250.3016307
1.450.307574
1.4750.3136947
1.50.32
1.5250.3264973
1.550.3331945
1.5750.3400999
1.60.3472222
1.6250.3545706
1.650.3621548
1.6750.369985
1.70.3780718
1.7250.3864268
1.750.3950617
1.7750.4039894
1.80.4132231
1.8250.4227771
1.850.4326663
1.8750.4429066
1.90.4535147
1.9250.4645086
1.950.4759072
1.9750.4877305
20.5

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
QuantityValueUnit
Heat rate toward cold boundary150W
Entropy influx at hot boundary0.25W/K
Entropy outflux at cold boundary0.5W/K
Integrated entropy generation rate0.25W/K
Steady entropy storage rate0W/K

A steady state stores no additional entropy while producing entropy continuously and exporting it.

With heat-flux density j=−k dT/dx, local generation is j d(1/T)/dx=k(dT/dx)²/T²≥0. Integrating over area and length gives Qdot(1/Tc−1/Th). Boundary entropy outflux minus influx equals generation; neither heat rate alone nor a zero storage derivative measures irreversibility.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

A steady state stores no additional entropy while producing entropy continuously and exporting it.

With heat-flux density j=−k dT/dx, local generation is j d(1/T)/dx=k(dT/dx)²/T²≥0. Integrating over area and length gives Qdot(1/Tc−1/Th). Boundary entropy outflux minus influx equals generation; neither heat rate alone nor a zero storage derivative measures irreversibility.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Declare the system and evidence

A steady slab has conductivity k=2 W/(m K), area 0.5 m² and length L, with hot boundary Th and cold boundary 300 K. There is no volumetric heating, material motion or time-dependent storage. Under the supplied constant-conductivity/local-equilibrium assumptions, T(x)=Th+(300−Th)x/L.

Derive the state or rate account

Fourier heat-flux density is j=−k dT/dx, positive toward the cold end. Entropy-flux density is j/T. Its derivative in a steady constant-j slab is j d(1/T)/dx=k(dT/dx)²/T², a nonnegative entropy-generation density with units W/(m³ K). A squared gradient alone is insufficient without conductivity and absolute temperature.

Audit the constraint and model

Integrating over volume gives generation rate Qdot(1/300−1/Th). This equals cold-side entropy outflux Qdot/300 minus hot-side influx Qdot/Th. The entropy storage rate is zero at steady state, even though generation is positive. Heat energy flow is constant while entropy flow increases toward the cold end.

Predict a changed boundary

Doubling length at fixed boundary temperatures halves heat rate and total entropy generation, while the local generation density at corresponding fractional positions falls by four. The doubled volume restores one factor. Conversely, equal boundary temperatures give zero gradient, heat rate and generation; this limiting case is analytic even though the controls keep a positive contrast.

Worked original case

At Th=600 K, L=2 m, Qdot=150 W. Entropy influx is 0.25 W/K and outflux 0.5 W/K, so generation is 0.25 W/K. Local density is 0.125 W/(m³ K) at the hot end and 0.5 at the cold end; multiplying either endpoint by the whole volume would not integrate the varying density.

Investigation

Derive the local expression, integrate it independently and reconcile both boundary fluxes. Change length while retaining boundary temperatures, and distinguish no storage from no production.

Changed-case practice

Try the changed case before feedback

Use Th=450 K, L=1 m with the same k and area. Find total entropy generation rate in W/K.

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and boundaries

Original derivations and synthetic cases checked 31 October 2026 against the primary teaching reference. No source exercises or diagrams are reproduced. All thermal operations are mathematical proposals or supplied records, not physical heating, pressure-vessel or exposure instructions. Public feedback and retained prose do not certify independent reasoning or actual practical participation. Preserve first work, critique, revision and a new assigned response in the separate course workspace.

Apply

What does zero steady entropy storage imply here?

Public practice checks a specific claim. It does not establish independent mastery.

Which system boundary, reference path, constitutive law or independently calibrated record supports this entropy inference?

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