This tool solves the Rankine steam power cycle using the IAPWS-IF97 formulation for water and steam properties. Enter any two independent properties at each state point and it fills in the rest, exactly the way you would by hand. Energy balances around the boiler, turbine, condenser, and pump then link the four states into a complete cycle, including thermal efficiency, back-work ratio, and net work output.
Works for ideal isentropic cycles and actual cycles with pump and turbine isentropic efficiencies. Flags over-specified or conflicting states so you know immediately if the inputs are inconsistent. The same solver also covers the air-standard Otto cycle and the R-134a vapor-compression refrigeration cycle, described further down this page. Runs entirely in the browser. No account required.
The ideal Rankine cycle is the standard model for steam power plants. It consists of four processes:
Problem: Boiler at 8 MPa / 480 C, condenser at 8 kPa. Ideal isentropic cycle. Find thermal efficiency.
State 3 (turbine inlet): P = 8 MPa, T = 480 C. IAPWS-IF97 gives h3 = 3349.5 kJ/kg, s3 = 6.661 kJ/kg·K.
State 1 (pump inlet): sat. liquid at 8 kPa. h1 = 173.9 kJ/kg, s1 = 0.593 kJ/kg·K.
State 4 (turbine exit): s4 = s3 at 8 kPa gives x4 = 0.795, h4 = 2083.4 kJ/kg.
State 2 (pump exit): s2 = s1 at 8 MPa gives h2 = 181.9 kJ/kg.
Result: w_net = (h3 - h4) - (h2 - h1) = 1258.1 kJ/kg; q_in = 3167.6 kJ/kg; thermal efficiency = 39.7%.
| Quantity | Expression | Notes |
|---|---|---|
| Turbine work | w_t = h3 - h4 | Per unit mass flow |
| Pump work | w_p = h2 - h1 | Small for liquids |
| Heat input | q_in = h3 - h2 | Boiler |
| Heat rejected | q_out = h4 - h1 | Condenser |
| Thermal efficiency | eta = w_net / q_in | First law |
| Back-work ratio | bwr = w_p / w_t | Typically 0.5-2% |
Switch the Cycle control to Otto Cycle to solve the ideal air-standard Otto cycle, the standard model for spark-ignition engines. The four states are: 1 at bottom dead center before compression, 2 after isentropic compression, 3 after constant-volume heat addition, and 4 after isentropic expansion. Enter any two independent properties per state (P, T, v, or u), plus the compression ratio or a heat or work quantity on the process chips, and the solver propagates the rest through the ideal gas law, the isentropic relations, and closed-system energy balances.
Two property models are available. Cold air-standard uses constant specific heats with an editable ratio k (1.4 by default), matching the closed-form relations T2 = T1 r^(k-1) and eta = 1 - r^(1-k). Turning the toggle off switches to variable specific heats using tabulated air properties (u and relative specific volume vr), referenced to match standard air tables so intermediate values can be checked against a textbook table directly.
Problem: Compression ratio 8, air at 100 kPa and 290 K before compression, 800 kJ/kg of heat added. Variable specific heats. Find the state properties, net work, thermal efficiency, and MEP.
State 2: vr2 = vr1 / 8 gives T2 = 652 K, P2 = 1799 kPa.
State 3: u3 = u2 + 800 kJ/kg gives T3 = 1575 K, P3 = 4345 kPa.
State 4: vr4 = 8 vr3 gives T4 = 795 K.
Result: w_net = 418.3 kJ/kg; thermal efficiency = 52.3%; MEP = 574 kPa. Cold air-standard with k = 1.4 gives 56.5% for comparison.
| Quantity | Expression | Notes |
|---|---|---|
| Compression ratio | r = v1 / v2 | Also equals v4 / v3 |
| Heat input | q_in = u3 - u2 | Constant volume |
| Heat rejected | q_out = u4 - u1 | Constant volume |
| Thermal efficiency | eta = 1 - q_out / q_in | Cold air: 1 - r^(1-k) |
| Mean effective pressure | MEP = w_net / (v1 - v2) | Constant pressure that would produce w_net over one stroke |
Switch the Cycle control to Refrigeration (R-134a) to solve the vapor-compression cycle. Refrigerant properties come from a Helmholtz equation of state for R-134a (Tillner-Roth and Baehr), with enthalpy and entropy referenced to saturated liquid at -40 C, matching the convention in standard refrigerant tables. The four states are linked the same way as the Rankine cycle: enter any two properties at each state and the rest are computed.
The coefficient of performance is the useful effect over the compressor work. For cooling, COP = (h1 - h4) / (h2 - h1). For heating (a heat pump), COP = (h2 - h3) / (h2 - h1), which is always one greater than the cooling COP for the same cycle. Enter a mass flow rate to convert the refrigeration effect into cooling capacity in kW and tons of refrigeration.
Problem: R-134a cycle, evaporator at 140 kPa, condenser at 800 kPa. Saturated vapor enters the compressor, saturated liquid leaves the condenser, isentropic compressor. Find the COP for cooling.
State 1 (compressor inlet): sat. vapor at 140 kPa. h1 = 239.2 kJ/kg, s1 = 0.9446 kJ/kg·K.
State 2 (compressor exit): s2 = s1 at 800 kPa gives T2 = 39.0 C, h2 = 275.4 kJ/kg.
State 3 (condenser exit): sat. liquid at 800 kPa. h3 = 95.5 kJ/kg.
State 4 (evaporator inlet): throttle from State 3, so h4 = h3 = 95.5 kJ/kg.
Result: refrigeration effect = h1 - h4 = 143.7 kJ/kg; compressor work = h2 - h1 = 36.2 kJ/kg; COP_cooling = 3.97, COP_heating = 4.97.
IAPWS-IF97 is the International Association for the Properties of Water and Steam industrial formulation from 1997. It covers water and steam from 0 to 800 C and up to 100 MPa using region-based equations for enthalpy, entropy, and specific volume. This solver implements Regions 1 (compressed liquid) and 2 (superheated steam) along with the saturation curve (Region 4).
For a pure substance with one phase (liquid or vapor), specifying any two intensive properties (P, T, h, s, v) fully determines the thermodynamic state. In the two-phase region, P and T are no longer independent, so you need quality x as one of the two inputs instead. The solver detects which region a state is in and uses the appropriate equations.
Low exit quality means a large fraction of the steam is liquid droplets at the turbine exit, which erodes turbine blades. The practical lower limit is usually around x = 0.85. To increase exit quality, superheat the steam further, raise the boiler pressure (counterintuitively, higher pressure with sufficient superheat gives higher exit quality), or use a reheat stage.
Yes. Switch the Cycle control in the sidebar to Refrigeration (R-134a) and the tool solves the vapor-compression cycle using a Helmholtz equation of state for R-134a. The same fill-what-you-know approach applies: enter any two properties at each state and it propagates the rest, then reports COP for cooling and heating. Other refrigerants are not included yet.