Rankine cycle read off the steam tables
Thermodynamics · the Rankine cycle from the steam tables
A campus cogeneration set runs a simple Rankine cycle: boiler, turbine, condenser, feed pump, and nothing clever in between. Steam leaves the superheater at 4.00 MPa and 400 °C, expands through a single-casing turbine, and exhausts into a surface condenser held at 10 kPa by its air ejectors, where the cooling water condenses it back to saturated liquid before the feed pump sends it round again. The turbine's own test report puts its isentropic efficiency at 85 %, and the plant meters 18,000 kg/h of steam through it. Four state points fix the whole cycle, and the steam tables give them all. State 1 comes straight off the superheated page at 4.00 MPa and 400 °C: h₁ = 3,214.4 kJ/kg with an entropy of s₁ = 6.7712 kJ/(kg·K). The IDEAL exhaust is where that same entropy lands at 10 kPa, and the saturation page at 10 kPa reads s_f = 0.6492 and s_fg = 7.4997 kJ/(kg·K), so the ideal quality is x₂s = (6.7712 − 0.6492)/7.4997 = 0.8163, and with h_f = 191.81 and h_fg = 2,392.1 kJ/kg on the same row that puts h₂s = 191.81 + 0.8163 × 2,392.1 = 2,144.5 kJ/kg. State 3 is the condensate itself, saturated liquid at 10 kPa, so h₃ = 191.81 kJ/kg. State 4 is that water after the pump, and since liquid barely compresses, the pump work is v_f(p₁ − p₂) = 0.001010 × 3,990 = 4.03 kJ/kg, giving h₄ = 195.84 kJ/kg.
Every number in this problem is editable — change any value below and the whole chain recalculates.
- h₁ = 3,214.4 kJ/kg — Turbine inlet, 4.00 MPa and 400 °C (s₁ = 6.7712)
- h₂s = 2,144.5 kJ/kg — Ideal exhaust at 10 kPa and s₁ (x₂s = 0.8163)
- h₃ = 191.81 kJ/kg — Condensate, saturated liquid at 10 kPa
- h₄ = 195.84 kJ/kg — Feedwater after the pump (h₃ + 4.03 kJ/kg)
- η_isen = 0.85 — — Turbine isentropic efficiency, from its test report
- ṁ = 18,000 kg/h — Steam through the machine
- (a)the enthalpy the steam really leaves the turbine with
- (b)the work the turbine takes out of each kilogram
- (c)the thermal efficiency of the cycle
- (d)the shaft power at the coupling
The ideal expansion is a construction, not a place any steam has ever been: it is the vertical line down the Mollier chart from state 1 to the condenser pressure, and it defines the largest drop physically available, 1,069.9 kJ/kg. A real turbine takes 85 % of that and turns the rest into entropy — friction, leakage past the tips, moisture dragging on the blades — so the real end point sits ABOVE the ideal one on the chart at the same pressure. Solving here for h₂ is running the efficiency definition backwards, which is how it is used in practice: the efficiency is known from the machine's test, the end point is not.
Carried onward at full precision, not this rounded figure.
w = h₁ − h₂ is the steady-flow energy equation with everything that does not matter struck out: no heat crosses the casing worth counting, and the kinetic and potential terms are noise beside a megajoule per kilogram. So the whole of a turbine's output is the enthalpy it takes off the steam. The answer has to come back as 0.85 × 1,069.9 — it is the same statement part (a) made, read from the other end — and that redundancy is deliberate: a chain that closes on itself is a chain you can trust with the numbers you are less sure of.
Carried onward at full precision, not this rounded figure.
Now all four state points earn their keep. Net work is the turbine's 909.4 kJ/kg less the pump's 4.03, and heat added is what the boiler puts in between states 4 and 1 — measured from the FEEDWATER enthalpy h₄, not from the condensate h₃, because the pump has already added its share by the time the water reaches the drum. The pump term is what makes a steam cycle work at all: compressing a vapour between the same two pressures would cost hundreds of kilojoules, and pumping the liquid costs four.
Carried onward at full precision, not this rounded figure.
Specific work becomes power the moment a flow rate is attached: 18,000 kg/h is exactly 5.000 kg/s, so the coupling sees five times the per-kilogram figure. Read the answer as GROSS turbine output — the feed pump's 4.03 kJ/kg is 20 kW of it, and a generator and gearbox will take a few percent more before anything reaches a breaker. The trade cross-check to carry: 18,000 kg/h for 4,547 kW is a steam rate of about 8.7 lb per kWh, right in the middle of the 8–12 that condensing turbines of this size actually deliver.
Carried onward at full precision, not this rounded figure.
Therefore the steam really leaves the turbine at 2,305.0 kJ/kg and still 88 % dry, the machine takes 909.4 kJ out of every kilogram, the cycle converts 30.0 % of its heat input into net work, and the coupling develops 4,547 kW — of which the feed pump takes 20 kW back, leaving 4,527 kW net against the 15.09 MW the boiler is supplying.
Why this order
Every number in this chain came off a table, and the chain's real subject is the order in which a reader is allowed to look things up. State 1 is a direct reading, pressure and temperature in, enthalpy and entropy out. State 2s cannot be read directly at all: the condenser page has no column for "the steam that arrived from 4 MPa", so the entropy has to be carried down from state 1 and the lever rule run on the s_f and s_fg columns to recover a quality, which is then run on the h columns to recover an enthalpy. That two-column round trip is the single most useful manoeuvre in applied thermodynamics, and it is worth doing slowly once: entropy is the quantity that is conserved in the ideal machine, so entropy is the thread you follow between pages. State 3 is a reading again. State 4 is the only place the incompressible approximation appears, and it earns 4.03 kJ/kg against a boiler input of 3,018.6 — a back-work ratio of 0.44 %, which is why nobody has ever built a Rankine cycle around a vapour compressor.
Two sanity checks belong beside the answer. The first is the exhaust quality: x₂ = (2,305.0 − 191.81)/2,392.1 = 0.883, so the last stage is running in 12 % moisture. Turbine designers hold a floor near 88 % because water droplets at blade-tip speeds erode the trailing edges, and the fact that this cycle sits right on that floor is exactly why real plants reheat — take the steam back to the boiler mid-expansion and the exhaust comes out drier and the efficiency rises with it. The second check is Carnot: between 400 °C at the throttle and 45.8 °C in the condenser, the ceiling is 1 − 318.96/673.15 = 52.6 %, and this cycle reaches 30.0 %, or 57 % of the ceiling. That gap is not waste to be scolded — it is the honest cost of boiling water at constant temperature and rejecting heat at constant temperature in a real machine, and the entire history of steam plant design is the attempt to shave it: higher throttle pressures, reheat, and feedwater heaters that bleed steam mid-expansion to warm state 4 toward state 1. The condenser deserves the last word. Dropping it from 10 kPa to 5 kPa is worth about two efficiency points, which is far more than any plausible improvement at the hot end, and it costs nothing but colder cooling water — the reason power stations sit on rivers.
Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.