JJ207 Thermodynamic Topic 2 First Law of Thermodynamics

JJ207-THERMOD JJ207THERMODYNAMICS YNAMICS 1

Topic 2- First First Law o o T!"r#o$%&a#ics T!"r#o$%&a#ics a&$ its proc"ss"s

TO'IC 2 FIRST LAW OF THERMODYNAMICS THERMODYNAMI CS AND ITS PROCESSES At the end of the topic you will be able to: 

Describe forms of energy.



Describe energy transfer by heat and energy energy transfer by work.



Classify the mechanical forms of work.



Distinguish the mode of energy transfer between closed and open systems.



Defined of a closed system.



Describe the non-flow energy equation. Explain the specific heat internal energy energy enthalpy of ideal gases solids and



  

 

.

liquids. Calculate the energy balance for close systems. Definition of an open system. Describe control !olumes" steadily flow processes and engineering de!ices unsteadily-flow processes. Calculate mass and !olume flow rate and the continuity equation Determine and explain the steady flow energy equation #$egligible change in kinetic or potential energy% leading to the conce pt of enthalpy - typical applications such as turbine compressor boiler boiler no&&le diffuser and heat h eat exchanger.



2.0

Calculate the energy balance for open systems.

INTRODUCTION ' KBD/JKM/PUO

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he *irst +aw of ,hermodynamics is the most basic and fundamental law of nature. hile its name makes it sound intimidating it is actually the most intuiti!e law of

nature as well. ,he *irst +aw of ,hermodynamics pro!ides a method method for accounting

for all energy inputs outputs and stores within a system. ne of the most fundamental laws of

natu nature re is the the conse conser! r!at atio ion n of energ energy y prin princi cipl ple. e. /t simp simply ly stat states es that that duri during ng an ener energy gy interaction energy can change from one form to another but the total amount of energy remains constant. ,hat is energy cannot be created or destroyed.

2.1

The Firs irst La Law ! ! The Therr"#$% #$%a a"i&s i&s

Energy can exist in many forms such as thermal, kinetic, potential, electric, chemical,…

Fi'(re 2.1

0ictures showing types of energy

,he first thermodynamic law is the formulation of a more general law of physics #the law of cons conser er!a !attion ion of ener energy gy%% for ther hermody odynam namic proc proces essses. es. ,he fir first law of thermodynamics is simply a statement of conser!ation of energy principle and it asserts that total energy is a thermodynamic thermodynamic property. Energy can neither be created nor destroyed" destroyed" it can only change forms. ,his principle is based on experimental experimental obser!ations and is known as the First Law of Thermodynamics. ,he Thermodynamics. ,he *irst *irst +aw of ,hermodynami ,hermodynamics cs can therefore therefore be stated stated as follows:

Energy cannot be created or destroyed; destroyed; it can only be transformed from one form into another. ))The First Law ! Ther"#$%a"i&s

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During a transformation the change in the internal energy of a system is equal with the sum of work and heat exchanged by the system with its surroundings. dQ – dW = dU here: dQ are dQ are positi!e if they are transferred into the system and negati!e if they are released by the system. ,o apply the first law of thermodynamics to a cyclic process we ha!e to remember that internal energy is a system !ariable. /n a cyclic process the system returns at the same state hence there is no change in its internal energy. /n this case work produced by the system equals the heat exchanged by the system. hen a system undergoes a thermodynamic cycle then the net heat supplied to the system from its surroundings is equal to the net work done by the systems on its surroundings. /n symbols

Σ dQ dQ = Σ dW dW

#2.(%

where Σ represents the sum of a complete cycle. 2.2

Fr"s ! E%er'$

Energy is the ability to do work. /t is one of the basic human needs and is an essential component in any de!elopment programme. /n this lesson we are going to look at the forms that energy exists namely: heat, light, sond, electrical, electrical, chemical, chemical, nclear nclear and mechanical mechanical . ,hese forms of energy may be transformed from one form to the other usually with losses. a*

Heat e% e%er'$  also also referr referred ed to as therm thermal al energy energy is really really the effect effect of mo!ing mo!ing

molecules. 3atter is made up of molecules which are in continual motion and in a solid solid !ibrate !ibrate about a mean position. position. ,he motion of any molecule increases increases when the energy of the substance is increased. ,his may cause an increase in the temperature of the substance substance or lead to a change of state. ,he higher the temperature temperature the greater the internal energy of the substance. +*

Li'ht e%er'$  is a type of wa!e motion. ,hat is light is a form of energy caused by

light wa!es. /t enables us to see as ob4ects are only !isible when they reflect light into our eyes. &*

S(%# e%er'$, is also a type of wa!e motion. 5ound energy may be con!erted into

electrical energy for transmission and later the electrical energy can be con!erted

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back into sound energy at the recei!ing end. An example of such transformations could be seen in the microphone and the loudspeaker. #*

E-e&tri&a- e% e%er'$, is really the effect of mo!ing electrical charges from one point to

another another in a conduct conductor or.. Electr Electrica icall charges charges mo!ing mo!ing throug through h a conduct conductor or is called called electricity. Electrical energy may be easily changed into other forms of energy to suit our particular particular needs. +ightning +ightning is an example example of electrical energy. energy. Electric current is the means by which electrical energy is most easily transported to places where it is needed and con!erted into other forms. e*

energy stored stored within within chemic chemical al compou compounds nds.. A chemical chemical Che"i&a- e% e%er'$, is the energy compound is formed by the rearrangement of atoms that is accompanied by energy loss or gain. ,his energy is the chemical energy gained or lost in the formation of the compound. compound. *ood biomass biomass fuel and explosi!es explosi!es ha!e a store store of chemical energy. energy. ,he energy from food is released by chemical chemical reactions in our bodies in the form of heat. *uels like coal oil and natural gas contain chemical energy that may be con!erted into other forms of energy like heat and light. ,he chemical energy present in a gi!en fuel is determined by its calorific !alue 7 the heat liberated when ( 8g of the fuel is burnt. 9atteries and explosi!es also contain chemical energy that could be con!erted into other forms of energy some beneficial others harmful.

!*

N(&-ear e%er'$, also known as atomic energy is energy stored in the nucleus of an

atom. /t is this energy that holds the nucleus together and could be released released when the nuclei are combined #fusion% or split #fission% apart. $uclear energy can be used for peaceful purpose as well as destructi!e purposes #as in the atomic bomb%. Considering peaceful purposes nuclear energy is used to generate electricity in nuclear power plants produce steam for dri!ing machines powering some submarines and spacecrafts. /n these applications the nuclei of uranium atoms are split in a process called fission. $uclear energy is also the source of the suns energy. ,he sun combines the nuclei of hydrogen atoms into helium atoms in a process called fusion. '*

Me&ha%i&a- e%er'$, is the kind of energy that can do mechanical work directly.

$aturally occurring sources of mechanical energy include winds waterfalls and tides. ,here are two kinds of mechanical energy energy namely kinetic energy and potential energy: 

i%eti& e%er'$, is the energy a body possesses by !irtue of its motion. A mo!ing

body of mass m kg and !elocity C m;s possesses kinetic energy. energy. ,hus the magnitude of the kinetic energy of an ob4ect depends both the mass and the !elocity of the ob4ect. *lowing water and winds ha!e kinetic energy. 

Pte%tia- e%er'$, is the energy of a body due to its position or shape. ,his form of

energy could be considered as energy stored in a body to be released when it ' KBD/JKM/PUO

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begins to mo!e or change its position or shape. =uite often potential energy changes to kinetic energy. A ball at the top of a slope water behind a dam a compressed spring and a stretched elastic band possess potential energy. energy.

E%er'$ &%/ersi% . ne important property of energy is its ability to change from one form

to another form. *or example chemical energy from fossil fuels #coal oil and natural gas% can be con!erted into heat energy when burned. ,he heat energy may be con!erted into kinetic energy in a gas turbine and finally into electrical energy by a generator. ,he electric energy may subsequently be con!erted into light sound or kinetic energy in our homes through !arious household appliances. During any energy con!ersion the amount of energy input is the same as the energy output. ,his concept is known as the law of conser!ation of energy and energy and sometimes referred to as the First the First Law of Thermodynamics. Thermodynamics. ,his law states: energy cannot be created nor destroyed but can be transformed from one form to another. ,hus the total energy of an isolated system is always constant and when energy of one form is expended an equal amount of energy in another form is produced.

2. 2.

E%er E% er'$ '$ Tra%s! a%s!er er +$ Heat eat a%# a%# Wr

>istorically heat was considered to be a fluid that can spontaneously flow from a hot body to a cold body. >eat is a form of energy energy which crosses the boundary of a system during a change change of state state produc produced ed by the diffe differen rence ce in temper temperatu ature re betwee between n the syste system m and its its surroundings. ,he unit of heat is is taken as the amount of heat energy equi!alent to one 4oule or $m. ,he 4oule is defined as the the work done when the point of application of a force of one newton is displaced through a distance of o f one meter in the direction of the force. Work transfer is defined as a product of the force and the distance mo!ed in the

direction of the force. hen a boundary of a close close system mo!es in the direction of the the force acting acting on it then the system system does work on its surroundings. surroundings. hen the boundary boundary is mo!ed inward inwardss the work work is done done on the system system by its surrou surroundi ndings. ngs. ,he units units of work are for example $m or ?. /f work is done on unit mass mass of a fluid then the work done per kg of fluid has the units units of $m;kg or ?;kg. Consider the fluid expanding behind the piston of an engine. ,he force * #in the absence of friction% will be gi!en by: F = "#

#2.2%

where " is " is the pressure exerted on the piston and # is # is the area of the piston /f d$ is the disp displa lacem cement ent of the the pist piston on and and " " can be assumed constant o!er this displacement then the work done W will be gi!en by W = F x x d$ ' KBD/JKM/PUO

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= "# x d$ = " x #d$ = " x d% = "&% ' – % ( ) )

#2.)%

where d% = #d$ = change in !olume.

F

0E55FE

d$

ork transfer Fi'(re 2. ork hen two systems at different temperatures are in contact with each other energy will transf transfer er betwee between n them them until until they they reach reach the same temper temperatu ature re #that #that is when they they are in equilibrium with each other%. ,his energy is called heat or thermal energy energy and the term heat flow refers to an energy transfer as a consequence of a temperature difference. 2. 2.

Si'% Si'% C%/ C%/e% e%ti ti% % !r !r Wr Tra%s! a%s!er er

/t is con!enient to consider a con!ention of sign in connection with work transfer and the usual con!ention adopted is: •

if work energy is transferred from the system to the surroundings it is donated as positi!e.

•

if work energy is transferred from the surroundings to the system it is donated as 5FF$D/$G5 negati!e. 9F$DA

8 2

55,E3

 !e

8 ( - !e

Fi'(re 2. 5ign Con!ention for work transfer 2.3

Si'% C%/e%ti% !r Heat Tr Tra%s!er a%s!er

,he sign con!ention usually adopted ado pted for heat energy transfer is such that : ' KBD/JKM/PUO

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•


if heat energy flows into the system from the surroundings it is said to be 4siti/e .

•

if heat energy flows from the system to the surroundings it is said to be %e'ati/e . /t is incorrect to speak of heat in a system since heat energy exists only

when it flows across the boundary. nce in the system it is con!erted to other types of energy. energy. 5FF$D/$G5

55,E3

>EA, E$EG Q( 

>EA, E$EG Q2 -!e

!e

9F$DA Fi'(re 2.3 5ign con!ention for heat transfer 2.5

S4e&i!i& !i& heat &a4a&it$

,he specific heat capacity #symbol * % is the amount of heat expressed in k? which must be transferred to or from ( kg of a gi!en material to cause its temperature to change by (HC. >ence the unit of measurement is k?;kg.HC or in thermodynamics scale the unit is k?;kg.8. ,he word specific used in this and o ther terms we will use later refers to (kg.

2.6

I%ter%a- E%er'$

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/nternal energy is the sum of all the energies a fluid possesses and stores within itself. ,he molecules of a fluid may be imagined to be in motion thereby possessing kinetic energy of tran transl slat atio ion n and rota rotati tion on as well well as the the ener energy gy of !ibr !ibrat atio ion n of the the atom atomss with within in the the molecu molecules les..

/n additi addition on the fluid fluid also posses possesses ses interna internall potent potential ial energy energy due to interinter-

molecular forces. 5uppose we ha!e ( kg of gas in a closed container as shown in *igure 2.@. *or simplicity simplicity we shall assume that the !essel is at rest with respect to the earth and is located on a base hori&on. ,he gas in the !essel has neither macro kinetic energy nor potential energy. >owe!er the molecules of the gas are in motion and possess a molecular or JinternalJ kinetic energy. ,he term is usually shortened to i%ter%a- e%er'$ . /f we are to study thermal effects then we can no longer ignore this form of energy. ,he symbol for internal energy is F and in the /nternational 5ystem #5/% it is measured in 4oules #?% or kilo4oules #k?%. (k? K (LLL?. Also we shall denote the specific #per kg% internal energy as  ?;kg. $ow suppose that by rotation of an impeller within the !essel we add work dW dW to the closed system and we also introduce an amount of heat dQ dQ. ,he gas in the !essel still has &ero macro kinetic energy and &ero potential energy. energy. ,he energy that has been added has simply simply caused an increase in the internal energy. ,he change in internal energy is determined only by the net energy that has been transferred across the boundary and is independent of the form of that energy #work or heat% or the process path of the energy transfer. /n molecular simulations molecules can of course be seen so the changes occurring as a system gains or loses internal energy are apparent in the changes in the motion motion of the molecules. molecules. /t can be obser!ed that the molecules molecules mo!e faster faster when the internal energy is increased. /nternal energy is therefore a thermodynamic property of state. Equation 2.1 is sometimes known as the the non-flow energy equation and is a statement statement of the *irst +aw of ,hermodynamics. dU

or

U 2

=

−

dQ - dW

U (

=

Q(2

−

W (2

#2.1%

dW

dQ

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raise the internal energy of a close system Fi'(re 2.6 Added work and heat raise

2.7 2.7

E%th E% thaa-4 4$ 8H 8H* a%# a%# S4e&i 4e&i!i !i&& E E%t %tha ha-4 -4$ $ 8h* 8h*

A more useful system !ariable for this course is enthalpy. Enthalpy #>% unit k? is the sum of internal energy and the product of pressure and !olume of the system. + = U  "⋅ % /n a thermodynamic system with fixed !olume and pressure enthalpy has the same meaning as internal energy and it is also measured in kilo4oules #k?%. ,he reference condition regarding enthalpy differs from one substance to another. *or water #>2% the reference condition is defined at a temperature of approximately L.L(HC and a pressure of approximately L.<( k0a#a%. ,his is the only combination of pressure and temperature at which ice water and !apour can stably exist together. ,he condition is referred to as the triple point. E!ery substance has its own triple point. At the triple point enthalpy of the substance is considered to be L. ,he enthalpy contained in one kg of a substance is known as specific enthalpy #h%. 5pecific enthalpy enthalpy depends on the material material its pressure pressure temperature temperature and state. 5pecific 5pecific enthalpy data for light and hea!y water can be found in steam tables. 5pecific enthalpy #h% is measured in k?;kg. h =   "⋅ ! ,herefore taking the !alue of #h% and multiplying it by the mass one can obtain the !alue of enthalpy #>%. + = m⋅ h

E9a"4-e 2.1

W ot K #% M

Qin K (L k?

55,E3

W inK -2 k?

Qot K -) k?

,he figure abo!e shows a certain process which undergoes a complete cycle of operations. Determine the !alue of the work work output for a complete complete cycle W ot ' KBD/JKM/PUO ot .

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S-(ti% t E9a"4-e 2.1 ΣQ = Qin  Qot

Qin is (L k?

K #(L%  #-)% K @ k?

Qot is 7) k?

ΣW = W in in  W ot ot

= #-2%  #W #W ot ot %

W in is 72 k? W ot is !e

>ence ΣQ - ΣW K K L ΣW K K ΣQ

#-2%  #W ot ot % K @ W ot = I k? k? ot = E9a"4-e 2.2

A system system is allowed to do work amounting to 6LL k$m whilst heat energy amounting to BLL k? is transferred into it. it. *ind the change of internal energy and state whether it is an increase or decrease. d ecrease.

S-(ti% t E9a"4-e 2.2

U 2 7 U ( K Q(2 7 W (2 (2 now W (2 (2 K 6LL k$m K 6LL k? Q(2 K BLL k? U 2 7 U ( K BLL 7 6LL

∴

K )LL k? #5ince U 2 > U ( the internal energy has increased% E9a"4-e 2.

During a complete cycle operation a system is sub4ected to the following: >eat transfer is )LL k? supplied and (6L k? re4ected. ork ork done by the system is 2LL k?. k ?. Calculate the work transferred from the surrounding to the system. ' KBD/JKM/PUO

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S-(ti% t E9a"4-e 2. ΣQ = Qin  Qot K K #)LL%  #-(6L% K (6L k? ΣW = W in = #W inin%  #2LL% in  W ot ot =

>ence ΣQ - ΣW K K L ΣW K K ΣQ

#W inin%  #2LL% K (6L W inin = - 6L k?

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STEADY FLOW PROCESSES

/n heat engine it is the steady flow processes which are generally of most interest. ,he conditions which must be satisfied by all of these processes are: i.

,he ,he mass mass of flu fluid id flow flowin ing g past past any any sect sectio ion n in the the syst system em mus mustt be cons consta tant nt wit with h res respe pect ct to time.

ii. ii.

,he ,he prope propert rtie iess of the flu fluid id at any any parti particu cula larr secti section on in the the syst system em mus mustt be const constan antt

iii. iii.

with respect to time. All transf transfer er of work work energy energy and and heat heat which which take takess place place must must be done done at a unifo uniform rm rate. rate.

A typical example of a steady flow process is a steam boiler operating under a constant load as shown diagrammatically in *ig. 2.I. /n order to maintain the water le!el in the boiler the feed pump supplies water at exactly the same rate as that at which steam is drawn off from the boiler. ,o maintain the production of steam at this rate at a steady pressure the furnace will need to supply heat energy at a steady rate. Fnder these conditions the properties of the working fluid at any section within the system must be constant with respect to time. 5,EA3 F, A,E +ENE+

9F$D A,E /$

*F$ACE

Fi'(re 2.: 5team 9oiler

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2.10 2.10


E;(a E; (ati ti% % ! C%t C%ti% i%(i (it$ t$

,his is an equation which is often used in con4unction with the steady flow energy e nergy equation. /t is based on the fact that if a system is in a steady state then the mass of fluid passing any section during a specified time must be constant. Consider a mass flow rate rate #in  % kg;s flowing through a system in which all conditions are steady as illustrated symbol m in *ig.2.(L. (

2 C2

C(

2

AEA A2

AEA A( (

Fi'(re 2.10 3ass flowing through a system

+et A( and A2 represent the flow areas in m2 at the inlet and outlet respecti!ely. +et !( and !2 represent the specific !olumes in m);kg at the inlet and outlet respecti!ely. +et C( and C2 represent the !elocities in m;s at the inlet and outlet respecti!ely. ) ,hen mass flowing flowin g per sec K !olume flowing per pe r sec m ;s

!olume per kg K K

i.e.

2.11

 m

=

#(* ( -g !(

s

#2 * 2 -g !2 #(* ( !(

s

K

m);kg

at inlet at outlet

#2 * 2 !2

#2.6%

STEADY FLOW ENER
26



,his equation is a mathematical statement on the principle of Conser!ation of Energy as applied to the flow of a fluid through a thermodynamic system. ,he !arious forms of energy which the fluid can ha!e are as follows:

a*

Pte%tia- e%er'$

/f the fluid is at some height  height  abo!e abo!e a gi!en datum le!el then as a result of its mass it possesses potential energy with respect to that datum. ,hus for unit mass of fluid in the close !icinity of the earth 0otential energy K g K g  ≈ I.B(  I.B( 

+*

i%eti& e%er'$

A fluid in motion possesses kinetic energy. /f the fluid flows with !elocity *  then for unit mass of fluid

8inetic energy K &*

* 2 2

I%ter%a- e%er'$

All fluids store energy. ,he store of energy within any fluid can be increased or decreased as a result of !arious processes carried out on or by the fluid. ,he energy stored within a fluid which results from the internal motion of its atoms and molecules is called its internal energy and it is usually designated by the letter U . /f the internal energy of the unit mass of fluid is discussed this is then called the s"ecific the s"ecific internal energy and energy and is designated by . #*

F-w r r #is4-a&e"e%t e% e%er'$

/n order to enter or lea!e the system any entering or lea!ing !olume of fluid must be displaced with an equal !olume ahead of itself. ,he displacing mass must do work on the mass being displaced since the mo!ement of any mass can only be achie!ed at the expense of work. ,hus if the !olume of unit mass of liquid #its specific !olume% at entry is !( and its pressure is / is / ( then in order to enter a system it must displace an equal specific !olume !( inside the system. ,hus work to the !alue / !alue / (!( is done on the specific !olume inside the system by the specific !olume outside the system. ,his work is called flow or displacement work and at entry it is energy recei!ed by the system. ' KBD/JKM/PUO

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5imilarly at exit in order to lea!e the flow work must be done by the fluid inside the system on the fluid outside the system. ,hus if the pressure at exit is / is / ' and the specific !olume is !' the equation is then *low or displacement work re4ected K / K / '!' e*

Heat re re&ei/e# r r re>e&te#

During its passage through the system the fluid can ha!e direct reception or re4ection of heat energy through the system boundary. ,his is designated by Q. ,his must be taken in its algebraic sense. ,hus Q is positi!e when heat is recei!ed. Q is negati!e when heat is re4ected. Q K L if heat is neither recei!ed nor re4ected. !*

E9ter%a- wr #%e

During its passage through the system the fluid can do external work or ha!e external work done on it. ,his is usually designated by W . ,his also must be taken in its algebraic sense. ,hus if External work is done by the fluid then W is is positi!e. External work is done on the fluid then W is is negati!e. /f no external work is done on or by the fluid then W K K L. *igure 2.(( illustrates some thermodynamic system into which is flowing a fluid with pressure / pressure / ( specific !olume !( specific internal energy ( and !elocity * (. ,he entry is at height  height  ( abo!e some datum le!el. /n its passage through the system external work W may may be done on or by the fluid and also heat energy Q may be recei!ed or re4ected by the fluid from or to the surroundings.

,he fluid then lea!es the system with pressure / pressure / ' specific !olume !' specific internal energy ' and !elocity * '. ,he exit is at height  height  ' abo!e some datum le!el.

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/ (,(, !( U (,(,* (

/ ' ,!' ' ,* '

Fi'(re 2.11 A schematic of a steady flow system

,he application of the principle of energy conser!ation to the system is ,otal ,otal energy entering the system K ,otal energy lea!ing the system or for unit mass of substance 2

g. ( + ( + / (!( +

* ( 2

2

+ 0 = g. +  + / ! + 2

2

2

2

* 2 2

+w

#2.<%

,his is called the stea#$ !-w e%er'$ e;(ati%. ,his equation is not applicable to all energy forms. /n such cases the energy forms concerned are omitted from the energy equation. /n equation 2.< it was stated that the particular combination of properties of the form h =   /! is called specific enthalpy and is designated as h. ,hus the steady flow energy equation is written as 2

g. ( + h( +

* ( 2

2

+ 0 = g. + h + 2

2

* 2 2

+w

or in the easy way the equations becomes

 * − * 0 − w = ( h − h ) +   2 2

2

2

$ote:

(

2

(

   + ( . − . ) g  2

(

#2.@%

0 K specific heat #k?;kg% w K specific work #k?;kg% ' KBD/JKM/PUO

2B



 #kg;s% the equation may be written as and with the flow rate m

  * − *  ( h − h ) +  Q − W = m  2  2

2

2

2

(

(

    + ( . − . ) g    2

(

#2.B%

$ote: = K heat flow #k?;s or k%  K work #k?;s or k%

2.12

APPLICATION APPLICATION OF STEADY FLOW E=UATION E=UATION

,he steady flow energy equation may be applied to any apparatus through which a fluid is flowing pro!ided that the conditions stated pre!iously are applicable. 5ome of the most common cases found in engineering practise are dealt with in detail as below. a* ?i-ers

/n a boiler operating under steady conditions water is pumped into the boiler along the feed line at the same rate as which steam lea!es the boiler along the steam main and heat energy is supplied from the furnace at a steady rate. 2 2

9F$D

55,E3

(

A,E /$

5,EA3 F,

(

= *F$ACE

Fi'(re 2.12.1 5team 9oiler

,he steady flow energy equation gi!es

  * − *  ( h − h ) +  Q − W = m   2 2

2

2

(

2

(

    + ( . − . ) g    2

(

#2.I%

/n applying this equation to the boiler the following points should be noted : ' KBD/JKM/PUO

2I



i.

= is the the am amount ount of heat heat ener energy gy pass passin ing g int into o the the flu fluid id per per seco second nd

ii.

 is is &ero &ero si since nce a boi boiler ler has has no mo! mo!iing par parts ts cap capab ablle of af affect ecting a work transfer

iii. iii.

,he ,he kine kineti ticc ener energy gy is is smal smalll as com compa pare red d to the the oth other er ter terms ms and and may may usually be neglected

i!. i!. !.

,he ,he pot poten enti tial al energ energy y is is gener general ally ly smal smalll eno enough ugh to be negl neglect ected ed..  #kg;s% is the rate of the flow of fluid. m

>ence the equation is reduced to

 ( h2 Q=m

+*

−h )

(

#2.(L%

C%#e%sers

/n principle a condenser is a boiler re!erse. /n a boiler heat energy is supplied to con!ert the liquid into !apour whereas in a condenser heat energy en ergy is remo!ed in order to condense the !apour into a liquid. /f the condenser is in a steady state then the amount of liquid usually called condensate lea!ing the condenser must be equal to the amount of !apour entering the condenser. NA0F

A,E F, 55,E3

A,E /$

9F$DA C$DE$5A,E

Fi'(re 2.12.2 Condenser


 

 ( h2 Q − W = m

 * − * − h ) +   2 2

2

(

2

(

    + ( . − . ) g    2

(

0oints to note ' KBD/JKM/PUO

)L



i.

Q is the amount of heat energy en ergy per second transferred from the system

ii.

W is is &ero in the boiler

iii. iii.

,he ,he kin kinet etic ic ener energy gy term term may may be be negl neglec ecte ted d as as in in the the boil boiler er

i!. i!.

,he ,he pote potent ntia iall ener energy gy is gener general ally ly smal smalll enou enough gh to to be be negl neglec ecte ted d  is the rate of the flow of fluid. m

!.

,hus the equation is reduced to

 ( h2 Q=m

−h )

(

#2.((%

E9a"4-e 2.

A boiler operates at a constant pressure of (6 bar and e!aporates fluid at the rate of (LLL kg;h. At entry to the boiler the fluid has an enthalpy of (<6 k?;kg and on lea!ing the boiler the enthalpy of the fluid is 22LL k?;kg. k?;kg. Determine the heat energy supplied to the boiler.

S-(ti% t E9a"4-e 2.

2 5,EA3 F, 2 55,E3 9F$DA

( A,E /$

( =


 

 * − * − h ) +   2 2

 ( h2 Q − W = m

2

(

2

(

    + ( . − . ) g    2

(

Q K heat energy per hour entering system W K K work energy per hour lea!ing system K L  K fluid flow m flow rate K (LLL kg;h K L.2@B kg;s h' K 22LL k?;kg h( K (<6 k?;kg * (1 * ' K neglected  (1  ' K neglected ,hus the steady flow energy equation becomes

 ( h2 Q=m

−h ) (

' KBD/JKM/PUO

)(


=

L.2@B

=

-g s

6<6.@)


( 22LL − (<6)

-2 -g

k? s

E9a"4-e 2.3

/f <6 O of the heat energy energy supplied to the boiler in in example 2.1 is used in e!aporating the fluid determine the rate of fuel consumption required to maintain this rate of e!aporation if ( kg of fuel produces )2LLL k? of o f heat energy. S-(ti% t E9a"4-e 2.3 6<6.@)

>eat energy from fuel required per seconds K

L.<6

K [email protected] k?;s P k >eat energy obtained from the fuel K )2 ) 2 LLL k?;kg

*uel required K

[email protected] k? )2LLL s

x

kg k?

K L.L2@2 kg;s

E9a"4-e 2.5

*luid enters a condenser at the the rate of )6 kg;min with a specific specific enthalpy of 22LL k?;kg and lea!es with a specific enthalpy of 266 k?;kg. Determine the rate of heat energy loss from the system.

S-(ti% t E9a"4-e 2.5

 K )6 kg;min kg;min K L.6B kg;s

m


  * − *  ( h − h ) +  Q − W = m   2 2

2

2

(

2

(

    + ( . − . ) g    2

(

' KBD/JKM/PUO

)2



*or a condenser  K L and the term representing the chang e in kinetic and potential energy may be neglected. ,herefore the equation is reduced to

 ( h2 Q=m

−h ) (

*rom the abo!e equation

Q

=

L.6B

-g s

#266 − 22LL%

-2 -g

K - ( (2B.( k?;s ACTI@ITY 1*

2.(% 2.(%

/n an air condi conditi tion onin ing g syste system m air air is cooled cooled by passi passing ng it o!er o!er a chil chille led d water water coil coil condenser. ater enters the coil with an enthalpy of 12 k?;kg and lea!es the coil with an enthalpy of BL k?;kg. ,he water flow rate is 2LL kg;h. *ind the rate of heat absorption by the water in kilowatts.

2.2%

/n a stead steady y flow flow syste system m a substa substance nce flow flowss at the the rate rate of 1 kg;s. kg;s. /t ente enters rs at a pres pressur suree of <2L k$;m2 a !elocity !elocity of )LL m;s internal energy energy 2(LL k?;kg and specific !olume ) L.)@ m ;kg. /t lea!es the system at a pressure of ()L k$;m 2 a !elocity of (6L m;s internal internal energy (6LL k?;kg and specific specific !olume (.2 m);kg. During its passage through the system the substance has a loss by heat transfer of )L k?;kg to the surroundings. Determine the power of the system in kilowatts stating whether it is from or to the system. $eglect any change in potential energy.

A%swer t ACTI@ITY 1*

2 .( %

Data:

 K 2LL kg;h K m

2LL )
=

L.L6< kg;s

h( K 12 k?;kg" h2 K BL k?;kg Q KM

,he diameter of the water tube in a cooler is normally constant. ,herefore there is no change in water !elocity and kinetic energy. /n general the change in potential energy is also negligible. ,he equation of steady flow is therefore reduced to

 ( h2 Q=m

−h ) (

' KBD/JKM/PUO

))



K L.L6<#12 7 BL% K - 2.() k?;s or k ,he rate of heat absorption by the water is 2.() k

2.2%

9y ne negle glecting the ch change in in po potential ene enerrgy fo for a un unit ma mass of of su substance nce th the steady flow energy equation becomes:

 * − * 0 − w = ( h − h ) +   2 2

2

2

2

(

(

   + ( . − . ) g  2

#(%

(

0 is written negati!e since )L k?;kg are lost to the surroundings. *rom equation #(%

5pecific work w K #(

−

( % + # / (!( − / 2 ! 2 % + #

* (2

−

* 22

2

%+0

orki in kilo4oules #k?%

5pecificwork w K #2(LL-(6LL%  #<2LxL.)@ -()Lx(.2%  #

)LL 2

−

(6L 2

2 $(L )

% #-)L%

K <@<.@6 k?;kg. ,he substance flows at the rate of

 K 1 kg;s

m

 ×w ∴utput #since W is is positi!e%  K m K 1 x <@<.@6 K 2@L@ k?;s or k

&*

T(r+i%e

' KBD/JKM/PUO

)1



A turbine is a de!ice which uses a pressure drop to produce work energy ene rgy which is used to dri!e an external load. *+F/D /$ (

=out SYSTEM

out

?OUNDARY

2

*+F/D F,

Fi'(re 2.12. ,urbine


  * − *  ( h − h ) +  Q − W = m   2 2

2

2

(

2

(

    + ( . − . ) g    2

(

0oints to note : i.

,he ,he a!er a!erag agee !elo !eloci city ty of of flow flow of of flu fluid id thr throu ough gh a tur turbi bine ne is is norm normal ally ly hig high h and and the fluid passes quickly through the turbine. /t may be assumed that because of this heat energy does not ha!e time to flow into or out of the fluid during its passage through the turbine and hence Q K L .

ii. ii.

Alth Althou ough gh !el !eloc ocit itie iess are are high high the the dif diffe fere rence nce betw betwee een n them them is is not not lar large ge and and the the term representing the change in kinetic energy may be neglected.

iii. iii.

0ote 0otent ntia iall ener energy gy is gener general ally ly sma small ll enoug enough h to to be be negl neglect ected ed..

i!.

W is is the amount of external work energy produced per second.

,he steady flow energy equation becomes beco mes

or

 ( h2 - W = m

−h )

 ( h( W = m

−h )

(

2

' KBD/JKM/PUO

#2.(2%

)6



E9a"4-e 2.6

A fluid flows through a turbine at the rate of 16 kg;min. Across the turbine the specific enthalpy drop of the fluid is 6BL k?;kg and the turbine loss 2(LL k?;min in the form of heat energy. Determine the power produced by the turbine assuming that changes in kinetic and potential p otential energy may be neglected.

S-(ti% t E9a"4-e 2.6


  * − *  ( h − h ) +  Q − W = m   2 2

2

2

2

(

(

    + ( g. − g. )    2

(

Q K heat energy flow into system K -2(LL k?;min K -)6 k?;s W K K work energy flow from system k?;min  K fluid flow m flow rate K 16 kg;min K L.@6 kg;s h' 7 h h( K -6BL k?;kg * (1 * ' K neglected  (1  ' K neglected ,herefore the steady flow energy equation becomes

 ( h2 Q − W = m

−h ) (

-)6 k?;s 7 W K K L.@6 kg;s #-6BL k?;kg% W K K #-)6%  #1)6% k?;s K 1LL k?;s K 1LL k #*

N-e

A no&&le utilises a pressure drop to produce an increase in the kinetic energy of the fluid. ( 2 55,E3 *+F/D /$

9F$DA *+F/D F,

2 (

Fi'(re 2.12. $o&&le ' KBD/JKM/PUO

)<




  * − *  ( h − h ) +  Q − W = m  2  2

2

2

(

2

(

    + ( . − . ) g    2

(

0oints to note : i. ,he ,he a!e a!era rage ge !el !eloc ocit ity y of flow flow thr throu ough gh a no& no&&l &lee is is hig high h hen hence ce the the flu fluid id spends only a short time in the no&&le. *or this reason it may be assumed that there is insufficient time for heat energy to flow into or ii. ii.

out of the fluid during its passage through the no&&le i.e. Q K L. 5inc 5incee a no&& no&&le le has no mo! mo!in ing g part parts s no work work ener energy gy will will be tra trans nsfe ferr rred ed

iii. iii.

to or from the fluid as it passes through the no&&le i.e. W K K L. 0ote 0otent ntia iall ener energy gy is gene genera rall lly y sma small ll eno enough ugh to be neg negle lect cted ed..

>ence the equation becomes

  * − *  ( h − h ) +  L=m  2  2

2

2

      

2

(

−

h2 ) + * (2

(

2

* 2 2

= ( h − h ) + *

or

2

(

* 2

2

(

=

2( h(

#2.()%

ften C( is negligible compared with C2. /n this case the equation equation becomes * 2

=

2( h(

−h ) 2

#2.(1%

E9a"4-e 2.7

*luid with a specific enthalpy of 2BLL k?;kg enters a hori&ontal no&&le with negligible !elocity at the rate of (1 kg;s. At the outlet from the no&&le the specific enthalpy and specific !olume of the fluid are 226L k?;kg and (.26 m);kg respecti!ely. Assuming Assuming an adiabatic flow determine the required outlet area of the no&&le.

S-(ti% t E9a"4-e 2.7

' KBD/JKM/PUO

)@




  * − *  ( h − h ) +  Q − W = m  2  2

2

2

2

(

(

    + ( g. − g. )    2

(

hen applied to the no&&le this becomes

  * − *  ( h − h ) +  L=m   2 2

2

2

(

2

(

      

5ince the inlet C( is negligible this may be written as * 2

2( h(

=

−

h2 )

(

K

2 2BLL

−

226L

) × (L

)

K (L1I m;s Applying the equation of continuity at outlet gi!es

*rom

 m

=

#2

* 2 #2

=

=

!2



 × !2 m * 2 (1 × (.26 (L1I

K L.L(<
ACTI@ITY 2 TEST YOUR UNDERSTANDIN< UNDERSTANDIN< 9E*E F C$,/$FE /,> ,>E $EQ,

/$0F,RS ' KBD/JKM/PUO

)B


2.)


5team 5team enters enters a turbin turbinee with with a !eloci !elocity ty of (< m;s and specif specific ic enthal enthalpy py

2IIL

k?;kg. ,he steam lea!es the turbine with a !elocity of )@ m;s and specific enthalpy 26)L k?;kg. ,he heat lost to the surroundings as the steam passes through the turbine is 26 k?;kg. ,he steam flow rate is )21LLL kg;h. Determine the work output from the turbine in kilowatts. 2.1

/n a tur turbo bo 4et 4et engi engine ne the the mome moment ntum um of the the gase gasess lea! lea!in ing g the the no&&l no&&lee produ produce cess the propulsi!e force. ,he enthalpy and !elocity of the gases at the no&&le entrance are (2LL k?;kg and 2LL m;s respecti!ely. ,he enthalpy of the gas at exit is ILL k?;kg. /f the heat loss from the no&&le is negligible determine the !elocity of the gas 4et at exit from the no&&le.

' KBD/JKM/PUO

)I



ANSWER TO ACTI@ITY 2* 2.)

$eglect $eglecting ing the the chan changes ges in potent potential ial energy energy the the stea steady dy flow flow energy energy equat equation ion is

  * − * Q − W = ( h − h ) +   2  2

2

      

2

2

(

(

= is negati!e since heat is lost from the steam to the surroundings

 * − * ∴specific W K K ( h − h ) +   2 2

(

(

2

K #2III-26)L% 

2

2

   - Q 

#(< 2

−

)@ 2 %

2 $(L )

−

26

K 1)1.11) k?;kg ,he steam flow rate K )21LLL;)
K )ILII.I@ k?;s or k ≈ )I(LL k ≈ )I.( 3

2.

,he steady energy flow equation for no&&le gi!es

  * − *  ( h − h ) +  L=m   2 2

2

2

2

(

(

      

n simplification * 2

=

K

2( h(

−

h2 ) + * (2

2((2LL × (L )

−

ILL × (L ) ) + 2LL 2

K BLL m;s

' KBD/JKM/PUO

1L


e*


Thrtt-i%' 4r&ess A throttling process is one in which the fluid is made to flow through a restriction  e.g. a partially opened !al!e or orifice causing a considerable drop in the pressure of the fluid.

2 (

Fi'(re 2.12.3 ,hrottling process


  * − *  ( h − h ) +  Q − W = m   2 2

2

2

(

2

(

    + ( . − . ) g    2

(

0oints to note: i.

ii. ii.

5in 5ince thro throtttlin tling g take takess plac placee o!er o!er a !ery ery smal smalll dista istan nce ce the a!a a!ailab ilable le are area through which heat energy can flow is !ery small and it is normally assumed that no energy is lost by heat transfer i.e. Q K L. 5inc 5incee ther theree are are no mo! mo!in ing g part parts s no no ener energy gy can can be be tran transf sfer erre red d in the the for form m of work energy i.e. W K K L.

iii.

,he difference between * ( and * ' will not be great and consequently the term representing the change in kinetic energy is normally neglected.

i!. i!.

,he ,he pote potent ntia iall ener energy gy is gener general ally ly smal smalll enou enough gh to to be be negl neglec ecte ted. d.

,he steady flow energy equation becomes

 &h' – h( ) LK m ) or

h' = h(

#2.(6%

i.e. during a throttling process the enthalpy remains constant. E9a"4-e 2.:

A fluid flowing along a pipeline undergoes undergoes a throttling throttling process from from (L bar to to ( bar in passing through a partially open !al!e. 9efore throttling the specific !olume of the fluid is L.) m);kg and after throttling is (.B m );kg. Determine the change in specific internal energy during the throttling process.

' KBD/JKM/PUO

1(



S-(ti% t E9a"4-e 2.:

*or a throttling process the steady flow energy equation becomes

 &h' 3 h( ) LK m ) or

h' = h(

*rom h =   /!, ' = h' 3 / '!' (= h( 3 / (!( ,herefore the change in specific internal energy ' – ( K # h' 3 / '!' % - # h( – / (!(% K # h' 3 h( ) – & / '!' 3 / (!( ) K L 7 # ( x (L2 x (.B 7 (L x (L2 x L.) % k$;m2 x m);kg K (2L k$m;kg K (2L k?;kg !*

P("4

,he action of a pump is the re!erse of that of a turbine i.e. it uses external work energy to produce a pressure rise. /n applying the steady flow energy equation to a pump exactly the same arguments are used as for turbine and the equation becomes

 ( h2 - W = m

−h)

(

5ince h' 4 h( , , W will will be found to be negati!e.

#2.(<%

F,+E, 2 =

55,E3 

9F$DA (

/$+E,

Fi'(re 2.12.5 0ump ' KBD/JKM/PUO

12



E9a"4-e 2.10

A pump deli!ers fluid at the rate of 16 kg;min. At the inlet to the pump the specific enthalpy of the fluid is 1< k?;kg and at the outlet from the pump the specific enthalpy of the fluid is (@6 k?;kg. /f (L6 k?;min of heat energy are lost to the surroundings by the pump determine the power required to dri!e the pump if the efficiency of the dri!e is B6 O. S-(ti% t E9a"4-e 2.10

 K 16 kg;min ,he flow rate of fluid m K L.@6 kg;s ,he steady flow energy is

 

 ( h2 Q − W = m

 * − * − h ) +   2 2

2

(

2

(

    + ( . − . ) g    2

(

Q K - (L6 k?;min K - (.@6 k?;s W K K work energy flow #k?;s% h( K 1< k?;kg h' K (.2@ k?;kg m K L.@6 kg;s ,he kinetic and potential energy may be neglected 5ubstituting the data abo!e with the steady flow energy equation gi!es  &h' – h( ) Q 3 W K K m ) - (.@6 7 W K K L.@6 #(@6 7 1<% W K K -(.@6 7 #L.@6 x (2I% K - IB.6 k?;s K - IB.6 k #$.9. ,he negati!e sign indicates work energy required by the pump% 5ince the efficiency of the dri!e is B6 O 0ower required by the compressor K - IB.6 IB. 6 x

(LL B6

K - ((1.B k

' KBD/JKM/PUO

1)



TEST YOUR UNDERSTANDIN< UNDERSTANDIN< 9E*E F C$,/$FE /,> ,>E $EQ,

/$0F,RS 2.

A rotary air pump is required to deli!er ILL kg of air per hour. ,he enthalpy at the inlet and exit of the pump are )LL k?;kg and 6LL k?;kg respecti!ely. ,he air !elocity at the entrance and exit are (L m;s and (6 m;s respecti!ely. ,he rate of heat loss from the pump is 26LL . Determine the power required to dri!e the pump.

2.

/n Acti!ity 2 for question $o. 2.1 if the diameter of the no&&le at exit is 6LL mm find the mass flow rate of gas. ,he gas density at the no&&le inlet and exit are L.B( kg;m) and L.)I kg;m) respecti!ely. Also Also determine the diameter of the no&&le at the inlet.

2.

Data : m

=

ILL K L.26 kg;s )
h( K )LL k?;kg h' K 6LL k?;kg * (K (L m;s * 'K (6 m;s Q K 26LL  K 2.6 k W K K M ,he steady flow energy equation gi!es

  * − *  ( h − h ) +  Q − W = m   2 2

2

2

2

(

(

    + ( . − . ) g    2

(

$eglecting the change in 0otential energy since it is negligible

  * − *  ( h − h ) +  Q − W = m  2  2

2

2

2

(

(

3W K K L.26 T# 6LL- )LL%  #

    +   (6 2

−

(L 2

2 $(L )

%U  2.6

W K K - 62.6 k

' KBD/JKM/PUO

11


2 .1

L.6

Data : A2 K π

1


2

K L.(I< m2

L.B( kg;m) ) ρ ' K L.)I kg;m ρ ( K

 K M m d K M 3ass flow rate at exit

 K # m K #' * ' ρ ' K <(.2 kg;s *rom the mass balance 3ass entering the no&&le K mass lea!ing the no&&le K m

 K # K #( * ( ρ ( K A2 C2 ρ2

m

n substitution #( x 2LL x L.B( K <(.2 n simplification #( K L.)@B m2 or d ( K L.
SELF ASSESMENT*

ou ou are approaching app roaching success. Tr$ a-- the ;(esti%s in this self-assessment section and check your answers with those gi!en in the *eedback to 5elf-Assessment on the next page. /f you face any problem discuss it with your lecturer. lecturer. Good luck. (

5team fl flows thr through ugh a turbine st stage at the the rat rate of 16L 16LL k?; k?;h. ,he ,he ste steam !elocities at inlet and outlet are (6 m;s and (BL m;s respecti!ely. ,he rate of heat energy flow from the turbine casing to the surroundings is 2) k?;kg of steam flowing. /f the specific enthalpy of the steam decreases by 12L k?;kg in passing through the turbine stage calculate the power de!eloped.

2

A rota rotary ry pump pump draw drawss
' KBD/JKM/PUO

6LLL 6LLL . .

16



$eglecting the changes in kinetic and potential energy energy determine the power required to dri!e the pump. )

A no&&l no&&lee is is supp suppli lied ed with with steam steam ha!i ha!ing ng a spe speci cifi ficc ent enthal halpy py of 2@BL 2@BL k?;k k?;kg g at at the the rate of I.( kg;min. kg;min. At outlet from the no&&le the !elocity of the steam is (L@L m;s. Assuming that the inlet !elocity of the steam is negligible and that the process is adiabatic determine: a%

the the spe speci cifi ficc ent entha halp lpy y of of the the ste steam am at the the no& no&&l &lee exi exitt

b%

the outlet area required if the final specific !olume of the steam is (B.@6 m);kg.

1

*luid at (L (L.)6 bar bar ha! ha!ing a spe specific !ol !olume of L. L.(B m);kg is throttled to a pressure of ( bar. /f the specific !olume of the fluid after throttling is L.(L@ m);kg calculate the change in specific internal energy during the process.

>a!e you tried tried the questionsMMMMM /f VYESW check your answers now. (.

1@< k

2.

)6) k

).

22LB k?;kg" 2<
1.

(@6.@ k?;kg

' KBD/JKM/PUO

1<

JJ207 Thermodynamic Topic 2 First Law of Thermodynamics

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