The Rankine cycle

The Rankine cycle is the cycle of choice to produce electricity and is shown in Fig. 9.1 in a pressure (P)-enthalpy (H) diagram. Water is the usual working fluid in the cycle. Although there are more details to the cycle than shown, such as a reheat component, this will be ignored here so the reader may understand the basics of the cycle and the concept behind its design. It differs from the basic Carnot cycle in important aspects which will be discussed within the text below.

The four components to the cycle are the pump, boiler, turbine and condenser. We will begin with the pump and work around the cycle explaining how the steps between each component occur and the reason why they are designed that way. The state a is saturated liquid, al­though the liquid may be cooled slightly below the saturation state in practice. A saturated liquid means it is just at its boiling temperature for the given pressure, yet, it is in the liquid state. If the temperature were increased very slightly then a bubble of vapor would be produced. If the temperature is decreased then it is technically called a compressed

liquid. The normal boiling point of water is 100 0 C at 1 atm. pressure so water at ambient conditions is a compressed liquid since the pressure required to boil water at 20 0C is very low (2.3kPa и 0.0023 atm). The reason one desires a liquid at state a is because the pump does not work well at all if it is a mixture of vapor and liquid. Furthermore, it does not pay to use energy to decrease the temperature of the water below the saturation temperature (boiling point) since this is wasted energy. Thus, the target is to have saturated liquid water at state a.

In the Rankine cycle the liquid is pumped reversibly and adiabati – cally to state b at a higher pressure and a required power input of Wp. Reference to the Second Law of Thermodynamics (SLOT) in Chapter 3 shows that this means Sb = Sa, where Si is the specific entropy of state i. This part of the cycle is isentropic, which is important in deter­mining state b. A Carnot cycle would operate identically to drive the pump, as can be seen in Fig. 9.2, where the two cycles are graphed on a temperature (T)-entropy diagram (S).

The high pressure at state b is required so that the (compressed) liquid water can be forced through the pipes that circulate within the furnace that make up the boiler. The boiler operates at constant pressure for the Rankine cycle and heat at a rate of Qb is supplied to boil the water. This is the heat flow supplied by burning coal or from the Sun. A Carnot cycle operates differently for this part of the process and heat transfer operates at constant temperature. This difference, constant pressure boiling of water rather than constant temperature, is the reason why the two cycles look so dissimilar in the T-S diagram. The only way the Carnot cycle can have constant temperature boiling of water is for the line b’-c’ to be within the vapor-liquid envelope. If it is not, then the temperature will rise as energy is supplied to either liquid water or water vapor (steam). As may be familiar to the reader, when water boils it occurs at constant temperature for an equilibrium process and all energy supplied to the water goes to achieving the phase transition. Before the phase transition, liquid water will increase in temperature while after the completion of the phase transition the vapor will increase in temperature. The line b-b’-c’-c in the Rankine cycle is boiling of water at constant pressure and so this line is an isobar which ends at state c, which is superheated vapor (compare Figures 9.1 and 9.2).

Подпись: Fig. 9.2 Comparison of the ideal Rankine and Carnot cycles that operate between the same temperatures on a temperature-entropy diagram, whose shape is similar to that for water. The Rankine cycle operates in the cycle a- b-c-d and the Carnot cycle in the cycle a'-b'-c'-d'. The high pressure, superheated vapor is forced through a turbine that makes electricity at a rate Wt along the isentropic (reversible, adiabatic) line c-d. The line c’-d’ for the Carnot cycle is a similar line in that it is isentropic too, however, it ends in the vapor-liquid envelope. This is undesirable since the state d should not have any liquid droplets in it which could damage the turbine blades should they hit them. So ideally, state d should be right at the saturated vapor line; in reality, the design usually has it slightly to the right of the line to ensure no water droplets are formed should the cycle vary in conditions during normal operation. In addition to state d’ having water droplets which could damage the rotating blades, state a’ in the Carnot cycle is a vapor-liquid mixture which is very difficult to pump, it is much easier

to pump pure, liquid water. So, one can ascertain that the Carnot cycle is useful as a thermodynamic teaching tool, yet, its utility is negligible for power cycles due to its inherent engineering defects.

The final part of the process is condensation of the water vapor to liquid and to achieve state a. This occurs at constant pressure for the Rankine cycle and constant temperature for the Carnot cycle. If state d and state a are saturated vapor and liquid, respectively, then both cy­cles follow the same path since as long as the pressure is kept constant in the Rankine cycle it will be isothermal too. The heat extracted at a rate Qc by the condenser is used to condense water vapor to liquid. Again, for a conservative design, state d may very well be a superheated vapor just to the right of the vapor envelope and state a may be compressed liquid to the left of the liquid envelope. If this is the case the line d-a will look similar to c-b.

Now it is possible to analyze the Rankine cycle and to do this the First and Second Laws of Thermodynamics, the FLOT and SLOT, are written below, since they will be required throughout this chapter.

m [H2 – #1] = Q + Wm FLOT (3.12)

m [§2 – Si] > Q SLOT (3.13)

The change in kinetic and potential energies in the FLOT have been ignored. They can contribute, especially changes in kinetic energy, how­ever, for the purposes of this chapter they will be ignored as are frictional energy losses in the pumping of liquids long distances in a pipe. Before delving into the various components that constitute the power cycle let’s perform a FLOT analysis of the entire cycle to arrive at

0 = Wp – Wt + Qb – Qc [=] W (9.1)

Подпись: Пя Подпись: Wnet Q b Подпись: (9.2)

where the signs of the heat and work terms have been explicitly written and the symbol [=] means ‘has dimensions of.’ The net work is defined as Wnet = Wt – Wp and, as the definition implies, is the work obtained from the cycle ignoring the other loss terms (see Table 9.1) and is equivalent to the electrical output. The efficiency of the Rankine cycle пЯ is written as

representing the rate of energy one obtains from the cycle to that which is supplied. As mentioned above, a coal-fired power plant has an efficiency of about 40%. To understand the scale of these power plants and the coal that must be supplied to create electricity consider this example.

Example 9.1

Determine how much coal is required to operate the coal-fired power plant in Table 9.1 and the flow rate of water required to condense the water vapor coming from the turbine. Assume the cooling water comes from a nearby river and it can be returned from the power plant if it increases in temperature at most by 10 ° C.

The power plant is large with a capacity of 500 MW and an efficiency of 39%, so, the rate of primary energy supply, from coal, should be Qb = 500MW/0.39 = 1.28 GW. The energy content of coal is approximately 2 x 104 kJ/kg making 64 kg/s or 5.5 x 106 kg/day required. A rail car can hold about 100 tonnes (105 kg) making 55 cars required per day. Each car is approximately 15 m long meaning that the delivery train would be 825 m long, or almost a kilometer, for each day! If Qb were to be gathered from the Sun a rather large area would have to be used. Assuming the irradiance is 500 W/m2 then an area of 1.6 x 1.6 km2 would have to be used, assuming 100% collection efficiency.

The cooling water requirement is Qc = 0.525 x 1.28 GW = 672 MW. This is the rate of energy given to the river water pumped to the power plant and subsequently returned to the river. One can write the FLOT for the river water going to and from the condenser as mw Cpw AT = Q c, where mw is the river water flow rate to/from the condenser, Cpw, the heat capacity of water, and AT, the water temperature rise which must be at most 10 °C. Rearranging this equation and substituting in numbers gives mw = 1.6 x 104 kg/s.

The Susquehanna River, a large river in the east of the United States, has a flow rate of about 800 m3/s or 8 x 106 kg/s. Injecting this much energy into the river will produce a temperature rise of 0.02 °C, which is negligible. However, care should be taken to ensure that there is little local temperature rise, since this will reduce oxygen in the water and promote conditions for algal blooms and invasive species propagation. Brunner Island in the river hosts three units that produce 1.49 GW, so, the temperature rise is expected to be three times this amount, still negligible. Yet, cooling towers have been installed on the island to reduce thermal discharge to the river, obviously local heating effects are severe or the company would not have done this. These challenges will be the same for a solar powered unit since this process also requires a heat sink.

The details of the Rankine power cycle can now be considered and an analysis of each process unit operation is shown in Table 9.2. 2 As explained in the side note the pump work is best obtained by

Wp = – к [Pb – Pa] = – [Pb – Pa] (9.3)

Pa

Подпись:where Va is the specific volume of water in state a, Pa and Pb, pressures for state a and state b, respectively, and pa, the water density at

Table 9.2 Conditions and equations governing the components to the Rankine cycle.

Component

Process line

FLOT

SLOT

Pump

Rev. & Adia. (a-b)

m [Hb – Ha] = Wp

in [Sb – Sa] = 0

Boiler

Const. P (b-c)

hi [Hc – Hb] = Qь

not useful

Turbine

Rev. & Adia. (c-d)

in [Hc – Hd] = Wt

m [Sd – .Sc] = 0

Condenser

Const. P (d-a)

n [Hd — Ha] = Qc

not useful

state a. One can use thermodynamic tables for compressed liquids to find Ha and Hb, yet, eqn (9.3) is accurate enough for most conditions.

The process design is typically performed by stating the pressure level in the condenser and boiler and the power output, really, the electricity required. Or the boiler may operate at a given temperature and by knowing this and the condenser pressure together with the electrical output the entire process can be designed.

Clearly, the design entails finding the enthalpy of each state, which can be obtained through a variety of software packages. There are also tables of data as well as graphs. Appendix B has a pressure-enthalpy diagram which is very useful for visualizing the process as shown in Fig.

9.1 and will be used here.

Example 9.2

There are several small solar thermal energy generated electricity (STEGE) power plants in the Mojave Desert in California, United States, one of which is called SEGS VI (Solar Electric Generating System VI), which has generated 30 MW of electricity continuously since 1989. A simplified Rankine cycle for this power plant is that the steam at the exit from the boiler is at 375 °C and a pressure of 10 MPa. The pressure at the exit from the turbine is 10 kPa, this is absolute pressure and so is below atmospheric pressure and is a vacuum, while the exit state from the condenser will be saturated liquid. Determine the working fluid mass flow rate, the amount of heat rejected to the condenser, the heat required to make steam in the boiler and the efficiency of the process.

The FLOT and SLOT equations in Table 9.2 will be used to analyze this process, which is shown in Fig. 9.3. The place to start is at state c, which is superheated steam at a temperature of 375 °C and a pressure of 10 MPa, allowing one to determine the enthalpy and entropy: Hc = 3010 kJ/kg and Sc = 6.078 kJ/kg-K, respectively. Since the turbine operates isentropically state d will have Sd = 6.078 kJ/kg-K and Pd = 10 kPa.

image567Reference to the figure shows that state d will be a mixture of liquid and vapor, which is undesirable, since as previously mentioned the liquid droplets can ruin the turbine. We will use this as the exit state in this example for simplicity, a representation of the cycle which more closely resembles the power cycle is also given in Fig. 9.3. It involves reheat cycles and high and low pressure turbines. Although states c2, c5 and Сб also appear to be within the vapor-liquid envelope, they are most likely not. This is due to the fact that a real turbine will not be reversible (nor adiabatic), which requires the entropy of the exit state to increase. This tends to put these states outside the envelope thereby avoiding formation of liquid droplets and is most easily visualized on a T-S diagram.

Now to determine the enthalpy of state d we need to find the fraction of vapor which is determined by

Подпись: (9.4)

image569

Sd = [1 – Xd]Sl + XdSv

where xd is the mass fraction of vapor (often called the quality), S, the enthalpy of saturated liquid and Sv, the enthalpy of saturated vapor. This is a simple mixing rule for vapor and liquid which is very accurate. At a pressure of 10 kPa one can determine Si = 0.6492 kJ/kg-K and Sv = 8.1501 kJ/kg-K, making xd = 0.724. Now we can also determine Hl = 191.8 kJ/kg and Hv = 2585 kJ/kg and, by using a similar mixing rule to that in eqn (9.4) for the enthalpy, one finds Hd = 1925 kJ/kg. The values for the saturated entropy and vapor were determined with steam tables (or one can do this with convenient software that is available) which is an accurate way of finding their value. However, to the accuracy of the design calculations applied in this chapter one can estimate them from the pressure-enthalpy diagram for water in the appendix that is reproduced in Fig. 9.3.

The above enthalpy values allow the work (electricity) per unit mass flow rate generated by the turbine to be calculated,

W = Hc – Hd = 1085 J

m kg

State a should be determined now to find the heat load on the con­denser since state d has already been determined. There is no pressure drop through the condenser (this is an assumption) making Pa = 10 kPa and since it is saturated liquid, the enthalpy is Ha = 191.8 kJ/kg and Va = 0.00101 m3/kg. Before considering other parts of the cycle it is best to determine the power required to drive the pump, which will in turn allow the net work to be found and the mass flow rate of the working fluid m. Reference to eqn (9.3) and Table 9.2 yields

W = Hb – Ha = Va[Pb – Pa] = 1 J

m kg

and the enthalpy at state b can be determined via Hb = Ha + Va [Pb – Pa] = 192.8 kJ/kg.

The net work per unit mass flow rate Wnet /rn is Wt/m – Wp/m = 1084 kJ/kg, so, the mass flow rate of the working fluid can be determined,

. Wnet 30 MW o^kg

m = ———— =————- ;— = 27.7 —

Wnet /m 1084 kJ/kg s

The condenser heat load Q c is

Qc = m [Hd – Ha] = 48.0 MW while the heat load on the boiler Qb is

Qb = m [Hc – Hb] = 78.0 MW

image570 Подпись: 30.0 MW 78.0 MW Подпись: 38.5%

Finally, the efficiency of the process can be determined from eqn (9.2)

One can ascertain that the Rankine power cycle is a reasonably efficient cycle, with almost 40% of the heat source energy content being con­verted into electricity,3 however, the waste of energy is apparent when the condenser heat load is determined. Unfortunately, this cannot be avoided.

Подпись:Finally, the heat load on the boiler, 78.0 MW, is the amount of power that must be supplied by the Sun since this Rankine cycle is repre­sentative of that for the SEGS VI power plant in the Mojave Desert. Assuming there is 1000 W/m2 of insolation available, an area of at least

78.0 m2, or in terms of linear dimensions about 280 x280 m2, is needed. In reality, approximately twice this will be required since 1000 W/m2 is a substantial power density and in fact the insolation is gathered over

188.0 m2 (и 430 x 430 m2) at SEGS VI, which is approximately twice the above estimate.

STEGE is typically performed by raising the temperature of a heat transfer fluid with the Sun’s energy and having a thermal reservoir to dampen out times when the Sun is obscured by clouds. This can be done by heating molten salt stored on-site that can be used at will to boil water; other technologies are also available. Variance in the high pressure steam properties (state c) would put strain on turbine operation if the quality of the steam entering it were to significantly change and the heat transfer fluid system is designed to minimize this. Furthermore, the heat transfer fluid reduces the pressures obtained in the field since it does not boil, which is desirable.

The heat transfer fluid is placed in thermal contact with water through a heat exchanger to subsequently generate steam in another part of the plant. This causes some process inefficiency and an effort does exist to directly generate steam to eliminate this deleterious effect. The heat transfer fluid is considered integral to STEGE and will be used in the designs discussed within this chapter as will become clear later. Before that is considered the manner in which solar radiation is concentrated is discussed.

Updated: August 21, 2015 — 2:57 pm