Heat Transfer Calculator

Heat Transfer is evaluated from Heat Transfer Mode, Mass and Specific Heat. The calculation reports Heat Transfer, Heat Transfer Rate and Power / Rate.

Results

Thanks — we’ve logged this for review.

About the Heat Transfer Calculator

Heat Transfer is treated here as a quantitative relation between Heat Transfer Mode, Mass, Specific Heat and Temperature Difference and Heat Transfer, Heat Transfer Rate, Power / Rate and Power / Rate.

The calculator uses a multi formula configuration. Each reported value is read as a direct evaluation of the stored rules with the declared field formats and units.

Formula basis:
Sensible heat: Q(BTU) = m(lb) x cp(BTU/lb· degF) x ΔT( degF)
Conduction: Q(BTU/hr) = k x A(ft^2) x ΔT( degF) / d(in)
Convection: Q(BTU/hr) = h x A(ft^2) x ΔT( degF)
R-value: R = d(in) / k
1 BTU/hr = 0.2931 W

Interpret the outputs in the order shown by the result fields. Optional inputs affect only the outputs that depend on those variables.

Formula & How It Works

The calculation applies the following relations exactly as recorded in the metadata:

Sensible heat: Q(BTU) = m(lb) x cp(BTU/lb· degF) x ΔT( degF)
Conduction: Q(BTU/hr) = k x A(ft^2) x ΔT( degF) / d(in)
Convection: Q(BTU/hr) = h x A(ft^2) x ΔT( degF)
R-value: R = d(in) / k
1 BTU/hr = 0.2931 W

Each output field is produced by substituting the supplied inputs into the relevant relation and then applying the declared rounding or text format.

Worked Examples

Example 1: Heating Water for a Shower

Inputs

mode: Sensible Heat: Q = mcΔT mass_lb: 60 cp: 1 delta_T_F: 80

Heat Transfer: 4,800 BTU. Power / Rate: 1.407 kW. Power / Rate: 1,406.7 W

With Heat Transfer Mode = Sensible Heat: Q = mcΔT, Mass = 60, Specific Heat = 1 and Temperature Difference = 80 as the stated inputs, the result is Heat Transfer = 4,800 BTU, Power / Rate = 1.407 kW and Power / Rate = 1,406.7 W. Each value corresponds to the declared output fields.

Example 2: Heat Loss Through an Insulated Wall

Inputs

mode: Conduction: Q = kAΔT/d delta_T_F: 50 k_val: 0.27 area_ft2: 120 thickness_in: 3.5

Heat Transfer: 462.86 BTU. Heat Transfer Rate: 462.86 BTU/hr. Power / Rate: 0.136 kW. Power / Rate: 135.7 W. R-Value: 12.96 hr·ft^2· degF/BTU

With Heat Transfer Mode = Conduction: Q = kAΔT/d, Temperature Difference = 50, Thermal Conductivity = 0.27 and Surface Area = 120 as the stated inputs, the result is Heat Transfer = 462.86 BTU, Heat Transfer Rate = 462.86 BTU/hr and Power / Rate = 0.136 kW. Each value corresponds to the declared output fields.

Example 3: Convective Cooling — Electronics Heat Sink

Inputs

mode: Convection: Q = hAΔT delta_T_F: 40 area_ft2: 0.25 h_coeff: 3.5

Heat Transfer: 35 BTU. Heat Transfer Rate: 35 BTU/hr. Power / Rate: 0.01 kW. Power / Rate: 10.3 W

With Heat Transfer Mode = Convection: Q = hAΔT, Temperature Difference = 40, Surface Area = 0.25 and Convection Coefficient = 3.5 as the stated inputs, the result is Heat Transfer = 35 BTU, Heat Transfer Rate = 35 BTU/hr and Power / Rate = 0.01 kW. Each value corresponds to the declared output fields.

Example 4: Concrete Slab — Thermal Mass in Building

Inputs

mode: Sensible Heat: Q = mcΔT mass_lb: 12000 cp: 0.2 delta_T_F: 20

Heat Transfer: 48,000 BTU. Power / Rate: 14.067 kW. Power / Rate: 14,067.4 W

With Heat Transfer Mode = Sensible Heat: Q = mcΔT, Mass = 12,000, Specific Heat = 0.2 and Temperature Difference = 20 as the stated inputs, the result is Heat Transfer = 48,000 BTU, Power / Rate = 14.067 kW and Power / Rate = 14,067.4 W. Each value corresponds to the declared output fields.

Common Use Cases

Calculate heat loss through a wall by conduction
Find sensible heat to raise water temperature
Determine convective heat transfer from a surface