Thermal Engineering

Brayton Cycle – Turbine Engine

In 1872, an American engineer, George Bailey Brayton advanced the study of heat engines by patenting a constant pressure internal combustion engine, initially using vaporized gas but later using liquid fuels such as kerosene. This heat engine is known as “Brayton’s Ready Motor”. It means, the original Brayton engine used a piston compressor and piston expander instead of a gas turbine and gas compressor.

Today, modern gas turbine engines and airbreathing jet engines are also a constant-pressure heat engines, therefore we describe their thermodynamics by the Brayton cycle. In general, the Brayton cycle describes the workings of a constant-pressure heat engine.

It is the one of most common thermodynamic cycles that can be found in gas turbine power plants or in airplanes. In contrast to Carnot cycle, the Brayton cycle does not execute isothermal processes, because these must be performed very slowly. In an ideal Brayton cycle, the system executing the cycle undergoes a series of four processes: two isentropic (reversible adiabatic) processes alternated with two isobaric processes.

Since Carnot’s principle states that no engine can be more efficient than a reversible engine (a Carnot heat engine) operating between the same high temperature and low temperature reservoirs, a gas turbine based on the Brayton cycle must have lower efficiency than the Carnot efficiency.

A large single-cycle gas turbine typically produces for example 300 megawatts of electric power and has 35–40% thermal efficiency. Modern Combined Cycle Gas Turbine (CCGT) plants, in which the thermodynamic cycle of consists of two power plant cycles (e.g. the Brayton cycle and the Rankine cycle), can achieve a thermal efficiency of around 55%.

Brayton Cycle – Processes

Brayton Cycle – Processes

In a closed ideal Brayton cycle, the system executing the cycle undergoes a series of four processes: two isentropic (reversible adiabatic) processes alternated with two isobaric processes:

• 

  Isentropic compression (compression in a compressor) – The working gas (e.g. helium) is compressed adiabatically from state 1 to state 2 by the compressor (usually an axial-flow compressor). The surroundings do work on the gas, increasing its internal energy (temperature) and compressing it (increasing its pressure). On the other hand the entropy remains unchanged. The work required for the compressor is given by W_C = H₂ – H₁.

• Isobaric heat addition (in a heat exchanger) – In this phase (between state 2 and state 3) there is a constant-pressure heat transfer to the gas from an external source, since the chamber is open to flow in and out. In an open ideal Brayton cycle, the compressed air then runs through a combustion chamber, where fuel is burned and air or another medium is heated (2 → 3). It is a constant-pressure process, since the chamber is open to flow in and out. The net heat added is given by Q_add = H₃ – H₂
• Isentropic expansion (expansion in a turbine) – The compressed and heated gas expands adiabatically from state 3 to state 4 in a turbine. The gas does work on the surroundings (blades of the turbine) and loses an amount of internal energy equal to the work that leaves the system. The work done by turbine is given by W_T = H₄ – H₃. Again the entropy remains unchanged.
• Isobaric heat rejection (in a heat exchanger) – In this phase the cycle completes by a constant-pressure process in which heat is rejected from the gas. The working gas temperature drops from point 4 to point 1. The net heat rejected is given by Q_re = H₄ – H₁

During a Brayton cycle, work is done on the gas by the compressor between states 1 and 2 (isentropic compression). Work is done by the gas in the turbine between stages 3 and 4 (isentropic expansion). The difference between the work done by the gas and the work done on the gas is the net work produced by the cycle and it corresponds to the area enclosed by the cycle curve (in pV diagram).

As can be seen, it is convenient to use enthalpy or specific enthalpy and to express the first law in terms of enthalpy in analysis of this thermodynamic cycle. This form of the law simplifies the description of energy transfer. At constant pressure, the enthalpy change equals the energy transferred from the environment through heating:

Isobaric process (Vdp = 0):

dH = dQ     →     Q = H₂ – H₁

At constant entropy, i.e. in isentropic process, the enthalpy change equals the flow process work done on or by the system:

Isentropic process (dQ = 0):

dH = Vdp     →     W = H₂ – H₁

See also: Why power engineers use enthalpy? Answer: dH = dQ + Vdp

Isentropic Process

An isentropic process is a thermodynamic process, in which the entropy of the fluid or gas remains constant. It means the isentropic process is a special case of an adiabatic process in which there is no transfer of heat or matter. It is a reversible adiabatic process. The assumption of no heat transfer is very important, since we can use the adiabatic approximation only in very rapid processes.

Isentropic Process and the First Law

For a closed system, we can write the first law of thermodynamics in terms of enthalpy:

dH = dQ + Vdp

or

dH = TdS + Vdp

Isentropic process (dQ = 0):

dH = Vdp     →     W = H₂ – H₁     →     H₂ – H₁ = C_p (T₂ – T₁)    (for ideal gas)

Isentropic Process of the Ideal Gas

The isentropic process (a special case of adiabatic process) can be expressed with the ideal gas law as:

pV^κ = constant

or

p₁V₁^κ = p₂V₂^κ

in which κ = c_p/c_v is the ratio of the specific heats (or heat capacities) for the gas. One for constant pressure (c_p) and one for constant volume (c_v). Note that, this ratio κ  = c_p/c_v is a factor in determining the speed of sound in a gas and other adiabatic processes.

Isobaric Process

An isobaric process is a thermodynamic process, in which the pressure of the system remains constant (p = const). The heat transfer into or out of the system does work, but also changes the internal energy of the system.

Since there are changes in internal energy (dU) and changes in system volume (∆V), engineers often use the enthalpy of the system, which is defined as:

H = U + pV

Isobaric Process and the First Law

The classical form of the first law of thermodynamics is the following equation:

dU = dQ – dW

In this equation dW is equal to dW = pdV and is known as the boundary work. In an isobaric process and the ideal gas, part of heat added to the system will be used to do work and part of heat added will increase the internal energy (increase the temperature). Therefore it is convenient to use the enthalpy instead of the internal energy.

Isobaric process (Vdp = 0):

dH = dQ → Q = H₂– H₁

At constant entropy, i.e. in isentropic process, the enthalpy change equals the flow process work done on or by the system.

Isobaric Process of the Ideal Gas

The isobaric process can be expressed with the ideal gas law as:

or

On a p-V diagram, the process occurs along a horizontal line (called an isobar) that has the equation p = constant.

See also: Charles’s Law

See also:

Brayton Cycle

Brayton Cycle – Turbine Engine

In 1872, an American engineer, George Bailey Brayton advanced the study of heat engines by patenting a constant pressure internal combustion engine, initially using vaporized gas but later using liquid fuels such as kerosene. This heat engine is known as “Brayton’s Ready Motor”. It means, the original Brayton engine used a piston compressor and piston expander instead of a gas turbine and gas compressor.

Today, modern gas turbine engines and airbreathing jet engines are also a constant-pressure heat engines, therefore we describe their thermodynamics by the Brayton cycle. In general, the Brayton cycle describes the workings of a constant-pressure heat engine.

It is the one of most common thermodynamic cycles that can be found in gas turbine power plants or in airplanes. In contrast to Carnot cycle, the Brayton cycle does not execute isothermal processes, because these must be performed very slowly. In an ideal Brayton cycle, the system executing the cycle undergoes a series of four processes: two isentropic (reversible adiabatic) processes alternated with two isobaric processes.

Since Carnot’s principle states that no engine can be more efficient than a reversible engine (a Carnot heat engine) operating between the same high temperature and low temperature reservoirs, a gas turbine based on the Brayton cycle must have lower efficiency than the Carnot efficiency.

A large single-cycle gas turbine typically produces for example 300 megawatts of electric power and has 35–40% thermal efficiency. Modern Combined Cycle Gas Turbine (CCGT) plants, in which the thermodynamic cycle of consists of two power plant cycles (e.g. the Brayton cycle and the Rankine cycle), can achieve a thermal efficiency of around 55%.

Brayton Cycle – pV, Ts diagram

The Brayton cycle is often plotted on a pressure volume diagram (pV diagram) and on a temperature-entropy diagram (Ts diagram).

When plotted on a pressure volume diagram, the isobaric processes follow the isobaric lines for the gas (the horizontal lines), adiabatic processes move between these horizontal lines and the area bounded by the complete cycle path represents the total work that can be done during one cycle.

The temperature-entropy diagram (Ts diagram) in which the thermodynamic state is specified by a point on a graph with specific entropy (s) as the horizontal axis and absolute temperature (T) as the vertical axis. Ts diagrams are a useful and common tool, particularly because it helps to visualize the heat transfer during a process. For reversible (ideal) processes, the area under the T-s curve of a process is the heat transferred to the system during that process.

See also:

Brayton Cycle

Thermal Efficiency of Brayton Cycle

In general the thermal efficiency, η_th, of any heat engine is defined as the ratio of the work it does, W, to the heat input at the high temperature, Q_H.

The thermal efficiency, η_th, represents the fraction of heat, Q_H, that is converted to work. Since energy is conserved according to the first law of thermodynamics and energy cannot be be converted to work completely, the heat input, Q_H, must equal the work done, W, plus the heat that must be dissipated as waste heat Q_C into the environment. Therefore we can rewrite the formula for thermal efficiency as:

This is very useful formula, but here we express the thermal efficiency using the first law in terms of enthalpy.

To calculate the thermal efficiency of the Brayton cycle (single compressor and single turbine) engineers use the first law of thermodynamics in terms of enthalpy rather than in terms of internal energy.

The first law in terms of enthalpy is:

dH = dQ + Vdp

In this equation the term Vdp is a flow process work. This work,  Vdp, is used for open flow systems like a turbine or a pump in which there is a “dp”, i.e. change in pressure. There are no changes in control volume. As can be seen, this form of the law simplifies the description of energy transfer.

There are expressions in terms of more familiar variables such as temperature and pressure:

dH = C_pdT + V(1-αT)dp

Where C_p is the heat capacity at constant pressure and α is the coefficient of (cubic) thermal expansion. For ideal gas αT = 1 and therefore:

dH = C_pdT

At constant pressure, the enthalpy change equals the energy transferred from the environment through heating:

Isobaric process (Vdp = 0):

dH = dQ     →     Q = H₃ – H₂   →   H₃ – H₂ = C_p (T₃ – T₂)

At constant entropy, i.e. in isentropic process, the enthalpy change equals the flow process work done on or by the system:

Isentropic process (dQ = 0):

dH = Vdp     →     W = H₄ – H₃     →     H₄ – H₃ = C_p (T₄ – T₃)

The enthalpy can be made into an intensive, or specific, variable by dividing by the mass. Engineers use the specific enthalpy in thermodynamic analysis more than the enthalpy itself.

Now, let assume the ideal Brayton cycle that describes the workings of a constant pressure heat engine. Modern gas turbine engines and airbreathing jet engines also follow the Brayton cycle. This cycle consist of four thermodynamic processes:

1. 

   isentropic compression – ambient air is drawn into the compressor, where it is pressurized (1 → 2). The work required for the compressor is given by W_C = H₂ – H₁.

2. isobaric heat addition – the compressed air then runs through a combustion chamber, where fuel is burned and air or another medium is heated (2 → 3). It is a constant-pressure process, since the chamber is open to flow in and out. The net heat added is given by Q_add = H₃ – H₂
3. isentropic expansion – the heated, pressurized air then expands on turbine, gives up its energy. The work done by turbine is given by W_T = H₄ – H₃
4. isobaric heat rejection – the residual heat must be rejected in order to close the cycle. The net heat rejected is given by Q_re = H₄ – H₁

As can be seen, we can fully describe and calculate such cycles (similarly for Rankine cycle) using enthalpies.

Thermal Efficiency – Brayton Cycle

The thermal efficiency of such simple Brayton cycle, for ideal gas can now be expressed in terms of the temperatures:

where

• W_T the work done by the gas in the turbine
• W_C the work done on the gas in the compressor
• c_p is the heat capacity ratio

Pressure Ratio – Brayton Cycle – Gas Turbine

The thermal efficiency in terms of the compressor pressure ratio (PR = p₂/p₁), which is the parameter commonly used:

In general, increasing the pressure ratio is the most direct way to increase the overall thermal efficiency of a Brayton cycle, because the cycle approaches the Carnot cycle.

According to Carnot’s principle higher efficiencies can be attained by increasing the temperature of the gas.

But there are also limits on the pressure ratios that can be used in the cycle. The highest temperature in the cycle occurs at the end of the combustion process, and it is limited by the maximum temperature that the turbine blades can withstand. As usual, metallurgical considerations (about 1700 K) place an upper limits on thermal efficiency.

Consider the effect of compressor pressure ratio on thermal efficiency when the turbine inlet temperature is restricted to the maximum allowable temperature. There are two Ts diagrams of Brayton cycles having the same turbine inlet temperature but different compressor pressure ratios on the picture. As can be seen for a fixed-turbine inlet temperature, the net work output per cycle (W_net = W_T – W_C) decreases with the pressure ratio (Cycle A). But the Cycle A has the greater efficiency.

On the other hand, the Cycle B has a larger net work output per cycle (enclosed area in the diagram) and thus the greater net work developed per unit of mass flow. The work produced by the cycle times a mass flow rate through the cycle is equal to the power output produced by the gas turbine.

Therefore with less work output per cycle (Cycle A), a larger mass flow rate (thus a larger system) is needed to maintain the same power output, which may not be economical. This is the key consideration in design of gas turbine, since here engineers must balance the thermal efficiency and the compactness. In most common designs, the pressure ratio of a gas turbine ranges from about 11 to 16.

Thermal Efficiency Improvement – Brayton Cycle

There are several methods, how can be the thermal efficiency of the Brayton cycle improved. Assuming that the maximum temperature is limited by metallurgical consideration, these methods are:

Increasing pressure ratio

In general, increasing the pressure ratio is the most direct way to increase the overall thermal efficiency of a Brayton cycle, since the thermodynamic efficiency is primarily dependent on the pressure ratio, PR.

As was discussed, increasing the pressure ratio increases the compressor discharge temperature. Since the turbine inlet temperature is limited by the maximum temperature that the turbine blades can withstand, the pressure ration influences the heat amount that can be added to the flow. Moreover with an increase of the pressure ratio, the diameter of the compressor blades becomes progressively smaller in higher pressure stages of the compressor. Because the gap between the blades and the engine casing increases in size as a percentage of the compressor blade height as the blades get smaller in diameter, a greater percentage of the compressed air can leak back past the blades in higher pressure stages. This causes a leak back and in result it decreases the isentropic compressor efficiency (will be discussed later). Finally, from the formula for the thermal efficiency in term of pressure ratio can be seen, there is smaller gain as the pressure ratio increases (due to the exponent).

Heat regeneration

Significant increases in the thermal efficiency of gas turbine power plants can be achieved through reducing the amount of fuel that must be burned in the combustion chamber. This can be done by transferring heat from the turbine exhaust gas, which is normally well above the ambient temperature, to the compressor discharge air flow known as heat regeneration. Especially at a low or moderate pressure ratio, there is a high temperature increase in the combustion chamber and the turbine exhaust gas might still contain significant amount of heat at higher temperature than the compressor outlet gas (after the last compression stage but before the combustor). For this purpose a heat exchanger called a regenerator is used. Sometimes engineers use the term economiser that are heat exchangers intended to reduce energy consumption, especially in case of preheating of a fluid.

This heat regenerator allows the air exiting the compressor to be preheated before it enters the combustion chamber, thereby reducing the amount of fuel that must be burned in the combustor. This form of heat recycling is only possible if the gas turbine is run with low pressure ratio.

As was stated the temperature difference between turbine outlet and compressor outlet is crucial and determines the amount of heat that can be recovered. In case of negative difference (i.e. T₂ > T₄), the heat regeneration is not possible. There are two main ways, how to change this difference:

• to increase the turbine outlet temperature (T₄) through reheat of the flow during expansion phase (i.e. use of a multistage turbine with a reheat combustor or with a reheater)
• to decrease the compressor outlet temperature (T₂) through intercooling of the flow during compression phase (i.e. use of a multistage compressor with an intercooler)

Therefore reheat and intercooling are complementary with heat regeneration. By itself, they would not necessarily increase the thermal efficiency, however, when intercooling or reheat is used in conjunction with heat regeneration, a significant increase in thermal efficiency can be achieved.

It must be noted, transferring heat from the turbine outlet to the compressor inlet would reduce efficiency, as hotter inlet air means more volume, thus more work for the compressor. Engineers must also take into consideration pressure losses generated by the heat exchanger that slightly reduce the power of the gas turbine.

Regeneration vs. Recuperation of Heat

In general, the heat exchangers used in regeneration may be classified as either regenerators or recuperators.

• Regenerator is a type of heat exchanger where heat from the hot fluid is intermittently stored in a thermal storage medium before it is transferred to the cold fluid. It has a single flow path in which the hot and cold fluids alternately pass through.

• Recuperator is a type of heat exchanger has separate flow paths for each fluid along their own passages and heat is transferred through the separating walls. Recuperators (e.g. economisers) are often used in power engineering, to increase the overall efficiency of thermodynamic cycles. For example, in a gas turbine engine. The recuperator transfers some of the waste heat in the exhaust to the compressed air, thus preheating it before entering the combustion chamber. Many recuperators are designed as counterflow heat exchangers.

Reheat - Reheaters

As was discussed, the maximum temperature is limited by metallurgical consideration, but in order to deliver more of the heat at a temperature close to the peak of the cycle the gas can be reheated in a reheater. This involves splitting the turbine, i.e. use of a multistage turbine with a reheat combustor or with a reheater. High pressure and low pressure stages of the turbine may be on the same shaft to drive a common generator, but they will have separate cases. With a reheater, the flow is extracted after a partial expansion (point a), run back through the heat exchanger to heat it back up to the peak temperature (point b), and then passed to the lower pressure stage of the turbine. The expansion then completed in this stage from point b to point 4.

With this arrangement the net work per unit of mass flow can be increased. Despite the increase in net work with reheat, the cycle thermal efficiency would not necessarily increase because a greater total heat addition would be required. On the other hand, the temperature at the exit of the turbine (low pressure stage) is higher with reheat than without reheat, so there is the potential for heat regeneration. Therefore reheat and regeneration are complementary, they are used usually together in order to increase the thermal efficiency of gas turbine.

Compression with Intercooling

Significant increases in the thermal efficiency of gas turbine power plants can be achieved also through intercooling. Intercooling can be applied between compressor stages to reduce compression work, W_C, hence increasing overall power of the gas turbine.

For this purpose a heat exchanger known as an intercooler is usually used between stages of a multi-stage compression process. In general, intercoolers are heat exchangers that are used in many applications, including air compressors, air conditioners, refrigerators, and gas turbines. Intercoolers are widely known also in automotive use as a turbocharger or supercharger, but here they increase intake air charge density, hence the power of an engine.

In gas turbine power plant the thermal efficiency is of the highest importance and intercooling with heat regeneration are widely used. This involves splitting the compressor, i.e. use of a multistage compressor with an intercooler or intercoolers. High pressure and low pressure stages of the compressor may be on the same shaft even with turbine or a generator, but it is not a rule. With an intercooler, the flow is extracted after a partial compression (point c), run through the heat exchanger (intercooler) to cool it to the ambient temperature (point d), and then passed to the high stage of compressor. The compression is then completed in the second compressor from point d to point 2.

With this arrangement the net work per unit of mass flow (↑W_net = W_T – ↓W_C) can be increased by reducing the compression work (↓W_C).  Despite the increase in net work with intercooling, the cycle thermal efficiency would not necessarily increase because the temperature of the air entering the combustor would be reduced and a greater total heat addition would be required to achieve the desired turbine inlet temperature. On the other hand, the temperature at the exit of the compressor (high pressure stage) is lower with intercooling than without intercooling, so there is the potential for heat regeneration (Q_regen increases). Note that, the heat regeneration requires lower compressor outlet temperature than the turbine outlet temperature (simply due to 2nd law) and this temperature difference determines the amount of heat available for heat regeneration.

Therefore reheat and intercooling are complementary with heat regeneration. By itself, they would not necessarily increase the thermal efficiency, however, when intercooling or reheat is used in conjunction with heat regeneration, a significant increase in thermal efficiency can be achieved.

Some large compressors with higher pressure ratio have several stages of compression with intercooling between stages. Engineers must also take into consideration pressure losses generated by all heat exchangers that slightly increase compression work. The certain gas turbine design (the number of intercoolers, reheaters and regenerators) is an engineering problem and depends on certain purpose of the gas turbine.

Reheat, Intercooling and Regeneration  in Brayton Cycle

As was discussed reheat and intercooling are complementary with heat regeneration. By itself, they would not necessarily increase the thermal efficiency, however, when intercooling or reheat is used in conjunction with heat regeneration, a significant increase in thermal efficiency can be achieved and the net work output is also increased. This requires a gas turbine with two stages of compression and two turbine stages.

Isentropic Efficiency – Turbine, Compressor

Most steady-flow devices (turbines, compressors, nozzles) operate under adiabatic conditions, but they are not truly isentropic but are rather idealized as isentropic for calculation purposes. We define parameters η_T,  η_C, η_N, as a ratio of real work done by device to work by device when operated under isentropic conditions (in case of turbine). This ratio is known as the Isentropic Turbine/Compressor/Nozzle Efficiency.

See also: Irreversibility of Natural Processes

These parameters describe how efficiently a turbine, compressor or nozzle approximates a corresponding isentropic device. This parameter reduces the overall efficiency and work output. For turbines, the value of η_T is typically 0.7 to 0.9 (70–90%).

See also:

Brayton Cycle

Pressure Ratio – Brayton Cycle – Gas Turbine

The thermal efficiency in terms of the compressor pressure ratio (PR = p₂/p₁), which is the parameter commonly used:

In general, increasing the pressure ratio is the most direct way to increase the overall thermal efficiency of a Brayton cycle, because the cycle approaches the Carnot cycle.

According to Carnot’s principle higher efficiencies can be attained by increasing the temperature of the gas.

But there are also limits on the pressure ratios that can be used in the cycle. The highest temperature in the cycle occurs at the end of the combustion process, and it is limited by the maximum temperature that the turbine blades can withstand. As usual, metallurgical considerations (about 1700 K) place an upper limits on thermal efficiency.

Consider the effect of compressor pressure ratio on thermal efficiency when the turbine inlet temperature is restricted to the maximum allowable temperature. There are two Ts diagrams of Brayton cycles having the same turbine inlet temperature but different compressor pressure ratios on the picture. As can be seen for a fixed-turbine inlet temperature, the net work output per cycle (W_net = W_T – W_C) decreases with the pressure ratio (Cycle A). But the Cycle A has the greater efficiency.

On the other hand, the Cycle B has a larger net work output per cycle (enclosed area in the diagram) and thus the greater net work developed per unit of mass flow. The work produced by the cycle times a mass flow rate through the cycle is equal to the power output produced by the gas turbine.

Therefore with less work output per cycle (Cycle A), a larger mass flow rate (thus a larger system) is needed to maintain the same power output, which may not be economical. This is the key consideration in design of gas turbine, since here engineers must balance the thermal efficiency and the compactness. In most common designs, the pressure ratio of a gas turbine ranges from about 11 to 16.

See also:

Brayton Cycle

Brayton Cycle with Reheat, Regeneration and Intercooling

Reheat, Intercooling and Regeneration  in Brayton Cycle

As was discussed reheat and intercooling are complementary with heat regeneration. By itself, they would not necessarily increase the thermal efficiency, however, when intercooling or reheat is used in conjunction with heat regeneration, a significant increase in thermal efficiency can be achieved and the net work output is also increased. This requires a gas turbine with two stages of compression and two turbine stages.

Thermal Efficiency Improvement – Brayton Cycle

There are several methods, how can be the thermal efficiency of the Brayton cycle improved. Assuming that the maximum temperature is limited by metallurgical consideration, these methods are:

See also:

Brayton Cycle

Brayton Cycle – Problem with Solution

Let assume the closed Brayton cycle, which is the one of most common thermodynamic cycles that can be found in modern gas turbine engines. In this case assume a helium gas turbine with single compressor and single turbine arrangement. One of key parameters of such engines is the maximum turbine inlet temperature and the compressor pressure ratio (PR = p₂/p₁) which determines the thermal efficiency of such engine.

In this turbine the high-pressure stage receives gas (point 3 at the figure) from a heat exchanger:

• p₃ = 6.7 MPa;
• T₃ = 1190 K (917°C))
• the isentropic turbine efficiency is η_T = 0.91 (91%)

and exhaust it to another heat exchanger, where the outlet pressure is (point 4):

• p₄ = 2.78 MPa
• T_4,is = ?

Thus the compressor pressure ratio is equal to PR = 2.41.  Moreover we know, that the compressor receives gas (point 1) at the figure:

• p₁ = 2.78 MPa;
• T₁ = 299 K (26°C)
• the isentropic compressor efficiency η_K = 0.87 (87%).

The heat capacity ratio, , for helium is equal to =c_p/c_v=1.66

Calculate:

1. the heat added by the heat exchanger (between 2 → 3)
2. the compressor outlet temperature of the gas (T_2,is)
3. the real work done on this compressor, when the isentropic compressor efficiency is η_K = 0.87 (87%)
4. the turbine outlet temperature of the gas (T_4,is)
5. the real work done by this turbine, when the isentropic turbine efficiency is η_T = 0.91 (91%)
6. the thermal efficiency of this cycle

Solution:

1) + 2)

From the first law of thermodynamics, the net heat added is given by Q_add,ex = H₃ – H₂ [kJ] or Q_add = C_p.(T₃-T_2s), but in this case we do not know the temperature (T_2s) at the outlet of the compressor. We will solve this problem in intensive variables. We have to rewrite the previous equation (to include η_K) using the term (+h₁ – h₁) to:

Q_add = h₃ – h₂ = h₃ – h₁ – (h_2s – h₁)/η_K  [kJ/kg]

Q_add = c_p(T₃-T₁) – (c_p(T_2s-T₁)/η_K)

Then we will calculate the temperature, T_2s, using p, V, T Relation (from Ideal Gas Law) for adiabatic process between (1 → 2).

In this equation the factor for helium is equal to =c_p/c_v=1.66. From the previous equation follows that the compressor outlet temperature, T_2s, is:

Q_add = c_p(T₃-T₁) – (c_p(T_2s-T₁)/η_K) = 5200.(1190 – 299) – 5200.(424-299)/0.87 = 4.633 MJ/kg – 0.747 MJ/kg = 3.886 MJ/kg

3)

The work done on the gas by the compressor in the isentropic compression process is:

W_C,s = c_p (T_2s – T₁) = 5200 x (424 – 299) = 0.650 MJ/kg

The real work done on the gas by the compressor in the adiabatic compression is then:

W_C,real = c_p (T_2s – T₁). η_C = 5200 x (424 – 299) / 0.87 = 0.747 MJ/kg

4)

The turbine outlet temperature of the gas, T_4,is, can be calculated using the same p, V, T Relation as in 2) but between states 3 and 4:

From the previous equation follows that the outlet temperature of the gas, T_4,is, is:

5)

The work done by gas turbine in the isentropic expansion is then:

W_T,s = c_p (T₃ – T_4s) = 5200 x (1190 – 839) = 1.825 MJ/kg

The real work done by gas turbine in the adiabatic expansion is then:

W_T,real = c_p (T₃ – T_4s) . η_T = 5200 x (1190 – 839) x 0.91 = 1.661 MJ/kg

6)

As was derived in the previous section, the thermal efficiency of an ideal Brayton cycle is a function of pressure ratio and κ:

therefore

η_th = 0.295 = 29.5%

The thermal efficiency can be also calculated using the work and the heat (without η_K):

η_th,s = (W_T,s – W_C,s) / Q_add,s = (1.825 – 0.650) / 3.983  =  0.295 = 29.5%

Finally, the thermal efficiency including isentropic turbine/compressor efficiency is:

η_th,real = (W_T,real – W_C,real) / Q_add = (1.661 – 0.747) / 3.886  =  0.235 = 23.5%

See also:

Brayton Cycle

Ericsson Cycle

The Ericsson cycle is named after a Swedish-American inventor John Ericsson, who designed and built many unique heat engines based on various thermodynamic cycles. He is credited with inventing two unique heat engine cycles and developing practical engines based on these cycles.

His first thermodynamic cycle “the first Ericsson cycle” is now called the “Brayton cycle“, in fact it is the closed Brayton cycle, which is commonly applied to modern closed cycle gas turbine engines.

The second Ericsson cycle is what is now called the Ericsson cycle. The second Ericsson cycle is similar to the Brayton cycle, but uses external heat and incorporates the multiple use of an intercooling and reheat. In fact, it is like a Brayton cycle with an infinite number of reheat and intercooler stages in the cycle. Compared to the Brayton cycle which uses adiabatic compression and expansion, an ideal Ericsson cycle consists of isothermal compression and expansion processes, combined with isobaric heat regeneration between them. Applying intercooling, heat regeneration and sequential combustion significantly increases thermal efficiency of a turbine, in fact, the thermal efficiency of the ideal Ericsson cycle equals to the Carnot efficiency.

It is assumed (in an ideal case) that each intercooler return the working fluid to the ambient temperature T₁ and each reheater reheats the working fluid to the temperature T₃. The regenerator is 100% efficient and allows the heat input for process 2 → 3 to be obtained from the heat rejected in process 4 → 1. Since there is no need of heat transfer (Q_add) in process 2 → 3, all the heat added externally would occur in the reheaters and all the heat rejected to the surroundings would take place in the intercoolers. As can be seen from the picture, in this case all the heat added would occur when the working fluid is at its highest temperature, T₃, and all the heat rejected would take place when the working fluid is at its lowest temperature, T₁. Since irreversibilities are presumed absent and all the heat is supplied and rejected isothermally, the thermal efficiency of an ideal Ericsson cycle can be calculated from these temperatures:

where:

• η_Carnot is the efficiency of Carnot cycle, i.e. it is the ratio = W/Q_H of the work done by the engine to the heat energy entering the system from the hot reservoir.
• T_C is the absolute temperature (Kelvins) of the cold reservoir,
• T_H is the absolute temperature (Kelvins) of the hot reservoir.

Although the thermodynamic processes of the Ericsson cycle differ from those of the Carnot cycle, both cycles have the same value of thermal efficiency when operating between the temperatures T_H and T_C.

See also:

Thermodynamic Cycles

Rankine Cycle – Steam Turbine Cycle

In 1859, a Scottish engineer, William John Macquorn Rankine advanced the study of heat engines by publishing the “Manual of the Steam Engine and Other Prime Movers”. Rankine developed a complete theory of the steam engine and indeed of all heat engines. Together with Rudolf Clausius and William Thomson (Lord Kelvin), he was a contributor to the thermodynamics, particularly focusing on the first of the three thermodynamic laws.

The Rankine cycle was named after him and describes the performance of steam turbine systems, though the theoretical principle also applies to reciprocating engines such as steam locomotives. In general, the Rankine cycle is an idealized thermodynamic cycle of a constant pressure heat engine that converts part of heat into mechanical work. In this cycle the heat is supplied externally to a closed loop, which usually uses water (in a liquid and vapor phase) as the working fluid. In contrast to the Brayton cycle, the working fluid in the Rankine cycle undergo the phase change from a liquid to vapor phase and vice versa.

While many substances could be used as the working fluid in the Rankine cycle (inorganic or even organic), water is usually the fluid of choice due to its favorable properties, such as its non-toxic and unreactive chemistry, abundance, and low cost, as well as its thermodynamic properties. For example, water has the highest specific heat of any common substance –  4.19 kJ/kg K. Moreover it has very high heat of vaporization, which makes it an effective coolant and medium in thermal power plants and other energy industry. In case of the Rankine cycle, the Ideal Gas Law almost cannot be used (steam do not follow pV=nRT), therefore all important parameters of water and steam are tabulated in so called “Steam Tables“.

One of the major advantages of the Rankine cycle is that the compression process in the pump takes place on a liquid. By condensing the working steam to a liquid (inside a condenser) the pressure at the turbine outlet is lowered and the energy required by the feed pump consumes only 1% to 3% of the turbine output power and these factors contribute to a higher efficiency for the cycle.

In contrast to Carnot cycle, the Rankine cycle does not execute isothermal processes, because these must be performed very slowly. In an ideal Rankine cycle, the system executing the cycle undergoes a series of four processes: two isentropic (reversible adiabatic) processes alternated with two isobaric processes.

Since Carnot’s principle states that no engine can be more efficient than a reversible engine (a Carnot heat engine) operating between the same high temperature and low temperature reservoirs, a steam turbine based on the Rankine cycle must have lower efficiency than the Carnot efficiency.

In modern nuclear power plants the overall thermal efficiency is about one-third (33%), so 3000 MWth of thermal power from the fission reaction is needed to generate 1000 MWe of electrical power. Higher efficiencies can be attained by increasing the temperature of the steam. But this requires an increase in pressures inside boilers or steam generators. However, metallurgical considerations place an upper limits on such pressures. In comparison to other energy sources the thermal efficiency of 33% is not much. But it must be noted that nuclear power plants are much more complex than fossil fuel power plants and it is much easier to burn fossil fuel ,than to generate energy from nuclear fuel.

Rankine Cycle – Processes

In an ideal Rankine cycle, the system executing the cycle undergoes a series of four processes: two isentropic (reversible adiabatic) processes alternated with two isobaric processes:

• 

  Isentropic compression (compression in centrifugal pumps) – The liquid condensate is compressed adiabatically from state 1 to state 2 by centrifugal pumps (usually by condensate pumps and then by feedwater pumps). The liquid condensatei s pumped from the condenser into the higher pressure boiler. In this process, the surroundings do work on the fluid, increasing its enthalpy (h = u+pv) and compressing it (increasing its pressure). On the other hand the entropy remains unchanged. The work required for the compressor is given by W_Pumps = H₂ – H₁.

• Isobaric heat addition (in a heat exchanger – boiler) – In this phase (between state 2 and state 3) there is a constant-pressure heat transfer to the liquid condensate from an external source, since the chamber is open to flow in and out.  The feedwater (secondary circuit) is heated from to the boiling point (2 → 3a) of that fluid and then evaporated in the boiler (3a → 3). The net heat added is given by Q_add = H₃ – H₂
• Isentropic expansion (expansion in a steam turbine) – Steam from the boiler expands adiabatically from state 3 to state 4 in a steam turbine to produce work and then is discharged to the condenser (partially condensed). The steam does work on the surroundings (blades of the turbine) and loses an amount of enthalpy equal to the work that leaves the system. The work done by turbine is given by W_T = H₄ – H₃. Again the entropy remains unchanged.
• Isobaric heat rejection (in a heat exchanger) – In this phase the cycle completes by a constant-pressure process in which heat is rejected from the partially condensed steam. There is heat transfer from the vapor to cooling water flowing in a cooling circuit. The vapor condenses and the temperature of the cooling water increases. The net heat rejected is given by Q_re = H₄ – H₁

During a Rankine cycle, work is done on the fluid by the pumps between states 1 and 2 (isentropic compression). Work is done by the fluid in the turbine between stages 3 and 4 (isentropic expansion). The difference between the work done by the fluid and the work done on the fluid is the net work produced by the cycle and it corresponds to the area enclosed by the cycle curve (in pV diagram). The working fluid in a Rankine cycle follows a closed loop and is reused constantly.

As can be seen, it is convenient to use enthalpy and the first law in terms of enthalpy in analysis of this thermodynamic cycle. This form of the law simplifies the description of energy transfer. At constant pressure, the enthalpy change equals the energy transferred from the environment through heating:

Isobaric process (Vdp = 0):

dH = dQ     →     Q = H₂ – H₁

At constant entropy, i.e. in isentropic process, the enthalpy change equals the flow process work done on or by the system:

Isentropic process (dQ = 0):

dH = Vdp     →     W = H₂ – H₁

See also: Why power engineers use enthalpy? Answer: dH = dQ + Vdp

Isentropic Process

An isentropic process is a thermodynamic process, in which the entropy of the fluid or gas remains constant. It means the isentropic process is a special case of an adiabatic process in which there is no transfer of heat or matter. It is a reversible adiabatic process. The assumption of no heat transfer is very important, since we can use the adiabatic approximation only in very rapid processes.

Isentropic Process and the First Law

For a closed system, we can write the first law of thermodynamics in terms of enthalpy:

dH = dQ + Vdp

or

dH = TdS + Vdp

Isentropic process (dQ = 0):

dH = Vdp     →     W = H₂ – H₁    

Isobaric Process

An isobaric process is a thermodynamic process, in which the pressure of the system remains constant (p = const). The heat transfer into or out of the system does work, but also changes the internal energy of the system.

Since there are changes in internal energy (dU) and changes in system volume (∆V), engineers often use the enthalpy of the system, which is defined as:

H = U + pV

Isobaric Process and the First Law

The classical form of the first law of thermodynamics is the following equation:

dU = dQ – dW

In this equation dW is equal to dW = pdV and is known as the boundary work. In an isobaric process and the ideal gas, part of heat added to the system will be used to do work and part of heat added will increase the internal energy (increase the temperature). Therefore it is convenient to use the enthalpy instead of the internal energy.

Isobaric process (Vdp = 0):

dH = dQ → Q = H₂– H₁

At constant entropy, i.e. in isentropic process, the enthalpy change equals the flow process work done on or by the system.

Rankine Cycle – pV, Ts diagram

The Rankine cycle is often plotted on a pressure volume diagram (pV diagram) and on a temperature-entropy diagram (Ts diagram).

When plotted on a pressure volume diagram, the isobaric processes follow the isobaric lines for the gas (the horizontal lines), adiabatic processes move between these horizontal lines and the area bounded by the complete cycle path represents the total work that can be done during one cycle.

The temperature-entropy diagram (Ts diagram) in which the thermodynamic state is specified by a point on a graph with specific entropy (s) as the horizontal axis and absolute temperature (T) as the vertical axis. Ts diagrams are a useful and common tool, particularly because it helps to visualize the heat transfer during a process. For reversible (ideal) processes, the area under the T-s curve of a process is the heat transferred to the system during that process.

Thermal Efficiency of Rankine Cycle

In general the thermal efficiency, η_th, of any heat engine is defined as the ratio of the work it does, W, to the heat input at the high temperature, Q_H.

The thermal efficiency, η_th, represents the fraction of heat, Q_H, that is converted to work. Since energy is conserved according to the first law of thermodynamics and energy cannot be be converted to work completely, the heat input, Q_H, must equal the work done, W, plus the heat that must be dissipated as waste heat Q_C into the environment. Therefore we can rewrite the formula for thermal efficiency as:

This is very useful formula, but here we express the thermal efficiency using the first law in terms of enthalpy.

Typically most of nuclear power plants operates multi-stage condensing steam turbines. In these turbines the high-pressure stage receives steam (this steam is nearly saturated steam – x = 0.995 – point C at the figure; 6 MPa; 275.6°C) from a steam generator and exhaust it to moisture separator-reheater (point D). The steam must be reheated in order to avoid damages that could be caused to blades of steam turbine by low quality steam. The reheater heats the steam (point D) and then the steam is directed to the low-pressure stage of steam turbine, where expands (point E to F). The exhausted steam then condenses in the condenser and it is at a pressure well below atmospheric (absolute pressure of 0.008 MPa), and is in a partially condensed state (point F), typically of a quality near 90%.

In this case, steam generators, steam turbine, condensers and feedwater pumps constitute a heat engine, that is subject to the efficiency limitations imposed by the second law of thermodynamics. In ideal case (no friction, reversible processes, perfect design), this heat engine would have a Carnot efficiency of

= 1 – T_cold/T_hot = 1 – 315/549 = 42.6%

where the temperature of the hot reservoir is 275.6°C (548.7K), the temperature of the cold reservoir is 41.5°C (314.7K). But the nuclear power plant is the real heat engine, in which thermodynamic processes are somehow irreversible. They are not done infinitely slowly. In real devices (such as turbines, pumps, and compressors) a mechanical friction and heat losses cause further efficiency losses.

To calculate the thermal efficiency of the simplest Rankine cycle (without reheating) engineers use the first law of thermodynamics in terms of enthalpy rather than in terms of internal energy.

The first law in terms of enthalpy is:

dH = dQ + Vdp

In this equation the term Vdp is a flow process work. This work,  Vdp, is used for open flow systems like a turbine or a pump in which there is a “dp”, i.e. change in pressure. There are no changes in control volume. As can be seen, this form of the law simplifies the description of energy transfer. At constant pressure, the enthalpy change equals the energy transferred from the environment through heating:

Isobaric process (Vdp = 0):

dH = dQ     →     Q = H₂ – H₁

At constant entropy, i.e. in isentropic process, the enthalpy change equals the flow process work done on or by the system:

Isentropic process (dQ = 0):

dH = Vdp     →     W = H₂ – H₁

It is obvious, it will be very useful in analysis of both thermodynamic cycles used in power engineering, i.e. in Brayton cycle and Rankine cycle.

The enthalpy can be made into an intensive, or specific, variable by dividing by the mass. Engineers use the specific enthalpy in thermodynamic analysis more than the enthalpy itself. It is tabulated in the steam tables along with specific volume and specific internal energy. The thermal efficiency of such simple Rankine cycle and in terms of specific enthalpies would be:

It is very simple equation and for determination of the thermal efficiency you can use data from steam tables.

In modern nuclear power plants the overall thermal efficiency is about one-third (33%), so 3000 MWth of thermal power from the fission reaction is needed to generate 1000 MWe of electrical power. The reason lies in relatively low steam temperature (6 MPa; 275.6°C). Higher efficiencies can be attained by increasing the temperature of the steam. But this requires an increase in pressures inside boilers or steam generators. However, metallurgical considerations place an upper limits on such pressures. In comparison to other energy sources the thermal efficiency of 33% is not much. But it must be noted that nuclear power plants are much more complex than fossil fuel power plants and it is much easier to burn fossil fuel ,than to generate energy from nuclear fuel. Sub-critical fossil fuel power plants, that are operated under critical pressure (i.e. lower than 22.1 MPa), can achieve 36–40% efficiency.

Causes of Inefficiency

As was discussed, an efficiency can range between 0 and 1. Each heat engine is somehow inefficient. This inefficiency can be attributed to three causes.

• Irreversibility of Processes. There is an overall theoretical upper limit to the efficiency of conversion of heat to work in any heat engine. This upper limit is called the Carnot efficiency. According to the Carnot principle, no engine can be more efficient than a reversible engine (a Carnot heat engine) operating between the same high temperature and low temperature reservoirs. For example, when the hot reservoir have T_hot of 400°C (673K) and T_cold of about 20°C (293K), the maximum (ideal) efficiency will be: = 1 – T_cold/T_hot = 1 – 293/673 = 56%. But all real thermodynamic processes are somehow irreversible. They are not done infinitely slowly. Therefore, heat engines must have lower efficiencies than limits on their efficiency due to the inherent irreversibility of the heat engine cycle they use.
• Presence of Friction and Heat Losses. In real thermodynamic systems or in real heat engines, a part of the overall cycle inefficiency is due to the losses by the individual components. In real devices (such as turbines, pumps, and compressors) a mechanical friction, heat losses and losses in the combustion process cause further efficiency losses.
• Design Inefficiency. Finally, last and also important source of inefficiencies is from the compromises made by engineers when designing a heat engine (e.g. power plant). They must consider cost and other factors in the design and operation of the cycle. As an example consider a design of the condenser in the thermal power plants.  Ideally the steam exhausted into the condenser would have no subcooling. But real condensers are designed to subcool the liquid by a few degrees of Celsius in order to avoid the suction cavitation in the condensate pumps. But, this subcooling increases the inefficiency of the cycle, because more energy is needed to reheat the water.

Thermal Efficiency Improvement – Rankine Cycle

There are several methods, how can be the thermal efficiency of the Rankine cycle improved. Assuming that the maximum temperature is limited by the pressure inside the reactor pressure vessel, these methods are:

Isentropic Efficiency – Turbine, Pump

In previous chapters we assumed that the steam expansion is isentropic and therefore we used T_4,is  as the outlet temperature of the gas. These assumptions are only applicable with ideal cycles.

Most steady-flow devices (turbines, compressors, nozzles) operate under adiabatic conditions, but they are not truly isentropic but are rather idealized as isentropic for calculation purposes. We define parameters η_T,  η_P, η_N, as a ratio of real work done by device to work by device when operated under isentropic conditions (in case of turbine). This ratio is known as the Isentropic Turbine/Pump/Nozzle Efficiency. These parameters describe how efficiently a turbine, compressor or nozzle approximates a corresponding isentropic device. This parameter reduces the overall efficiency and work output. For turbines, the value of η_T is typically 0.7 to 0.9 (70–90%).

See also: Isentropic Process

Rankine Cycle – Problem with Solution

Let assume the Rankine cycle, which is the one of most common thermodynamic cycles in thermal power plants. In this case assume a simple cycle without reheat  and without with condensing steam turbine running on saturated steam (dry steam). In this case the turbine operates at steady state with inlet conditions of  6 MPa, t = 275.6°C, x = 1 (point 3). Steam leaves this stage of turbine at a pressure of 0.008 MPa, 41.5°C and x = ??? (point 4).

Calculate:

1. the vapor quality of the outlet steam
2. the enthalpy difference between these two states (3 → 4), which corresponds to the work done by the steam, W_T.
3. the enthalpy difference between these two states (1 → 2), which corresponds to the work done by pumps, W_P.
4. the enthalpy difference between these two states (2 → 3), which corresponds to the net heat added in the steam generator
5. the thermodynamic efficiency of this cycle and compare this value with the Carnot’s efficiency

1)

Since we do not know the exact vapor quality of the outlet steam, we have to determine this parameter. State 4 is fixed by the pressure p₄ =  0.008 MPa and the fact that the specific entropy is constant for the isentropic expansion (s₃ = s₄ = 5.89 kJ/kgK for 6 MPa). The specific entropy of saturated liquid water (x=0) and dry steam (x=1) can be picked from steam tables. In case of wet steam, the actual entropy can be calculated with the vapor quality, x, and the specific entropies of saturated liquid water and dry steam:

s₄ = s_v x + (1 – x ) s_l              

where

s₄ = entropy of wet steam (J/kg K) = 5.89 kJ/kgK

s_v = entropy of “dry” steam (J/kg K) = 8.227 kJ/kgK (for 0.008 MPa)

s_l = entropy of saturated liquid water (J/kg K) = 0.592 kJ/kgK (for 0.008 MPa)

From this equation the vapor quality is:

x₄ =  (s₄ – s_l) / (s_v – s_l) = (5.89 – 0.592) / (8.227 – 0.592) = 0.694 = 69.4%

2)

The enthalpy for the state 3 can be picked directly from steam tables, whereas the enthalpy for the state 4 must be calculated using vapor quality:

h_{3, v} = 2785 kJ/kg

h_{4, wet} = h_4,v x + (1 – x ) h_4,l  = 2576 . 0.694 + (1 – 0.694) . 174 = 1787 + 53.2 = 1840 kJ/kg

Then the work done by the steam, W_T, is

W_T = Δh = 945 kJ/kg

3)

Enthalpy for state 1 can be picked directly from steam tables:

h_{1, l} = 174 kJ/kg

State 2 is fixed by the pressure p₂ =  6.0 MPa and the fact that the specific entropy is constant for the isentropic compression (s₁ = s₂ = 0.592 kJ/kgK for 0.008 MPa). For this entropy s₂ = 0.592 kJ/kgK and p₂ =  6.0 MPa we find h_{2, subcooled} in steam tables for compressed water (using interpolation between two states).

h_{2, subcooled} = 179.7 kJ/kg

Then the work done by the pumps, W_P, is

W_P = Δh = 5.7 kJ/kg

4)

The enthalpy difference between (2 → 3), which corresponds to the net heat added in the steam generator, is simply:

Q_add = h_{3, v}  – h_{2, subcooled} = 2785 – 179.7 =  2605.3 kJ/kg

Note that, there is no heat regeneration in this cycle. On the other hand most of the heat added is for the enthalpy of vaporization (i.e. for the phase change).

5)

In this case, steam generators, steam turbine, condensers and feedwater pumps constitute a heat engine, that is subject to the efficiency limitations imposed by the second law of thermodynamics. In the ideal case (no friction, reversible processes, perfect design), this heat engine would have a Carnot efficiency of

η_Carnot = 1 – T_cold/T_hot = 1 – 315/549 = 42.6%

where the temperature of the hot reservoir is 275.6°C (548.7 K), the temperature of the cold reservoir is 41.5°C (314.7K).

The thermodynamic efficiency of this cycle can be calculated by the following formula:

thus
η_th = (945 – 5.7) / 2605.3 = 0.361 = 36.1%

See also:

Thermodynamic Cycles

Theory of Rankine Cycle

The Rankine cycle was named after him and describes the performance of steam turbine systems, though the theoretical principle also applies to reciprocating engines such as steam locomotives. In general, the Rankine cycle is an idealized thermodynamic cycle of a constant pressure heat engine that converts part of heat into mechanical work. In this cycle the heat is supplied externally to a closed loop, which usually uses water (in a liquid and vapor phase) as the working fluid. In contrast to the Brayton cycle, the working fluid in the Rankine cycle undergo the phase change from a liquid to vapor phase and vice versa.

While many substances could be used as the working fluid in the Rankine cycle (inorganic or even organic), water is usually the fluid of choice due to its favorable properties, such as its non-toxic and unreactive chemistry, abundance, and low cost, as well as its thermodynamic properties. For example, water has the highest specific heat of any common substance –  4.19 kJ/kg K. Moreover it has very high heat of vaporization, which makes it an effective coolant and medium in thermal power plants and other energy industry. In case of the Rankine cycle, the Ideal Gas Law almost cannot be used (steam do not follow pV=nRT), therefore all important parameters of water and steam are tabulated in so called “Steam Tables“.

One of the major advantages of the Rankine cycle is that the compression process in the pump takes place on a liquid. By condensing the working steam to a liquid (inside a condenser) the pressure at the turbine outlet is lowered and the energy required by the feed pump consumes only 1% to 3% of the turbine output power and these factors contribute to a higher efficiency for the cycle.

In an ideal Rankine cycle, the system executing the cycle undergoes a series of four processes: two isentropic (reversible adiabatic) processes alternated with two isobaric processes:

• 

  Isentropic compression (compression in centrifugal pumps) – The liquid condensate is compressed adiabatically from state 1 to state 2 by centrifugal pumps (usually by condensate pumps and then by feedwater pumps). The liquid condensatei s pumped from the condenser into the higher pressure boiler. In this process, the surroundings do work on the fluid, increasing its enthalpy (h = u+pv) and compressing it (increasing its pressure). On the other hand the entropy remains unchanged. The work required for the compressor is given by W_Pumps = H₂ – H₁.

• Isobaric heat addition (in a heat exchanger – boiler) – In this phase (between state 2 and state 3) there is a constant-pressure heat transfer to the liquid condensate from an external source, since the chamber is open to flow in and out.  The feedwater (secondary circuit) is heated from to the boiling point (2 → 3a) of that fluid and then evaporated in the boiler (3a → 3). The net heat added is given by Q_add = H₃ – H₂
• Isentropic expansion (expansion in a steam turbine) – Steam from the boiler expands adiabatically from state 3 to state 4 in a steam turbine to produce work and then is discharged to the condenser (partially condensed). The steam does work on the surroundings (blades of the turbine) and loses an amount of enthalpy equal to the work that leaves the system. The work done by turbine is given by W_T = H₄ – H₃. Again the entropy remains unchanged.
• Isobaric heat rejection (in a heat exchanger) – In this phase the cycle completes by a constant-pressure process in which heat is rejected from the partially condensed steam. There is heat transfer from the vapor to cooling water flowing in a cooling circuit. The vapor condenses and the temperature of the cooling water increases. The net heat rejected is given by Q_re = H₄ – H₁

During a Rankine cycle, work is done on the fluid by the pumps between states 1 and 2 (isentropic compression). Work is done by the fluid in the turbine between stages 3 and 4 (isentropic expansion). The difference between the work done by the fluid and the work done on the fluid is the net work produced by the cycle and it corresponds to the area enclosed by the cycle curve (in pV diagram). The working fluid in a Rankine cycle follows a closed loop and is reused constantly.

As can be seen, it is convenient to use enthalpy and the first law in terms of enthalpy in analysis of this thermodynamic cycle. This form of the law simplifies the description of energy transfer. At constant pressure, the enthalpy change equals the energy transferred from the environment through heating:

Isobaric process (Vdp = 0):

dH = dQ     →     Q = H₂ – H₁

At constant entropy, i.e. in isentropic process, the enthalpy change equals the flow process work done on or by the system:

Isentropic process (dQ = 0):

dH = Vdp     →     W = H₂ – H₁

See also: Why power engineers use enthalpy? Answer: dH = dQ + Vdp

Isentropic Process

An isentropic process is a thermodynamic process, in which the entropy of the fluid or gas remains constant. It means the isentropic process is a special case of an adiabatic process in which there is no transfer of heat or matter. It is a reversible adiabatic process. The assumption of no heat transfer is very important, since we can use the adiabatic approximation only in very rapid processes.

Isentropic Process and the First Law

For a closed system, we can write the first law of thermodynamics in terms of enthalpy:

dH = dQ + Vdp

or

dH = TdS + Vdp

Isentropic process (dQ = 0):

dH = Vdp     →     W = H₂ – H₁    

Isobaric Process

An isobaric process is a thermodynamic process, in which the pressure of the system remains constant (p = const). The heat transfer into or out of the system does work, but also changes the internal energy of the system.

Since there are changes in internal energy (dU) and changes in system volume (∆V), engineers often use the enthalpy of the system, which is defined as:

H = U + pV

Isobaric Process and the First Law

The classical form of the first law of thermodynamics is the following equation:

dU = dQ – dW

In this equation dW is equal to dW = pdV and is known as the boundary work. In an isobaric process and the ideal gas, part of heat added to the system will be used to do work and part of heat added will increase the internal energy (increase the temperature). Therefore it is convenient to use the enthalpy instead of the internal energy.

Isobaric process (Vdp = 0):

dH = dQ → Q = H₂– H₁

At constant entropy, i.e. in isentropic process, the enthalpy change equals the flow process work done on or by the system.

Thermal Efficiency of Rankine Cycle

In general the thermal efficiency, η_th, of any heat engine is defined as the ratio of the work it does, W, to the heat input at the high temperature, Q_H.

The thermal efficiency, η_th, represents the fraction of heat, Q_H, that is converted to work. Since energy is conserved according to the first law of thermodynamics and energy cannot be be converted to work completely, the heat input, Q_H, must equal the work done, W, plus the heat that must be dissipated as waste heat Q_C into the environment. Therefore we can rewrite the formula for thermal efficiency as:

This is very useful formula, but here we express the thermal efficiency using the first law in terms of enthalpy.

Typically most of nuclear power plants operates multi-stage condensing steam turbines. In these turbines the high-pressure stage receives steam (this steam is nearly saturated steam – x = 0.995 – point C at the figure; 6 MPa; 275.6°C) from a steam generator and exhaust it to moisture separator-reheater (point D). The steam must be reheated in order to avoid damages that could be caused to blades of steam turbine by low quality steam. The reheater heats the steam (point D) and then the steam is directed to the low-pressure stage of steam turbine, where expands (point E to F). The exhausted steam then condenses in the condenser and it is at a pressure well below atmospheric (absolute pressure of 0.008 MPa), and is in a partially condensed state (point F), typically of a quality near 90%.

In this case, steam generators, steam turbine, condensers and feedwater pumps constitute a heat engine, that is subject to the efficiency limitations imposed by the second law of thermodynamics. In ideal case (no friction, reversible processes, perfect design), this heat engine would have a Carnot efficiency of

= 1 – T_cold/T_hot = 1 – 315/549 = 42.6%

where the temperature of the hot reservoir is 275.6°C (548.7K), the temperature of the cold reservoir is 41.5°C (314.7K). But the nuclear power plant is the real heat engine, in which thermodynamic processes are somehow irreversible. They are not done infinitely slowly. In real devices (such as turbines, pumps, and compressors) a mechanical friction and heat losses cause further efficiency losses.

To calculate the thermal efficiency of the simplest Rankine cycle (without reheating) engineers use the first law of thermodynamics in terms of enthalpy rather than in terms of internal energy.

The first law in terms of enthalpy is:

dH = dQ + Vdp

In this equation the term Vdp is a flow process work. This work,  Vdp, is used for open flow systems like a turbine or a pump in which there is a “dp”, i.e. change in pressure. There are no changes in control volume. As can be seen, this form of the law simplifies the description of energy transfer. At constant pressure, the enthalpy change equals the energy transferred from the environment through heating:

Isobaric process (Vdp = 0):

dH = dQ     →     Q = H₂ – H₁

At constant entropy, i.e. in isentropic process, the enthalpy change equals the flow process work done on or by the system:

Isentropic process (dQ = 0):

dH = Vdp     →     W = H₂ – H₁

It is obvious, it will be very useful in analysis of both thermodynamic cycles used in power engineering, i.e. in Brayton cycle and Rankine cycle.

The enthalpy can be made into an intensive, or specific, variable by dividing by the mass. Engineers use the specific enthalpy in thermodynamic analysis more than the enthalpy itself. It is tabulated in the steam tables along with specific volume and specific internal energy. The thermal efficiency of such simple Rankine cycle and in terms of specific enthalpies would be:

It is very simple equation and for determination of the thermal efficiency you can use data from steam tables.

In modern nuclear power plants the overall thermal efficiency is about one-third (33%), so 3000 MWth of thermal power from the fission reaction is needed to generate 1000 MWe of electrical power. The reason lies in relatively low steam temperature (6 MPa; 275.6°C). Higher efficiencies can be attained by increasing the temperature of the steam. But this requires an increase in pressures inside boilers or steam generators. However, metallurgical considerations place an upper limits on such pressures. In comparison to other energy sources the thermal efficiency of 33% is not much. But it must be noted that nuclear power plants are much more complex than fossil fuel power plants and it is much easier to burn fossil fuel ,than to generate energy from nuclear fuel. Sub-critical fossil fuel power plants, that are operated under critical pressure (i.e. lower than 22.1 MPa), can achieve 36–40% efficiency.

Rankine Cycle – Equations and Calculation

Let assume the Rankine cycle, which is the one of most common thermodynamic cycles in thermal power plants. In this case assume a simple cycle without reheat  and without with condensing steam turbine running on saturated steam (dry steam). In this case the turbine operates at steady state with inlet conditions of  6 MPa, t = 275.6°C, x = 1 (point 3). Steam leaves this stage of turbine at a pressure of 0.008 MPa, 41.5°C and x = ??? (point 4).

Calculate:

1. the vapor quality of the outlet steam
2. the enthalpy difference between these two states (3 → 4), which corresponds to the work done by the steam, W_T.
3. the enthalpy difference between these two states (1 → 2), which corresponds to the work done by pumps, W_P.
4. the enthalpy difference between these two states (2 → 3), which corresponds to the net heat added in the steam generator
5. the thermodynamic efficiency of this cycle and compare this value with the Carnot’s efficiency

1)

Since we do not know the exact vapor quality of the outlet steam, we have to determine this parameter. State 4 is fixed by the pressure p₄ =  0.008 MPa and the fact that the specific entropy is constant for the isentropic expansion (s₃ = s₄ = 5.89 kJ/kgK for 6 MPa). The specific entropy of saturated liquid water (x=0) and dry steam (x=1) can be picked from steam tables. In case of wet steam, the actual entropy can be calculated with the vapor quality, x, and the specific entropies of saturated liquid water and dry steam:

s₄ = s_v x + (1 – x ) s_l              

where

s₄ = entropy of wet steam (J/kg K) = 5.89 kJ/kgK

s_v = entropy of “dry” steam (J/kg K) = 8.227 kJ/kgK (for 0.008 MPa)

s_l = entropy of saturated liquid water (J/kg K) = 0.592 kJ/kgK (for 0.008 MPa)

From this equation the vapor quality is:

x₄ =  (s₄ – s_l) / (s_v – s_l) = (5.89 – 0.592) / (8.227 – 0.592) = 0.694 = 69.4%

2)

The enthalpy for the state 3 can be picked directly from steam tables, whereas the enthalpy for the state 4 must be calculated using vapor quality:

h_{3, v} = 2785 kJ/kg

h_{4, wet} = h_4,v x + (1 – x ) h_4,l  = 2576 . 0.694 + (1 – 0.694) . 174 = 1787 + 53.2 = 1840 kJ/kg

Then the work done by the steam, W_T, is

W_T = Δh = 945 kJ/kg

3)

Enthalpy for state 1 can be picked directly from steam tables:

h_{1, l} = 174 kJ/kg

State 2 is fixed by the pressure p₂ =  6.0 MPa and the fact that the specific entropy is constant for the isentropic compression (s₁ = s₂ = 0.592 kJ/kgK for 0.008 MPa). For this entropy s₂ = 0.592 kJ/kgK and p₂ =  6.0 MPa we find h_{2, subcooled} in steam tables for compressed water (using interpolation between two states).

h_{2, subcooled} = 179.7 kJ/kg

Then the work done by the pumps, W_P, is

W_P = Δh = 5.7 kJ/kg

4)

The enthalpy difference between (2 → 3), which corresponds to the net heat added in the steam generator, is simply:

Q_add = h_{3, v}  – h_{2, subcooled} = 2785 – 179.7 =  2605.3 kJ/kg

Note that, there is no heat regeneration in this cycle. On the other hand most of the heat added is for the enthalpy of vaporization (i.e. for the phase change).

5)

In this case, steam generators, steam turbine, condensers and feedwater pumps constitute a heat engine, that is subject to the efficiency limitations imposed by the second law of thermodynamics. In the ideal case (no friction, reversible processes, perfect design), this heat engine would have a Carnot efficiency of

η_Carnot = 1 – T_cold/T_hot = 1 – 315/549 = 42.6%

where the temperature of the hot reservoir is 275.6°C (548.7 K), the temperature of the cold reservoir is 41.5°C (314.7K).

The thermodynamic efficiency of this cycle can be calculated by the following formula:

thus
η_th = (945 – 5.7) / 2605.3 = 0.361 = 36.1%

See also:

Rankine Cycle

Rankine Cycle – Processes

In an ideal Rankine cycle, the system executing the cycle undergoes a series of four processes: two isentropic (reversible adiabatic) processes alternated with two isobaric processes.

Rankine Cycle – Processes

• 

  Isentropic compression (compression in centrifugal pumps) – The liquid condensate is compressed adiabatically from state 1 to state 2 by centrifugal pumps (usually by condensate pumps and then by feedwater pumps). The liquid condensatei s pumped from the condenser into the higher pressure boiler. In this process, the surroundings do work on the fluid, increasing its enthalpy (h = u+pv) and compressing it (increasing its pressure). On the other hand the entropy remains unchanged. The work required for the compressor is given by W_Pumps = H₂ – H₁.

• Isobaric heat addition (in a heat exchanger – boiler) – In this phase (between state 2 and state 3) there is a constant-pressure heat transfer to the liquid condensate from an external source, since the chamber is open to flow in and out.  The feedwater (secondary circuit) is heated from to the boiling point (2 → 3a) of that fluid and then evaporated in the boiler (3a → 3). The net heat added is given by Q_add = H₃ – H₂
• Isentropic expansion (expansion in a steam turbine) – Steam from the boiler expands adiabatically from state 3 to state 4 in a steam turbine to produce work and then is discharged to the condenser (partially condensed). The steam does work on the surroundings (blades of the turbine) and loses an amount of enthalpy equal to the work that leaves the system. The work done by turbine is given by W_T = H₄ – H₃. Again the entropy remains unchanged.
• Isobaric heat rejection (in a heat exchanger) – In this phase the cycle completes by a constant-pressure process in which heat is rejected from the partially condensed steam. There is heat transfer from the vapor to cooling water flowing in a cooling circuit. The vapor condenses and the temperature of the cooling water increases. The net heat rejected is given by Q_re = H₄ – H₁

During a Rankine cycle, work is done on the fluid by the pumps between states 1 and 2 (isentropic compression). Work is done by the fluid in the turbine between stages 3 and 4 (isentropic expansion). The difference between the work done by the fluid and the work done on the fluid is the net work produced by the cycle and it corresponds to the area enclosed by the cycle curve (in pV diagram). The working fluid in a Rankine cycle follows a closed loop and is reused constantly.

As can be seen, it is convenient to use enthalpy and the first law in terms of enthalpy in analysis of this thermodynamic cycle. This form of the law simplifies the description of energy transfer. At constant pressure, the enthalpy change equals the energy transferred from the environment through heating:

Isobaric process (Vdp = 0):

dH = dQ     →     Q = H₂ – H₁

At constant entropy, i.e. in isentropic process, the enthalpy change equals the flow process work done on or by the system:

Isentropic process (dQ = 0):

dH = Vdp     →     W = H₂ – H₁

See also: Why power engineers use enthalpy? Answer: dH = dQ + Vdp

Isentropic Process

An isentropic process is a thermodynamic process, in which the entropy of the fluid or gas remains constant. It means the isentropic process is a special case of an adiabatic process in which there is no transfer of heat or matter. It is a reversible adiabatic process. The assumption of no heat transfer is very important, since we can use the adiabatic approximation only in very rapid processes.

Isentropic Process and the First Law

For a closed system, we can write the first law of thermodynamics in terms of enthalpy:

dH = dQ + Vdp

or

dH = TdS + Vdp

Isentropic process (dQ = 0):

dH = Vdp     →     W = H₂ – H₁    

Isobaric Process

An isobaric process is a thermodynamic process, in which the pressure of the system remains constant (p = const). The heat transfer into or out of the system does work, but also changes the internal energy of the system.

Since there are changes in internal energy (dU) and changes in system volume (∆V), engineers often use the enthalpy of the system, which is defined as:

H = U + pV

Isobaric Process and the First Law

The classical form of the first law of thermodynamics is the following equation:

dU = dQ – dW

In this equation dW is equal to dW = pdV and is known as the boundary work. In an isobaric process and the ideal gas, part of heat added to the system will be used to do work and part of heat added will increase the internal energy (increase the temperature). Therefore it is convenient to use the enthalpy instead of the internal energy.

Isobaric process (Vdp = 0):

dH = dQ → Q = H₂– H₁

At constant entropy, i.e. in isentropic process, the enthalpy change equals the flow process work done on or by the system.

Rankine Cycle – pV, Ts diagram

The Rankine cycle is often plotted on a pressure volume diagram (pV diagram) and on a temperature-entropy diagram (Ts diagram).

When plotted on a pressure volume diagram, the isobaric processes follow the isobaric lines for the gas (the horizontal lines), adiabatic processes move between these horizontal lines and the area bounded by the complete cycle path represents the total work that can be done during one cycle.

The temperature-entropy diagram (Ts diagram) in which the thermodynamic state is specified by a point on a graph with specific entropy (s) as the horizontal axis and absolute temperature (T) as the vertical axis. Ts diagrams are a useful and common tool, particularly because it helps to visualize the heat transfer during a process. For reversible (ideal) processes, the area under the T-s curve of a process is the heat transferred to the system during that process.

See also:

Rankine Cycle