## Theory of Steam Turbines – Thermodynamics

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.

Today, the **Rankine cycle** is the fundamental operating cycle of **all thermal power plants** where an operating fluid is continuously evaporated and condensed. It is the one of most common **thermodynamic cycles, **because in most of the places in the world the turbine is steam-driven.

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}**= H**_{2}**– H**_{1}**.****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}**= H**_{3 }**– H**_{2}**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}**= H**_{4}**– H**_{3}**.**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}**= H**_{4 }**– H**_{1}

During a Rankine cycle, work is done on the fluid by the pumps between states 1 and 2 (**i****sentropic compression**). Work is done by the fluid in the turbine between stages 3 and 4 (**i****sentropic 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 = H**_{2}** – H**_{1}

**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 = H**_{2}** – H**_{1}

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 = H**_{2}** – H**_{1}** **

## 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 = H**_{2}**– H**_{1}

**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**

**, that is converted**

_{H}**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**

**into the environment. Therefore we can rewrite the formula for thermal efficiency as:**

_{C}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,

**, is used for**

*Vdp***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 = H**_{2}** – H**_{1}

**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 = H**_{2}** – H**_{1}

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***is typically 0.7 to 0.9 (70–90%).**

_{T}See also: Isentropic Process

## Steam Turbine – 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:

- the vapor quality of the outlet steam
- the enthalpy difference between these two states (3 → 4), which corresponds to the work done by the steam, W
_{T}. - the enthalpy difference between these two states (1 → 2), which corresponds to the work done by pumps, W
_{P}. - the enthalpy difference between these two states (2 → 3), which corresponds to the net heat added in the steam generator
- 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 **p _{4} = 0.008 MPa** and the fact that the specific entropy is constant for the isentropic expansion (s

_{3}= s

_{4}= 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:

*s*_{4}* = s*_{v}* x + (1 – x ) s*_{l}* *

*where*

*s*_{4}* = 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:

x_{4} = (*s*_{4}* – s** _{l}*) / (

*s*

_{v}*– s*

*) = (5.89 – 0.592) / (8.227 – 0.592) = 0.694 = 69.4%*

_{l}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 p_{2} = 6.0 MPa and the fact that the specific entropy is constant for the isentropic compression (s_{1} = s_{2} = 0.592 *kJ/kgK for 0.008 MPa*). For this entropy s_{2} = **0.592 kJ/kgK** and p

_{2}= 6.0 MPa we find

*h***in steam tables for compressed water (using interpolation between two states).**

_{2, subcooled}*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%**

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