Energy content per kg of fuel (Natural Gas approx 50,000 kJ/kg)
Heat Rate (Metric):–
Heat Rate (Imperial Approx):–
Energy Input:–
Thermal Efficiency:–
Understanding Gas Turbine Heat Rate
In power generation, Heat Rate is the inverse of efficiency. It represents the amount of thermal energy required to produce one unit of electrical energy. While typical efficiency is expressed as a percentage, engineers often prefer Heat Rate to quantify the fuel cost of generation.
The lower the Heat Rate, the more efficient the gas turbine is. A lower Heat Rate means less fuel is consumed to generate the same amount of electricity, reducing operational costs and emissions.
The Calculation Formulas
To determine the efficiency of a thermodynamic cycle, we compare the useful work output (electricity) to the energy input (fuel). The calculator above uses the standard metric approach.
Note: 3600 is the conversion factor because 1 kWh equals 3600 kJ.
Key Parameters Explained
Gross Power Output (MW): The total electricity generated by the turbine generator terminals before auxiliary loads are subtracted.
Fuel Mass Flow (kg/hr): The rate at which fuel (usually natural gas or distillate oil) is supplied to the combustion chamber.
Lower Heating Value (LHV): The amount of heat released by combusting a specified quantity of fuel (initially at 25°C) and returning the temperature of the combustion products to 150°C. For natural gas, this is typically around 47,000 – 50,000 kJ/kg.
Heat Rate vs. Efficiency Table
Below is a quick reference converting common Heat Rates to Thermal Efficiency:
Heat Rate (kJ/kWh)
Heat Rate (Btu/kWh)
Thermal Efficiency (%)
8,000
~7,582
45.0%
9,000
~8,530
40.0%
10,285
~9,750
35.0%
12,000
~11,374
30.0%
Improving Gas Turbine Performance
Several factors can degrade heat rate over time, including compressor fouling, turbine blade degradation, and seal leakage. Regular maintenance cycles, including offline compressor washing and hot gas path inspections, are critical for maintaining the design heat rate. Additionally, ambient conditions significantly affect performance; gas turbines are generally more efficient on colder days due to higher air density.
function calculateHeatRate() {
// 1. Get input values
var powerMW = document.getElementById('powerOutput').value;
var fuelFlow = document.getElementById('fuelFlow').value;
var lhv = document.getElementById('heatingValue').value;
// 2. Validate inputs
if (powerMW === "" || fuelFlow === "" || lhv === "") {
alert("Please fill in all fields (Power Output, Fuel Flow, and LHV).");
return;
}
var powerVal = parseFloat(powerMW);
var fuelVal = parseFloat(fuelFlow);
var lhvVal = parseFloat(lhv);
if (isNaN(powerVal) || powerVal <= 0) {
alert("Please enter a valid positive number for Power Output.");
return;
}
if (isNaN(fuelVal) || fuelVal <= 0) {
alert("Please enter a valid positive number for Fuel Flow.");
return;
}
if (isNaN(lhvVal) || lhvVal kJ/h. But standard definition relates Input Energy per hour to Output Power.
var heatRateMetric = energyInputKJ_hr / powerKW;
// Thermal Efficiency (%)
// Since 1 kWh = 3600 kJ
// Efficiency = Output / Input = 3600 / Heat Rate (kJ/kWh)
var efficiency = (3600 / heatRateMetric) * 100;
// Convert to Imperial for reference (1 kJ/kWh ≈ 0.94781712 Btu/kWh)
var heatRateImperial = heatRateMetric * 0.94781712;
// Energy Input in GJ/hr for display
var energyInputGJ = energyInputKJ_hr / 1000000;
// 4. Update the UI
document.getElementById('resHeatRateMetric').innerText = heatRateMetric.toFixed(2) + " kJ/kWh";
document.getElementById('resHeatRateImperial').innerText = heatRateImperial.toFixed(0) + " Btu/kWh";
document.getElementById('resEnergyInput').innerText = energyInputGJ.toFixed(2) + " GJ/hr";
document.getElementById('resEfficiency').innerText = efficiency.toFixed(2) + "%";
// Show results
document.getElementById('results').style.display = "block";
}