PE Architectural Engineering · Complete public sample

Parallel heat-flow paths

Distinguish adjacent envelope areas from layers stacked along the same heat-flow path.

Understand the governing concept.

Layers through the thickness of one assembly are in series: the same heat passes through each layer, and their thermal resistances add on a consistent area basis. A window beside an opaque wall is a parallel path: both see the same indoor-to-outdoor temperature difference, and their heat-transfer rates add.

For adjacent areas, calculate each area multiplied by its own U-factor, then sum these UA products. A small high-U area can transfer as much heat as a much larger well-insulated area. Averaging R-values across adjacent areas hides that effect; use total UA or an area-weighted U-factor.

By the end of this refresher

Explain the basis or assumption controlling this decision.
Apply the method within its stated limits.
Check the result and choose a related study task.

\dot Q=\left(\sum_i U_iA_i\right)\Delta T,\qquad U_{\mathrm{equiv}}=\frac{\sum_i U_iA_i}{\sum_i A_i}

Sum UA for parallel areas with the same temperature difference; do not average their R-values.

Define the basis before substituting a value
Symbol	Meaning	Unit or basis
$\dot Q$	Steady heat-transfer rate	Btu/h
$U$	Whole-assembly conductance per area	Btu/(h·ft²·°F)
$A$	Area of a parallel path	ft²
$\Delta T$	Temperature difference across all paths	°F
$R$	Consistent total resistance per area, 1/U	h·ft²·°F/Btu

When the method applies

This steady one-dimensional conduction estimate uses supplied whole-assembly U-factors or consistent total R-values; it excludes solar gains and air leakage.
Thermal bridges, two-dimensional details, and differently exposed surfaces require an appropriate model or their stated correction data.

A wall cutaway shows the physical opaque assembly. — **Connect the physical model and the decision**Connect the equipment to the calculation

Opaque wall and glazing are parallel whole-path conductances with the same 40-degree-F temperature difference, and each has UA of 90 Btu per hour degree F. — **Connect the physical model and the decision**Connect the equipment to the calculation

Original teaching case

Work the decision, then cross-check it.

An exterior elevation has 1,800 ft² of opaque wall with total R = 20 h·ft²·°F/Btu and 200 ft² of glazing with U = 0.45 Btu/(h·ft²·°F). Both see a 40°F temperature difference. Ignore air leakage and solar gain.

Opaque-wall U is 1/20 = 0.05; its UA is 0.05 × 1,800 = 90 Btu/(h·°F).
Glazing UA is 0.45 × 200 = 90 Btu/(h·°F).
Total UA is 180 Btu/(h·°F); heat transfer is 180 × 40 = 7,200 Btu/h.
Equivalent U for the full 2,000 ft² elevation is 180/2,000 = 0.09 Btu/(h·ft²·°F).

\dot Q=[(1800/20)+(200)(0.45)](40)=7200\ \mathrm{Btu/h}

Decision

The glazing causes half the conductive heat transfer despite representing only 10% of the area.

Independent check

Separate path heat rates are 3,600 Btu/h each. Their sum matches 0.09 × 2,000 × 40 = 7,200 Btu/h.

A fresh application

Use the concept under changed conditions.

Choose a response before inspecting the reasoning. This is a learning check, separate from scored diagnostic and mock attempts.

Connect this refresher to your program.

Read module M1 in the study studio →
Connect this prerequisite to the complete program explanation and its conditions.
Move to related program practice →
Use new questions after focused repair; keep learning checks separate from diagnostic and mock results.

The Architectural handbook and specified codes and standards are supplied references. Build system understanding, a unit basis and coordination judgment before selecting a formula or code provision. This topic is classified as understanding and application; no individual reference omission has been verified.

Original ReasonSpan^™ teaching material. AI-assisted checks are complete; independent specialist review remains pending.