Temp Loss Thru Air Ducts
Instructions:
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Review the methodology to ensure it aligns with your project's requirements.
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Enter the duct size: for round ducts, enter inside diameter (in); for rectangular ducts, enter height and width (in).
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Enter the Exterior Duct Conditions:
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Still air: h_outside= 1 BTU/(hr·ft²·°F)
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Moving air: h_outside= 1.75 BTU/(hr·ft²·°F)
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Outdoors (wind): h_outside= 4 BTU/(hr·ft²·°F)
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Enter the Air Flow Rate, Q, in CFM.
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Enter the Duct entering Temperature, Tair, in °F.
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Enter the Exterior duct temperature,Toutside, in °F.
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Enter the Duct Insulation Value, Rinsulation, duct, in hr·ft²·°F/BTU.
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Enter the Duct Length in feet.
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Click the <Calculate> button
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The results are displayed in the Exit Air Temperature table.
Assumptions
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Steady-state heat transfer conditions
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Uniform air temperature across duct cross-section
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Constant air properties (c_p, ρ) along duct length
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One-dimensional heat transfer in radial direction
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Negligible heat generation within system
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Fully developed turbulent flow
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Constant outside air temperature along duct length
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Negligible thermal resistance of thin metal duct walls
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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
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Duct Shape: Rectangle or Round
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Duct Dimensions:
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Rectangle: Height and width (inches)
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Round: Diameter (inches)
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Exterior Conditions: Convection coefficient for air outside the duct
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Flow Rate (Q): Volumetric air flow rate (CFM)
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T_air: Temperature of air entering the duct (°F)
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T_outside: Ambient air temperature surrounding the duct (°F)
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R_insulation: Thermal resistance of duct insulation (hr·ft²·°F/BTU)
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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]
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Convective resistance on inside duct surface
R_duct = (thickness / k_duct) [hr·ft²·°F/BTU]
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Conductive resistance through duct wall
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Assumed negligible (R_duct = 0) for thin sheet metal construction
R_insulation = User input [hr·ft²·°F/BTU]
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Conductive resistance through insulation layer
R_outside = 1 / h_outside [hr·ft²·°F/BTU]
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Convective resistance on outside surface
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Still air: h_outside = 1 BTU/(hr·ft²·°F)
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Moving air: h_outside = 1.75 BTU/(hr·ft²·°F)
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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
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Steady-state heat transfer conditions
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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
Key References
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ASHRAE Handbook - Fundamentals (2021). American Society of Heating, Refrigerating and Air-Conditioning Engineers, Atlanta, GA.
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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.
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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.
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Holman, J.P. (2010). Heat Transfer, 10th ed., McGraw-Hill, New York.
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Cengel, Y.A., and Boles, M.A. (2015). Thermodynamics: An Engineering Approach, 8th ed., McGraw-Hill Education, New York.
Applications
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HVAC system design and optimization
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Energy loss assessment in air distribution systems
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Insulation requirement determination
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Temperature maintenance verification for process air
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Building energy modeling and analysis