## Thermal Efficiency of Steam Turbine

**thermal efficiency**,

*η***, of any heat engine is defined as the ratio of the work it does,**

_{th}**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%.

**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**.

**Thermal Efficiency of Steam Turbine**

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 – Steam Turbine

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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