Steam Turbine Power
Calculate the mechanical power generated by a steam turbine, P = ṁ × (h₁ − h₂), multiplying the steam mass flow (kg/s) by the enthalpy drop between turbine inlet and outlet (kJ/kg). The result, in kW, is the shaft power delivered to the generator, accounting for the expansion of high-pressure, high-temperature steam down to condenser pressure. It is the core calculation in sizing thermal power and cogeneration plants: the larger the enthalpy drop, the more power per kg of steam. Enter the steam flow and the inlet and outlet enthalpies.
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Steam turbine power
The steam turbine converts the energy of steam into mechanical shaft work, which drives the electric generator. The power produced is simply the steam flow rate multiplied by the energy each kilogram delivers as it crosses the turbine: P = ṁ × (h₁ − h₂). The mass flow rate ṁ (kg/s) is how much steam passes per second; (h₁ − h₂) is the enthalpy drop, the difference between the enthalpy of the steam at the inlet (high pressure and temperature) and at the outlet (low pressure, heading to the condenser), in kJ/kg. The product gives the power in kW. Everything conspires in favour of a larger enthalpy drop: hotter inlet steam at the highest feasible pressure, and the lowest possible exhaust pressure (vacuum in the condenser) — each one widens the available enthalpy jump and therefore the power per kg of steam. In practice the figure is further multiplied by the mechanical and electrical efficiency. It is the starting calculation in the design of thermal power stations and cogeneration plants. Enter the steam flow rate and the inlet and outlet enthalpies.
Related Tools
Specific Steam Consumption
Calculate the specific steam consumption (steam rate) of a turbine, SSC = 3600 ÷ Δh, dividing 3600 (s/h) by the available enthalpy drop in the turbine (kJ/kg). The result, in kg/kWh, gives how many kilograms of steam are needed to generate one kilowatt-hour. The lower the specific consumption, the more efficient the conversion: larger enthalpy drops (hotter steam and greater expansion) cut the steam needed per kWh. It is a practical indicator to compare turbines and estimate the steam flow required for a given power. Enter the available enthalpy drop.
Rankine Cycle Efficiency
Calculate the thermal efficiency of a Rankine cycle, η = (w_turbine − w_pump) ÷ q_boiler × 100%, dividing the net work (turbine work minus pump work) by the heat added in the boiler, all in kJ/kg. The Rankine cycle is the basis of steam power plants: water is pumped, heated and vaporized in the boiler, expands through the turbine producing work, then condenses. The result, in %, measures how much boiler heat becomes useful work; real cycles run 30–45%. Enter the turbine work, the pump work and the boiler heat.
Solids Mass Flow (Dredge)
Calculate the mass flow of solids transported by a dredge or pipeline, ṁ_s = Q·C_v·ρ_s, from the total slurry flow Q (m³/s), the solids volumetric concentration C_v (fraction) and the solids density ρ_s (kg/m³). Solids mass flow is the MASS of useful material transported per unit time (kg/s, or tonnes per hour), the production indicator used when TONNAGE matters — the typical case of ore transport by pipeline (measured in t/h of dry ore) and mineral processing. It is the product of three factors: the slurry flow (pump capacity), the solids concentration (how 'loaded' the slurry is) and the solids density (iron ores, for example, are very dense, ~5000 kg/m³, so little volumetric concentration already gives high tonnage). Mass flow, integrated over time, gives the total transported tonnage, the basis of billing and operational mass balance. Optimizing it — maximizing tonnage per unit pumping energy — is the central goal of pipeline operation, which moves hundreds of millions of tonnes of ore per year over long distances far more energy-efficiently than trucks or trains. Enter the slurry flow, the volumetric concentration and the solids density.
Compressor Work (Isentropic)
Compute the specific compression work in an ideal refrigeration cycle, W = h₂ − h₁, the enthalpy difference between the compressor outlet and inlet (isentropic compression, at constant entropy). It is the energy the compressor adds to the refrigerant per kilogram — the cycle's 'electricity bill'. Together with the refrigerating effect, it defines the COP (COP = refrigerating effect/work). Enter the outlet and inlet enthalpies.
Axle Load Equivalency Factor
Computes how many passes of the standard axle are equivalent to one pass of the real axle, using the power law of pavement design: factor = (axle load ÷ standard axle load) raised to the damage exponent. This factor is what converts a traffic count into the number N of standard axle repetitions, which in Brazil is the 8.2 tf, or 80 kN, single axle with dual wheels. The exponent amplifies overload brutally: an axle 20% heavier than the standard does not consume 20% more pavement but 2.07 times as much, which is why a single overloaded truck weighs more on the life of the road than thousands of cars, whose factor is practically zero. The exponent is an input rather than fixed at 4, the AASHTO value known as the fourth power law, because rigid pavement and fatigue cracking models work with exponents between 3 and 5 and the result shifts by a whole level depending on the choice. Enter the axle load, the standard axle load and the damage exponent.
Steam Quality
Calculate the quality (title) of a wet steam, x = (h − h_f) ÷ (h_g − h_f), from the mixture enthalpy and the enthalpies of saturated liquid (h_f) and saturated vapour (h_g) at the same pressure. The result (between 0 and 1, or ×100%) is the vapour mass fraction in the liquid-vapour mixture: x = 0 is saturated liquid, x = 1 is dry saturated vapour, and intermediate values are wet steam. Quality is essential in steam cycles to know the state at the turbine outlet (very low quality erodes the blades) and at the condenser inlet. Enter the mixture enthalpy, h_f and h_g.
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