Compressor Heat Recovery Calculator

The calculator separates available heat, technically recoverable heat and actually utilized heat so that theoretical potential is not presented as guaranteed savings.

1. Purpose and Engineering Decision

Most electrical energy entering an air compressor ultimately appears as heat. That fact does not mean the same percentage is always recoverable or useful. Recovery depends on compressor type, cooling arrangement, heat-exchanger design, operating load, available temperature, duct or water-side losses, seasonal coincidence and the temperature required by the receiving process.

2. Three Heat Quantities Must Stay Separate

The calculator must not label all compressor input power as recoverable heat. The user enters an explicit recoverable fraction or verified recovery curve from the compressor/OEM recovery package.

3. Core Thermal-Potential Equation

Q_recoverable (kWth) = P_input (kW) x f_recoverable

P_input should preferably be measured or verified total package electrical input for the operating state being modeled. f_recoverable is a decimal fraction such as 0.70, 0.80 or another verified value.

4. Useful Heat After Coincidence and Distribution Losses

Q_useful = Q_recoverable x f_distribution x f_utilization

f_distribution represents duct, piping, heat-exchanger or other transfer losses. f_utilization represents the fraction of recovered heat that the site can actually absorb during the compressor operating period. Keeping both factors visible prevents seasonal or process mismatch from being hidden inside one optimistic number.

5. Input Model

6. Hot-Air Recovery Mode

Air-cooled compressor packages can often provide a stream of heated cooling air.

Q = m_dot_air x cp_air x DeltaT

V_dot_air = Q / (rho_air x cp_air x DeltaT)

The output is a thermal balance, not a duct-design result. Fan static pressure, duct resistance, fire/smoke requirements, recirculation, summer bypass and room ventilation must be engineered separately.

7. Hot-Water Recovery Mode

Where the compressor or recovery package can transfer heat to water, the required water mass flow follows the same sensible-heat balance.

m_dot_water = Q / (cp_water x DeltaT)

Water flow (L/min) approx = Q_kW x 60 / (4.186 x DeltaT_C)

The calculator must check that the selected outlet temperature is technically achievable with the actual compressor-side recovery temperature. If the requested water temperature exceeds the verified capability of the recovery system, the result should be flagged rather than treated as valid.

8. Reference Test Case A - Thermal Potential

A compressor package draws 160 kW at the selected operating point. The verified heat-recovery arrangement is modeled at a 78% recoverable fraction, 92% distribution efficiency and 80% utilization.

The example illustrates why useful heat can be materially lower than theoretical compressor heat generation even before annual operating hours are considered.

9. Reference Test Case B - Hot-Air Flow

Use 100 kWth of useful heat to warm process/make-up air from 25 C to 45 C. Assume air density 1.15 kg/m3 and cp = 1.006 kJ/(kg.K).

V_dot = 100 / (1.15 x 1.006 x 20) = 4.32 m3/s

Reference result: approximately 15,559 m3/h of air. This is the thermal airflow requirement only; actual duct sizing requires a separate velocity/pressure-loss design.

10. Reference Test Case C - Hot-Water Flow

Use 90 kWth to heat water from 30 C to 50 C. Use cp = 4.186 kJ/(kg.K) and density approximately 1 kg/L.

Water flow = 90 x 60 / (4.186 x 20) = 64.5 L/min

The result indicates the water flow required to absorb 90 kWth across a 20 C rise, provided the recovery heat exchanger can maintain the selected outlet temperature.

11. Annual Useful Energy

Annual Useful Heat (kWhth/y) = Q_useful x Operating Hours

For plants with multiple operating states, annual heat should be summed by load band because compressor input kW, recovery fraction and thermal demand coincidence may change with production.

Annual Useful Heat = Sum(Q_useful,i x Hours_i)

Using Test Case A for 6,000 h/y gives approximately 551,117 kWhth/y of useful recovered heat under the stated utilization assumptions.

12. Displaced Electricity Value

If recovered heat directly displaces electric resistance heating, the avoided electrical input is approximately equal to useful thermal kWh divided by the heater efficiency.

Avoided Electricity kWh = Useful Heat kWhth / eta_electric_heater

Annual Value = Avoided Electricity kWh x Electricity Tariff

13. Displaced Fuel / Boiler Value

When recovered heat displaces a fuel-fired heater or boiler, compare useful thermal energy at the process with the fuel input that would otherwise have been required.

Avoided Fuel Energy = Useful Heat / eta_existing_heater

The fuel price input must use a compatible energy basis, such as currency per kWh of fuel input. If fuel is purchased per litre, kg or standard volume, a separate verified fuel-energy conversion is required.

14. Reference Test Case D - Annual Economic Value

Assume Test Case A's useful heat of 91.9 kWth operates 6,000 h/y and displaces electric heating at 95% efficiency with an electricity tariff of Rs 9/kWh.

The result is an energy-only value. Demand charges, taxes, maintenance changes, backup requirements and financing should be handled separately unless deliberately added to the project model.

15. Simple Payback

Simple Payback (years) = Installed Recovery Cost / Annual Net Savings

Annual net savings may include avoided energy less estimated recovery-system fan, pump, maintenance or other incremental operating costs.

16. Seasonal Utilization

Space-heating recovery may be valuable in cool periods and unwanted in hot periods. Process hot-water demand may be much more stable.

17. Temperature-Match Gate

18. Confidence Model

19. Validation and Guardrails

20. Output Specification