THERMODYNAMIC POWER CYCLES - I

N S Senanayake

ME4001Applied Thermodynamics

Content

Introduction Power cycles Efficiency definition Introduction to Carnot Cycle and Carnot Efficiency Vapour power cycles Mollier charts (h – s) Carnot vapour power cycle Simple Rankine Power Cycle Rankine Cycle with super heat steam Comparison of efficiency for Carnot, Simple

Rankine cycle and Rankine cycle with Super heat steam

Introduction

In any thermal power generation plant, heat energy converts into mechanical work. Then it is converted to electrical energy by rotating a generator which produces electrical energy. Heat is derived from various energy sources like fossil fuels, nuclear fusion of radioactive elements and geothermal energy.

Potential energy stored in water also can produce mechanical work, and then mechanical work to electrical energy, this conversion of energy is commonly known as hydropower generation. Kinetic energy of wind and solar radiation are the other sources of energy that can be used to produce electrical energy by conversion.

Power cycles

• Carnot vapour power cycle• Rankine Cycle

vapour power cycle

• Carnot gas power cycle• Otto cycle• Diesel cycle• Dual cycle• Brayton cycle

Gas power cycle

Depending on the types of processes involved power cycles can also be classified as follows:

Efficiency

The overall efficiency of a thermal power plant is the combination of two efficiencies.

Combustion efficiency or the source efficiencyThe proportion of the available energy transferred as heat to the working fluid.

Thermodynamic Cycle efficiency

This indicates the proportion of the heat supplied to system which is finally converted to useful mechanical work. As the efficiency can never be 100% according the Second Law of thermodynamics, our concern is to raise the efficiency as much as possible. The efficiency depends on the nature of thermodynamic processes that constitute a cycle.

Carnot Cycle

Carnot Cycle is a reversible thermodynamic cycle comprising of 4 processes.

1 -2 Reversible Isothermal heat addition

2 – 3 Reversible adiabatic expansion (isentropic)

3 – 4 Reversible Isothermal heat rejection

4 -1 Reversible Adiabatic compression (isentropic)

The temperature –entropy diagram (T-S diagram) is the best form of cycle that gives the maximum efficiency

Carnot efficiency

Corollary 1 of Carnot's Theorem:All reversible heat engines operating between the same two heat reservoirs must have the same efficiency.

Corollary 2 of Carnot's Theorem:The efficiency of a reversible heat engine is a function only of the respective temperatures of the hot and cold reservoirs. It can be evaluated by replacing the ratio of heat transfers QL and QH by the ratio of temperatures TL and TH of the respective heat

reservoirs.

Carnot efficiency cont..

H

L

H

LH

H

Rth Q

Q

Q

QQ

Q

W

1

Thermal efficiency for a reversible heat engine can be written as follows.

H

L

H

Rth T

T

Q

W 1

H

L

H

L

T

T

Q

QAccording to Carnot

theorem

Vapor power cycles

Introduction to vapor power cycles

In converting heat to work it is necessary to have a substance called working fluid to “store” energy and release it in the form of mechanical work. Such a fluid should not upset the system and its components that are needed to bring about this heat and work transfer. You are aware of that it is the water (or steam) we mostly use as the working substance in thermal power plant. The thermodynamic processes undergone by the working substance with the efficiency will be discussed in detail in the next session.

Although the Carnot Cycle is applicable to vapor power cycles, because of the constant temperature heat absorption and rejection (latent heat), there are practical difficulties to exactly match the Carnot Cycle.

Enthalpy changes

In all thermodynamic power cycles we are concerned with changes in enthalpy rather than the enthalpy as a property. Therefore it is convenient to use enthalpy-entropy diagram, commonly known as Mollier diagram of steam (h-s chart) in our calculations to determine heat and work transfer, rather than using tables of thermodynamic properties.

The enthalpy-entropy diagram

Carnot vapour power cycle

Carnot cycle is an ideal cycle having the highest thermodynamics efficiency

TH= T1= T2

TL = T3= T4

Carnot vapour power cycle…

Processes : all processes are reversible

1 – 2 : isothermal heat addition in the boiler

2 – 3 : adiabatic expansion in the turbine

3 – 4 : isothermal heat rejection in the condenser

4 – 1 : adiabatic compression in the feed pump

q12 = h2 – h1

- w23 = h3 – h2

q34 = h4 – h3

- w41 = h1 – h4

Carnot vapour power cycle…

addedHeat

Net Workth

12

1432

hh

hhhhth

4123 wwNet work 3412 qqheatNet

H

Lth T

T1

q

q1

q

qq

q

ww

12

34

12

3412

12

4123

Thermal efficiency can also be expressed as:

Thermal efficiency

Net work done

T

qs sTq

12BA Area,)( 1212 ssTq H

43BA Area,)( 2134 ssTq L

3412 qq heat Net

43BA Area,)( 1234 ssTq L

1234 AreaTT heat Net 12LH ss

For cycle net heat = net work

Net work

Example 1

Steam power plant operates between a boiler pressure of 4MPa and condenser pressure of 50kPa. The steam at 300oC enters the turbine. Assuming the cycle to be Carnot cycle, Determine the thermal efficiency.

Also find the turbine work, feed pump work and heat added.

solution

Thermal efficiency

H

Lth T

T1

From steam tables, the saturation temperature at 50kPa is 81.32oC.

38.0273300

27332.811

th

Turbine work

- w23 = h3 – h2

h2 = 2961.7kJ/kg

State 2 is in the superheated region as T2 is greater than sat. temperature at 4MPa, which is 250.35oC

h3 = hf3 + x3hfg3

hf3 = 340.54kJ/kg

hfg3 = 2304.7kJ/kg

x3 = ?

s3 = sf3 + x3sfg3

sf3 =1.0912kJ/kg

sfg3 = 6.5019kJ/kg

s3 = s2 = 6.3639kJ/kg

From steam tables at 50kPa

To find out dryness at state 3, consider the entropy

From steam tables at 50kPa

6.3639 = 1.0912 + x3 (6.5019)

x3 = 0.81

h3 = hf3 + x3hfg3

h3 = 340.54 + 0.81(2304.7) = 2207.35kJ/kg

Turbine work, w23 = (h2 – h3) = 2961.7 – 2207.35 =754.35kJ/kg

Heat added

q12 = h2 – h1

h1 =1344.8kJ/kg

h2 = 2961.7kJ/kg

For saturated liquid at 300oC

q12 = 2961.7 – 1344.8 = 1616.9kJ/kg

Feed pump work - w41 = h1 – h4

h1 =1344.8kJ/kg

h4 = hf4 + x4hfg4

hf4 = 340.54kJ/kg

hfg4 = 2304.7kJ/kg x4 = ?

From steam tables at 50kPa

To find out dryness at state 4, consider the entropy

sf4 =1.0912kJ/kg

sfg4 = 6.5019kJ/kg

s4 = s1 = 3.2548kJ/kg

From steam tables at 50kPa

3.2548 = 1.0912 + x4 (6.5019)

x4 = 0.33

h4 = hf4 + x4hfg4

h4 = 340.54 + 0.33 (2304.7) = 1101.09kJ/kg

- w41 = h1 – h4 = 1344.8 – 1101.09 = 243.71kJ/kg

Net work output = w23 + w41 = 754.35kJ/kg - 243.71kJ/kg = 510.65kJ/kg

31.01616.9

65.510

addedheat

net workth

Net work RatioWork

Rankine cycle

Rankine cycle is ideal cycle for vapour power cycles

The Rankine Cycle is similar to the Carnot Cycle, but the Rankine Cycle is much more practical because the working fluid typically exists as a single phase (liquid or vapor) for the two pressure change processes.

Processes in simple Rankine cycle

5 to 1: water is heated until it reaches saturation (phase change / boiling point) in a constant-pressure process.

1 to 2: Once saturation is reached, further heat transfer takes place at constant pressure, until the working fluid is completely vaporized.

2 to 3: The vapour is expanded isentropically through a turbine to produce work. The vapour (steam) pressure falls as it passes through the turbine and exits at low pressure.

3 to 4: The working fluid is routed through a condenser, where it condenses (phase change) into liquid (water).

4 to 5: The working fluid is pumped back into the boiler.

Rankine cycle on h- s

Rankine cycle with superheat steam

To ensure that steam leaving the turbine is sufficiently dry, in practical steam power plant superheat steam is used instead of saturated steam to the turbine input. However, we need to ensure that wet steam enters the condenser to get the maximum turbine output.

Advantages of super heating

• Average temperature at which heat is supplied increased by superheating the steam and therefore ideal cycle efficiency is increased.

• The specific steam consumption is markedly reduced as the net work per unit mass of steam is high.

• Since the requirement of steam per unit of energy (kg/kWh) is less, the complexity added by introducing the superheating system is compensated by a smaller boiler.

There is an additional reason for using superheat steam in practice. If you closely look at the T-S diagram below, higher boiler pressures tends to give wetter steam at turbine exhaust, which is not desirable. Therefore superheating becomes necessary to get dried steam at the turbine exhaust.

Effect of super heating

Example 2

Steam turbine plant operates on Rankine Cycle with steam entering the turbine at 40bar, 350oC and leaving at 0.05bar. Steam leaving the turbine condenses to saturated liquid inside condenser. Feed pump pumps saturated liquid into the boiler.

Determine the net work per kg of steam and the cycle efficiency assuming all processes to be ideal.

Solution

From steam tables

h4 = hf4 = 137.75kJ/kg

At 0.05bar (5kPa) and saturated liquids4 = sf4 = 0.4762kJ/kg

v4 = vf4 = 0.001005m3/kg

h2 = 3093.3kJ/kgAt 40bar (4MPa) and 350oC

s2 = 6.5843kJ/kg

To find h3

s3 = sf3 + x3sfg3sf3 = 0.4762kJ/kg

sfg3 = 7.9176kJ/kg at 0.05 bar

s3 = s2

6.5843 = 0.4762 + x3 (7.9176)

x3 = 0.77

h3 = hf3 + x3hfg3

hf3 = 137.75kJ/kg

hfg3 = 2423.0kJ/kg at 0.05 bar

h3 = 137.75 + 0.77(2423.0) =2003.46kJ/kg

To find h1

Consider the pump workw14 = h1 – h4 = vf4 (p2 – p1)

h1 = vf4 (p2 – p1) + h4 = 0.001005(4000 – 0.005)+137.75=141.78kJ/kg

Turbine work (w23) = h2 – h3 =(3093.3 - 2003.46) = 1089.84kJ/kg

Feed pump work (w41) = h1 – h4 =(141.78 – 137.75) = 4.03kJ/kg

Net work = w23 – w41

Net work = 1089.84 – 4.03 = 1085.81kJ/kg

3678.02951.52

1085.81

addedheat

Net work Efficiency

Heat added= h2 – h 1 =3093.3 - 141.78 =2951.52kJ/kg

Exercise

Calculate the ideal efficiency and specific steam consumption for the following cases with 30bar and 0.04bar turbine and condenser pressure respectively

Carnot cycle Simple Rankine cycle Rankine Cycle with super heat steam

at450oC

Comparison of different cycles

Performance indicator Cycle Name

Carnot Simple Rankine

Rankine with super

heatIdeal efficiency, h 0.404 0.35 0.375

Specific steam consumption (kg/kWh)

4.97 3.84 2.98

Isentropic efficiency - Turbine

The desired output from a turbine is the work output. Hence, the definition of isentropic efficiency of a turbine is the ratio of the actual work output of the turbine to the work output of the turbine if the turbine undergoes an isentropic process between the same inlet and exit pressures.     

WorkIsentropic

WorkTurbineActualT

The isentropic efficiency of turbine can be written as

   

12

12

hh

hh

s

aT

h1 = enthalpy at the inlet

h2a = enthalpy of actual process at the exit

h2s = enthalpy of isentropic process at the exit 

12

12

TT

TT

s

aT

Isentropic efficiency - compressor

The isentropic efficiency of a compressor or pump

is defined as the ratio of the work input to

an isentropic process, to the work input to the

actual process between the same inlet and exit

pressures.

WorkActual

WorkCompressorIsentropicC

The isentropic efficiency of compressor can be written as

   12

12

hh

hh

a

sC

h1 = enthalpy at the inlet

h2a = enthalpy of actual process at the exit

h2s = enthalpy of isentropic process at the exit 

12

12 )(

hh

ppv

aC

Example

If the isentropic efficiency for turbine is 0.8 redo the example 1

3a

solution

From steam tables h2 = 3093.3kJ/kgAt 40bar (4MPa) and 350oC

s2 = 6.5843kJ/kg

h4 = hf4 = 137.75kJ/kg

At 0.05bar (5kPa) and saturated liquids4 = sf4 = 0.4762kJ/kg

v4 = vf4 = 0.001005m3/kg

To find h3

s3 = sf3 + x3sfg3sf3 = 0.4762kJ/kg

sfg3 = 7.9176kJ/kg at 0.05 bar

s3 = s2

6.5843 = 0.4762 + x3 (7.9176)

x3 = 0.77

h3 = hf3 + x3hfg3 hf3 = 137.75kJ/kg

hfg3 = 2423.0kJ/kg at 0.05 bar

h3 = 137.75 + 0.77(2423.0) =2003.46kJ/kg

32

32

hh

hh aT

84.089,1

3.3093

46.20033.3093

3.30938.0 33 aa hh

43.221,23 ah

To find h1Consider the pump work

h1 = vf4 (p2 – p1) + h4 = 0.001005(4000 – 0.005)+137.75=141.78kJ/kg

Turbine work (w23) = h2 – h3a =(3093.3 – 2221.43) = 871.87kJ/kg

Feed pump work (w41) = h1 – h4 =(141.78 – 137.75) = 4.03kJ/kg

Net work = w23 – w41

Net work = 871.87 – 4.03 = 730.09kJ/kg

2473.02951.52

730.09

addedheat

Net work Efficiency

Heat added= h2 – h 1 =3093.3 - 141.78 =2951.52kJ/kg