The thermal cycle with the highest theoretical efficiency is the Carnot cycle, consisting of two isotherms and two adiabats.

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1 Carnot to Reality The thermal cycle with the highest theoretical efficiency is the Carnot cycle, consisting of two isotherms and two adiabats. η = W/Q in Carnot's theorem states that for a heat engine operating with a hot thermal reservoir (temperature TH) and a cold thermal reservoir (temperature Tc), none can have an efficiency greater than that of the following cycle: 1. Isothermal, reversible expansion of the working fluid at TH 2. Adiabatic, reversible expansion of the working fluid, which cools it to Tc 3. Isothermal, reversible compression of the working fluid at Tc 4. Adiabatic, reversible expansion of the working fluid, which heats it back up to TH In practice, there are difficulties in creating a Carnot heat engine. In order to achieve the theoretical Carnot efficiency, all of the processes in the cycle must be perfectly reversible, and thus the cycle time, τ, approaches infinity and there is zero power output. If one optimizes the cycle for power, the efficiency is much lower, but achievable and useful. Compare the Carnot formula with Curzon and Ahlborn s modification below. See poster by Lam, Liao and CEW on Hennings 3 rd Floor.

2 Steam Engines: the Rankine Cycle This is how electricity is generated from coal, oil, nuclear fuels. Complications due to two phases, liquid water and steam. Water 3 Steam Water+Steam 4 Process 1-2: The working fluid is pumped from low to high pressure. As the fluid is a liquid at this stage, the pump requires little input energy. Process 2-3: The high pressure liquid enters a boiler where it is heated at constant pressure by an external heat source to become a dry saturated vapour. Process need not stop at the saturation line, may go to 3. The input energy required can be easily calculated using steam tables, e.g.

3 Process 3-4: The dry saturated vapour expands through a turbine, generating power. This decreases the temperature and pressure of the vapour, and some condensation may occur. The output in this process can be easily calculated using the Enthalpy-entropy chart or the steam tables. This line may cross the saturation line as in 3-4. Process 4-1: The wet vapour then enters a condenser where it is condensed at a constant pressure to become a saturated liquid. The working fluid is NOT an ideal gas! Hence we cannot calculate the efficiency from a PV diagram; if we have the pressures everywhere and the temperature at point 3 or 3, we can use steam tables. The change in enthalpy is the heat absorbed or expelled under constant pressure conditions. As heat is absorbed and expelled at constant pressure in the Rankine cycle, we can write: η = 1 Q out Q in = 1 H 4 H 1 H 3 H 2 1 H 4 H 1 H 3 H 1 The last approximation arises because the pump adds very little energy to the water. To estimate the efficiency we need, at minimum, the enthalpies for points 1, 3, 4. Example: Critical point for water: 374 C (647 K) and 22 MPa; above this temperature or above this pressure, liquid and gaseous water are not separate phases. The green lines are isovolumetric.

4 Example: P1 = P4 = 20 kpa; P2 = P3 = 6 MPa with T3 = 500 C Row 1: Row 3: Process 1->2 is adiabatic (isentropic), so enter the known P and S into: Process 3->4 is adiabatic (isentropic), so you know S = kj/kg/k, but the steam is wet. Request a table of values for a range of dryness and a saturation pressure of 20 kpa at: See that S = kj/kg/k occurs at 85.5% dryness Note that the entropy of wet steam is the sum of the entropies of the water content and that of evaporation for the steam.

5 Now we can calculate the heat added and expelled, and the work done, from the change in enthalpies: Point H (kj/kg) Work done Heat (kj/kg) (kj/kg) (out) (by pump) (in) (on turbine) η = 1 H 4 H 1 H 3 H 2 = W turbine W pump Q in = 34.7% Note the Carnot efficiency for these temperatures is 61%; the Curzon-Ahlborn formula gives 38%. Questions 1. The turbines of a nuclear power station operate on the same cycle as the example given above, and they deliver 1 GW of power to the electrical generators. At what rate (in kg/s) does steam pass through the turbines? 2. Vary the parameters of the cycle in the example above, and see what happens to the efficiency: a. Lower the maximum pressure by 1 MPa b. Raise the maximum pressure by 1 MPa c. Lower the maximum temperature by 100 C d. Raise the maximum temperature by 100 C