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How to Calculate Mass Flow Rate of Steam in a Boiler
Determining the mass flow rate of steam is critical for boiler operators and engineers to ensure the system is meeting process demands and operating efficiently. The mass flow rate ($\dot{m}$) represents the quantity of steam (in kilograms) produced over a specific period (usually per hour or per second).
The calculation relies on the First Law of Thermodynamics, specifically the energy balance across the boiler. By knowing the total energy input (boiler power) and the energy required to convert 1 kg of feedwater into steam at a specific pressure, we can determine how much steam is being generated.
The Calculation Formula
The standard formula for calculating the steam mass flow rate is derived from the heat load equation:
$\dot{m} = \frac{Q}{h_{steam} – h_{feedwater}}$
Where:
- $\dot{m}$ = Mass flow rate of steam (kg/s)
- $Q$ = Boiler Heat Output or Thermal Power (kW or kJ/s)
- $h_{steam}$ = Specific Enthalpy of the steam exiting the boiler (kJ/kg)
- $h_{feedwater}$ = Specific Enthalpy of the water entering the boiler (kJ/kg)
Understanding the Inputs
1. Boiler Heat Output ($Q$): This is the net thermal energy transferred to the water/steam. It is often rated in Kilowatts (kW) or Megawatts (MW). If your boiler is rated in Boiler Horsepower (BHP), you can convert it roughly using 1 BHP $\approx$ 9.81 kW.
2. Specific Enthalpy of Steam ($h_g$): This value depends on the pressure and temperature of the steam. You can find this in steam tables. For saturated steam at 10 bar absolute, the specific enthalpy is approximately 2778 kJ/kg.
3. Feedwater Enthalpy ($h_f$): This is the energy content of the water before it is heated. For liquid water, specific enthalpy can be approximated using the specific heat capacity ($C_p \approx 4.186$ kJ/kg°C) and the temperature ($T$):
$h_{feedwater} \approx 4.186 \times T_{feedwater}$.
Example Calculation
Let's assume you have a boiler with an output of 1,000 kW. The steam is produced at 10 bar (enthalpy $\approx$ 2778 kJ/kg), and the feedwater enters at 80°C.
- Calculate Feedwater Enthalpy: $80 \times 4.186 = 334.88$ kJ/kg.
- Calculate Energy Added per kg: $2778 – 334.88 = 2443.12$ kJ/kg.
- Calculate Mass Flow (kg/s): $1000 / 2443.12 \approx 0.409$ kg/s.
- Convert to kg/hr: $0.409 \times 3600 \approx 1472.4$ kg/h.
function calculateSteamFlow() {
// Get input values
var powerKw = document.getElementById('boilerPower').value;
var enthalpySteam = document.getElementById('steamEnthalpy').value;
var tempFeed = document.getElementById('feedTemp').value;
// Validate inputs
if (powerKw === "" || enthalpySteam === "" || tempFeed === "") {
alert("Please fill in all fields to calculate the flow rate.");
return;
}
// Parse values to floats
var Q = parseFloat(powerKw);
var h_s = parseFloat(enthalpySteam);
var t_w = parseFloat(tempFeed);
// Constant: Specific heat of water (kJ/kg°C)
var Cp_water = 4.186;
// Calculate Feedwater Enthalpy (h_w)
// h = Cp * T
var h_w = t_w * Cp_water;
// Calculate Delta Enthalpy (Energy added to convert water to steam)
var delta_h = h_s – h_w;
// Validation: Cannot have negative or zero energy addition (Physically impossible for a boiler)
if (delta_h <= 0) {
alert("Error: Steam enthalpy must be greater than feedwater enthalpy. Please check your inputs.");
document.getElementById('results').classList.remove('visible');
return;
}
// Calculate Mass Flow Rate
// Formula: m_dot (kg/s) = Q (kW) / delta_h (kJ/kg)
var flow_kg_per_sec = Q / delta_h;
// Convert to kg/hour
var flow_kg_per_hour = flow_kg_per_sec * 3600;
// Update the UI
document.getElementById('flowRateHour').innerHTML = flow_kg_per_hour.toFixed(2) + " kg/h";
document.getElementById('flowRateSec').innerHTML = flow_kg_per_sec.toFixed(4) + " kg/s";
document.getElementById('feedEnthalpy').innerHTML = h_w.toFixed(2) + " kJ/kg";
document.getElementById('energyAdded').innerHTML = delta_h.toFixed(2) + " kJ/kg";
// Show results
document.getElementById('results').classList.add('visible');
}