Evaporation Rate of Boiling Water Calculator

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Boiling Water Evaporation Calculator

Calculate the rate of steam generation based on heat input.

The power rating of your stove, burner, or heating element in Kilowatts.
Typical induction: ~85%, Gas: ~40-50%, Electric Coil: ~75%.
Enter volume to calculate time until dry.
Standard Atmosphere (1 atm / 100°C) High Altitude (0.8 atm / ~94°C) Pressure Cooker (2 atm / ~120°C)
Affects the Latent Heat of Vaporization.
Effective Heating Power:
Evaporation Rate (Mass):
Evaporation Rate (Volume):
Water Consumed per Minute:
Time to Boil Dry:

Understanding the Evaporation Rate of Boiling Water

When water reaches its boiling point ($100^\circ\text{C}$ at sea level), any additional thermal energy added to the system does not increase the temperature. Instead, this energy is consumed by the phase change from liquid to gas (steam). This energy is known as the Latent Heat of Vaporization.

This calculator determines how fast water boils away based on the power of the heat source and the efficiency of the heat transfer. This is critical for industrial boilers, culinary sciences, distillation processes, and HVAC humidification calculations.

The Physics Formula

The rate of evaporation during boiling is governed by the conservation of energy. The formula used is:

Rate (kg/s) = P_eff / h_fg
  • $P_{eff}$ (Effective Power): The actual heat energy entering the water in Watts (Joules/second). This is calculated as $\text{Source Power} \times \text{Efficiency}$.
  • $h_{fg}$ (Latent Heat): The specific latent heat of vaporization of water. At standard atmospheric pressure, this is approximately $2,260 \text{ kJ/kg}$.

Why Efficiency Matters

Not all energy from a heat source reaches the water.

Heat Source Typical Efficiency Notes
Induction Cooktop 80% – 90% Direct magnetic heating of the vessel. Very efficient.
Electric Coil 70% – 80% Heat loss to the surrounding air and stove surface.
Gas Burner 35% – 50% Significant heat escapes around the sides of the pot.

Example Calculation

Imagine you have a 3 kW electric kettle (3000 Watts) that is 90% efficient.

  1. Effective Power: $3 \text{ kW} \times 0.90 = 2.7 \text{ kW}$ (or $2.7 \text{ kJ/s}$).
  2. Latent Heat: We assume standard pressure ($2260 \text{ kJ/kg}$).
  3. Calculation: $2.7 / 2260 \approx 0.00119 \text{ kg/s}$.
  4. Hourly Rate: $0.00119 \times 3600 \text{ seconds} \approx 4.3 \text{ kg/hr}$.

Since 1 kg of water equals roughly 1 liter, the kettle will boil away approximately 4.3 liters of water per hour.

Factors Affecting Evaporation

1. Atmospheric Pressure: At higher altitudes, atmospheric pressure is lower. This lowers the boiling point of water, slightly altering the specific latent heat required (it actually increases slightly as temperature drops, requiring more energy per kg to vaporize).

2. Surface Area: While surface area dictates evaporation below boiling point, during boiling, the rate is primarily driven by the power input. However, a wider pot may lose more heat to the environment via radiation/convection, reducing efficiency.

function calculateEvaporation() { // 1. Get Input Values var powerKW = document.getElementById('heatPower').value; var efficiency = document.getElementById('efficiency').value; var volumeL = document.getElementById('waterVolume').value; var latentHeat = document.getElementById('pressureCondition').value; // 2. Validation if (!powerKW || powerKW < 0) { alert("Please enter a valid heating power."); return; } if (!efficiency || efficiency 100) { alert("Please enter a valid efficiency percentage (0-100)."); return; } // 3. Convert Types var p_kw = parseFloat(powerKW); var eff_percent = parseFloat(efficiency); var h_fg = parseFloat(latentHeat); // kJ/kg var vol_initial = parseFloat(volumeL); // 4. Calculate Effective Power // Power in kW is equivalent to kJ/s var effective_power_kw = p_kw * (eff_percent / 100); // 5. Calculate Evaporation Rate // Rate (kg/s) = Power (kJ/s) / Latent Heat (kJ/kg) var rate_kg_per_sec = effective_power_kw / h_fg; // Convert to hourly rates var rate_kg_per_hour = rate_kg_per_sec * 3600; // Assume water density roughly 1kg/L at boiling point (actually ~0.96, but 1 is standard for general estimation) // For precision in this context, we usually equate kg to Liters for water. var rate_liters_per_hour = rate_kg_per_hour; var rate_ml_per_min = (rate_liters_per_hour * 1000) / 60; // 6. Calculate Time to Dry (if volume provided) var timeDisplay = ""; var showTime = false; if (!isNaN(vol_initial) && vol_initial > 0) { showTime = true; // Time (hours) = Volume / Rate var hours_to_dry = vol_initial / rate_liters_per_hour; // Format time nicely var total_minutes = hours_to_dry * 60; var h = Math.floor(total_minutes / 60); var m = Math.round(total_minutes % 60); if (h > 0) { timeDisplay = h + " hr " + m + " min"; } else { timeDisplay = m + " min"; } } // 7. Update UI document.getElementById('results-area').style.display = 'block'; document.getElementById('resEffectivePower').innerHTML = effective_power_kw.toFixed(2) + " kW"; document.getElementById('resKgPerHour').innerHTML = rate_kg_per_hour.toFixed(2) + " kg/hr"; document.getElementById('resLitersPerHour').innerHTML = rate_liters_per_hour.toFixed(2) + " L/hr"; document.getElementById('resMLPerMin').innerHTML = rate_ml_per_min.toFixed(0) + " mL/min"; var timeRow = document.getElementById('timeRow'); if (showTime) { timeRow.style.display = 'flex'; document.getElementById('resTimeDry').innerHTML = timeDisplay; } else { timeRow.style.display = 'none'; } }

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