Steam Power Plants and the Rankine Cycle

The steam power plant is the workhorse of the Industrial and Power Plant Engineering paper. It runs on the Rankine cycle, and almost every item reduces to an energy balance on one of four devices: boiler, turbine, condenser, pump. Learn to read enthalpies off the given data and apply the first law to each device, and this becomes one of the most reliable scoring areas on the MELE. Work everything in SI units: enthalpy in kJ/kg, pressure in kPa, mass flow in kg/s.

The ideal Rankine cycle

The ideal Rankine cycle has four processes, each at a fixed condition that makes the energy balance simple:

1. Boiler (1 to 2): heat is added at constant pressure, turning feedwater into high-pressure superheated steam. Heat added qin = h2 minus h1.
2. Turbine (2 to 3): steam expands isentropically (constant entropy), producing work wt = h2 minus h3.
3. Condenser (3 to 4): heat is rejected at constant pressure, condensing the steam to saturated liquid. Heat rejected qout = h3 minus h4.
4. Pump (4 to 1): the liquid is compressed back to boiler pressure. Pump work is small: wp = v x (P1 minus P4), with v the saturated-liquid specific volume, about 0.001 m3/kg.

The thermal efficiency is net work over heat added:

eta = (wt minus wp) / qin = [(h2 minus h3) minus wp] / (h2 minus h1)

Worked example: turbine inlet h2 = 3,350 kJ/kg, turbine exit h3 = 2,100 kJ/kg, condenser exit (saturated liquid) hf = 200 kJ/kg. Neglecting the small pump work, qin = 3,350 minus 200 = 3,150 kJ/kg and wnet = 3,350 minus 2,100 = 1,250 kJ/kg, so eta = 1,250/3,150 = 0.397, or about 39.7%.

The pump work itself: for water at 10 kPa raised to 6,000 kPa, wp = 0.001 x (6,000 minus 10) = about 5.99 kJ/kg, tiny next to the turbine's thousand-plus kJ/kg. That is why steam plants use a liquid pump instead of compressing vapor: pumping a liquid costs almost nothing.

Reheat and regeneration

Two modifications push Rankine efficiency higher:

• Reheat: expand the steam partway in a high-pressure turbine, send it back to the boiler to be reheated at constant pressure, then finish the expansion in a low-pressure turbine. This raises the average temperature of heat addition and, just as important, keeps the steam drier at turbine exit, protecting the last blades from erosion.
• Regeneration: bleed (extract) some steam partway through the turbine and use it to preheat the feedwater in a feedwater heater before it reaches the boiler. Less external heat is then needed to raise the feedwater, so efficiency climbs. The cost is a little lost turbine work from the bled steam.

Both changes raise efficiency by raising the average temperature at which heat is added, the single lever behind every vapor-cycle improvement.

Boilers and steam generation

The boiler (steam generator) turns feedwater into steam by burning fuel. Its performance is the boiler efficiency: the fraction of the fuel's heating value that ends up in the steam.

boiler efficiency = ms x (h_steam minus h_feed) / (mf x HHV)

where ms is steam flow, mf is fuel flow, and HHV is the fuel's higher heating value.

Worked example: a boiler makes 8 kg/s of steam at h_steam = 3,200 kJ/kg from feedwater at h_feed = 700 kJ/kg, burning 0.8 kg/s of fuel of HHV = 30,000 kJ/kg. Boiler efficiency = 8 x (3,200 minus 700) / (0.8 x 30,000) = 20,000/24,000 = 0.833, or 83.3%. The rest goes up the stack as hot flue gas and as radiation loss.

The factor of evaporation and the equivalent evaporation (referred to steam from and at 100 degrees C) are related boiler-rating measures, but boiler efficiency is the one the MELE leans on most.

Steam properties

Steam states come from the steam tables. Below the saturation curve, steam is wet, described by its quality x, the mass fraction that is vapor:

h = hf + x x hfg

where hf is saturated-liquid enthalpy and hfg is the latent heat. At x = 0 the water is saturated liquid; at x = 1 it is saturated (dry) vapor; above the saturation temperature it is superheated. Keeping quality high at turbine exit is exactly why reheat is used.

Condensers and feedwater heaters

The condenser rejects the cycle's waste heat to cooling water, condensing the turbine exhaust to liquid at low pressure (a vacuum), which maximizes the turbine's pressure drop and thus its work.

heat rejected = ms x (h3 minus hf) = mw x cp x (delta T)

where mw is cooling-water flow and cp = 4.187 kJ/kg K.

Worked example: 5 kg/s of exhaust steam at h3 = 2,300 kJ/kg condenses to hf = 180 kJ/kg. Heat rejected = 5 x (2,300 minus 180) = 10,600 kW. If the cooling water may rise only 10 degrees C, mw = 10,600 / (4.187 x 10) = about 253 kg/s.

A feedwater heater raises the feedwater temperature using bled steam. For an open (direct-contact) heater, an energy balance on 1 kg of feedwater fixes the fraction m of steam that must be extracted:

m x h_extraction + (1 minus m) x hf_in = 1 x hf_out

Worked example: extraction steam h_extraction = 2,700 kJ/kg, feedwater in hf_in = 200 kJ/kg, mixed out hf_out = 760 kJ/kg. Then m = (760 minus 200)/(2,700 minus 200) = 560/2,500 = 0.224, so about 22.4% of the steam is bled to the heater.

Exam-day strategy

• Tag each Rankine device with its energy balance before plugging numbers: boiler qin = h2 minus h1, turbine wt = h2 minus h3, condenser qout = h3 minus hf, pump wp = v x delta P.
• Pump work is almost always negligible next to turbine work, but compute it if the item gives you the pressures.
• Reheat keeps steam dry and raises efficiency; regeneration preheats feedwater and raises efficiency. Both work by raising the mean temperature of heat addition.
• For a condenser or feedwater heater, always start from a first-law balance on a stated mass flow.
• Boiler efficiency is heat into the steam divided by heat released by the fuel: ms(delta h) over mf(HHV).