6.5 Air Distribution, Duct Design, Fans & Hydronic Water Networks

Q: A ventilation fan running at 900 RPM delivers 12,000 CFM at a static pressure of 2.0 inches water gauge, consuming 6.0 Brake Horsepower (BHP). If the fan sheave is changed to increase fan speed to 1,080 RPM on the identical duct system, what will be the new brake horsepower?

10.37 BHP. Applying Fan Law 3 for speed variation on a fixed system resistance: BHP_2 / BHP_1 = (N_2 / N_1)^3. The speed ratio is N_2 / N_1 = 1,080 / 900 = 1.20. Computing: BHP_2 = 6.0 * (1.20)^3 = 6.0 * 1.728 = 10.368 BHP, or 10.37 BHP. The value 7.2 BHP assumes linear scaling (6.0 * 1.20), and 8.64 BHP assumes quadratic scaling (6.0 * 1.20^2).

Q: A chilled water cooling coil requires a water flow rate of 80 GPM. If the coil is designed for a chilled water temperature supply of 44°F and return of 56°F, what is the cooling heat removal capacity of this coil?

480,000 BTU/hr (40.0 TR). Apply the standard hydronic heat transfer equation for water: q = 500 * GPM * Delta T. Here Delta T = 56°F - 44°F = 12°F. Substituting: q = 500 * 80 * 12 = 40,000 * 12 = 480,000 BTU/hr. Converting to tons of refrigeration: 480,000 / 12,000 = 40.0 TR. The value 400,000 BTU/hr incorrectly uses a 10°F Delta T, and 360,000 BTU/hr incorrectly uses a 9°F Delta T.

Q: A cooling tower operates with entering hot water at 96°F and leaving cold water at 86°F. If the ambient outdoor weather conditions are 92°F dry-bulb and 76°F wet-bulb, what is the cooling tower Approach and Effectiveness?

Approach = 10°F, Effectiveness = 50.0%. Cooling tower Range is T_hot - T_cold = 96°F - 86°F = 10°F. The Approach is the difference between leaving cold water and entering wet-bulb temperature: Approach = T_cold - T_wb = 86°F - 76°F = 10°F. The tower Effectiveness is eta = Range / (Range + Approach) = 10 / (10 + 10) = 10 / 20 = 0.50, or 50.0%. Comparing against dry-bulb (86 - 92) is a fundamental error.

Q: Air flows through a supply air duct at a velocity of 2,400 Feet Per Minute (FPM) under standard atmospheric conditions. What is the velocity pressure in inches of water gauge?

0.36 in. w.g.. Velocity pressure in US Customary units for standard air is VP = (V / 4005)^2. Substituting V = 2,400 FPM: VP = (2400 / 4005)^2 = (0.59925)^2 = 0.3591 in. w.g., which rounds to 0.36 in. w.g. The value 0.60 is the linear ratio V/4005, while 0.24 and 0.48 represent incorrect coefficient scalings.

Key Takeaways

Air distribution aerodynamics balances total, static, and velocity pressures ($TP = SP + VP$), with velocity pressure given by $VP = (V/4005)^2$ at standard air density.
Fan Laws govern operational scaling: airflow scales linearly with RPM ($CFM \propto N$), static pressure quadratically ($SP \propto N^2$), and brake horsepower cubically ($BHP \propto N^3$).
The standard hydronic heat transfer equation for water is $\dot{q} = 500 \times GPM \times \Delta T\text{ [BTU/hr]}$ ($4.184 \times \dot{V}_{L/s} \times \Delta T\text{ [kW]}$).
Control valve sizing requires balancing $C_v$ ($Q = C_v \sqrt{\Delta P}$) and valve authority ($N = \frac{\Delta P_{\text{valve}}}{\Delta P_{\text{valve}} + \Delta P_{\text{branch}}} \ge 0.50$) to maintain linear temperature modulation.
Cooling tower thermal performance is dictated by Range ($T_{\text{hot,in}} - T_{\text{cold,out}}$) and Approach ($T_{\text{cold,out}} - T_{wb,\text{amb}}$), with total makeup water compensating for evaporation, drift, and blowdown ($M = E \frac{COC}{COC - 1}$).

Last updated: August 2026

Air Distribution, Duct Design, Fans & Hydronic Water Networks

Fluid distribution networks—both air ductwork and hydronic piping loops—transport thermal energy between central plant equipment and conditioned building zones. Mastering duct sizing, fan performance scaling, chilled water hydronics, control valve authority, and cooling tower thermodynamics is vital for success on the NCEES PE Mechanical exam.

1. Air Distribution Fundamentals & Duct Aerodynamics

Air flowing through a duct system obeys the conservation of energy expressed by Bernoulli's equation in terms of total, static, and velocity pressures:

+-----------------------------------------------------------------------------------------+
|                        DUCT SYSTEM PRESSURE RELATIONSHIPS                               |
|                                                                                         |
|   Total Pressure (TP) = Static Pressure (SP) + Velocity Pressure (VP)                   |
|                                                                                         |
|   - Static Pressure (SP): Potential energy exerted outward equally in all directions    |
|     on duct walls; measured perpendicular to airflow.                                   |
|                                                                                         |
|   - Velocity Pressure (VP): Kinetic energy of moving air stream; always positive;       |
|     measured parallel to airflow using a pitot tube.                                    |
|                                                                                         |
|   - Standard Air Formula:   VP = \left(\frac{V}{4005}\right)^2 \quad [\text{in. w.g.}]  |
|     Where V is velocity in Feet Per Minute (FPM).                                       |
|                                                                                         |
|   - Velocity from VP:       V = 4005 \sqrt{VP} \quad [\text{FPM}]                       |
|                                                                                         |
|   - SI Units:               VP = \frac{1}{2} \rho V^2 \quad [\text{Pa}]                 |
+-----------------------------------------------------------------------------------------+

Duct Design Methodologies

Equal Friction Method: Sizes all duct sections for a constant friction loss per unit length (typically $0.08\text{ to }0.10\text{ in. w.g. per } 100\text{ ft}$ of equivalent duct run). Simple and widely used, but requires manual balancing dampers on shorter branch runs.
Static Regain Method: Sizes duct cross-sections such that the increase in static pressure from velocity reduction at each branch takeoff exactly balances the downstream friction loss of the next section.
Velocity Reduction Method: Arbitrarily steps down velocities along successive downstream branches.

Huebscher Formula for Equivalent Circular Duct Diameter

To convert a rectangular duct with dimensions $a$ and $b$ (inches) carrying a specified airflow to a circular duct of diameter $D_e$ having the exact same friction loss per foot:

$D_e = 1.30 \frac{(a \cdot b)^{0.625}}{(a + b)^{0.25}}$

2. Fan Types, Fan Laws & System Resistance Curves

+-----------------------------------------------------------------------------------------+
|                        CENTRIFUGAL VS. AXIAL FAN CLASSIFICATIONS                        |
|                                                                                         |
|   CENTRIFUGAL FANS (Air turned 90°):                                                    |
|   - Forward Curved (FC): Squirrel-cage, low speed/low SP; overloading horsepower curve. |
|   - Backward Inclined (BI) / Airfoil (AF): High efficiency (up to 85%), high SP;        |
|     self-limiting non-overloading power curve.                                          |
|   - Radial: Rugged, industrial material handling and sawdust collection.                |
|                                                                                         |
|   AXIAL FANS (Straight-through flow):                                                   |
|   - Propeller: High volume, low static pressure (< 0.5 in. w.g.); wall exhaust fans.    |
|   - Tubeaxial: Propeller inside cylinder; medium static pressure (< 2.0 in. w.g.).      |
|   - Vaneaxial: Straightening guide vanes convert swirl into static regain; up to 90% eta|
+-----------------------------------------------------------------------------------------+

The Affinity Laws for Fans (Fan Laws)

+-----------------------------------------------------------------------------------------+
|                              FAN LAWS FORMULATION SUMMARY                               |
|                                                                                         |
|   LAW 1: SPEED VARIATION (Constant Diameter D and Density \rho)                          |
|   - Airflow Rate:      \frac{CFM_2}{CFM_1} = \frac{N_2}{N_1}                             |
|   - Static Pressure:   \frac{SP_2}{SP_1} = \left(\frac{N_2}{N_1}\right)^2                |
|   - Brake Horsepower:  \frac{BHP_2}{BHP_1} = \left(\frac{N_2}{N_1}\right)^3             |
|                                                                                         |
|   LAW 2: DENSITY VARIATION (Constant Diameter D and Speed N)                            |
|   - Airflow Rate:      CFM_2 = CFM_1                                                    |
|   - Static Pressure:   \frac{SP_2}{SP_1} = \frac{\rho_2}{\rho_1}                       |
|   - Brake Horsepower:  \frac{BHP_2}{BHP_1} = \frac{\rho_2}{\rho_1}                      |
+-----------------------------------------------------------------------------------------+

Fan Static & Total Efficiency Formulas

$\text{Air Horsepower (AHP)} = \frac{CFM \times SP}{6356} \quad [\text{HP}]$ $\text{Fan Static Efficiency (}\eta_{fs}\text{)} = \frac{CFM \times SP}{6356 \times BHP}$ $\text{Fan Total Efficiency (}\eta_t\text{)} = \frac{CFM \times TP}{6356 \times BHP}$

System Effect Factors (SEF)

Abrupt elbows, transitions, or obstructions located immediately adjacent to fan inlets or outlets disrupt the uniform velocity profile, introducing severe uncalculated pressure losses known as System Effects. SEF must be added to calculated duct friction when selecting fan operating points.

3. Hydronic Distribution Networks (Chilled & Heating Water)

+-----------------------------------------------------------------------------------------+
|                        HYDRONIC PIPING SYSTEM ARCHITECTURES                             |
|                                                                                         |
|   PRIMARY-SECONDARY DECOUPLED:                                                          |
|   - Primary Loop: Constant flow chillers protected by dedicated primary pumps.          |
|   - Secondary Loop: Variable-speed distribution pumps modulating with zone 2-way valves.|
|   - Decoupler / Common Pipe: Hydraulic bridge with zero pressure drop separating loops. |
|                                                                                         |
|   VARIABLE PRIMARY FLOW (VPF):                                                          |
|   - Single variable-speed pump set modulates flow across variable-flow chillers.        |
|   - Requires minimum-flow bypass valve to prevent evaporator freezing during low loads. |
+-----------------------------------------------------------------------------------------+

The Fundamental Hydronic Equation

$\dot{q} = \dot{m} c_p \Delta T = (GPM \times 8.33\text{ lbm/gal} \times 60\text{ min/hr}) \times 1.0\text{ BTU/lbm}\cdot^\circ\text{F} \times \Delta T$ $\dot{q} = 500 \times GPM \times \Delta T \quad [\text{BTU/hr}]$

In SI Units (Water $\rho = 1000\text{ kg/m}^3, c_p = 4.184\text{ kJ/kg}\cdot\text{K}$): $\dot{q} = 4.184 \times \dot{V}_{L/s} \times \Delta T \quad [\text{kW}]$

For Ethylene/Propylene Glycol mixtures: $\dot{q} = 500 \times SG \times c_p \times GPM \times \Delta T$.

Hydronic Control Valves & Valve Authority ($N$)

Valve Flow Coefficient ($C_v$): The flow rate of $60^\circ\text{F}$ water in GPM that produces a $1.0\text{ psi}$ pressure drop across a fully open valve: $Q = C_v \sqrt{\Delta P} \implies \Delta P = \left(\frac{Q}{C_v}\right)^2$
Valve Authority ($N$): The ratio of the pressure drop across the fully open control valve to the total pressure drop of the valve plus its associated piping branch: $N = \frac{\Delta P_{\text{valve, 100\%}}}{\Delta P_{\text{valve, 100\%}} + \Delta P_{\text{piping branch}}}$ $N = \frac{Δ P _{valve, 100%}}{Δ P _{valve, 100%} + Δ P _{piping branch}}$
- If $N < 0.25$, the valve loses control authority, distorting linear heat output and causing severe actuator hunting.
- Target design rule: $N \ge 0.50$ (the valve drop should equal or exceed the piping branch drop).

4. Cooling Towers & Condenser Water Systems

Cooling towers reject waste heat from water-cooled chillers to the ambient atmosphere via direct evaporative contact with ambient air.

+-----------------------------------------------------------------------------------------+
|                        COOLING TOWER TEMPERATURE RELATIONSHIPS                          |
|                                                                                         |
|   Hot Water In (T_hot,in) = 95°F                                                        |
|         |                                                                               |
|         | <--- Cooling Tower Range = T_hot,in - T_cold,out = 95°F - 85°F = 10°F         |
|         v                                                                               |
|   Cold Water Out (T_cold,out) = 85°F                                                    |
|         |                                                                               |
|         | <--- Cooling Tower Approach = T_cold,out - T_wb,ambient = 85°F - 78°F = 7°F   |
|         v                                                                               |
|   Ambient Wet-Bulb Temp (T_wb) = 78°F                                                   |
+-----------------------------------------------------------------------------------------+

Key Cooling Tower Performance Metrics:

Range: The temperature drop of the condenser water across the tower: $\text{Range} = T_{\text{hot, entering}} - T_{\text{cold, leaving}}$
Approach: The temperature difference between the cold water leaving the tower basin and the ambient entering wet-bulb temperature: $\text{Approach} = T_{\text{cold, leaving}} - T_{wb,\text{ambient}}$ (Note: The leaving water temperature can NEVER be cooled below the ambient wet-bulb temperature).
Cooling Tower Effectiveness ($\eta_{\text{tower}}$): $\eta_{\text{tower}} = \frac{\text{Range}}{\text{Range} + \text{Approach}} = \frac{T_{\text{hot}} - T_{\text{cold}}}{T_{\text{hot}} - T_{wb}}$

Cooling Tower Water Consumption & Makeup Water

Water lost through heat rejection must be continuously replenished by fresh municipal makeup water ($M$):

+-----------------------------------------------------------------------------------------+
|                        COOLING TOWER WATER BALANCE                                      |
|                                                                                         |
|   Total Makeup Water (M) = Evaporation Loss (E) + Blowdown (B) + Drift Loss (D)         |
|                                                                                         |
|   - Evaporation Rate (E):   E \approx 0.0008 \times GPM \times \text{Range}             |
|     Rule of thumb: ~1% of circulating flow per 10°F Range (or ~3 GPM per 100 TR).       |
|                                                                                         |
|   - Drift Loss (D):         D \approx 0.0002 \text{ to } 0.002 \times GPM               |
|                                                                                         |
|   - Cycles of Concentration (COC): Ratio of dissolved solids in basin to makeup:       |
|     COC = \frac{\text{TDS}_{\text{basin}}}{\text{TDS}_{\text{makeup}}} = \frac{M}{B + D}|
|                                                                                         |
|   - Blowdown Rate (B):      B = \frac{E}{COC - 1} - D \approx \frac{E}{COC - 1}         |
|                                                                                         |
|   - Total Makeup (M):       M = E \left(\frac{COC}{COC - 1}\right)                      |
+-----------------------------------------------------------------------------------------+

5. Step-by-Step Worked Engineering Problem

Problem Statement

A central chilled water plant features a $500\text{ TR}$ water-cooled centrifugal chiller operating with a condenser heat rejection rate of $15,000\text{ BTU/hr}\cdot\text{TR}$.

Condenser water circulation rate $= 1,500\text{ GPM}$.
Entering cooling tower hot water temperature $= 95.0^\circ\text{F}$.
Ambient air conditions: $90.0^\circ\text{F } T_{db}$ and $78.0^\circ\text{F } T_{wb}$.
Leaving cooling tower cold water temperature $= 85.0^\circ\text{F}$.
Water treatment operates at $COC = 4.0$ Cycles of Concentration. Drift loss is $0.001 \times GPM$.

Calculate:

The cooling tower Range, Approach, and Effectiveness.
The evaporation rate ($E$) in $\text{GPM}$.
The required blowdown rate ($B$) in $\text{GPM}$.
The total municipal makeup water requirement ($M$) in $\text{GPM}$.
If the cooling tower fan motor is upgraded from $1,200\text{ RPM}$ to $1,400\text{ RPM}$, calculate the percentage increase in fan Brake Horsepower ($BHP$).

+-----------------------------------------------------------------------------------------+
|                          CENTRAL PLANT CALCULATION STEPS                                |
|                                                                                         |
|   Range    = 95.0°F - 85.0°F = 10.0°F                                                   |
|   Approach = 85.0°F - 78.0°F = 7.0°F                                                    |
+-----------------------------------------------------------------------------------------+

Step 1: Range, Approach & Effectiveness

$\text{Range} = 95.0 - 85.0 = 10.0^\circ\text{F}$ $\text{Approach} = 85.0 - 78.0 = 7.0^\circ\text{F}$ $\eta_{\text{tower}} = \frac{\text{Range}}{\text{Range} + \text{Approach}} = \frac{10.0}{10.0 + 7.0} = \frac{10.0}{17.0} = 0.5882 \implies 58.8\%$

Step 2: Evaporation Loss Rate ($E$)

Total Heat Rejected: $\dot{Q}{\text{cond}} = 500\text{ TR} \times 15,000\text{ BTU/hr}\cdot\text{TR} = 7,500,000\text{ BTU/hr}$. Using latent heat of vaporization ($h{fg} \approx 1,000\text{ BTU/lbm}$): $E = \frac{\dot{Q}_{\text{cond}}}{h_{fg} \times 8.33\text{ lbm/gal} \times 60\text{ min/hr}} = \frac{7,500,000}{1000 \times 500} = 15.0\text{ GPM}$ (Verification with rule of thumb: $E = 0.0008 \times 1500 \times 10 = 12.0\text{--}15.0\text{ GPM}$).

Step 3: Blowdown Rate ($B$)

$\text{Drift Loss } D = 0.001 \times 1,500\text{ GPM} = 1.5\text{ GPM}$ $B = \frac{E}{COC - 1} - D = \frac{15.0\text{ GPM}}{4.0 - 1} - 1.5 = \frac{15.0}{3} - 1.5 = 5.0 - 1.5 = 3.5\text{ GPM}$

Step 4: Total Makeup Water ($M$)

$M = E + B + D = 15.0 + 3.5 + 1.5 = 20.0\text{ GPM}$ (Verification: $M = E \times \frac{COC}{COC-1} = 15.0 \times \frac{4}{3} = 20.0\text{ GPM}$).

Step 5: Fan Law Scaling for Brake Horsepower

Speed ratio: $\frac{N_2}{N_1} = \frac{1400}{1200} = 1.1667$. $\frac{BHP_2}{BHP_1} = \left(\frac{N_2}{N_1}\right)^3 = (1.1667)^3 = 1.5879$ $\text{Percentage Increase} = (1.5879 - 1) \times 100\% = 58.8\%$

6. Common Exam Traps & PE Pro-Tips

Trap 1 — Fan Law 3 Cubed Power Scaling: Increasing fan speed by just $20%$ ($1.20$) increases electrical power draw by $72.8%$ ($1.20^3 = 1.728$). Never assume linear power scaling!
Trap 2 — Fan Laws Do Not Apply to Duct Resistance Alterations: The Fan Laws relate performance changes on the same system resistance curve when fan speed is modified. If a damper is closed or duct geometry changes, the system curve itself shifts, and you must find the intersection of the new system curve with the fan curve.
Trap 3 — Cooling Tower Approach Limit: Remember that no cooling tower, regardless of size or airflow, can cool water below the ambient wet-bulb temperature. If an exam question claims leaving water is colder than ambient wet-bulb, the scenario violates the Second Law of Thermodynamics.

Test Your Knowledge

A ventilation fan running at 900 RPM delivers 12,000 CFM at a static pressure of 2.0 inches water gauge, consuming 6.0 Brake Horsepower (BHP). If the fan sheave is changed to increase fan speed to 1,080 RPM on the identical duct system, what will be the new brake horsepower?

7.2 BHP

10.37 BHP

8.64 BHP

12.0 BHP

Test Your Knowledge

A chilled water cooling coil requires a water flow rate of 80 GPM. If the coil is designed for a chilled water temperature supply of 44°F and return of 56°F, what is the cooling heat removal capacity of this coil?

360,000 BTU/hr (30.0 TR)

400,000 BTU/hr (33.3 TR)

480,000 BTU/hr (40.0 TR)

520,000 BTU/hr (43.3 TR)

Test Your Knowledge

A cooling tower operates with entering hot water at 96°F and leaving cold water at 86°F. If the ambient outdoor weather conditions are 92°F dry-bulb and 76°F wet-bulb, what is the cooling tower Approach and Effectiveness?

Approach = 4°F, Effectiveness = 71.4%

Approach = 10°F, Effectiveness = 40.0%

Approach = 6°F, Effectiveness = 62.5%

Approach = 10°F, Effectiveness = 50.0%

Test Your Knowledge

Air flows through a supply air duct at a velocity of 2,400 Feet Per Minute (FPM) under standard atmospheric conditions. What is the velocity pressure in inches of water gauge?

0.36 in. w.g.

0.60 in. w.g.

0.24 in. w.g.

0.48 in. w.g.

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6.6 Energy Recovery Devices, HVAC Acoustics & Vibration Isolation

Continue learning

PE Mechanical

1Chapter 1: Exam Overview, Reference Handbook & Mechanical Engineering Fundamentals

2Chapter 2: Mechanical Design & Analysis (Stress, Failure Theories & Components)

3Chapter 3: Mechanical Power Transmission, Mechanisms & Vibrations

4Chapter 4: Thermal & Fluid Systems

5Chapter 5: Power Plants & Energy Conversion Systems

6Chapter 6: HVAC & Refrigeration Systems

7Chapter 7: Materials, Manufacturing & Processing

8Chapter 8: Measurements, Controls, Codes & Engineering Ethics

PE Mechanical

6.5 Air Distribution, Duct Design, Fans & Hydronic Water Networks

Key Takeaways

Air Distribution, Duct Design, Fans & Hydronic Water Networks

1. Air Distribution Fundamentals & Duct Aerodynamics

Duct Design Methodologies

Huebscher Formula for Equivalent Circular Duct Diameter

2. Fan Types, Fan Laws & System Resistance Curves

The Affinity Laws for Fans (Fan Laws)

Fan Static & Total Efficiency Formulas

System Effect Factors (SEF)

3. Hydronic Distribution Networks (Chilled & Heating Water)

The Fundamental Hydronic Equation

Hydronic Control Valves & Valve Authority ($N$)

4. Cooling Towers & Condenser Water Systems

Key Cooling Tower Performance Metrics:

Cooling Tower Water Consumption & Makeup Water

5. Step-by-Step Worked Engineering Problem

Problem Statement

Step 1: Range, Approach & Effectiveness

Step 2: Evaporation Loss Rate ($E$)

Step 3: Blowdown Rate ($B$)

Step 4: Total Makeup Water ($M$)

Step 5: Fan Law Scaling for Brake Horsepower

6. Common Exam Traps & PE Pro-Tips