Heat Transfer Analysis: Parallel and Series-Parallel Thermal Resistance Networks

Fundamentals of Parallel Thermal Resistance Networks

Thermal resistance analysis extends the electrical circuit analogy to heat transfer across multiple materials. When heat conducts simultaneously through adjacent materials situated side by side along the path of heat flow, the arrangement is defined as a parallel network.

In a parallel configuration, the heat transfer rate divides among the parallel branches while the temperature drop across each parallel component remains identical:

$Q = Q_1 + Q_2$

The rate of heat transfer through any thermal network is expressed by the general thermal resistance equation:

$Q = \frac{\Delta T}{R_{\text{total}}}$

Where:

• $Q$ is the total heat transfer rate, expressed in watts ($\text{W}$) or kilowatts ($\text{kW}$).

• $\Delta T$ is the overall temperature difference driving the heat flow ($\Delta T = T_{\text{high}} - T_{\text{low}}$), expressed in degrees Celsius ($^\circ\text{C}$) or Kelvin ($\text{K}$).

• $R_{\text{total}}$ is the total equivalent thermal resistance of the network, expressed in degrees Celsius per watt ($^\circ\text{C/W}$) or Kelvin per watt ($\text{K/W}$).

For two conductive layers arranged in parallel between isothermal surfaces at temperatures $T_1$ and $T_2$ (where $T_1 > T_2$):

$\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2}$

Algebraic rearrangement yields the equivalent resistance for two parallel components:

$R_{\text{total}} = R_1 \parallel R_2 = \frac{R_1 R_2}{R_1 + R_2}$

The conductive thermal resistance of an individual layer is determined by its thickness, thermal conductivity, and cross-sectional area:

$R_1 = \frac{L_1}{k_1 A_1}$

$R_2 = \frac{L_2}{k_2 A_2}$

Where:

• $L$ is the material thickness along the direction of heat transfer in meters ($\text{m}$).

• $k$ is the thermal conductivity in watts per meter-degree Celsius ($\text{W/(m}\cdot^\circ\text{C)}$) or watts per meter-Kelvin ($\text{W/(m}\cdot\text{K)}$).

• $A$ is the cross-sectional heat transfer area normal to the direction of heat transfer in square meters ($\text{m}^2$).

In parallel arrangements, the individual material layers often possess different cross-sectional areas ($A_1 \neq A_2$) due to differing vertical dimensions or heights ($H_1 \neq H_2$), even though their thicknesses ($L$) along the heat flow path may be identical.

Combined Series-Parallel Thermal Circuits

Engineering systems frequently combine series layers and parallel layers within composite structures, incorporating both conductive solid layers and convective fluid boundaries.

To analyze a combined series-parallel network:

1. Construct the complete thermal circuit diagram identifying all thermal nodes (temperatures) and resistance elements.

2. Formulate the total equivalent thermal resistance expression by reducing parallel groups into single equivalent resistances, then summing all components in series.

3. Apply the general resistance equation using the global temperature boundary conditions.

Consider a composite wall where parallel materials 1 and 2 are bounded on the left by surface temperature $T_1$ and interface temperature $T_2$, connected in series to solid material 3 (spanning from $T_2$ to $T_3$), which transfers heat via convection from outer surface $T_3$ to an ambient fluid at $T_\infty$ with convection heat transfer coefficient $h$:

$R_{\text{total}} = (R_1 \parallel R_2) + R_3 + R_{\text{conv}}$

Expanding each individual resistance component:

$R_1 \parallel R_2 = \frac{R_1 R_2}{R_1 + R_2}$

$R_1 = \frac{L_1}{k_1 A_1}$

$R_2 = \frac{L_2}{k_2 A_2}$

$R_3 = \frac{L_3}{k_3 A_3}$

$R_{\text{conv}} = \frac{1}{h A_{\text{conv}}}$

The surface area for convection ($A_{\text{conv}}$) must correspond strictly to the physical surface interfacing with the moving fluid. Because material 3 covers the combined outer face of the composite structure, its area equals the total exposed frontal area:

$A_3 = A_1 + A_2$

$A_{\text{conv}} = A_3$

Once the total resistance is established, the total heat transfer rate across the system is given by:

$Q = \frac{T_1 - T_\infty}{R_{\text{total}}}$

Intermediate local temperatures can be solved using local energy balances across specific subsections of the network:

$Q = \frac{T_1 - T_2}{R_1 \parallel R_2}$

$Q = \frac{T_1 - T_3}{(R_1 \parallel R_2) + R_3}$

$Q = \frac{T_3 - T_\infty}{R_{\text{conv}}}$

Analysis of Series-Parallel Composite Wall Example

Consider steady, one-dimensional heat conduction across a composite slab composed of four distinct material blocks designated A, B, C, and D. Heat flows horizontally in the positive x-direction from the left surface to the right surface without surface convection.

Given Parameters:

• Left surface temperature: $T_1 = 370\,^\circ\text{C}$

• Right surface temperature: $T_4 = 66\,^\circ\text{C}$

• Cross-sectional area of material C: $A_C = 0.1\,\text{m}^2$

• Area relationships: $A_B = A_D = \frac{1}{2} A_C = 0.05\,\text{m}^2$

• Area of material A: Material A shares the full height of material C, yielding $A_A = A_C = 0.1\,\text{m}^2$

• Thickness of layer A: $L_A = 2.5\,\text{cm} = 0.025\,\text{m}$

• Thickness of parallel layers B and D: $L_B = L_D = 7.5\,\text{cm} = 0.075\,\text{m}$

Thermal Circuit Architecture:

• Node $T_1$: Outer surface of material A.

• Node $T_2$: Interface between material A and the parallel block containing materials B and D.

• Node $T_3$: Interface between the parallel block (B and D) and material C.

• Node $T_4$: Outer surface of material C.

The equivalent total thermal resistance of the composite system is:

$R_{\text{total}} = R_A + (R_B \parallel R_D) + R_C$

$R_{\text{total}} = R_A + \frac{R_B R_D}{R_B + R_D} + R_C$

Individual conductive resistances are computed after converting all linear dimensions to meters:

$R_A = \frac{L_A}{k_A A_A} = \frac{0.025}{k_A (0.1)}$

$R_B = \frac{L_B}{k_B A_B} = \frac{0.075}{k_B (0.05)}$

$R_D = \frac{L_D}{k_D A_D} = \frac{0.075}{k_D (0.05)}$

$R_C = \frac{L_C}{k_C A_C}$

The total heat transfer rate through the composite assembly is:

$Q = \frac{T_1 - T_4}{R_{\text{total}}}$

To determine the unknown interface temperature $T_3$, utilize the heat flow through material C alone:

$Q = \frac{T_3 - T_4}{R_C}$

$T_3 = T_4 + Q \cdot R_C$

Periodic Composite Structures: Brick Wall with Plaster and Insulation

A representative recurring cross-section (unit cell) approach is utilized when walls consist of repetitive geometric patterns in the vertical direction.

Overall Wall Specifications:

• Wall height: $H_{\text{wall}} = 3\,\text{m}$

• Wall width: $W_{\text{wall}} = 5\,\text{m}$

• Total wall face area: $A_{\text{wall}} = H_{\text{wall}} \times W_{\text{wall}} = 3\,\text{m} \times 5\,\text{m} = 15\,\text{m}^2$

Internal Layer and Material Data:

• Horizontal brick core dimensions: $16\,\text{cm}$ vertical height by $22\,\text{cm}$ horizontal thickness ($L_{\text{brick}} = 0.22\,\text{m}$).

• Thermal conductivity of brick: $k_{\text{brick}} = 0.72\,\text{W/(m}\cdot^\circ\text{C)}$.

• Plaster separation: Bricks are separated horizontally by $3\,\text{cm}$ thick plaster layers (allocating $1.5\,\text{cm}$ above and $1.5\,\text{cm}$ below each brick within a symmetric unit cell).

• Lateral plaster layers: $2\,\text{cm}$ thick vertical plaster layers on both sides of the brick core ($L_{P,1} = L_{P,4} = 0.02\,\text{m}$).

• Thermal conductivity of plaster: $k_{\text{plaster}} = 0.22\,\text{W/(m}\cdot^\circ\text{C)}$.

• Rigid foam insulation layer: $3\,\text{cm}$ thick ($L_{\text{foam}} = 0.03\,\text{m}$) situated on the indoor side.

• Thermal conductivity of foam: $k_{\text{foam}} = 0.026\,\text{W/(m}\cdot^\circ\text{C)}$.

Thermal Boundaries:

• Indoor air temperature: $T_{\infty,1} = 20\,^\circ\text{C}$

• Outdoor air temperature: $T_{\infty,2} = -10\,^\circ\text{C}$

• Inner convection heat transfer coefficient: $h_1 = 10\,\text{W/(m}^2\cdot^\circ\text{C)}$

• Outer convection heat transfer coefficient: $h_2 = 25\,\text{W/(m}^2\cdot^\circ\text{C)}$

• Assumptions: Steady-state, one-dimensional heat flow, negligible radiation effects.

Unit Cell Characterization: The repeating pattern recurs every $25\,\text{cm}$ ($0.25\,\text{m}$) in the vertical direction. Assuming a unit depth of $W = 1\,\text{m}$ into the page:

$H_{\text{cell}} = 0.25\,\text{m}$

$A_{\text{cell}} = 0.25\,\text{m} \times 1.0\,\text{m} = 0.25\,\text{m}^2$

Layer Subdivision of the Unit Cell (Indoor to Outdoor):

1. Indoor convection: $R_{\text{conv},1}$

2. Rigid foam layer ($F$): $L_F = 0.03\,\text{m}$, $A_F = 0.25\,\text{m}^2$

3. Inner plaster layer ($P_1$): $L_{P,1} = 0.02\,\text{m}$, $A_{P,1} = 0.25\,\text{m}^2$

4. Parallel core layer ($L = 0.22\,\text{m}$) consisting of three parallel branches:

   • Top plaster section ($P_2$): Height $= 1.5\,\text{cm} = 0.015\,\text{m}$, area $A_{P,2} = 0.015\,\text{m} \times 1.0\,\text{m} = 0.015\,\text{m}^2$

   • Central brick ($B$): Height $= 16\,\text{cm} = 0.16\,\text{m}$, area $A_B = 0.16\,\text{m} \times 1.0\,\text{m} = 0.16\,\text{m}^2$

   • Bottom plaster section ($P_3$): Height $= 1.5\,\text{cm} = 0.015\,\text{m}$, area $A_{P,3} = 0.015\,\text{m} \times 1.0\,\text{m} = 0.015\,\text{m}^2$

5. Outer plaster layer ($P_4$): $L_{P,4} = 0.02\,\text{m}$, $A_{P,4} = 0.25\,\text{m}^2$

6. Outdoor convection: $R_{\text{conv},2}$

Unit Cell Resistance Equations:

$R_{\text{conv},1} = \frac{1}{h_1 A_{\text{cell}}} = \frac{1}{(10)(0.25)}$

$R_F = \frac{L_F}{k_{\text{foam}} A_{\text{cell}}} = \frac{0.03}{(0.026)(0.25)}$

$R_{P,1} = \frac{L_{P,1}}{k_{\text{plaster}} A_{\text{cell}}} = \frac{0.02}{(0.22)(0.25)}$

$\frac{1}{R_{\text{parallel}}} = \frac{1}{R_{P,2}} + \frac{1}{R_B} + \frac{1}{R_{P,3}}$

Where:

$R_{P,2} = \frac{L_{\text{brick}}}{k_{\text{plaster}} A_{P,2}} = \frac{0.22}{(0.22)(0.015)}$

$R_B = \frac{L_{\text{brick}}}{k_{\text{brick}} A_B} = \frac{0.22}{(0.72)(0.16)}$

$R_{P,3} = \frac{L_{\text{brick}}}{k_{\text{plaster}} A_{P,3}} = \frac{0.22}{(0.22)(0.015)}$

$R_{P,4} = \frac{L_{P,4}}{k_{\text{plaster}} A_{\text{cell}}} = \frac{0.02}{(0.22)(0.25)}$

$R_{\text{conv},2} = \frac{1}{h_2 A_{\text{cell}}} = \frac{1}{(25)(0.25)}$

Total Unit Cell Resistance:

$R_{\text{total, cell}} = R_{\text{conv},1} + R_F + R_{P,1} + R_{\text{parallel}} + R_{P,4} + R_{\text{conv},2}$

Heat Transfer Rate of the Unit Cell:

$Q_{\text{cell}} = \frac{T_{\infty,1} - T_{\infty,2}}{R_{\text{total, cell}}} = \frac{20 - (-10)}{R_{\text{total, cell}}} = \frac{30}{R_{\text{total, cell}}}$

Total Heat Transfer Rate Across Entire Wall: The total heat transfer rate across the entire wall is calculated by scaling the unit cell heat transfer rate by the ratio of the total wall area to the unit cell area:

$N_{\text{cells}} = \frac{A_{\text{wall}}}{A_{\text{cell}}} = \frac{15\,\text{m}^2}{0.25\,\text{m}^2} = 60$

$Q_{\text{total}} = N_{\text{cells}} \times Q_{\text{cell}} = 60 \times Q_{\text{cell}}$