Calculate heat rejection rate, mass or volumetric flow, specific heat, density, or temperature change with flexible SI and US units.

Sensible heat with constant average properties and no phase change. The fluid helper fills editable reference values; use actual operating properties.

Heat Rejection Formula

The heat rejection rate is thermal power carried away by a moving fluid. This constant-average-property sensible model assumes no phase change; volumetric flow and density must refer to the same state. For this calculator, when specific heat capacity is entered in kJ/(kg·K), density in kg/m³, volumetric flow rate in m³/hr, and temperature change in K or °C, the governing equation is:

Q = (cₚ ρ F Δ T) / (3600)
Variable Meaning Typical Unit
Q Heat rejection rate kW
cp Specific heat capacity of the fluid kJ/(kg·K)
ρ Fluid density kg/m³
F Volumetric flow rate m³/hr
ΔT Temperature change across the fluid stream K or °C

This relationship is linear: if you double flow rate, specific heat, density, or temperature change, the calculated heat rejection doubles as well. The 3600 in the denominator converts the hourly flow basis into seconds so the final result is expressed in kW instead of kJ/hr.

How to Calculate Heat Rejection

  1. Identify the fluid properties: specific heat capacity cp and density ρ.
  2. Measure or estimate the volumetric flow rate F.
  3. Determine the temperature change across the fluid stream.
  4. Substitute the values into the formula.
  5. Interpret the result as the rate of heat being removed from the process or equipment.

If you know inlet and outlet temperatures, compute the temperature difference first:

Δ T = | Tₒᵤₜ - Tᵢₙ |

For most cooling calculations, heat rejection is reported as a positive magnitude, so the absolute difference is typically the most useful form.

Equivalent Forms

Sometimes you know mass flow rate instead of volumetric flow rate. In that case, use the flow conversion:

dotm = ρ dotV

The heat rejection equation then becomes:

Q = dotm cₚ Δ T

That form is especially convenient when ˙m is in kg/s and cp is in kJ/(kg·K), because Q comes out directly in kW.

Example

Assume the following inputs:

  • cp = 4.186 kJ/(kg·K)
  • ρ = 1000 kg/m³
  • F = 12 m³/hr
  • ΔT = 6 °C
Q = (4.186 × 1000 × 12 × 6) / (3600) = 83.72 kW

In this case, the fluid stream is rejecting 83.72 kW of heat. That means the system must continuously transfer about 83.72 kJ of thermal energy every second to maintain those operating conditions.

Unit Notes

Heat rejection calculations are only as good as the unit consistency. Keep these points in mind:

  • A temperature difference of 1 K is equal to a temperature difference of 1 °C.
  • If your temperature change is in °F, convert it before using SI-based properties.
  • Specific heat values in J/(kg·K) are 1000 times larger numerically than values in kJ/(kg·K).
  • Density and specific heat should match the actual fluid and operating temperature range as closely as possible.
Δ TK = Δ T^° C
Δ TK = (5) / (9)Δ T^° F
QBTU / hr = 3412.14 QkW

Where Heat Rejection Matters

This calculation is commonly used anywhere a fluid removes heat from equipment, structures, or processes, including:

  • HVAC chilled-water and condenser-water loops
  • Engine and powertrain cooling systems
  • Process heat exchangers
  • Hydronic systems
  • Refrigeration equipment
  • Liquid cooling for electronics and industrial machinery

Within this model and with other inputs fixed, larger flow, temperature change or heat capacity increases the calculated sensible duty. This does not establish exchanger capacity, attainable temperatures or phase-change duty.

Common Mistakes

  • Entering absolute temperature instead of temperature difference. The formula uses ΔT, not the inlet or outlet temperature alone.
  • Mixing unit systems. For example, using J/(kg·K) for specific heat with a formula expecting kJ/(kg·K).
  • Using the wrong density. Density can change significantly between liquids, gases, and operating conditions.
  • Ignoring flow basis. A value in m³/s should not be treated like m³/hr.
  • Assuming all fluids behave like water. Oils, glycol mixtures, refrigerants, and air can produce very different results.

Practical Interpretation of the Result

  • A larger Q means the system is removing heat more aggressively.
  • If the calculated heat rejection is below the required thermal load, the system will tend to run hotter.
  • If two systems have the same fluid and the same temperature change, the one with the higher flow rate rejects more heat.
  • If flow rate is fixed, increasing ΔT increases heat rejection proportionally.

Use this fluid energy balance as one input to performance checks and design calculations. The editable helper values are approximate near20°C and atmospheric pressure (dry air at1atm); supply actual average operating properties for other conditions. Zero flow or zero temperature difference gives zero forward sensible heat transfer. Inverse solves require a unique valid solution. Refrigeration tons here mean12,000BtuIT/h and do not select equipment.