Heat Exchanger Effectiveness

Infer sensible heat duty, cold outlet temperature and effectiveness from the hot-side temperature drop.

How this calculation works

Effectiveness compares the measured sensible heat transfer with the maximum available from the inlet temperature difference and the smaller heat-capacity rate. Enter both mass flows and specific heats, plus hot inlet, measured hot outlet and cold inlet temperatures. Multiplying mass flow by specific heat gives each stream's capacity rate in kW/K. The hot-side temperature drop establishes duty, and an equal cold-side gain predicts the cold outlet temperature. The minimum capacity rate and its ratio to the maximum are also reported. Inputs that imply excessive duty or an invalid hot outlet are rejected. A predicted cold outlet above the hot outlet requires a suitable counterflow arrangement. The balance assumes constant properties and no external losses; it does not size an exchanger or model phase change.

Inputs and units

Hot-stream mass flow (kg/s)
Hot-stream specific heat (kJ/(kg·K))
Cold-stream mass flow (kg/s)
Cold-stream specific heat (kJ/(kg·K))
Hot inlet temperature (°C): Must be above absolute zero
Measured hot outlet temperature (°C): Must be above absolute zero
Cold inlet temperature (°C): Must be above absolute zero

Method and formula

Ch = mh cph; Cc = mc cpc; Q = Ch(Thi − Tho); Qmax = min(Ch,Cc)(Thi − Tci); effectiveness = Q/Qmax; Tco = Tci + Q/Cc.

Worked example

Example inputs

Hot-stream mass flow: 2 kg/s
Hot-stream specific heat: 4.18 kJ/(kg·K)
Cold-stream mass flow: 3 kg/s
Cold-stream specific heat: 4.18 kJ/(kg·K)
Hot inlet temperature: 80 °C
Measured hot outlet temperature: 50 °C
Cold inlet temperature: 20 °C

Calculation steps

Calculate capacity rates: Ch = 2 × 4.18 = 8.36 kW/K; Cc = 3 × 4.18 = 12.54 kW/K.
The minimum capacity rate is 8.36 kW/K and the capacity-rate ratio is 8.36/12.54 = 0.666667.
Calculate hot-side duty: Q = 8.36 × (80 − 50) = 250.8 kW.
Calculate the maximum duty: Qmax = 8.36 × (80 − 20) = 501.6 kW; effectiveness = 100 × 250.8/501.6 = 50%.
Apply the cold-side energy balance: Tcold,out = 20 + 250.8/12.54 = 40 °C.

Example results

Heat duty: 250.8 kW
Effectiveness: 50 %
Predicted cold outlet: 40 °C
Minimum capacity rate: 8.36 kW/K
Capacity-rate ratio: 0.6666666667 1

Assumptions

Steady, adiabatic two-stream sensible exchanger with constant heat capacities
Hot-side measured heat loss is transferred to the cold stream

Limitations

Not an NTU sizing model; excludes phase change, external losses and fouling estimation
Measurement uncertainty can produce impossible apparent effectiveness

Sources

Portland State University: Heat Exchangers: Heat-capacity rates, energy balance and effectiveness definition