10.2 Duct Design Methodologies: Equal Friction, Static Regain & Velocity Reduction

Key Takeaways

  • Total Pressure in duct systems is the algebraic sum of Static Pressure and Velocity Pressure (Pt=Ps+PvP_t = P_s + P_v), where Velocity Pressure is calculated for standard air as Pv=(V4005)2P_v = \left(\frac{V}{4005}\right)^2.

  • The Equal Friction duct design method sizes all duct sections for a constant frictional resistance per unit length (typically 0.080.08 to 0.15 in. wg/100 ft0.15\text{ in. wg}/100\text{ ft}), causing duct velocities to naturally decrease downstream as flow rate drops.

  • The Static Regain duct design method sizes downstream duct transitions such that static pressure recovery from velocity reduction (ΔPsr=R(Pv1−Pv2)\Delta P_{sr} = R(P_{v1} - P_{v2})) balances downstream friction and fitting losses, yielding equal static pressure at all branch takeoffs.

  • Huebscher's circular equivalent diameter formula converts rectangular duct dimensions (a,b)(a, b) to an equivalent round duct diameter (DeD_e) that exhibits identical airflow and frictional pressure gradient: De=1.30(a⋅b)0.625(a+b)0.250D_e = \frac{1.30 (a \cdot b)^{0.625}}{(a + b)^{0.250}}.

  • Rectangular duct aspect ratio (AR=a/b\text{AR} = a/b) should be maintained below 3:13:1 (maximum 4:14:1) to minimize sheet metal weight, heat gain/loss surface area, and acoustic breakout.

Last updated: August 2026

10.2 Duct Design Methodologies: Equal Friction, Static Regain & Velocity Reduction

Duct design is the process of sizing air distribution ductwork, selecting routing geometries, and sizing terminal runouts to deliver design airflows to all building zones while balancing initial sheet metal fabrication cost, fan operating energy, space plenum constraints, and acoustic noise generation. On the PE Mechanical: HVAC and Refrigeration exam, duct design questions test pressure classifications, velocity pressure calculations, equivalent round duct conversions, Darcy-Weisbach friction factors, static regain derivations, and index run pressure summation.


1. Governing Fluid Mechanics in Air Distribution Systems

Airflow through closed conduits follows the steady-state conservation of mass and the extended Bernoulli energy equation. Total Pressure (PtP_t), Static Pressure (PsP_s), and Velocity Pressure (PvP_v) are related by:

Pt=Ps+PvP_t = P_s + P_v
   TOTAL PRESSURE (Pt) = STATIC PRESSURE (Ps) + VELOCITY PRESSURE (Pv)
  +-------------------------------------------------------------------+
  |                                                                   |
  |  Ps: Potential energy exerted equally in all directions on walls. | 
  |      Measured perpendicular to airflow direction.                 |
  |                                                                   |
  |  Pv: Kinetic energy directed parallel to fluid streamlines.       |
  |      Always POSITIVE in the direction of flow.                    |
  |                                                                   |
  +-------------------------------------------------------------------+

Velocity Pressure Equation (Standard Air)

Velocity pressure represents fluid kinetic energy per unit volume. For standard air density (ρ0=0.075 lbm/ft3\rho_0 = 0.075\text{ lbm/ft}^3 at 70∘F70^\circ\text{F} and 29.921 in. Hg29.921\text{ in. Hg}):

Pv=ρV22gc=0.075 lbm/ft3×(V ft/min/60)22×32.174 ft⋅lbm/(lbf⋅s2)×5.197 lbf/ft2 per in. wgP_v = \frac{\rho V^2}{2 g_c} = \frac{0.075\text{ lbm/ft}^3 \times (V\text{ ft/min} / 60)^2}{2 \times 32.174\text{ ft}\cdot\text{lbm}/(\text{lbf}\cdot\text{s}^2) \times 5.197\text{ lbf/ft}^2\text{ per in. wg}} Pv (in. wg)=(V4,005)2  ⟺  V (FPM)=4,005Pv (in. wg)P_v\text{ (in. wg)} = \left(\frac{V}{4,005}\right)^2 \iff V\text{ (FPM)} = 4,005 \sqrt{P_v\text{ (in. wg)}}

Where:

  • V=CFMAV = \frac{\text{CFM}}{A} = Mean air velocity in feet per minute (FPM\text{FPM})
  • AA = Internal cross-sectional area of duct in square feet (ft2\text{ft}^2)
  • For non-standard air density (ρ≠0.075 lbm/ft3\rho \ne 0.075\text{ lbm/ft}^3):
Pv=(V4,005)2×(ρ0.075)P_v = \left(\frac{V}{4,005}\right)^2 \times \left(\frac{\rho}{0.075}\right)

2. Duct Friction Loss: Darcy-Weisbach & Colebrook Relations

Frictional pressure loss across a straight duct of length LL and hydraulic diameter DhD_h is governed by the Darcy-Weisbach equation:

ΔPf (in. wg)=f(LDh)Pv=f(LDh)(V4,005)2\Delta P_f\text{ (in. wg)} = f \left(\frac{L}{D_h}\right) P_v = f \left(\frac{L}{D_h}\right) \left(\frac{V}{4,005}\right)^2

Where:

  • ff = Darcy friction factor (dimensionless)
  • Dh=4APwD_h = \frac{4 A}{P_w} = Hydraulic diameter (where PwP_w is wetted perimeter)
  • For circular duct: Dh=DD_h = D
  • For rectangular duct (a×ba \times b): Dh=4(a⋅b)2(a+b)=2aba+bD_h = \frac{4(a \cdot b)}{2(a + b)} = \frac{2 a b}{a + b}

Colebrook-White Equation for Air Ducts

For turbulent flow in commercial galvanized sheet metal ducts (roughness ε≈0.0003 ft=0.09 mm\varepsilon \approx 0.0003\text{ ft} = 0.09\text{ mm}):

1f=−2log⁡10(ε/Dh3.7+2.51Ref)\frac{1}{\sqrt{f}} = -2 \log_{10}\left(\frac{\varepsilon / D_h}{3.7} + \frac{2.51}{\text{Re} \sqrt{f}}\right) Reynolds Number: Re=V⋅Dhν=V (FPM)×Dh (ft)60×1.60×10−4 ft2/s=104.2×V (FPM)×Dh (ft)\text{Reynolds Number: } \text{Re} = \frac{V \cdot D_h}{\nu} = \frac{V\text{ (FPM)} \times D_h\text{ (ft)}}{60 \times 1.60 \times 10^{-4}\text{ ft}^2/\text{s}} = 104.2 \times V\text{ (FPM)} \times D_h\text{ (ft)}

3. Equivalent Circular Diameter for Rectangular Ducts (Huebscher Formula)

Rectangular ducts experience higher boundary-layer friction than round ducts of equal cross-sectional area due to corner eddy vortices and higher perimeter-to-area ratios. Huebscher's equivalent diameter formula calculates the round duct diameter (DeD_e) that provides the exact same volumetric flow rate (CFM\text{CFM}) and frictional pressure gradient (ΔP/100 ft\Delta P / 100\text{ ft}) as a rectangular duct of dimensions a×ba \times b:

De=1.30(a⋅b)0.625(a+b)0.250=1.30[(a⋅b)5(a+b)2]0.125D_e = \frac{1.30 (a \cdot b)^{0.625}}{(a + b)^{0.250}} = 1.30 \left[\frac{(a \cdot b)^5}{(a + b)^2}\right]^{0.125}
RECTANGULAR DUCT (a x b)               EQUIVALENT ROUND DUCT (De)
+-----------------------+              /-----------------\
|                       |  b (Height) |                   |
|                       |             |      De (Dia)     | (Same CFM & Same Friction)
+-----------------------+              \-----------------/
       a (Width)

Important Note on Area: The equivalent round duct area (Around=π4De2A_{\text{round}} = \frac{\pi}{4} D_e^2) is smaller than the rectangular duct area (Arect=a⋅bA_{\text{rect}} = a \cdot b). Consequently, the actual air velocity in the rectangular duct is lower than the velocity in the equivalent round duct: Vrect=CFMa⋅b/144<Vround=CFMπ4De2/144V_{\text{rect}} = \frac{\text{CFM}}{a \cdot b / 144} < V_{\text{round}} = \frac{\text{CFM}}{\frac{\pi}{4} D_e^2 / 144}

Duct Aspect Ratio (AR\text{AR})

Aspect Ratio: AR=Longer Side (a)Shorter Side (b)\text{Aspect Ratio: } \text{AR} = \frac{\text{Longer Side (}a\text{)}}{\text{Shorter Side (}b\text{)}}
  • Optimal: AR=1.0\text{AR} = 1.0 (Square duct): Minimizes sheet metal surface area, insulation cost, and thermal conduction losses.
  • Recommended Commercial Maximum: AR≤3:1\text{AR} \le 3:1.
  • Absolute Limit: AR≤4:1\text{AR} \le 4:1. Higher aspect ratios (≥5:1\ge 5:1) dramatically increase metal gauge thickness requirements, duct oil-canning drumming noise, and installation cost.

4. Comparison of Primary Duct Design Methodologies

+---------------------------------------------------------------------------------------------------------+
|                                      DUCT DESIGN METHODOLOGY MATRIX                                     |
+-------------------+------------------------------------+------------------------------------------------+
| Method            | Governing Principle                | Best Application & Balancing Characteristics   |
+-------------------+------------------------------------+------------------------------------------------+
| **Equal**         | Constant friction loss per unit    | Dominant standard for low-velocity commercial  |
| **Friction**      | length (e.g., 0.10 in. wg/100 ft). | supply & return. Non-self-balancing; requires   |
|                   | Velocity reduces downstream.       | manual balancing dampers on shorter runs.      |
+-------------------+------------------------------------+------------------------------------------------+
| **Static**        | Velocity reduction after branches  | High-velocity long trunk VAV systems.          |
| **Regain**        | converts Pv to Ps, balancing       | Self-balancing; maintains uniform static       |
|                   | downstream section friction loss.  | pressure at all VAV terminal inlets.           |
+-------------------+------------------------------------+------------------------------------------------+
| **Velocity**      | Arbitrary step-down of velocities  | Simple exhaust/intake systems; outdated for    |
| **Reduction**     | from main trunk to branch runouts. | complex distribution networks.                 |
+-------------------+------------------------------------+------------------------------------------------+
| **T-Method**      | Life-cycle cost optimization of    | Complex industrial systems balancing lifetime  |
|                   | energy vs capital metal cost.      | fan kW cost vs upfront sheet metal fabrication.|
+-------------------+------------------------------------+------------------------------------------------+

Detailed Breakdown of Methodologies

A. Equal Friction Method

  1. Select Design Friction Rate: Based on noise criteria and energy standards, select an initial design friction rate (ΔP100\Delta P_{100}). Common values:
    • Quiet commercial offices/libraries: 0.080.08 to 0.10 in. wg/100 ft0.10\text{ in. wg} / 100\text{ ft}
    • Standard commercial: 0.100.10 to 0.15 in. wg/100 ft0.15\text{ in. wg} / 100\text{ ft}
    • Industrial / mechanical rooms: 0.200.20 to 0.30 in. wg/100 ft0.30\text{ in. wg} / 100\text{ ft}
  2. Check Maximum Velocity: Sizing the initial fan discharge trunk using equal friction may yield excessive velocity. If the resulting velocity exceeds recommended acoustic limits (1,5001,500 to 1,800 FPM1,800\text{ FPM} for commercial mains), size the first section for maximum allowable velocity, then size subsequent sections by equal friction.
  3. Size Downstream Sections: Using the constant ΔP100\Delta P_{100}, determine diameter DeD_e and rectangular dimensions for each section as airflow drops at branch takeoffs.
  4. Identify Index Run (Critical Path): The duct run with the highest cumulative total pressure loss from fan discharge to the most hydraulically remote terminal unit establishes the required Fan Static Pressure (FSP).

B. Static Regain Method

As air leaves a branch takeoff, the downstream trunk carries reduced airflow. If the downstream duct is sized such that air velocity decreases (V2<V1V_2 < V_1), velocity pressure decreases (Pv2<Pv1P_{v2} < P_{v1}). By the Bernoulli theorem, a fraction (RR) of this kinetic energy converts back into static pressure:

Static Regain: ΔPsr=R⋅(Pv1−Pv2)=R[(V14,005)2−(V24,005)2]\text{Static Regain: } \Delta P_{sr} = R \cdot (P_{v1} - P_{v2}) = R \left[\left(\frac{V_1}{4,005}\right)^2 - \left(\frac{V_2}{4,005}\right)^2\right]

Where RR is the Static Regain Recovery Factor (typically 0.750.75 for standard commercial duct transitions with included expansion angles ≤15∘\le 15^\circ).

STATIC REGAIN MECHANISM ACROSS TRUNK TRANSITION:

Section 1 (High V1, High Pv1)           Section 2 (Lower V2, Lower Pv2)
-------------------------\              -------------------------------
Airflow: Q1 = 5,000 CFM   \  Transition  Airflow: Q2 = 3,000 CFM
Velocity: V1 = 1,800 FPM   \   (R=0.75)  Velocity: V2 = 1,200 FPM
----------------------------\           -------------------------------
                             \---------
Total Pressure:    Pt1 =======================> Drops due to fitting loss (Pt2 < Pt1)
Velocity Pressure: Pv1 =======================> Drops (Pv2 < Pv1)
Static Pressure:   Ps1 ====[ Ps Regain ]======> Rises! (Ps2 = Ps1 + Delta_Psr - Delta_P_loss)

In ideal static regain design, the static regain is sized to exactly offset the downstream duct friction and fitting losses:

ΔPsr=ΔPf,2+ΔPfitting,2\Delta P_{sr} = \Delta P_{f,2} + \Delta P_{\text{fitting},2}

This produces identical static pressure at every terminal box takeoff along the main trunk, eliminating duct static imbalance.


5. Maximum Design Air Velocities & Noise Criteria (NC)

High duct velocity causes turbulent vortex shedding, casing vibration, and aerodynamic noise that radiates through ceiling plenums or breaks into occupied spaces.

System Component / ZoneMaximum Design Velocity (FPM)Typical NC Rating Range
Main Supply Trunk (Mechanical Room)1,800 to 2,500 FPMNC 45 to 55
Main Supply Trunk (Above Occupied Spaces)1,200 to 1,500 FPMNC 30 to 40
Branch Ducts to VAV Boxes1,000 to 1,200 FPMNC 30 to 35
Branch Runouts to Diffusers600 to 800 FPMNC 25 to 30
Diffuser Neck / Grille Face400 to 600 FPMNC 20 to 30
Main Return Air Ducts1,000 to 1,400 FPMNC 35 to 45
Louvers / Outside Air Intakes400 to 600 FPM (Prevents rain ingestion)N/A

6. Worked Example: Equal Friction Sizing & Index Run Pressure Loss

Problem: A commercial rooftop air handling unit supplies 4,500 CFM4,500\text{ CFM} of standard air through a supply duct system designed using the Equal Friction method at a rate of ΔP100=0.10 in. wg/100 ft\Delta P_{100} = 0.10\text{ in. wg} / 100\text{ ft}. The index run consists of three sequential duct sections:

  • Section 1: Carries 4,500 CFM4,500\text{ CFM} over a straight length of 60 ft60\text{ ft}. Due to ceiling plenum depth, maximum rectangular duct height is restricted to b=14 in.b = 14\text{ in.}
  • Section 2: After a takeoff of 1,500 CFM1,500\text{ CFM}, carries 3,000 CFM3,000\text{ CFM} over a straight length of 40 ft40\text{ ft}. Maximum duct height is b=12 in.b = 12\text{ in.}
  • Section 3: After another takeoff of 1,500 CFM1,500\text{ CFM}, carries 1,500 CFM1,500\text{ CFM} over a straight length of 50 ft50\text{ ft} with a round spiral duct.
  • Fittings along Index Run: Section 1 includes a fan discharge transition (Co=0.15C_o = 0.15) and one 90∘90^\circ radius elbow (Co=0.20C_o = 0.20). Section 2 includes one 45∘45^\circ elbow (Co=0.12C_o = 0.12). Section 3 terminates in a VAV terminal box requiring 0.50 in. wg0.50\text{ in. wg} minimum inlet static pressure, a flexible duct runout (10 ft10\text{ ft} at 0.05 in. wg0.05\text{ in. wg}), and a ceiling diffuser (0.08 in. wg0.08\text{ in. wg} drop).

Find:

  1. Sizing for Section 1: Equivalent diameter DeD_e, required width aa, actual rectangular area AA, actual duct velocity V1V_1, and velocity pressure Pv1P_{v1}.
  2. Sizing for Section 2: Equivalent diameter DeD_e, required width aa, actual velocity V2V_2, and velocity pressure Pv2P_{v2}.
  3. Sizing for Section 3: Round duct diameter D3D_3, velocity V3V_3, and velocity pressure Pv3P_{v3}.
  4. Total static pressure loss along the index run and required fan static pressure (FSP).

(Friction Chart lookup values at ΔP100=0.10 in. wg/100 ft\Delta P_{100} = 0.10\text{ in. wg}/100\text{ ft}: 4,500 CFM→De=21.6 in.4,500\text{ CFM} \to D_e = 21.6\text{ in.}; 3,000 CFM→De=18.5 in.3,000\text{ CFM} \to D_e = 18.5\text{ in.}; 1,500 CFM→De=14.2 in.1,500\text{ CFM} \to D_e = 14.2\text{ in.}).

Step-by-Step Solution:

Step 1: Size Section 1 (4,500 CFM4,500\text{ CFM}, b=14 in.b = 14\text{ in.}, De=21.6 in.D_e = 21.6\text{ in.}). Using Huebscher formula inversion or ductulator tables for De=21.6 in.D_e = 21.6\text{ in.} with b=14 in.b = 14\text{ in.}, solve for width aa:

a≈30.0 in.a \approx 30.0\text{ in.}

Let's check DeD_e for 30 in.×14 in.30\text{ in.} \times 14\text{ in.}:

De=1.30(30×14)0.625(30+14)0.250=1.30(420)0.625(44)0.250=1.30(43.916)2.576=57.092.576=22.16 in.≈22 in.D_e = \frac{1.30 (30 \times 14)^{0.625}}{(30 + 14)^{0.250}} = \frac{1.30 (420)^{0.625}}{(44)^{0.250}} = \frac{1.30 (43.916)}{2.576} = \frac{57.09}{2.576} = 22.16\text{ in.} \approx 22\text{ in.}

Using 30 in.×14 in.30\text{ in.} \times 14\text{ in.}, Cross-sectional Area:

A1=30×14144=420144=2.917 ft2A_1 = \frac{30 \times 14}{144} = \frac{420}{144} = 2.917\text{ ft}^2 V1=4,500 CFM2.917 ft2=1,543 FPMV_1 = \frac{4,500\text{ CFM}}{2.917\text{ ft}^2} = 1,543\text{ FPM} Pv1=(1,5434,005)2=(0.3853)2=0.1485 in. wgP_{v1} = \left(\frac{1,543}{4,005}\right)^2 = (0.3853)^2 = 0.1485\text{ in. wg}

Step 2: Size Section 2 (3,000 CFM3,000\text{ CFM}, b=12 in.b = 12\text{ in.}, De=18.5 in.D_e = 18.5\text{ in.}). For De=18.5 in.D_e = 18.5\text{ in.} with b=12 in.b = 12\text{ in.}, required width is a=26.0 in.a = 26.0\text{ in.}

A2=26×12144=312144=2.167 ft2A_2 = \frac{26 \times 12}{144} = \frac{312}{144} = 2.167\text{ ft}^2 V2=3,000 CFM2.167 ft2=1,385 FPMV_2 = \frac{3,000\text{ CFM}}{2.167\text{ ft}^2} = 1,385\text{ FPM} Pv2=(1,3854,005)2=(0.3458)2=0.1196 in. wgP_{v2} = \left(\frac{1,385}{4,005}\right)^2 = (0.3458)^2 = 0.1196\text{ in. wg}

Step 3: Size Section 3 (1,500 CFM1,500\text{ CFM}, Round Duct De=14.2 in.D_e = 14.2\text{ in.}). Select standard commercial round duct: D3=14.0 in.D_3 = 14.0\text{ in.}

A3=π4(1412)2=0.7854×1.361=1.069 ft2A_3 = \frac{\pi}{4} \left(\frac{14}{12}\right)^2 = 0.7854 \times 1.361 = 1.069\text{ ft}^2 V3=1,500 CFM1.069 ft2=1,403 FPMV_3 = \frac{1,500\text{ CFM}}{1.069\text{ ft}^2} = 1,403\text{ FPM} Pv3=(1,4034,005)2=(0.3503)2=0.1227 in. wgP_{v3} = \left(\frac{1,403}{4,005}\right)^2 = (0.3503)^2 = 0.1227\text{ in. wg}

Step 4: Compute Total Pressure Losses along Index Run.

  1. Frictional Losses (Straight Ducts): ΔPfriction=(60 ft+40 ft+50 ft100)×0.10 in. wg=150100×0.10=0.150 in. wg\Delta P_{\text{friction}} = \left(\frac{60\text{ ft} + 40\text{ ft} + 50\text{ ft}}{100}\right) \times 0.10\text{ in. wg} = \frac{150}{100} \times 0.10 = 0.150\text{ in. wg}
  2. Fitting Dynamic Losses: Sec 1 Transition: ΔP=0.15×0.1485=0.0223 in. wg\text{Sec 1 Transition: } \Delta P = 0.15 \times 0.1485 = 0.0223\text{ in. wg} Sec 1 90∘ Elbow: ΔP=0.20×0.1485=0.0297 in. wg\text{Sec 1 } 90^\circ\text{ Elbow: } \Delta P = 0.20 \times 0.1485 = 0.0297\text{ in. wg} Sec 2 45∘ Elbow: ΔP=0.12×0.1196=0.0144 in. wg\text{Sec 2 } 45^\circ\text{ Elbow: } \Delta P = 0.12 \times 0.1196 = 0.0144\text{ in. wg} ΣΔPfittings=0.0223+0.0297+0.0144=0.0664 in. wg\Sigma \Delta P_{\text{fittings}} = 0.0223 + 0.0297 + 0.0144 = 0.0664\text{ in. wg}
  3. Terminal Devices & Runout: ΔPterminals=0.50 (VAV Box)+0.05 (Flex)+0.08 (Diffuser)=0.630 in. wg\Delta P_{\text{terminals}} = 0.50\text{ (VAV Box)} + 0.05\text{ (Flex)} + 0.08\text{ (Diffuser)} = 0.630\text{ in. wg}
  4. Total Fan Static Pressure (FSP) Required: FSP=ΔPfriction+ΣΔPfittings+ΔPterminals=0.150+0.0664+0.630=0.8464 in. wg≈0.85 in. wg\text{FSP} = \Delta P_{\text{friction}} + \Sigma \Delta P_{\text{fittings}} + \Delta P_{\text{terminals}} = 0.150 + 0.0664 + 0.630 = 0.8464\text{ in. wg} \approx 0.85\text{ in. wg}

7. NCEES Reference Handbook Navigation & Exam Tips

  • Huebscher Formula: Look in the Mechanical Engineering Reference Handbook under Fluid Mechanics / HVAC Duct Design for De=1.30(ab)0.625(a+b)0.250D_e = \frac{1.30 (ab)^{0.625}}{(a+b)^{0.250}}.
  • Static Regain Equation: Remember ΔPsr=R(Pv1−Pv2)\Delta P_{sr} = R(P_{v1} - P_{v2}). When velocity drops, static pressure increases, offsetting friction.
  • Velocity Pressure Constant: Commit Pv=(V/4005)2P_v = (V / 4005)^2 to memory for standard air (0.075 lbm/ft30.075\text{ lbm/ft}^3).
Loading diagram...
Duct Static, Velocity & Total Pressure Gradients
Test Your Knowledge

Standard air flows through a 24 in. x 12 in. rectangular duct at a volumetric flow rate of 3,200 CFM. What are the air velocity and velocity pressure inside the duct?

A

Velocity = 1,200 FPM, Velocity Pressure = 0.090 in. wg

B

Velocity = 1,600 FPM, Velocity Pressure = 0.400 in. wg

C

Velocity = 2,000 FPM, Velocity Pressure = 0.250 in. wg

D

Velocity = 1,600 FPM, Velocity Pressure = 0.160 in. wg

Test Your Knowledge

Using Huebscher's formula, what is the circular equivalent diameter (D_e) of a rectangular duct measuring 36 inches wide by 18 inches high?

A

27.4 inches

B

24.0 inches

C

30.5 inches

D

22.5 inches

Test Your Knowledge

An expanding duct transition reduces air velocity from 1,800 FPM in upstream Section 1 down to 1,200 FPM in downstream Section 2. Assuming standard air and a static regain recovery factor of R = 0.75, what is the static pressure regain (Delta_P_sr) across the transition?

A

0.112 in. wg

B

0.084 in. wg

C

0.056 in. wg

D

0.150 in. wg

Test Your Knowledge

Why is the Equal Friction duct design method considered 'non-self-balancing' in commercial HVAC applications?

A

Because friction loss is forced to zero in the primary supply riser

B

Because velocity pressure increases continuously toward the terminal diffusers

C

Because shorter branch duct runs closer to the fan have less cumulative friction loss than the index run, resulting in excess static pressure that must be throttled with manual balancing dampers

D

Because equal friction can only be applied to round ducts and fails in rectangular geometries

Sections you finish are checked off in the contents.