Lab 3: Conduction and thermal networks

OpenFLUME's thermal network runs alongside the fluid network. It consists of solid nodes (lumped masses), ambient nodes (fixed-temperature reservoirs), and conduction, convection, and radiation conductors. Students build three small thermal models, predict each by hand, and then check the energy balance of the shipped conjugate example, Heated pipe with radiating wall (conjugate HT).

Parts A to C reproduce cases 1 to 3 of the thermal network validation report.

Learning objectives

By the end of the lab, students should be able to:

1. Build solid and ambient nodes and connect them with conduction, convection, and radiation conductors.
2. Solve a composite wall as thermal resistances in series, and predict the surface and interface temperatures.
3. Find the nonlinear equilibrium temperature of a body that loses heat by both convection and radiation.
4. Integrate a lumped-capacitance cooldown by hand with backward Euler and predict the solver's discrete answer exactly.
5. Close the energy balance of a conjugate (solid–fluid) model.
6. Discretize a fin into a conductor ladder and measure the discretization error (extension).

Prerequisites

• Heat transfer: thermal resistance, Newton's law of cooling, radiation between a small body and large surroundings, lumped capacitance.
• Lab 1 or the Getting started page.

Time

About 45 minutes of pre-lab work and 2 hours in the lab.

Theory

Elements (user manual §3.4–3.6, Appendix B).

Element	What it is	Key fields
Solid node	Lumped mass. In steady state its temperature floats to balance its heat flows. In a transient it also stores m cp dT/dt.	Temperature (initial value), Mass, cp, Heat Input
Ambient node	Fixed-temperature reservoir of unlimited capacity	Temperature
Conduction	Q = (k A / L)(T_from − T_to), between two thermal nodes	k, Area, Length
Convection	Q = h A (T_solid − T_fluid), between exactly one fluid node and one thermal node	Area, h
Radiation	Q = ε σ A F (T_from⁴ − T_to⁴), between two thermal nodes	Emissivity, Area, View Factor

σ = 5.670374419×10⁻⁸ W/(m²·K⁴) (src/core/solver/thermal.ts). A conductor's heat rate is positive from its from end to its to end, and a Heat Input is positive into the node (manual §1.4).

Every model needs a fluid network: at least one fluid node, one boundary node, and one branch (manual §3.1). Convection also needs a fluid node at one end. For the pure-conduction problems below, follow the validation report: two fluid boundary nodes at the required temperatures, joined by a Flow Source branch with zero mass flow, act as fixed-temperature fluid reservoirs.

Series resistances. In steady state the same Q passes through every element of a chain:

R_conv = 1/(h A),   R_cond = L/(k A),   Q = (T_hot − T_cold) / ΣR

Radiation–convection equilibrium. A body with heat input q, losing heat to surroundings at T∞, settles where

q = h A (T − T∞) + ε σ A F (T⁴ − T∞⁴)

This is a quartic in T. Solve it with Newton's method, using dQ/dT = h A + 4 ε σ A F T³.

Lumped capacitance. m c dT/dt = −h A (T − T∞) gives T = T∞ + (T₀ − T∞) exp(−t/τ) with τ = m c/(h A). Backward Euler replaces this with T_(n+1) − T∞ = (T_n − T∞) / (1 + Δt/τ), so after n steps

T_n = T∞ + (T₀ − T∞) (1 + Δt/τ)^(−n)

The solver applies this formula exactly, so your hand result should match to round-off.

Pre-lab hand calculation

Part A: Composite wall

A furnace wall of area A = 1 m² separates hot gas at 500 K (film coefficient 40 W/(m²·K)) from room air at 300 K (10 W/(m²·K)). It has two layers: brick, 10 cm thick with k = 1.2 W/(m·K), and insulation, 5 cm thick with k = 0.05 W/(m·K).

1. Compute the four resistances, ΣR, and Q.
2. Compute the hot-face, interface, and cold-face temperatures.
3. What fraction of the total resistance is the insulation?
4. Predict Q if the insulation is doubled to 10 cm.

Part B: Radiation–convection equilibrium

A small solid with heat input q convects to 300 K air (h = 15 W/(m²·K), A = 0.05 m²) and radiates to 300 K surroundings (ε = 0.8, A = 0.05 m², F = 1).

1. For q = 500 W, estimate T assuming convection only. Why is this estimate far too high?
2. Solve the quartic with Newton's method for q = 500 W and q = 1000 W. Report T and the fraction of q that leaves by radiation.

Part C: Lumped cooldown

A 1 kg solid with c = 500 J/(kg·K), initially at 400 K, cools by convection (hA = 10 W/K) to 300 K fluid.

1. Compute τ, T(50 s), and T(150 s) from the exact solution.
2. Compute the backward-Euler values at the same times for Δt = 2 s and Δt = 0.5 s.

Part D: Heated pipe with radiating wall

Read the model description in the examples gallery. Water at 300 K flows from a 200 kPa inlet through two pipes (each L = 0.5 m, D = 30 mm) and an outlet orifice (A = 1×10⁻⁴ m², Cd = 0.6) to 100 kPa. Two wall nodes, one per pipe, each receive 5000 W. Each wall convects to its fluid node with h = 1000 W/(m²·K) over the pipe's inner surface π D L, radiates to a 300 K ambient (ε = 0.8, A = 0.5 m², F = 1), and conducts to the other wall (k = 50, A = 0.05 m², L = 0.1 m).

1. Estimate the water mass flow by assuming the orifice takes the whole 100 kPa: ṁ = Cd A sqrt(2 ρ ΔP).
2. If all 10 kW went into the water, what would the outlet temperature rise be? (Water preset: cp = 4182 J/(kg·K).)
3. Estimate the wall temperature by ignoring radiation and wall conduction: T_w ≈ T_fluid + 5000/(h A). Then estimate how much the walls radiate at that temperature, and correct your estimate once.

Procedure in OpenFLUME

Set Units ▾ to SI. The left rail places Boundary node, Solid node, and Ambient node. To connect nodes, drag from one node's handle to another and choose from Connect with; conductors are under Thermal ties. The canvas rejects illegal connections with a red banner (for example, convection must have exactly one fluid end; manual §6.4).

Part A: Composite wall

1. Click New. A new model is steady-state incompressible water, which is what this part needs.
2. Place two Boundary nodes. Label them Hot gas and Cold air, set both Pressure values to 101325 Pa, and set Temperature to 500 K and 300 K.
3. Connect Hot gas to Cold air and choose Flow Source. Set its Mass Flow to 0.
4. Place three Solid nodes: Hot face, Interface, Cold face. In steady state they need no mass or cp. Their Temperature is only a starting guess.
5. Make the chain:
   • Hot gas → Hot face: Convection, h 40, Area 1.
   • Hot face → Interface: Conduction, k 1.2, Area 1, Length 0.1.
   • Interface → Cold face: Conduction, k 0.05, Area 1, Length 0.05.
   • Cold face → Cold air: Convection, h 10, Area 1.
6. Click Run. Click each solid node to read its temperature, and each conductor to read its heat rate. Every conductor should carry the same Q.
7. Change the insulation Length to 0.1 and run again.

If you are short of time, or want to check your build, click New, open Text view, paste the model below, and click Apply:

// Fluid Network config v2
network "Lab 3 composite wall" {
settings: {"mode":"steady","tolerance":0.000001,"maxIterations":200,"relaxation":0.9}
fluid: {"model":"incompressible","preset":"water"}
node "hot" boundary @ (0, 0) data: {"pressure":101325,"temperature":500,"label":"Hot gas"}
node "cold" boundary @ (600, 0) data: {"pressure":101325,"temperature":300,"label":"Cold air"}
solid "s1" solid @ (150, 150) data: {"temperature":450,"label":"Hot face"}
solid "s2" solid @ (300, 150) data: {"temperature":400,"label":"Interface"}
solid "s3" solid @ (450, 150) data: {"temperature":320,"label":"Cold face"}
branch "hold": "hot" -> "cold" flowSource data: {"label":"Zero flow","massFlow":0}
conductor "h1": "hot" -> "s1" convection data: {"label":"Hot film","h":40,"area":1}
conductor "kA": "s1" -> "s2" conduction data: {"label":"Brick","k":1.2,"area":1,"length":0.1}
conductor "kB": "s2" -> "s3" conduction data: {"label":"Insulation","k":0.05,"area":1,"length":0.05}
conductor "h2": "s3" -> "cold" convection data: {"label":"Cold film","h":10,"area":1}
}

Part B: Radiation–convection equilibrium

1. Starting from the Part A model, delete the three solid nodes. Their conductors are deleted with them, or you can delete the conductors yourself. Set Hot gas to 300 K.
2. Place one Solid node with Heat Input 500 W, and one Ambient node with Temperature 300 K.
3. Connect the solid node to Hot gas with Convection (h 15, Area 0.05), and to the ambient node with Radiation (Emissivity 0.8, Area 0.05, View Factor 1).
4. Run, and record the solid temperature and both heat rates. Repeat with Heat Input 1000 W.

Part C: Lumped cooldown

1. Starting from Part B, delete the radiation conductor and set the solid's Heat Input to 0. Set the solid's Temperature to 400 K, Mass to 1 kg, and cp to 500 J/(kg·K). Set the convection h to 10 and Area to 1.
2. In Setup → Solver, set Mode to Transient, choose Fixed dt, and set Time Step to 2 s and End Time to 150 s.
3. Run. In the Results tab, plot the solid temperature against Time (the Node & solid temperature set includes it). Read T at 50 s and 150 s.
4. Change Time Step to 0.5 s and run again. Overlay the two runs with + Compare run….

Part D: Heated pipe with radiating wall

1. Open the example: Heated pipe with radiating wall (conjugate HT). Opening it replaces your model, so Save Part C first if you want to keep it.
2. Check the parameters in the property panel. Click Conv w1-f1: its Area is a formula, pipe('b_in').surfaceArea. Use the f(x) preview to read the value.
3. Click Run.
4. Record the mass flow in the outlet orifice; the temperatures of Wall 1, Wall 2, Fluid 1, and Fluid 2; and the heat rates of all five conductors.
5. Set Color by to Temperature and look at where the temperature differences are.
6. Close the energy balance. Compare the 10 kW input with the heat convected to the water plus the heat radiated. Then compare the convected heat with ṁ cp (T_Fluid2 − T_inlet).

Questions

1. Part A: which layer controls the heat loss? Where would you add insulation, and where would a better hot-side film coefficient matter?
2. Part B: why is the convection-only estimate so far off at 500 W? At what heat input does radiation carry about half the load?
3. Part C: your backward-Euler hand calculation and the solver should agree to round-off. Why is that a stronger check than agreement with the exact exponential?
4. Part C: the error at 150 s fell by a factor of about 4 when the step fell by a factor of 4. What order of accuracy does that imply?
5. Part D: heat flows between the two walls through Cond w1-w2. In which direction, and why? (Hint: compare the fluid temperatures next to each wall.)
6. Part D: about 8 % of the heat is radiated rather than carried by the water. Which single parameter would you change to raise that fraction, and by roughly how much would the wall temperature have to rise to double it?
7. In Part D the water warms by only about 2.6 K, yet the walls run about 100 K above the water. What does that tell you about where the thermal resistance is?

Extensions

• Fin. Model a pin fin as a ladder of solid nodes: aluminium-like k = 200 W/(m·K), D = 10 mm, L = 100 mm, base held at 400 K (an ambient node), convection h = 20 W/(m²·K) to 300 K fluid, adiabatic tip. Use N equal cells with a solid node at each cell centre. The base conductor has length Δx/2, the cell-to-cell conductors have length Δx, each cell has a convection conductor of area π D Δx to a fluid boundary node, and nothing is attached to the tip. Compare the base heat rate for N = 2, 5, 10 with the analytic fin result Q = sqrt(h P k A_c) θ_b tanh(mL). The Repeat… selection action (manual §3.13) can chain a unit cell, or you can place five cells by hand.
• Sweep. In Part B, use the Sweep tab with the solid node's Heat Input as the target, from 100 to 4000 W. Compare with Figure 4 of the validation report.
• Heat exchanger. Open the Water-water counterflow heat exchanger example. Section 7.9 of the user manual and case 5 of the validation report show that segment count controls accuracy (first order in 1/N).

Answer key

Hand values were computed in Python from the formulas above, including a Newton solve of the quartic and a direct solve of the four-unknown Part D energy balance and of the fin ladder. Solver values are OpenFLUME 0.3.1, run locally on the models described here. In every part, hand and solver agree to at least six significant figures. Parts A to C also match the validation report.

Part A

• Resistances (K/W): hot film 0.0250, brick 0.0833, insulation 1.0000, cold film 0.1000. ΣR = 1.2083 K/W, so Q = 165.52 W.
• Hot face 495.86 K, interface 482.07 K, cold face 316.55 K (report case 1).
• Insulation share: 82.8 %.
• Insulation doubled: ΣR = 2.2083 K/W, so Q = 90.57 W. The solver gives hot face 497.74 K, interface 490.19 K, cold face 309.06 K.

Part B

q [W]	Convection-only T [K]	T_eq [K]	Convection [W]	Radiation [W]	Radiated share
500	966.7	599.79	224.84	275.16	55 %
1000	1633.3	741.85	331.39	668.61	67 %

The report (case 2) gives 599.786 K and 741.854 K. The convection-only estimate ignores a loss that grows as T⁴. At 967 K radiation alone would remove about 1960 W, four times the input. Radiation and convection carry equal shares at q ≈ 370 W (T ≈ 548 K). The report's sweep runs from 33 % radiated at 100 W to 85 % at 4000 W. A simple fixed-point iteration using a linearized radiation coefficient oscillates about the root. Newton's method converges in a few steps.

Part C

τ = m c/(hA) = 50 s. Exact: T(50) = 336.788 K, T(150) = 304.979 K.

Δt [s]	T(50 s) backward Euler [K]	T(150 s) backward Euler [K]	Error at 150 s [K]
2	337.512	305.278	0.300
0.5	336.971	305.053	0.0747

The solver reproduces these backward-Euler values to round-off. The ratio 0.300/0.0747 = 4.01 for a 4× step reduction (report case 3) is first-order accuracy.

Part D (solver; a hand Newton solve of the same equations gives identical values)

• Orifice-only estimate: ṁ = 0.6e-4 × sqrt(2 × 998 × 10⁵) = 0.8477 kg/s. The solved flow is 0.8453 kg/s. The two pipes take 0.28 kPa each, and the orifice takes the remaining 99.44 kPa.
• All heat into the water: ΔT = 10 000 / (0.8453 × 4182) = 2.83 K.
• h A = 1000 × π × 0.03 × 0.5 = 47.12 W/K. With no radiation, T_w ≈ 300 + 5000/47.12 = 406 K. At 406 K each wall would radiate about 0.43 kW, so the convective load falls to about 4.57 kW and T_w falls to about 397–398 K, close to the solved 399 K.

Quantity	Value
Wall 1 / Wall 2	399.38 K / 399.97 K
Fluid 1 / Fluid 2	301.307 K / 302.605 K
Conv w1-f1 / Conv w2-f2	4621.5 W / 4588.4 W
Rad w1-amb / Rad w2-amb	393.3 W / 396.8 W
Cond w1-w2	−14.86 W (heat flows from Wall 2 to Wall 1)

• Balance: 9209.9 W convected + 790.1 W radiated = 10 000.0 W. Water: 0.84530 × 4182 × (302.605 − 300) = 9209.9 W.
• Q5: Fluid 2 is 1.3 K warmer than Fluid 1, so Wall 2 must run slightly hotter to push its heat into the water, and a little heat conducts back to Wall 1.
• Q6: Raise emissivity or radiating area. Radiation is proportional to T⁴ − 300⁴. Doubling it from about 395 W per wall needs T_w ≈ 455 K, a rise of about 55 K, unless εA is doubled instead.
• Q7: The water's capacity rate ṁ cp ≈ 3535 W/K is much larger than the wall film conductance hA ≈ 47 W/K. Almost all of the temperature difference is across the convective film.

Fin extension (m = sqrt(hP/(kA_c)) = 6.325 m⁻¹, mL = 0.632)

Analytic: Q = 5.561 W, tip temperature 382.87 K, fin efficiency 0.885.

N	Q base [W]	Error
2	5.475	−1.5 %
5	5.547	−0.25 %
10	5.557	−0.06 %
20	5.560	−0.02 %

The error falls about fourfold when N doubles. This cell-centred ladder is second order in Δx. OpenFLUME 0.3.1 gives the same values to five significant figures as the hand solve of the ladder equations.