Temp Loss through an Air Duct

Assumptions

• Steady-state heat transfer conditions

• Uniform air temperature across duct cross-section

• Constant air properties (c_p, ρ) along duct length

• One-dimensional heat transfer in radial direction

• Negligible heat generation within system

• Fully developed turbulent flow

• Constant outside air temperature along duct length

• Negligible thermal resistance of thin metal duct walls

• Negligible radiation heat transfer

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: Rectangle or Round

• Duct Dimensions:

  • Rectangle: Height and width (inches)

  • Round: Diameter (inches)

• Exterior Conditions: Convection coefficient for air outside the duct

• Flow Rate (Q): Volumetric air flow rate (CFM)

• T_air: Temperature of air entering the duct (°F)

• T_outside: Ambient air temperature surrounding the duct (°F)

• R_insulation: Thermal resistance of duct insulation (hr·ft²·°F/BTU)

• Length: Total duct length (ft)

Calculation Methodology

STEP 1: Cross-Sectional Area Calculation

For Rectangular Ducts: Formula: A = (Height × Width) / 144 [ft²]

Example: Height = 16 in, Width = 14 in A = (16 × 14) / 144 = 1.556 ft²

For Round Ducts: Formula: A = π × (Diameter/12)² / 4 [ft²]

Example: Diameter = 18 in A = π × (18/12)² / 4 = π × 1.5² / 4 = 1.767 ft²

STEP 2: Air Velocity Calculation

Formula: V = Q / A [fps]

Where:

Q = volumetric flow rate [CFM]

A = cross-sectional area [ft²]

Example (Rectangular):

Q = 1000 CFM

A = 1.556 ft²

V = Q/A

     = 1000 / 1.556

     = 642.7 fpm

     = 10.7 fps

Example (Round):

Q = 2000 CFM

A = 1.767 ft²

V = Q/A

     = 2000 / 1.767

     = 1131.6 fpm

     = 18.9 fps

STEP 3: Hydraulic Diameter Calculation

For Rectangular Ducts: Formula: D_h = 4A / P [inches]

Where:

A = cross-sectional area [in²]

P = perimeter [inches] = 2(Height + Width)

Example (Rectangular):

Height = 16 in

Width = 14 in

P = 2(W x H)

     = 2(16 + 14)

     = 60 inches

A = W x H

     = 16 × 14

     = 224 in² D_h

     = 4(224) / 60

     = 14.93 inches

For Round Ducts: Formula: D_h = Diameter [inches]

Example (Round):

Diameter = 18 in

D_h = 18 inches

STEP 4: Mass Flow Rate Calculation

Formula: ṁ = ρ × Q × 60 [lb/hr]

Where:

ρ = air density [lb/ft³] ≈ 0.075 lb/ft³ at standard conditions

Q = volumetric flow rate [CFM]

60 = conversion factor (min to hr)

Reference: ASHRAE Handbook - Fundamentals (2021), Chapter 1: Psychrometrics.

Example:

Q = 1000 CFM

ρ = 0.075 lb/ft³

ṁ = ρ × Q × 60

      = 0.075 × 1000 × 60

      = 4500 lb/hr

STEP 5: Inside Convection Coefficient

Formula: h_inside = 1.5 × V^0.8 / (D_h / 12)^0.2 [BTU/(hr·ft²·°F)]

Where:

V = air velocity [fps]

D_h = hydraulic diameter [inches]

The division by 12 converts the hydraulic diameter from inches to feet in the denominator

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.

Example (Rectangular):

V = 10.7 fps

D_h = 14.93 inches

h_inside = 1.5 × V^0.8 / (D_h / 12)^0.2

                     = 1.5 × (10.7)^0.8 / (14.93 / 12)^0.2

                     = 1.5 × 6.77 / 1.067

                     = 15.7 BTU/(hr·ft²·°F)

Example (Round):

V = 18.9 fps

D_h = 18 inches

h_inside = 1.5 × V^0.8 / (D_h / 12)^0.2

                     = 1.5 × (18.9)^0.8 / (18 / 12)^0.2 h_inside

                     = 1.5 × 10.05 / 1.0845

                     = 13.9 BTU/(hr·ft²·°F)

STEP 6: Thermal Resistance Network

R_inside = 1 / h_inside [hr·ft²·°F/BTU]

• Convective resistance on inside duct surface

R_duct = (thickness / k_duct) [hr·ft²·°F/BTU]

• Conductive resistance through duct wall

• Assumed negligible (R_duct = 0) for thin sheet metal construction

R_insulation = User input [hr·ft²·°F/BTU]

• Conductive resistance through insulation layer

R_outside = 1 / h_outside [hr·ft²·°F/BTU]

• Convective resistance on outside surface

• Still air: h_outside = 1 BTU/(hr·ft²·°F)

• Moving air: h_outside = 1.75 BTU/(hr·ft²·°F)

• Outdoors (wind): h_outside = 4 BTU/(hr·ft²·°F)

Reference: ASHRAE Handbook - Fundamentals (2021), Chapter 26: Heat, Air, and Moisture Control in Building Assemblies.

Formula: R_Total = R_inside + R_duct + R_insulation + R_outside

Example:

h_inside = 15.7 BTU/(hr·ft²·°F)

h_outside = 1.75 BTU/(hr·ft²·°F) (moving air)

R_insulation = 6 hr·ft²·°F/BTU

R_inside = 1 / 15.7 = 0.064 hr·ft²·°F/BTU

R_duct = 0

R_outside = 1 / 1.75 = 0.571 hr·ft²·°F/BTU

R_Total = R_inside + R_duct + R_insulation + R_outside

                   = 0.064 + 0 + 6 + 0.571

                   = 6.635 hr·ft²·°F/BTU

NOTE: When summing thermal resistances, maintain precision through intermediate calculations and round only the final result to avoid cumulative rounding errors.

STEP 7: Heat Transfer Calculation

Formula: Q_transfer = |T_air - T_outside| × (P × L) / R_Total [BTU/hr]

Where:

|T_air - T_outside| = absolute temperature difference driving heat transfer [°F]

P = perimeter [ft]

Rectangular: P = 2(Height + Width) / 12

Round: P = π × Diameter / 12

L = duct length [ft]

R_Total = total thermal resistance [hr·ft²·°F/BTU]

Sign Convention:

If T_air > T_outside: Heat loss from duct (Q_loss)

If T_air < T_outside: Heat gain to duct (Q_gain)

Reference: Fourier's Law of Heat Conduction applied to composite walls. Holman, J.P. (2010). Heat Transfer, 10th ed., McGraw-Hill, pp. 76-82.

Example (Rectangular - Heat Loss): T_air = 155°F, T_outside = -4°F ΔT = |155 - (-4)| = 159°F P = 60 inches = 5 ft L = 100 ft R_Total = 6.635 hr·ft²·°F/BTU Surface Area = P × L = 5 × 100 = 500 ft²

Q_loss = |T_air - T_outside| × (P × L) / R_Total

                  = 159 × 500 / 6.635

                  = 11,983 BTU/hr

Example (Round - Heat Gain):

T_air = 55°F

T_outside = 95°F

ΔT = |55 - 95| = 40°F

P = π × 18 / 12 = 4.71 ft

L = 100 ft

R_Total = 6.613 hr·ft²·°F/BTU

Surface Area = 4.71 × 100 = 471 ft²

Q_gain = |T_air - T_outside| × (P × L) / R_Total

                  = 40 × 471 / 6.613

                  = 2,848 BTU/hr

STEP 8: Temperature Change Calculation

Formula: ΔT_air = Q_transfer / (ṁ × c_p) [°F]

Where:

Q_transfer = heat transfer through duct walls [BTU/hr]

ṁ = mass flow rate [lb/hr]

c_p = specific heat of air = 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.

Example (Heat Loss):

Q_loss = 11,983 BTU/hr

ṁ = 4500 lb/hr

c_p = 0.240 BTU/(lb·°F)

ΔT_air = Q_transfer / (ṁ × c_p) [°F]

                = 11,983 / (4500 × 0.240)

                = 11.1°F (temperature drop)

Example (Heat Gain):

Q_gain = 2,848 BTU/hr

ṁ = 9000 lb/hr

c_p = 0.240 BTU/(lb·°F)

ΔT_air = Q_transfer / (ṁ × c_p) [°F]

                = 2,848 / (9000 × 0.240)

                = 1.32°F (temperature rise)

STEP 9: Exit Air Temperature

Formula:

For heat loss (T_air > T_outside): T_exit = T_air - ΔT_air [°F]

For heat gain (T_air < T_outside): T_exit = T_air + ΔT_air [°F]

Example (Heat Loss):

T_air = 155°F

ΔT_air = 11.1°F

T_exit = T_air - ΔT_air

               = 155 - 11.1

               = 143.9°F

Example (Heat Gain):

T_air = 55°F

ΔT_air = 1.32°F

T_exit = T_air - ΔT_air

               = 55 + 1.32

               = 56.3°F

STEP 10: Effective Thermal Conductivity Parameter

Formula: k = P / (R_Total × ṁ × c_p) [1/ft]

This represents a system-level thermal parameter combining geometry and thermal resistance, used for quick reference calculations.

Example:

P = 5 ft

R_Total = 6.635 hr·ft²·°F/BTU

ṁ = 4500 lb/hr

c_p = 0.240 BTU/(lb·°F)

k = P / (R_Total × ṁ × c_p)

    = 5 / (6.635 × 4500 × 0.240)

    = 6.98×10⁻⁴ /ft

Assumptions

1. Steady-state heat transfer conditions

2. Uniform air temperature across duct cross-section

3. Constant air properties (c_p, ρ) along duct length

4. One-dimensional heat transfer in radial direction

5. Negligible heat generation within system

6. Fully developed turbulent flow

7. Constant outside air temperature along duct length

8. Negligible thermal resistance of thin metal duct walls

9. Negligible radiation heat transfer

Key References

1. ASHRAE Handbook - Fundamentals (2021). American Society of Heating, Refrigerating and Air-Conditioning Engineers, Atlanta, GA.

2. 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.

3. 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.

4. Holman, J.P. (2010). Heat Transfer, 10th ed., McGraw-Hill, New York.

5. 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

• Insulation requirement determination

• Temperature maintenance verification for process air

• Building energy modeling and analysis