How to use it

Review the methodology below to make sure it aligns with your project's requirements, then:

Choose the duct shape and enter its size: height and width for a rectangular duct, or the inside diameter for a round one (inches).
Choose the exterior duct conditions:
- Still air: h_outside = 1 Btu/(h·ft²·°F)
- Moving air: h_outside = 1.75 Btu/(h·ft²·°F)
- Outdoors (wind): h_outside = 4 Btu/(h·ft²·°F)
Enter the air flow rate, Q, in CFM.
Enter the temperature of the air entering the duct, T_air, and of the air around the duct, T_outside, in °F.
Enter the duct insulation value, R_insulation, in h·ft²·°F/Btu.
Enter the duct length in feet.
The exit air temperature, the heat gained or lost, and the values behind them appear in the results, and update as you type.

Duct heat loss calculator methodology

Purpose

This calculator estimates the heat loss (or heat gain) from rectangular or round ducts carrying air through unconditioned spaces, and calculates the resulting exit air temperature.

Input parameters

Duct shape: rectangular or round
Duct dimensions: height and width (in) for rectangular ducts, diameter (in) for round
Exterior conditions: the convection coefficient for the air outside the duct
Flow rate (Q): volumetric air flow rate (CFM)
T_air: temperature of the air entering the duct (°F)
T_outside: ambient air temperature around the duct (°F)
R_insulation: thermal resistance of the duct insulation (h·ft²·°F/Btu)
Length: total duct length (ft)

Step 1: cross-sectional area

Rectangular ducts: A = (Height × Width) / 144 [ft²]. Example: height 16 in, width 14 in: A = (16 × 14) / 144 = 1.556 ft²
Round ducts: A = π × (Diameter/12)² / 4 [ft²]. Example: diameter 18 in: A = π × 1.5² / 4 = 1.767 ft²

Step 2: air velocity

V = Q / A, where Q is the flow rate [CFM] and A the cross-sectional area [ft²].

Rectangular example: Q = 1,000 CFM, A = 1.556 ft²: V = 1,000 / 1.556 = 642.7 fpm = 10.7 fps
Round example: Q = 2,000 CFM, A = 1.767 ft²: V = 2,000 / 1.767 = 1,131.8 fpm = 18.9 fps

Step 3: hydraulic diameter

Rectangular ducts: D_h = 4A / P [in], where A is the area [in²] and P the perimeter [in] = 2(Height + Width). Example: height 16 in, width 14 in: P = 2(16 + 14) = 60 in; A = 16 × 14 = 224 in²; D_h = 4(224) / 60 = 14.93 in
Round ducts: D_h = diameter. Example: D_h = 18 in

Step 4: mass flow rate

ṁ = ρ × Q × 60 [lb/h], where ρ is the air density [lb/ft³] ≈ 0.075 lb/ft³ at standard conditions, Q the flow rate [CFM], and 60 converts minutes to hours. Reference: ASHRAE Handbook—Fundamentals (2025), Chapter 1: Psychrometrics.

Example: ṁ = 0.075 × 1,000 × 60 = 4,500 lb/h

Step 5: inside convection coefficient

h_inside = 1.5 × V^0.8 / (D_h / 12)^0.2 [Btu/(h·ft²·°F)], where V is the air velocity [fps] and D_h the hydraulic diameter [in]; dividing by 12 converts the hydraulic diameter from inches to feet.

Reference: simplified empirical correlation for forced convection in ducts, adapted from McQuiston, F.C., Parker, J.D., and Spitler, J.D. (2005). Heating, Ventilating, and Air Conditioning: Analysis and Design, 6th ed., Wiley.

Rectangular example: V = 10.7 fps, D_h = 14.93 in: h_inside = 1.5 × 6.67 / 1.045 = 9.57 Btu/(h·ft²·°F)
Round example: V = 18.9 fps, D_h = 18 in: h_inside = 1.5 × 10.48 / 1.0845 = 14.5 Btu/(h·ft²·°F)

Step 6: thermal resistance network

R_inside = 1 / h_inside: convective resistance on the inside duct surface
R_duct = thickness / k_duct: conductive resistance through the duct wall, taken as negligible (R_duct = 0) for thin sheet metal
R_insulation: user input, the conductive resistance of the insulation
R_outside = 1 / h_outside: convective resistance on the outside surface (still air 1, moving air 1.75, outdoors 4 Btu/(h·ft²·°F))

Reference: ASHRAE Handbook—Fundamentals (2025), Chapter 26: Heat, Air, and Moisture Control in Building Assemblies.

R_total = R_inside + R_duct + R_insulation + R_outside

Example: h_inside = 9.57, h_outside = 1.75 (moving air), R_insulation = 6: R_inside = 1 / 9.57 = 0.104; R_outside = 1 / 1.75 = 0.571; R_total = 0.104 + 0 + 6 + 0.571 = 6.676 h·ft²·°F/Btu. When summing thermal resistances, keep full precision through the intermediate steps and round only the final result.

Step 7: heat transfer

Q_transfer = |T_air − T_outside| × (P × L) / R_total [Btu/h], where P is the perimeter [ft] (rectangular: 2(Height + Width) / 12; round: π × Diameter / 12) and L the duct length [ft]. If T_air > T_outside the duct loses heat; if T_air < T_outside it gains heat. Reference: Fourier's law of heat conduction applied to composite walls; Holman, J.P. (2010). Heat Transfer, 10th ed., McGraw-Hill, pp. 76–82.

Rectangular, heat loss: T_air = 155 °F, T_outside = −4 °F, ΔT = 159 °F; P = 60 in = 5 ft; L = 100 ft; R_total = 6.676; surface area = 5 × 100 = 500 ft²; Q_loss = 159 × 500 / 6.676 = 11,908 Btu/h
Round, heat gain: T_air = 55 °F, T_outside = 95 °F, ΔT = 40 °F; P = π × 18 / 12 = 4.71 ft; L = 100 ft; R_total = 6.640 (h_inside = 14.5); surface area = 471.2 ft²; Q_gain = 40 × 471.2 / 6.640 = 2,839 Btu/h

Step 8: temperature change

ΔT_air = Q_transfer / (ṁ × c_p) [°F], with c_p = 0.240 Btu/(lb·°F). Reference: first law of thermodynamics for steady flow; Cengel, Y.A., and Boles, M.A. (2015). Thermodynamics: An Engineering Approach, 8th ed., McGraw-Hill, pp. 230–235.

Heat loss: 11,908 / (4,500 × 0.240) = 11.0 °F drop
Heat gain: 2,839 / (9,000 × 0.240) = 1.31 °F rise

Step 9: exit air temperature

Heat loss (T_air > T_outside): T_exit = T_air − ΔT_air. Example: 155 − 11.0 = 144.0 °F
Heat gain (T_air < T_outside): T_exit = T_air + ΔT_air. Example: 55 + 1.31 = 56.3 °F

Step 10: effective thermal parameter

k = P / (R_total × ṁ × c_p) [1/ft], a system-level parameter combining geometry and thermal resistance. Example: k = 5 / (6.676 × 4,500 × 0.240) = 6.93 × 10⁻⁴ /ft

Steps 7 to 9 treat the temperature difference as constant along the duct. The calculator goes one step further: because the air's temperature, and so the driving temperature difference, changes along the duct, it integrates dT/dL = −k (T − T_outside) over the length, giving T_exit = T_outside + (T_air − T_outside) × e^−kL. For short runs and well-insulated ducts the two agree closely; for long or poorly insulated runs the exponential form is the more accurate.

Assumptions

Steady-state heat transfer
Uniform air temperature across the duct cross-section
Constant air properties (c_p, ρ) along the duct length
One-dimensional heat transfer in the radial direction
Negligible heat generation within the system
Fully developed turbulent flow
Constant outside air temperature along the duct length
Negligible thermal resistance of thin metal duct walls
Negligible radiation heat transfer

Key references

ASHRAE Handbook—Fundamentals (2025). American Society of Heating, Refrigerating and Air-Conditioning Engineers, Atlanta, GA.
Incropera, F.P., DeWitt, D.P., Bergman, T.L., and Lavine, A.S. (2007). Fundamentals of Heat and Mass Transfer, 6th ed., John Wiley & Sons, New York.
McQuiston, F.C., Parker, J.D., and Spitler, J.D. (2005). Heating, Ventilating, and Air Conditioning: Analysis and Design, 6th ed., John Wiley & Sons, New York.
Holman, J.P. (2010). Heat Transfer, 10th ed., McGraw-Hill, New York.
Cengel, Y.A., and Boles, M.A. (2015). Thermodynamics: An Engineering Approach, 8th ed., McGraw-Hill Education, New York.

Applications

HVAC system design and optimization
Energy loss assessment in air distribution systems
Determining insulation requirements
Verifying temperature maintenance for process air
Building energy modeling and analysis

Temp Loss Thru Air Ducts

Exit air temperature