Nine cycles in one tool: the Rankine steam power cycle, with or without reheat, on IAPWS-IF97 water and steam data, the air-standard Brayton (with or without a regenerator), Otto, Diesel, dual and Atkinson cycles, and the R-134a vapor-compression refrigeration cycle. In every mode you enter the properties you already know at each state point and the solver fills in the rest, then closes the energy balances around each device.
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 by the same steps as a hand solution. 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. Over-specified or conflicting states are flagged. The reheat Rankine, Brayton, Otto, Diesel, dual, Atkinson and R-134a refrigeration cycles are described further down this page.
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 Rankine Cycle with Reheat for the six-state reheat cycle: 1 to 2 pump, 2 to 3 boiler, 3 to 4 high-pressure turbine, 4 to 5 reheat at constant pressure, 5 to 6 low-pressure turbine, 6 to 1 condenser. Each turbine has its own isentropic efficiency, and the heat input is the boiler and the reheater together: q_in = (h3 - h2) + (h5 - h4).
States can be fixed from either side of an ideal device. Textbook problems often give the turbine exit quality and ask for the reheat pressure: enter x6 with the condenser pressure and the reheat temperature, and the solver takes s5 = s6, finds P5 from T5 and s5, and carries P4 = P5 back to the high-pressure turbine exit.
Problem: Steam enters the high-pressure turbine at 15 MPa and 600 C and is reheated to 600 C. The condenser is at 10 kPa and the low-pressure turbine exit quality is 0.896. Ideal cycle. Find the reheat pressure and the thermal efficiency (Cengel and Boles, Example 10-4).
State 6: x6 = 0.896 at 10 kPa gives h6 = 2335.1 kJ/kg, s6 = 7.3689 kJ/kg·K.
State 5: s5 = s6 at 600 C gives P5 = 4.01 MPa, h5 = 3674.7 kJ/kg. The reheat pressure is 4 MPa.
State 4: s4 = s3 = 6.6797 kJ/kg·K at 4.01 MPa gives h4 = 3154.8 kJ/kg (h3 = 3583.3 kJ/kg).
Result: q_in = 3376.4 + 519.9 = 3896.3 kJ/kg; w_net = 1753.0 kJ/kg; thermal efficiency = 45.0%.
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 Diesel Cycle to solve the ideal air-standard Diesel cycle, the model for compression-ignition engines. It differs from the Otto cycle in one process: heat is added at constant pressure while the piston starts moving down, so 1 to 2 is isentropic compression, 2 to 3 is heat addition at constant pressure, 3 to 4 is isentropic expansion and 4 to 1 is heat rejection at constant volume. Each state carries P, T, v, u and h. Enter the compression ratio on the Compression chip and either the cutoff ratio rc = v3 / v2 or the heat added on the Heat addition chip, or give temperatures and pressures directly, and the solver works forward or backward from whatever you know.
Heat added at constant pressure is q_in = h3 - h2, and part of it leaves as boundary work P2 (v3 - v2) during the same process. The results list that work separately, so net work is not just expansion minus compression. Cold air-standard gives the closed form eta = 1 - r^(1-k) (rc^k - 1) / (k (rc - 1)). With the toggle off, u, h and vr come from the same air table as the Otto cycle.
Problem: Compression ratio 16, cutoff ratio 2, air at 95 kPa and 27 °C before compression. Variable specific heats. Find the temperature after heat addition, the thermal efficiency and the MEP.
State 2: vr2 = vr1 / 16 gives T2 = 862 K, P2 = 4368 kPa.
State 3: P3 = P2 and v3 = 2 v2, so T3 = 2 T2 = 1724 K and q_in = h3 - h2 = 1019.0 kJ/kg.
State 4: vr4 = (16 / 2) vr3 gives T4 = 882 K, and q_out = u4 - u1 = 445.6 kJ/kg.
Result: w_net = 573.4 kJ/kg; thermal efficiency = 56.3%; MEP = 675 kPa. Cold air-standard with k = 1.4 gives 61.4%.
Switch the Cycle control to Atkinson Cycle to solve the ideal Atkinson cycle, in which the expansion stroke is longer than the compression stroke. 1 to 2 is isentropic compression, 2 to 3 is heat addition at constant volume, 3 to 4 is isentropic expansion all the way down to the starting pressure, and 4 to 1 is heat rejection at constant pressure. The expansion ratio re = v4 / v3 is larger than the compression ratio r = v1 / v2, which is where the efficiency gain over the Otto cycle comes from. Enter r and either re, the heat added, or a temperature, and the rest follows.
Heat is rejected at constant pressure, so q_out = h4 - h1, and the piston does work on the gas P1 (v4 - v1) during that process. The results list it as boundary work in. Cold air-standard gives eta = 1 - k (re - r) / (re^k - r^k). With the toggle off, the expansion uses the Pr column of the air table and the compression uses vr.
Problem: Compression ratio 10, air at 100 kPa and 27 °C before compression, 1000 kJ/kg added at constant volume. Variable specific heats. Find the expansion ratio and the thermal efficiency.
State 2: vr2 = vr1 / 10 gives T2 = 730 K, P2 = 2433 kPa.
State 3: u3 = u2 + 1000 kJ/kg gives T3 = 1852 K, P3 = 6173 kPa.
State 4: isentropic expansion to P4 = 100 kPa gives T4 = 666 K and v4 = 1.913 m³/kg, so re = 22.2.
Result: q_out = h4 - h1 = 377.3 kJ/kg; w_net = 622.7 kJ/kg; thermal efficiency = 62.3%. An Otto cycle with the same compression ratio and heat input reaches 55.1%.
| Quantity | Expression | Notes |
|---|---|---|
| Cutoff ratio (Diesel) | rc = v3 / v2 | Equals T3 / T2, since P3 = P2 |
| Expansion ratio (Atkinson) | re = v4 / v3 | Larger than r |
| Heat input | Diesel: h3 - h2; Atkinson: u3 - u2 | Constant P or constant v |
| Heat rejected | Diesel: u4 - u1; Atkinson: h4 - h1 | Constant v or constant P |
| Net work | w_net = q_in - q_out | Includes the boundary work of the constant-pressure process |
| Mean effective pressure | MEP = w_net / (v_max - v_min) | v_max is v1 for Diesel, v4 for Atkinson |
The dual (limited pressure) cycle splits heat addition between constant volume and constant pressure, which fits a real compression-ignition engine better than either the Otto or the Diesel cycle alone. The five states are 1 before compression, 2 after isentropic compression, 3 after constant-volume heat addition, 4 after constant-pressure heat addition, and 5 after isentropic expansion back to v1. Enter the compression ratio r = v1/v2, the pressure ratio rp = P3/P2 and the cutoff ratio rc = v4/v3, or the two heat amounts, or any state properties you have.
Setting rc = 1 gives the Otto cycle and rp = 1 gives the Diesel cycle. For cold air-standard the efficiency is eta = 1 - (rp rc^k - 1) / (r^(k-1) [(rp - 1) + k rp (rc - 1)]).
Problem: Compression ratio 16, air at 100 kPa and 300 K, pressure ratio 1.5 during constant-volume heat addition, cutoff ratio 1.6. Cold air-standard, k = 1.4. Find the heat input and thermal efficiency.
State 2: T2 = 300 × 160.4 = 909.4 K, P2 = 4850 kPa.
State 3: P3 = 1.5 P2 = 7275 kPa at v3 = v2, so T3 = 1364.1 K and q23 = cv(T3 - T2) = 326.3 kJ/kg.
State 4: v4 = 1.6 v3 at P4 = P3, so T4 = 2182.6 K and q34 = cp(T4 - T3) = 822.2 kJ/kg.
Result: isentropic expansion to v1 gives T5 = 868.9 K and q_out = 408.2 kJ/kg; w_net = 740.2 kJ/kg; thermal efficiency = 64.5%, matching the closed form.
Switch the Cycle control to Brayton Cycle to solve the air-standard gas turbine cycle, drawn as the closed loop used in textbooks: 1 to 2 compressor, 2 to 3 heat addition at constant pressure, 3 to 4 turbine, 4 to 1 heat rejection at constant pressure. Enter P, T or h at each state, plus the pressure ratio, a work or heat quantity, or the compressor and turbine isentropic efficiencies, and the solver runs the steady-flow energy balance around each device.
The same two property models as the Otto cycle apply. Cold air-standard uses constant specific heats with cp = kR/(k - 1), which gives the closed forms T2 = T1 rp^((k-1)/k) and eta = 1 - rp^((1-k)/k). With the toggle off, enthalpy and the isentropic relation come from tabulated air properties (h and s0, equivalent to the Pr column of a standard air table). When the compressor or turbine is not isentropic, the T–s and P–v diagrams also mark the ideal exit states 2s and 4s.
Problem: Pressure ratio 8, air enters the compressor at 100 kPa and 300 K, turbine inlet 1300 K. Variable specific heats. Find the back work ratio and thermal efficiency.
State 2: isentropic compression by a factor of 8 in pressure gives T2 = 539.8 K, h2 = 544.17 kJ/kg (h1 = 300.17 kJ/kg).
State 4: isentropic expansion from h3 = 1395.92 kJ/kg gives T4 = 770.4 K, h4 = 789.60 kJ/kg.
Result: compressor work 244.0 kJ/kg, turbine work 606.3 kJ/kg, back work ratio = 40.2%; q_in = 851.7 kJ/kg, thermal efficiency = 42.5%. With 80% compressor and 85% turbine efficiencies the efficiency drops to 26.6%. Cold air-standard with k = 1.4 gives 44.8% for the ideal case.
| Quantity | Expression | Notes |
|---|---|---|
| Pressure ratio | rp = P2 / P1 | Also P3 / P4 when heat transfer is isobaric |
| Compressor work | w_c = h2 - h1 | Actual: (h2s - h1) / eta_c |
| Turbine work | w_t = h3 - h4 | Actual: eta_t (h3 - h4s) |
| Thermal efficiency | eta = (w_t - w_c) / (h3 - h2) | Cold air: 1 - rp^((1-k)/k) |
| Back work ratio | bwr = w_c / w_t | Typically 40 to 60% |
When the turbine exhaust is hotter than the compressor exit, a regenerator can use it to preheat the compressed air and cut the fuel heat. The Regenerative Brayton mode has six states: 1 to 2 compressor, 2 to 3 regenerator cold side, 3 to 4 heat addition, 4 to 5 turbine, 5 to 6 regenerator hot side, 6 to 1 heat rejection. The regenerator effectiveness is ε = (h3 - h2) / (h5 - h2), and an energy balance across it gives h3 - h2 = h5 - h6. Enter ε, or a temperature on either side of the regenerator and ε is computed. If the turbine exit is colder than the compressor exit the panel says so, since regeneration is then not possible.
Problem: The Brayton cycle above with 80% compressor and 85% turbine efficiencies, plus a regenerator with an effectiveness of 80%. Variable specific heats. Find the thermal efficiency (Cengel and Boles, Example 9-7).
Compressor and turbine: h2 = 605.17 kJ/kg and h5 = 880.55 kJ/kg (h4 = 1395.92 kJ/kg).
Regenerator: h3 = h2 + 0.80(h5 - h2) = 825.48 kJ/kg, so the heat added drops to h4 - h3 = 570.4 kJ/kg.
Result: w_net = 515.4 - 305.0 = 210.4 kJ/kg; thermal efficiency = 36.9%, up from 26.6% without the regenerator.
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.
Each cycle's math is checked against an independent reference before release.
Limits: Rankine with at most one reheat and no regeneration (feedwater heaters), IF97 Regions 1, 2 and 4 only (no Region 3, the area around the critical point, and no steam above 800 C), and R-134a as the only refrigerant.
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 or add a reheat stage. Raising the boiler pressure at the same turbine inlet temperature does the opposite: it raises efficiency but lowers the exit quality, which is why high-pressure plants use reheat.
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.