thermodynamics vpm psm 2.doc

INTRODUCTION

A Power Plant primarily exists for the important purpose of electricity (electrical power) generation and operates on EITHER the Vapour Power Cycles OR the Gas Power Cycles.

They are concerned with the transfer of heat and work, in many cases with the ultimate objective of converting one form of energy into another as efficiently as practicably possible in accordance to the Principle of Conservation of Energy.

In every case they make use of the way a fluid behaves as it is compressed or expanded, heated or cooled.

The various Vapour Power Cycles are Carnot Cycle, Rankine Cycle, Reheat Cycle and Regeneration Cycle.

The Vapour used is none other but steam.

As for Gas Power Cycles, there are the Gas Turbine Cycle and the Reciprocating Engine Cycles (such as the Otto Cycle, Diesel Cycle and Mixed Cycle).

The gas used is normally air (known as the Joule or Brayton Cycle) or the products of combustion of fuel and air.

With the above, a Power Plant has little choice but involve a tremendous amount of Thermodynamics.

The major equipments in Palm Oil Mill’s Power Plant are Boilers, Turbines, Condensers and Diesel Generating Sets where the details of construction and operating and maintenance instructions of each of these engineering plants (Boilers, Turbines and so on) can be obtained from the Manufacturer’s Manual upon purchase.

__________________________Prepared by VPM PSM 2 ~ NWK of 33

I. Introduction

Thermodynamics is the science dealing with the relationships between the properties of a substance and the quantities of ‘work ‘ and ‘heat’ which cause a change of state.

Thermodynamics is one of the 3 important sciences (other than the equally essential “mechanics of machines” and “properties of materials”) required for the design of an engineering plant like steam turbines, boilers, reciprocating engines, combustion systems, heat-exchangers, condenser, compressors, turbo jets for air planes, rocket motors for rockets and space ships and others.

There are 3 branches of Thermodynamics and divided as follows :

Engineering Thermodynamics Chemical Thermodynamics Physical Thermodynamics

We will focus on Engineering Thermodynamics as it is more relevant and applicable to the various engineering plant (steam turbine, boilers and so on) currently used in the Palm Oil Mills.


II. Laws of Thermodynamics and its Corollaries

a) First Law of Thermodynamics states that :

“When any closed system is taken through a cycle, the net work delivered to the surroundings is proportional to the net heat taken form the surroundings”.

This is basically the Law of Conservation of Energy and its formula is :

Q – W = 0

b) Corollary of the First Law of Thermodynamics states that :

“There exists a property (internal energy, u) of a closed system such that a change in its value is equal to the difference between the heat supplied and the work done during any change of state”.

The formula will be the non-flow energy equation :

Q – W = (u2 – u1)

c) Second Law of Thermodynamics states that :

“It is impossible to construct a system which will operate in a cycle, extract heat from a reservoir, and do an equivalent amount of work on the surroundings”.

OR

“It is impossible to convert all the heat supplied to an engine into work ; some heat must be rejected, representing a waste of energy”.


Thus the Second Law implies that if a system is to undergo a cycle and produce work, it must operate between at least two reservoirs of different temperature, however small this difference may be.

d) Corollary of the Second Law of Thermodynamics states that :

“There exists a property (entropy, s) of a closed system such that

a change in its value is equal to 2 dQ for any reversible process

1 Tundergone by the system between state 1 and state 2”.

The formula will be :

dS = dQ T reversible

OrdQ = TdS

Note : A reversible adiabatic process is called an isentropic process.


Hot Reservoir Boiler

W (+ve)(=Q1-Q2)

Cold Reservoir BPV

Q1 (+ve)

Q2 (-ve)

Turbine

e) From the First Law of Thermodynamics, the Mechanical Equivalent of heat, Joule’s equivalent, is established :

Joule’s equivalent expresses the number of “work” units which are equivalent to one “heat” unit.

Note : a) 1 kcal = 4.1868 kJb) 1 kgf = 9.80665 Nc) 1 J = 1 Nmd) 1 kcal = 4.1868 x 1000 Nm/9.80665 N/kgf

= 426.935 kgf-me) 1 Btu = 778.169 ft-lbf


= 4.19 kJ /kcal= 427 kgf - m/kcal= 778 ft-lbf/Btu= 1400 ft-lbf/Chu

III. Thermodynamic Properties

A) Basically there are 6 Thermodynamics properties and classified as follows :

a) Observable by the senses :

(i) Pressure, p (bar)(ii) Specific Volume, v (m3/kg)(iii) Temperature, T (degrees Kelvin)

b) Not directly observable by the senses :

(i) Internal energy, u (kJ/kg)(ii) Enthalpy, h (kJ/kg)(iii) Entropy, s (kJ/kg Deg K)

Note : 1 bar = 100000 N/m2 = 0.1 N/mm2 = 1/1.01325 atm

T Deg K = (t Deg C) + 273.15T Deg R = (t Deg F) + 459.671 Deg F = 5/9 Deg C

B) In a CLOSED SYSTEM in thermodynamics equilibrium, two of the above six properties can determine the “state” (values of the remaining four properties are then fixed).

However these two properties must be independent of each other.

Density and specific volume are not independent properties because one is simply the reciprocal of the other ; this fact is recognized in that density is not included in the six properties.

Similarly, less obviously, pressure and temperature are NOT ALWAYS independent properties ; when a liquid (water) is in contact with its vapour (steam) in a closed vessel (Boiler), it is found that the temperature at which the liquid and vapour are in equilibrium is


always associated with a particular pressure and one cannot be changed without the other.

Pressure and specific volume CAN be used to determine the “state” of the system.

IV. Work and Heat Transfer

A) The ultimate objective of Work Transfer and Heat Transfer is to convert one form of energy into another.

This Transfer are achieved through a fluid while the fluid is compressed or expanded, heated or cooled in open or closed systems.

The FLUID may be a gas (e.g. air), a vapour (e.g. steam), a liquid, or a mixture of these substances provided that they do not react chemically with one another.

B) OPEN SYSTEMS are Flow processes and examples are :

a) Steam flowing through a Turbine.b) Water entering a Boiler and leaving as steam.

CLOSED SYSTEMS are NON-Flow processes such as :

a) Mixture of water and steam in a closed system.b) Gas expanding in a cylinder by displacing a piston.

Note : In OPEN SYSTEMS, the MECHANICAL properties like (1) velocity relative to the fixed boundary of the system and (2) position in the gravitational field i.e. height above some datum level have to be considered in additional to the thermodynamic state/properties.


C) Work Transfer and Heat Transfer basically and strictly speaking is actually ENERGY transfer.

Work is a MORE VALUABLE form of energy transfer than heat because :

Heat can never by converted continuously and completely into work.

WHERE AS

Work can always be converted continuously and completely into heat.

D) HEAT transferred is actually the energy transferred by virtue of a temperature difference.

~ One kilo calorie (kcal) is defined as the heat required to raise one kilogram (kg) of water by one degree Celsius/Centrigrade in temperature.

~ Similarly One British Thermal Unit (Btu) is the heat required to raise one pound (lb) of water by one degree Fahreinheit.

E) Summarising : Neither heat nor work are properties of a system, but both are transient quantities, only appearing at the boundary while a change of “state” occurs within the system.

Heat and Work is the process necessary to the system during a change of state.


V. Non-Flow Process (reversible & irreversible)

A) The non-flow energy equation states that :

“Any quantity of heat supplied to a closed system must equal the increase of internal energy plus the work done by the system”.

and is represented as :

Q – W = (U2 – U1)

where u is the internal energy of the system.

Note : Internal energy is just random molecular energy of the fluid.

B) It is often more convenient to work with specific quantities throughout an analysis and than multiply the final result by the mass of the system.

Hence the non-flow energy equation can be rewritten as :

Q – W = (u2 – u1)

where Q is the quantity of heat per unit mass of fluid.W is the quantity of work done per unit mass of

fluid.U is the specific internal energy.


C) For reversible processes, the equation is written as :

dQ – dW = du

Note : For irreversible process the equation becomes :

dQ – dW = dE

where E = e m

D) As in mechanics, “work” is said to be done when a force moves through a distance i.e. :

W = F x L , where F = pressure x area

Where F is in Newtons (N)L is in meters (m)W is in Nm or Joule (J)

For reversible work :

dW = pdV where dV = Area x lOr

W = 2 pdV1

Note : Reversible process is an IDEAL process, ie. one which may be approached but never actually achieved in practice.


E) Constant Volume process :

Q – W = u2 – u1

Where W is zero (or negative)

Hence Q = u2 – u1

The specific heat at constant volume, Cv, is defined as the heat required to raise the temperature of unit mass one degree during a reversible constant volume process.

Cv = du dT v

OR

u2 – u1 = Cv (T2 – T1)


F) Constant Pressure process

Q – W = (u2 – u1)

dQ – pdv = du

Since p is constant , pdv = d (pv)

dQ – d (pv) = du

or dQ = d ( u + pv)

where u + pv = h, enthalpy

dQ = dh

Integrating, we have, the energy equation for a REVERSIBLE constant pressure process like :

Note : Thus the heat added in a reversible constant pressure is equal to the increase in enthalpy.

WHERE AS

The heat added was equal to the increase of internal energy in the reversible constant volume process.


Q = h2 – h1

The specific heat at constant pressure (Cp) is the heat required to raise unit mass one degree during a reversible constant pressure process.

Cp = dh dT p

OR

h2 – h1 = Cp (T2 – T1)

===> Cp – Cv = R


G) Polytropic process (both volume and pressure change)

p v n = constant

where n is called the index of expansion or compression

(i) If n= 0, then p = constant(ii) If n = , then v = constant

Because p v n = constant becomes p 1/n v = constant

The energy equation for a reversible polytropic process is written as :

Q – (p2v2 - p1v1) = (u2 – u1) 1 – n

by integrating W = 2 pdv 1

where p1 v1 n = p2 v2 n = p v n

Since, p1 v1 n = constant, and differentiating, we have :

Vn d p + p n v n-1 dv = 0

Or dp = np


dv = v

H) Adiabatic Process

The term adiabatic is used to describe any process during which heat is prevented from crossing the boundary of the system ie. NO HEAT TRASFERRED.

Energy Equation : Q – W = (u2 – u1)

If Q = 0, then, - W = (u2 – u1)

OR- pdv = (u2 – u1)

Note : In an adiabatic expansion work is done by the system at the expense of its internal energy

WHERE AS

In an adiabatic Compressor, the internal energy is increased by an amount equal to the Work Done on the system.

pdv = - duOR

du = - pdv

It is defined that :

dQ = Tds reversible process

dQ = 0 adiabatic process

ds = 0 entropy is constant


I) Isothermal process = No change in temperature

dQ - pdv = duOr

Q - 2 pdv = (u2 – u1) 1

and d Q = Tds

If T is constant (isothermal), then :

Q = T (s2 – s1)And

W = Q - (u2 – u1)


VI. Flow Processes (Steady Flow & Non Steady Flow)

A) pv is the flow work, flow energy or pressure energy.

The steady flow energy equation is :

Q – W = (h2-h1) + ½ (C22 – C1

2) + G (Z2-Z1)

Assuming :

a. The mass flow at inlet is constant with respect to time, and equal to the mass flow at outlet.

b. The properties at any point within the open system do not vary with time.

c. The properties are constant over the cross-section of the flow at inlet and outlet.

d. Any heat or work crossing the boundary does so at a uniform rate.

Note : Excluding hydraulic machinery (Fluid Mechanics - Bernoulli’s Theorem), normally the potential energy term g (z2

– z1 ) is either zero or negligible compared with the other terms.


B) Boiler (and similarly for condenser)

The fluid entering as a liquid (water) and leaving as a vapour (steam) at a constant rate has NO work done on or by the fluid as it passes though the system (boiler), and hence W = 0.

Kinetic energies and its difference is negligible here as shown by the following example.

Example :

20.000 kg of steam are to be generated per hour at a pressure of 30 bar.

Assumption 1 : Feed water temperature : 1000 C : Feed water velocity : 2 m/s

: Steam main velocity : 45 m/s

Assumption 2 : Enthalpy of a compressed liquid is equal to the enthalpy of

saturated liquid at the same temperature.

From the Saturation Table for steam and water, we have :

h1 = 100.09 kcal/kg ( 1 kcal = 4.187 kJ)

h2 = 669.30 kcal/kg

The heat required per kg of steam is :

Q = ( h2 – h1 ) + ½ ( C22 – C1

2 )

= 569.21 kcal. + ½ ( C22 – C1

2)

½ ( C22 – C1

2 ) = ½ ( 452 - 22 ) m 2 kg S2 kg

= ½ ( 47 x 43 ) Nm__________________________Prepared by VPM PSM 2 ~ NWK of 33

x

kg

= 1010.5 J 1 kJ kg 1000 J

= 1.0105 kJ/kg

= 1.0105 kJ 1 kcal kg 4.187 kJ

= 0.2413 kcal/kg

Q(+ve) = h2 – h1 where h2 > h1

For condensers, the formula is still the same except that h2 < h1, now, giving :

Q(-ve) = h2 – h1

C) Nozzle ( and Diffuser)

Principle : A drop in pressure from inlet to outlet to accelerate the flow.

The flow through a nozzle occurs at very high speed, and there is little time for the fluid to gain or lose energy by a flow of heat through the walls of the nozzle as the fluid passes through it.

The process is therefore adiabatic.


x

x

Nozzle

Steady Flow Energy Equation :

Q – W = ( h2 – h1 ) + ½ ( C22 – C1

2 ) + g ( z2 – z1 )

½ ( C22 – C1

2 ) = h1 – h2

C1 < C2

P1 > P2

The Diffuser is the reverse of a nozzle.

D) Turbine ( and Compressor ) A turbine is a means of extracting work from a flow of fluid (steam) expanding from a high pressure to a low pressure.

The fluid (steam) is accelerated in a set of fixed nozzles and the resulting high speed jets of fluid then change their direction as they pass over a row of curved blades/buckets attached to a rotor.


A force is exerted on the blades equal to the rate of change of momentum of the fluid, and this produces a torque at the rotor shaft.

At the same time, the velocity of the fluid is reduced to somewhat near the value it possessed before entering the nozzles and assumed to be equal.

Since the average velocity of flow through a turbine is very high, the process is assumed to be adiabatic.

Hence the energy equation becomes :

W = ( h1 – h2 )

The rotary compressor is a reversed turbine, work being done on the fluid to raise its pressure with the help of a set of fixed diffusers.

E) Throttling

1) A flow of fluid is said to the throttled when some restriction is placed in

the flow.

Example : Partially closed valve in a pipe line.

Note : Throttling is applied to relatively low-speed flow, i.e. kinetic

energy difference is negligible.


Heat transfer is neglected as throttling takes place in such a shortlength of pipe that the surface area across which heat can flow is very small.

In addition, no work crosses the boundary of the system concerned.

Thus, the energy equation becomes :

h1 = h2

i.e. the throttling, process is an adiabatic steady flow process such that the enthalpy is the same at inlet and outlet.

2) Dryness Fraction, x

If the dryness fraction of a wet vapour is x, it means unit mass of wet vapour can be regarded as a mixture of x kg of saturated vapour and (1-x) kg of saturated liquid, each having different specific properties denoted by suffixes g (gas) and f (fluid) respectively but the same pressure and temperature.

The applicable formulas are :

u = (1-x) uf + x ug

h = (1-x) hf + x hg

s = (1-x) sf + x sg


Dryness fraction, x, describes the “quality” of the vapour.

3) Throttling Calorimeter

To determine, the dryness fraction of a wet vapour, the throttling calorimeter is used.

A steady stream of wet vapour is sampled from the main flow and passed through a small orifice in a well lagged chamber (throttling calorimeter).

The vapour pressure, initially p1, falls to p2 as a result of the throttling process.

Since h2 = h1 it is evident from the h-s diagram that if the pressure drop is sufficient and the vapour not too wet, the sampled vapour will be superheated after the expansion.

The thermodynamic state can therefore be determined by measuring the independent properties p2 and T2.

The enthalpy h2 can then be found from the superheat table.

If the initial pressure p1 is also measured, and the corresponding values of hf and hfg are taken from the saturation table, the dryness fraction is found from

x1 = h1 – hf1 = h2 – hf1

hfg1 hfg1

The process must be continued long enough for the calorimeter to reach a steady temperature before readings are taken.


F) Cycles consisting of steady-flow processes.

Q12 ( + )

W23 ( +ve )

W41 (-ve )

Q34 ( -ve )

Steam Power Plant :

Boiler : Q12 = m ( h2 – h1 )

Turbine : -W23 = m ( h3 – h2 )

Condenser : Q34 = m ( h4 – h3 )


Boiler

Turbine

Condenser

Feed Pump

Feed Pump : -W41 = m ( h1 – h4 )

VII. Vapour Power Cycles

A) One common method of producing mechanical power employs the transfer of heat from a reservoir to a working fluid which is taken through an engineering thermodynamic cycle.

The cycles considered here have two characteristics in common :

a) The working fluid is a condensable vapour (steam) which is the liquid phase (water) during part of the cycle and

b) The cycle consists of a succession of steady-flow processes, with each process carried out in a separate component specially designed for the purposes.

Each component constitutes an OPEN SYSTEM, and all the components are connected in series so that as the fluid circulates through the power plant, each fluid element passes through a cycle of mechanical and thermodynamic states.

To SIMPLIFY the analysis, we shall ASSUME that the change in kinetic and potential energy of the fluid between entry and exit of each component is NEGLIGIBLE compared with the change in ENTHALPY.

Hence, the energy equation can be written as

Q – W = h2 – h1

Instead of :

Q – W = (h2-h1) + ½ (c22 – C1

2) + g (z2-z1)


B) Criteria for the comparison of cycles.

a) The choice of power plant for a given purpose is determined largely by considerations of OPERATING COST and CAPITAL COST.

Operating cost is primarily a function of the overall efficiency of the plant, while the Capital Cost depends mainly on its size and complexity.

In general the efficiency can always be improved by adding to the complexity of the plant, so that a suitable compromise (strike a proper balance) between the operating and capital costs must be reached.

b) Overall thermal efficiency of a vapour cycle is measured by the proportion of latent energy in the fuel which is converted into useful mechanical work.

OR

Combustion Efficiency X Cycle Efficiency

Note : All real processes are irreversible, and irreversibilities reduce the cycle efficiency.

Cycles composed of reversible processes are called IDEAL Cycles.

c) Efficiency Ratio = Actual cycle efficiency Ideal cycle efficiency

Some cycles are more sensitive to irreversibilities than others. That is two cycles may have the same ideal cycle efficiencies, but after allowing for the same process efficiencies we find that their ACTUAL cycle efficiencies are markedly/very different.


Hence a high IDEAL cycle efficiency is not therefore, by itself, a good indication of whether or not the cycle will provide a power plant of high overall efficiency.

d) Another criterion is the work ratio (rw)

All power cycles consists of processes of both positive and negative work transfer, and rw is defined as the ratio of the net work to the positive work done in the cycle.

Note : Irreversibilities decrease the positive work and increase the negative work.

If the ideal negative work is only slightly less than the ideal positive work i.e. if rw is only slightly greater than zero, quite a small amount of component inefficiency is sufficient to reduce the net work output to zero and so reduce the actual cycle efficiency to zero.

On the other hand, as the work ratio (rw) approaches unity, the same amount of component inefficiency will have a much smaller effect on the net work output and hence in the actual cycle efficiency.

HENCE a high ideal cycle efficiency TOGETHER with a high work ratio provides a reliable indicator that the real power plant will have a good overall efficiency.

Most practical vapour cycles, however, have work ratios very near unity,the work ratio is then not very informative and a more direct criterion of size is necessary.

e) In general, the size of the components will depend on the AMOUNT OR VOLUME OF WORKING FLUID which has to be passed through them.


A more direct indication of relative sizes of steam plant is therefore provided by the specific steam consumption (s.s.c) i.e. the mass flow of steam required per unit power output.

Its unit is kg/kWh

C) CARNOT Cycle

a) This cycle consists of two reversible isothermal processes at Ta

and Tb respectively, connected by two reversible adiabatic (isentropic) processes.

The wider the range of temperature, the more efficient becomes the cycle and the lowest temperature of 25°C to 30°C can be assumed.

The corresponding pressure of 25°C to 30°C is 0.032 – 0.042 bar absolute.

Hence for condensing turbines, condenser pressures of 0.04 bar is used generally.


The maximum possible temperature of the working fluid (steam), in power plants is governed by the strength of the materials available for the highly stressed parts of the plant e.q. boiler tubes or turbine blades.

This metallurgical limit is around 600-650°C for the steam power plant.

b) Carnot cycle is not used in practice because :

(i) It has a low work ratio(ii) Practical difficulties associated with the compression as it

would be extremely difficult to control the condensation process so that it stopped at state 4, and then carry out the compression of a very wet vapour efficiently.

D) RANKINE CYLCE


Rankine Cycle with Saturated Steam

a) It is comparatively easy to condense the vapour completely, and compress the liquid to boiler pressure in a small feed pump (compared to a large compressor in the CARNOT cycle) as illustrated in the Rankine Cycle.

Rankine Cycle with Superheated Steam

b) To increase the ideal cycle efficiency, superheated steam is used and this requires a separate bank of SUPERHEATER tubes into the flow of the combustion gases.


This is one way of raising the temperature of the steam without raising the boiler pressure.

With superheated steam, s.s.c is greatly/markedly reduced, so that the added complexity of a superheater is compensated by a reduction in the size of other components.

c) In saturated (un-superheated) cycle, the turbine exhaust steam’s dryness fraction is around 0.72 for inlet pressure of 30 barg and exhaust pressure of 0.04 barg. Inspection of the T-s diagram above, it shows that with higher boiler pressures the turbine exhaust would become even wetter.

Wetness in steam is undesirable because droplets in the steam erode turbine bladings and reduce the turbine isentropic efficiency.

In practical designs, the dryness fraction at the turbine exhaust of 0.88 is aimed at but almost impossible to achieve.

For example, when the steam enters the turbine at 160 bar absolute and 600C (superheated steam, saturation temperature is only 345.75C), after isentropic expansion to 0.04 bar absolute, the dryness fraction is 0.772.

REHEAT cycle is the solution.

Example :Steam, initially dry saturated, expands isentropically from a pressure of 20 bar gauge to 3 bar gauge.

S2 = S1 = Sg1 = 1.5109 kcal/kg Deg Kx2 = S2 – Sf2


Sfg2

= 1.5109 – 0.4225 = 1.0884 = 88.8% 1.6483 – 0.4225 1.2258

If it is expanded from 30 bar gauge, then :

x = 1.4757 – 0.4225 1.6483 – 0.4225

= 1.0532 1.2258 = 85.8%

E) Reheat Cycle


In this reheat cycle, the expansion takes place in two turbines. The steam expands in the high pressure (e.q. 30 bar gauge) to some intermediate pressure (say 3 bar gauge), and is then passed back to yet another bank of tubes in the boiler where it is reheated at constant pressure, usually to the original superheat temperature. It then expands in the low-pressure turbine to the condenser pressure.

At the turbine outlet, the dryness fraction can be as high as 0.98.


thermodynamics vpm psm 2.doc

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