/

: Measuring units , examples force , pressure ; specific volume : 1- . 2- Force Pressure Specific volumeThermodynamic

Definition :- The field of science , which deal with the energies possessed by gases and vapours . It also includes the conversion of these energies in term of heat and mechanical work and their relationship with properties of system .Measuring units S.I units [International System of units]

Physical Qantity

Symbol

Unit

Symbol of unit

Length LmeterM

Time TsecondS

Mass MkilogramKg

Temperature TkelvinK0

Electric current AAmpereA

The secondary SI unitsForce FNewtonN

Pressure PPascal

BarPa =

Bar = 105 pa

Density

Work WJouleJ = N.m

Power PwattW =

British unit system Physical QuantitySymbolUnitSymbol of unit

LengthLInchIn

TimeTSecondS

MassMPoundT , bm

TemperatureTFahrenheitF0

ForceFPoundIbF

PressureP-IbF/in2 psi

Density

-Ibm/in3

WorkW-IbF. in

Powerhorse powerh.p

Length : L 1m = 100 cm = 102 cm = 1000 mm = 103 mm

1 in = 2.54 cm 1 foot = 1 ft = 12 in

Time : t : 1 hour = 60 minute = 3600 s

mass : m : 1 pound = 0.45 kg

1 ton = 1000 kg = 103 kg

Temperature : T : F0 = 1.8 0C + 32

Force : F : 1 MN = 106 N , 1 KN = 103 N

Pressure : 1 bar = 105 pa = 105

Work : W 1 KJ = 1000 J = 103 J

J = N.m

Power : 1 h.p = 746 watt

1 KW = 1.34 h.p

Watt = , KW =

Definitions

1- Force : It is an agent which consider as a measure of mechanical effect on the bodies .

2- Pressure : It,s the force per unit area

,

3- Density : () mass per unit volume

4- Specific volume () : the volume occupies by unit mass

5- power : It is work done per unit time Power =

Units watt , kw , h.p

6- Specific gravity Sp . gr =

Questions

1- Define pressure , Specific volume .

2- Convert 20 kpa to at m .

3- Convert 50 ft to m .

: Thermodynamic terms state , process , equilibrium in thermodynamic classification of system :1-

2-

Thermodynamic terms 1- State :- Determines the value of the thermodynamic properties for

material at a certain time .2- process :- It is that which changes the system from a certain state of

thermodynamic equilibrium into anther state .

3- Thermodynamic equilibrium :- means that the thermodynamic properties of amatter is the same and constant and do not change at all areas in the system . thermodynamic systems Thermodynamic system is defined as :-

A definite area where some thermodynamic process is taking place .

Any thermodynamic system has its "boundaries" and any thing outside the boundary is called "Surrounding"

Boundary : It is a surface that separtes between the system and its

surrounding .

Surrounding : It is a region outside the boundary of the system.

Working Substance : It is amatter that transfer the energy through the system as steam , g as . etc .

Classification of Systems

The thermodynamic system may be classified into the following three groups .

1- Closed System : In this system the working substance does not cross the boundaries of the system , but heat and work can cross it .

2- Open System : In this system the working substance crosses boundary of the system . Heat and work may also cross the boundary .

3- Isolated System : It is a system of fixed mass and no heat or work cross its boundary .

Isolated System Closed System Open system

Questions

1- Define : process , state .2- write the classification of systems .

: 1- Temperature , kind of temp. measuring and relations between them .

2- Pressure measurements and relation between them .

: 1- .

2- .

Temperature :- It is may be defined the degree of hot hess or the level of heat intensity of a body .

Measurement of Temperature

The temperature of a body is measured by a thermometer . There are two scales for measuring the temp. of a body .

1- Centigrade or Celsius Scale This scale is most used by engineers . The freezing point of water = 0 . The boiling point of water = 100

We use symbol (C) to describe temp.

2- Fahrenheit Scale

The freezing point of water = 32

The boiling point of water = 212

We use the symbol (F0) to describe temp.

* The relation between centigrade scale and Fahrenheit scale is given by

F0 = 1.8 C0 + 32

Ex : 1) Convert 37C0 to F0 2) Convert 50 F0 to Celsius scale

Absolute Temperature :

The absolute centigrade scale is called degree "Kelvin"

K0 = C0 + 273

Absolute Fahrenheit scale is called degree "Rankine"

R0 = F0 + 460

Questions

1- (20 C0) to Kelvin scale .

2- Convert (400 K0) to Rankine scale .

3- Convert (170 F0) to Kelvin scale .Pressure Pressure :- is the force exerted by the system on unit area .

Absolute Pressure :- is the guage pressure plus atmospheric pressure .

Gauge pressure :- A gauge for measuring pressure records the pressure above atmospheric pressure .

Vaccume Pressure :- It is the pressure of the system below atmospheric pressure .

Equations Pabs = Patm + Pg

The positive guage pressure

> 0

The negative guage pressure :- [ vaccume pressure ]

Vaccume Pressure = Patm - Pabs < 0

Manometer and Barometer

Manometer :- An instrument for measuring a pressure difference in terms of the height of a liquid .

= Density of liquid

g =

= Height of the liquid

Barometer : An instrument for measuring the atmospheric pressure .

Units of pressure

1 pat m = 76 cm . Hg = 760 mm Hg

1 pat m =

1 pat m =

1 pat m = 1.01325 bar

Examples

1- change a pressure about 1500 mm Hg to bar .

Sol.

760 mm Hg = 1.0132 bar

1500 mm Hg * bar

2- A compound gauge reads (65 cm Hg ) vaccume pressure . Compute absolute pressure 14 bar .

Sol.

76 cm Hg = 1.0132 bar

Pvacc = 65 cm Hg * = 0.867 barPvacc = 0.867 barPvacc = Pat m - Pabs0.867 = 1.0132 Pabs

Pab = 1.0132 0.867 = 0.147 bar

Home work

1- Convert pressure 5 Kpa to bar

2- Convert pressure 76 cm Hg to

3- Convert pressure 50 pa to mmHg

Example (3)

A manometer is used to measure the pressure in a tank . The fluid is an 0.1 with a specific gravity of (0.87) and the liquid height = 45.2 cm . If the barometric pressure = 98.4 kpa , the density of water = 1000 the gravity = 9.78 , Determine the absolute pressure with in the tank in kpa , at m .

Solution

Sp. Gr.

P1 = 98.4 kpa + 3.853 kpa

P1 = 102.253 kpa

P1 = 102.253 kpa * = 1.022 atm

Note

1 kpa = 1000 pa

1 atm = 102 kpa

Questions

1- Convert 500 cm Hg to

2- A barometer reads (735 mm Hg) at room temp.

Determine :- the atmospheric pressure 14 bar millibar , kpa ,

Hg = 13600

3- Convert 2 kpa to mm Hg

: Work and kinds of work energy and forms of energy

: 1- .

2- .

Work :

Its may be defined as the product of a force with its corresponding displacement

------- (1)

J J = N.m

Work is one of energies types , that we can transform it to anther types of energy such as (transform of mechanical work to electrical energy , kinetic energy , heat energy

W = Force * displacement

W = F *

Notes* If the work done by a thermodynamic system we say that is a positive work +W or W > 0 Wout = + w* If the work done on a thermodynamic system we say that is a negative w or (-w)

W < 0 ex [compress or , generator]

Work of non of closed system

By taking a small element with length of p and width of dv Area of element = dw = p . dv

Area = area of piston

dL = the dis placement traveled by a piston

F = p . A

F = exerted force

P = exerted pressure

L

W = F . L L = strok length (m)

Flow work (open system)

Flow w on k = F . L

WF = F . L

WF = P . A . L

The flow work (flow energy) per unit mass [specific flow work]

= flow work / unit mass J/kg

= specific volume m3 / kg

J / kg Types of Energies

1- Potential energy (P.E)

The energy that system possesses by virtue of its position relative to the surface of the earth .

P . E = m g Z

m = mass

Z = elevation , m

2- Kinetic energy K.E

The energy that a system possesses owing to its motion

= velocity m/sec

= Final velocity

= initial velocity

3- Internal energy It is the energy stored in the substance total internal energy is given

U = m Cv T

Questions

1- A gas has a mass of (2 kg) and density = is transported by pipe of height (30.25 m) from earth , the temp = 138 C0 the velocity .

Flow = 6m / sec and Cv = 0.674

Compute : 1- potertial energy

2- kinetic energy 3- Internal energy

4- total energy

:The First Law of Thermodynamic

: 1- .

The First Law of Thermodynamics

The concept of energy and hypothes is that it can be neither created nor destroyed this is principle of the conservation of energy . The first law of thermodynamics is merly one statement of this general principle with particular of this general principle with particular reference to heat energy and work .

When a system under goes at hermodynamics cycle then the net heat supplied to the system from its surrounding is equal to the net work done by the system on its surroundings .

: the sum for a complete cycle .

: Enthalpy

: 1- .

Enthalpy (H)

Enthalpy is an extensive property is defined by the relation .

H = U + P . V (J , kJ)

Specific enthalpy (h)

= specific volume m3 / kg

= specific internal energy J/kg

= pressure

Heat energy (Q)

It is the type of energy that transfer due to the different in temp . between the system and its surrounding .Q : heat energy (J , kJ)

q : specific heat energy J/kg kg/kg

Heat Sign

1- heat added input use

Q +

2- heat rejected or out put us

(-Q) Net work (Wnet) & Net heat (Qnet)

Wnet = W0 Win

W0 = out put workWin = input work

Qnet = Qin Qout

Qin = input heat

Qo = output heat

: Applying the first law on closed systems : 1- .

Energy equation

By applying the first law of thermodynamic which state that the energy can neither created nor destroid but transfer from one form to another we obtain . .

Q + U1 + P1V1 + K. E1 + P . E1 = W + U2 + P2V2 + K . E2 + PE2

H = U1 + P1V1

Q W = equation of energy

We can written thia equation

Q W = m [(h2 h1) +

1- Non Flow energy equation (N. F. E. E)

Closed system

For closed system PV , K.E , P.E = 0

The energy equation become :-

2- Steady state flow energy equation (S. F. E. E)

(open system)

For open system and steady state

By dividing the equation of energy by time we obtain

S. F. E E

= mass flow rate ()

= rate of heat transfer ( = watt , kw)

= rate of heat transfer (w , kw)Examples

Closed system

Ex. 1

The change in the internal energy of closed system increase to (120 KJ) while (150 KJ) of work , that go out of the system , Determine the amount of heat transfer a cross system boundaries ?

Is the heat added or rejected ?

Solution

From N. F. E.E

The sign (+270) therefore the heat added to the system

Note

If the change in the internal energy increase we use (+)

If the change in the internal energy decrease we use ()

Ex . 2A tank containing a fluid is stirred by a paddle wheel the work input to the paddle wheel is (5090 KJ) . The heat from the tank is (1500 KJ) . Determine the change in the internal energy .

Sol.

From N. F.E.E

Q = - 1500 KJ (The heat out)

W = - 5090 KJ (the work input to the system)

- 1500 (-5090) =

Questions

1- A mass of oxygen is compressed with out friction from initial state volume = 0.14 m3 pressure = 1.5 bar ; the temp = 15 C0 to final state the volume = 0.06 m3 and the pressure = 1.5 bar at this process the oxygen losses quantity of heat = 0.6 KJ to the surrounding Cv = 0.649 KJ/Kg.k R = 0.25 KJ/Kg.k0 .

Find : 1) the change of internal energy .

2) the final temp.

: Applying the first Law on opened systems , examples : 1-

.Application of energy equation on open system :- S. F. E. E

1) The Boiler

The boiler is a heat exchange which convert the liquid water to steam at constant pressure . P = C

h2 > h1 heat added

Q (+)

2) The Condenser

It is a heat exchanger work on condenser steam of water and converted it into a liquid by cooling the steam under constant pressure .

Q = m (h2 h1)

h1 > h2

Q (-1) rejected (heat out)

Ex. 1A boiler is generate steam with rate of 8 kg/s the enthalpy of liquid water is 271 and the enthalpy of steam is 3150 . For steady state conditions and by neglecting the change in K.E and P.E . Determine the rate of heat added to the steam in Boiler .

Solution

Q = m (h2 h1)

h1 = 271 KJ/kg h2 = 3150

m = 8 kg/s

Q = m (h2 h1)

= 8 (3150 271) = 23032 = 23032 KW

( T )

3- The Turbine

It is a mechanical device used to convert the kinetic energy of fluid into mechanical work

T1 > T2 h1 > h2 P1 > P2

- W = m (h2 h1)

W = (+) work output 4- The Compressor It is a mechanical device used to increase fluid pressure by using on external mechanical work .

h1 > h2 T2 > T1P2 > P1

- W = m (h2 h1)

W = (-) work input

Ex. Air enters a compressor at enthalpy (5000 J/kg) and leave with enthalpy (250 KJ / kg) the mass flow rate = 0.25 kg/s , Determine the work done .

Sol.

h1 = 5000 J / kg =

h2 = 250

m = 0.25 kg / s

w = m (h1 h2) = - 0.25 (5 250)w = 0.25 (-245) = - 61.25 KW

: Specific heat kinds of specific heat and relations between them : 1-

Specific Heat

Is the a mount of heat required to rise the temperature of a unit mass of substance one degree .

Types of Specific heat

1- Specific heat at constant pressure

*

= Specific heat at constant pressure.

Unit [ ]

= change in enthalpy

Unit [ ]

= change in temp.

Unit [ k0 ]

*

*

m = mass (kg)

T2 = Final temp. (k0) T1 = initial temp. (k0) Q = heat (KJ , J)

2- Specific heat at constant volume

V = constant closed system

= Specific heat at constant volume

Unit (

= change in internal energy

Unit []

= change in temp. (k0) = T2 - T1 For m = 1 kg

*

*The relation between ()

Cama () :- Is the adiabatic expone which represent the ratio between specific heat at constant pressure and specific heat at constant volume .

Examples

1- compute the constant pressure specific heat of steam if the change in enthalpy is (104.2) KJ and the change in temp. is (50 KJ) .

Solution

KJ / kg . k0

2- Example (2)

Compute the change of enthalpy as (1 kg) of a gasis heated from 300 k0 to 1500 k0 , .

Sol.

3- Example (3)

1- Compute the change of internal energy as (219) of a gas is heated .2- Compute the amount of heat added . ,

Sol.1-

The relation between (R and Cp & Cv)

(1)We know that (2)

(3)

And for ideal gas p.v = m . R. T

For m = 1 kg

p.v = R . T

(4)

Sub . equations (2) , (3) , (4) in eq. (1)

2 3 4 (1)

Home work

Prove that : -

The relation between R and ()

R = (1)

(2)

(2) (1)

(3)

(3) (2)

Home work

Prove that 1)

2)

: Gas Constant , the universal gas constant and specific , Examples : 1-

The general equation of Ideal Gas

constant ----- (1) (1) (m)

Let

----- (2)

constant gas R = characteristic

----- (3)

Units

= absolute pressure

V = volume (m3)

T = temp. (k0)

= mass (kg)

R =

R0 = universal gas constant

M = Molecular weight (k mole) ( )

Example :- A tank has a volume of (0.5 in3) and contains (10 kg) of an ideal gas having a molecular weight = 24 , the temp = 25 C0 compute the pressure

SolutionR0 = R . M

kpaEx . (5) :- one kg of a perfect gas accupies a volume of (0.85 m3) at (15C0) and at a constant pressure of (1 bar) . The gas is first heated at a constant volume and then at a constant pressure .

Compute : 1) the specific heat at constant volume (Cv).

2) the specific heat at constant pressure (Cp) Solution

P1 V1 = m . R . T1

1 (kg) * R * (15 +273) k0

1)

Questions

What is the mass of air contained a room (6m x 10m x 4m) If the pressure is (100 kpa) and the temp. is 25 C0 , Assume air to be an ideal gas

:Ideal gas, boyle,s law, chal,s law , examples : 1-

.

Ideal gas

The ideal gas is defined as the state of substance that follows well know Bolyle,s and charle,s laws .

:

Laws of Ideal gas The physical properties of a gas are controlled by the following variables :- pressure (P) exerted by the gas .

Volume () occupied by the gas .

Temperature (T) of the gas .

The behaviour of perfect gas is governed by the following laws :-

1) Boyle,s Law

The absolute pressure of a given mass of ideal gas varies inversely of its volume when the temp. remain constant

.

T = C

T = CUnit of pressure (Pa)

Charle,s Law

The volume of a given mass of ideal gas varies directly with the temp. when the absolute pressure remain constant .

P = constant

P = CUnits

V1 = initial volume

V2 = final volume

Gay Lussac Law

The absolute pressure of a given mass of ideal gas is proportional directly with the temp. at constant volume . .

V = constant

General Process

It is transition of a substance from state (1) [ P1 , V1 , T1] to state (2) [P2 , V2 , T2] a cross a certain path .

(1) (2) .

= constant

P = pressure ()

V = volume m3

T = temperature (k0)

Examples (1) An air compress or is compress (2.8 m3) of air from initial pressure of (1 bar) to final , pressure of (14 bar) calculate the final volume of air if temp. is constant .Solution :-

P1 V1 = P2 V2 T = C

1 bar * 2.8 m3 = 14 bar * V2

Examples (2)

(0.2 m3) of a gas at (50 C0) the gas is heated at a constant pressure until its volume reached (0.4 m3) compute the final temp.

Solution :-

0.2 T2 = 0.4 [ 50 + 273]

Examples (3)(2 kg) of a gas at initial pressure of (1.4 bar) and temp. = 40 C0 , R = 188.34 , Determine (a) the volume of the gas (b) If the gas is heated at constant pressure to have a volume of (1.5 m3) . Find the final temp.

Solution

P1 V1 = m R T1

m3

b) P = C

T2 = 557.6 k0 : Thermodynamic Processes and applications

: 1- .

Some Processes for closed systems

1- Constant pressure process (Isobaric)

Constant volume process (Isometric)

3- Constant temp. process [Isothermal]

T = C

Ex. 1The pressure of a gas = 1.5 bar at 18 C0 temp. compute

1- the volume of (1 kg) of the gas .

If the gas is heated at constant pressure until the volume become [1m3]

Compute : a- the a mount of added heat .

b- the work done .

Cp = 1.005 KJ/kg.k Cv = 0.718

Solution

R = Cp - CvR = 1.005 0.718 = 0.278

P1 V1 = m R T1

V1 = 0.556 m3Q12 = m Cp [T2 T1]

T2 = 526.2 k0

Q = 1 * 1.005 * [526.2 291] = 243.4 KJ

W = P [V2 V1] = 1.5 x 102 [1- 0.556] = 66.48 KJ

Adiabatic Process

Ex. 1 : Given [ the mass of the gas = 1 kg , the pressure = 1.5 bar ,

the temp = 18C0 ] Cv = 0.718 , Cp= 1.005

compute : 1- the volume of the gas , the gas is heated at constant pressure

until the volume become (1m3). 2- the added heat .

3- the work done .

Solution

1.005 0.718 = 0.278

P1 V1 = m R T11.5 * 102 * V1 = 1 kg * 0.278 * [18 +273] .

m3Q = m Cp [T2 T1]

T2

Q = 1 * 1.005 (526.2 291)

Q = 243.4 KJ

W = P [ V2 V1]

W = 1.5 * 102 [1 - 0.566]

W = 66.48 KJ

Ex . 2 : Given [ the pressure of air = 1.013 bar , the volume = 0.827 m3 and the temp. = 25 C0 , the air is compressed with constant temp. until the pressure becomes = 13.78 bar .

Compute :- the work done to compress the air .

Solution

----- (1) V2

P1 V1 = P2 V2 T = C V2

1.013 * 0.827 = 13.78 V2

m3 V2 (1)

KJ

.

Ex . 3 : Given [ the volume of gas = 0.12 m3 , the temp. = 20 C0 and the pressure = 1.013 bar is compressed adibatically until the volume become = 0.024 m3 Cv = 0.718 , Cp = 1.005 .Compute : 1) the mass of the gas .

2) the final pressure and temp.

3) the work done .

Solution

R = Cp Cv = 1.005 0.718 = 0.287

P1 V1 = m R T11.013 * 102 * 0.12 m3 = m * 0.287 * [20 + 273]

1) kg

1.013 (0.12)1.4 = P2 (0.024)1.42) P2 = 1.013 = 9.64 bar ] ] T2

k03)

W12 = - 85.282 KJ

: Reversible and irreversibility

: 1-

.

Reversible and Irreversible Process

:

:

Reversible process :

The process in which the system and surrounding can be restored to the initial state without producing any change in the thermodynamic properties :-

Conditions of reversible process

1- All the initial and final state of system should be in equilibrium with the each other .

2- The process should occur in very small

Reversible processes

1- Friction relative motion

2- Extension of spring

3- Slow frictionless adiabatic expansion

Irreversible process

It is state that both the system and surrounding cannot return to original state .

: The second law of thermodynamic

Results of the second law of thermodynamic

The second law of thermodynamic The first law of thermodynamic indicates that the net heat supplied in a cycle is equal to the net work done the gross heat supplied must be greater than the net work done some of heat must be rejected by the system .

Kelven blank definition :-

It is impossible for heat engine to produce net work in a complete cycle if it exchange heat only with bodies at a single fixed temperature .

The second law has been stated in several ways .

(1) The principle of Thomson (Lord Kelvin) states :

It is impossible by a cyclic process to take heat from a reservoir and to convert it into work without simultaneously transferring heat from a hot to a cold reservoir . This statement of the second law is related to equilibrium , i.e. work can be obtained from a system only when the system is not already at equilibrium . If a system is at equilibrium , no spontaneous process occurs and no work is produced .

Evidently , Kelvin's principle indicates that the spontaneous process is the heat flow from a higher to a lower temperature , and that only from such a spontaneous process can work be obtained . (2) The principle of Clausius States : It is impossible to devise an engine which , working in a cycle , shall produce no effect other than the transfer of heat from a colder to a hotter body . A good example of this principle is the operation of a refrigerator .

(3) The principle of Planck states : It is impossible to construct an engine which , working in a complete cycle , will produce no effect other than raising of a weight and the cooling of a heat reservoir .

(4) The Kelvin Planck Principle :

May be obtained by combining the principles of Kelvin and of Planck into one equivalent statement as the Kelvin Planck statement of the second law . It states : No process is possible whose sole result is the absorption of heat from a reservoir and the conversion this heat into work . : Heat engine , heat pump : .Heat engine

A heat engine is a system operating in a complete cycle and developing net work from supply heat .

The second law implies that a source of heat supply and sink for the rejection of heat are both necessary since some heat must be always be rejected bythe system .

By first law

Net heat supplied = network doneQ1 Q2 = W

Q1 > W the second law of thermodynamic

The thermal efficiency of heat engine

One good example in practice of heat engine is a simple steam cycle in this cycle heat is supplied in boiler work is developed in yurbine heat is rejected in a condenser and small amount of work is required for the pump .

Simple steam cycleHeat Pump

The heat pump is reversed heat engine in the heat pump (or refrigerator)cycle an amount of heat Q2 is supplied from the cold reservoir and an amount of heat Q1 is rejected to the hot reservoir and there must be a work done on the cycle W

Heat pump (refrigerator)

Q1 = Q2 = W

There for W > 0 , the heat pump requires an input energy in order to transfer heat from the cold chamber and reject it at higher temp .

Heat pump system

: Entropy , changes on closed systems and temp - Entropy plan : 1- .

2- .

Entropy It is define as thermodynamic property that express the amount of storage energy in the system also represent measure of reversible and Irreversible process .

Temp. Entropy Plane

For the reversible process the area under (T-S) plane = the heat (1)

dQ = T ds --------- (1)

From equation (1)

J / k0

1- The entropy at constant volume

2- The entropy at constant pressure

Ex : 1

Comput the entropy for reversible process at constant pressure the temp vary from 120 C0 to 270 C0

Solution

: Carnot Cycle

: 1-

.

Carnot Cycle A Carnot gas cycle operating in a given temperature range is shown in the T-s diagram in Fig. 1(a) . One way to carry out the processes of this cycle is through the use of steady state , steady flow devices as shown in Fig. 1(b) . The isentropic expansion process 2-3 and the isentropic compression process 4-1 can be simulated quite well by a well designed turbine and compressor respectively , but the isothermal expansion process 1-2 and the isothermal compression process 3-4 are most difficult to achieve . Because of these difficulties , a steady flow Carnot gas cycle is not practical .

The Carnot gas cycle could also be achieved in a cylinder piston apparatus (a reciprocating engine) as shown in Fig. 4.2 (b) . The Carnot cycle on the p-v diagram is as shown in Fig 4.2 (a) , in which processes 1-2 and 3-4 are isothermal while processes 2-3 and 4-1 are isentropic . We know that the Carnot cycle efficiency is given by the expression .

Fig. 1 . Steady flow Carnot engine

Fig. 2. Reciprocating Carnot engine

Fig. 3. Carnot cycle on P-v and T-s diagrams

Fig. 4. Working of Carnot engine

Since the working fluid is an ideal gas with constant specific heats , we have , for the isentropic process ,

Now , T1 = T2 and T4 = T3 , therefore

compression or expansion ratio

Carnot cycle efficiency may be written as ,

From the above equation , it can be observed that the Carnot cycle efficiency increases as "r" increases . This implies that the high thermal efficiency of a Carnot cycle is obtained at the expense of large piston displacement . Also, for isentropic processes we have ,

and

Since , T1 = T2 and T4 = T3 , we have

pressure ratio

Therefore , Carnot cycle efficiency may be written as ,

From the above equation , it can be observed that , the Carnot cycle efficiency can be increased by increasing the pressure ratio . This means that Carnot cycle should be operated at high peak pressure to obtain large efficiency .

Ex. 1

The highes theortical efficiency of gasoline engine based on the Carnot cycle is 30 y0 if this engine expels its gases into at m . which has temp of 300k . compute 1) the temp in the cylinder immediately after combustion .

2) if the engine absorbs (837 J) of heat from the hot reservoir during each cycle how much work can it perform in each cycle ,

Solution 1)

2)

Ex . 2

A Carnot engine is operated between two heat reservoirs at temp of 450 k0 and 350 k0 , if the engine receive (1000 J) of heat in each cycle .

Compute : 1) the amount of heat reject .

2) the efficiency of the engine .

3) the work done by the engine in each cycle .

Solution

1) , ,

2)

3) the work done = Qh Qc = 1000 777.7

= 222.3 J

: : 1-

.Otto Cycle : [ Constant Volume Cycle ]

The cycle consist of four reversible process

1-2 Adibatic Compression

V1 V2 T1 T2 .

Win

23 Combustion stroke

Constant volume , heat addition

T2 T3 Qh (added heat) .

Added heat Qh = m Cv (T3 T2)3-4 Power stroke Adibatic expansion

V2 V1 T3 T4 ( )

A standard dtt cycle Exhaust stroke

4-1 Constant volume , heat rejection

( )

( ) T4 T1 Qc .

Rejected heat Qc = m Cv [ T1 T4 ]

The Heat efficiency (eff.) of otto cycle

5-1

: ( ) 5-1 ( ) .

:

Diesel Cycle net work out put and its eff.

: 1-

.

Diesel Cycle

The Diesel cycle is a compression ignition (rather than spark ignition) engine . Fuel is sprayed into the cylinder at P2 (high pressure) when the compression is complete , and there is ignition without spark . An idealized Diesel engine cycle is shown in figure 1.

Figure 1. The ideal Diesel cycle

The thermal efficiency is given by :

=

This cycle can operate with a higher compression ratio than the Otto cycle because only air is compressed and there is no risk of auto ignition of the fuel . Although for a given compression ratio the Otto cycle has higher efficiency , because the Diesel engine can be operated to higher compression ratio , the engine can actually have higher efficiency than an Otto cycle when both are operated at compression ratios that might be achieved in practice . Muddy Points When and where do we use Cv and Cp ? Some definitions use dU=Cv dT is it ever dU = Cp dT ? (MP 3.8)

Explanation of the above comparison between Diesel and Otto. (MP 3.9)

Air standard diesel engine cycleThe term "compression ignition" is typically used in technical literature to describe the modern engines commonly called "Diesel engines" . This is in contrast to "spark ignition" for the typical automobile gasoline engines that operate on a cycle derived from the Otto cycle . Rudolph Diesel patented the compression ignition cycle which bears his name in the 1890s.

Design of a Diesel Cycle

The General Idea

The Diesel cycle is very similar to the Otto cycle in that both are closed cycles commonly used to model internal combustion engines . The difference between them is that the Diesel cycle is a compression ignition cycle use fuels that begin combustion when they reach a temperature and pressure that occurs naturally at some point during the cycle and , therefore , do not require a separate energy source (e.g. from a spark plug) to burn . Diesel fuels are mixed so as to combust reliably at the proper thermal state so that Diesel cycle engines run well .

(We might note that most fuels will start combustion on their own at some temperature and pressure . But this is often not intended to occur and can result in the fuel combustion occurring too early in the cycle . For instance , when a gasoline engine ordinarily an Otto cycle device is run at overly high compression ratios , it can start "dieseling" where the fuel ignites before the spark is generated . It is often difficult to get such an engine to turn off since the usual method of simply depriving it of a spark may not work .

Stages of Diesel Cycles

Diesel Cycles have four stages : compression , combustion , expansion , and cooling .

Compression

We start out with air at ambient conditions often just outside air drawn into the engine . In preparation for adding heat to the air , we compress it by moving the piston down the cylinder . It is in this part of the cycle that we contribute work to the air . In the ideal Diesel cycle , this compression is considered to be isentropic .

It is at this stage that we set the volumetric compression ratio , r which is the ratio of the volume of the working fluid before the compression process to its volume after .

Piston : moving from top dead center to bottom dead center . Combustion

Next , heat is added to the air by fuel combustion . This process begins just as the piston leaves its bottom dead center position . Because the piston is moving during this part of the cycle , we say that the heat addition is isochoric , like the cooling process .

Piston : starts at bottom dead center , begins moving up .

Expansion

In the Diesel cycle , fuel is burned to heat compressed air and the hot gas expands forcing the piston to travel up in the cylinder . It is in this phase that the cycle contributes its useful work , rotating the automobile's crankshaft . We make the ideal assumption that this stage in an ideal Diesel cycle is isentropic .

Piston : moving from bottom dead center to top dead center .

Cooling

Next , the expanded air is cooled down to ambient conditions . In an actual automobile engine , this corresponds to exhausting the air from the engine to the environment and replacing it with fresh air . Since this happens when the piston is at the top dead center position in the cycle and is not moving , we say this process is isochoric (no change in volume) .

Piston : at top dead center .

: Dual Cycle

: 1- .

Limited Pressure Cycle (or Dual Cycle) :

This cycle is also called as the dual cycle , which is shown in Fig. 1 Here the heat addition occurs partly at constant volume and partly at constant pressure . This cycle is a closer approximation to the behavior of the actual Otto and Diesel engines because in the actual engines , the combustion process does not occur exactly at constant volume or at constant pressure but rather as in the dual cycle .

Process 1-2 : Reversible adiabatic compression .

Process 2-3 : Constant volume heat addition .

Process 3-4 : Constant pressure heat addition .

Process 4-5 : Reversible adiabatic expansion .

Process 5-1 : Constant volume heat rejection .

Fig. 1 Dual cycle on p-v and T-s diagramsAir Standard Efficiency

Heat supplied = m Cv (T3 T2) + m Cp (T4 T3)

Heat rejected = m Cv (T5 T1)

Net work done = m Cv (T3 T2) + m Cp (T4 T3) m Cv (T5 T1)

Let ,

=

=

From the above equation , it is observed that , a value of rp > 1 results in an increased efficiency for a given value of rc and . Thus the efficiency of the dual cycle lies between that of the Otto cycle and the Diesel cycle having the same compression ratio . Mean Effective Pressure

=

=

= : Comparing between Fuel air and the air standard cycles

: 1-

.Fuel air cycle

The simple ideal air standard cycles overestimate the engine efficiency by a factor of about 2. A significant simplification in the air standard cycles is the assumption of constant specific heat capacities . Heat capacities of gases are strongly temperature dependent , as shown by figure (1) .

The molar constant volume heat capacity will also vary , as will the ratio of heat capacities :

If this is allowed for , air standard Otto cycle efficiency falls from 57 per cent to 49.4 per cent for a compression ratio of 8 .

When allowance is made for the presence of fuel and combustion products , there is an even greater reduction in cycle efficiency . This leads to the concept of a fuel air cycle which is the same as the ideal air standard Otto cycle , except that allowance is made for the real thermodynamic behaviour of the gases . The cycle assumes instantaneous complete combustion , no heat transfer , and reversible compression and expansion . Taylor (1966) discusses these matters in detail and provides results in graphical form . Figure (2) and (3) . INTRODUCTION TO INTERNAL COMBUSTION ENGINES

Figure (1) : Molar heat capacity at constant pressure of gases above 150C quoted as averages between 150C and abscissa temperature Show the variation in fuel air cycle efficiency as a function of equivalence ratio for a range of compression ratios . Equivalence ratio is defined as the chemically correct (stoichiometric)air / fuel ratio divided by the actual air / fuel ratio . The datum conditions at the start of the compression stroke are pressure (p1) 1.013 bar , temperature (T1) 1150C , mass fraction of combustion residuals (f) 0.05 , and specific humidity () 0.02 the mass fraction of water vapour .

The fuel 1- octane has the formula C8H16 , and structure

Figure (2) shows the pronounced reduction in efficiency of the fuel air cycle for rich mixtures . The improvement in cycle efficiency with increasing compression ratio is shown in figure 3 , where the ideal air standard Otto cycle efficiency has been included for comparison .

In order to make allowances for the losses due to phenomena such as heat transfer and finite combustion time , it is necessary to develop computer models .

Prior to the development of computer models , estimates were made for the various losses that occur in real operating cycles . Again considering the Otto

Figure (2) : Variation of efficiency with equivalence ratio for a constant volume fuel air cycle with 1 octane fuel for different compression ratios ( adapted from Taylor (1966))

Cycle , these are as follows :

(a) "Finite piston speed losses" occur since combustion takes a finite time and cannot occur at constant volume . This leads to the rounding of the indicator diagram and Taylor (1960) estimates these losses as being about 6 per cent .

Figure (3) : Variation of efficiency with compression ratio for a constant volume fuel air cycle with 1 octane fuel for different equivalence ratios (adapted from Taylor (1966))

(b) "Heat losses" , in particular between the end of the compression stroke and the beginning of the expansion stroke . Estimates of up to 12 per cent have been made by both Taylor (1966) and Ricardo and Hempson (1968) . However, with no heat transfer the cycle temperatures would be raised and the fuel air cycle efficiencies would be reduced slightly because of increasing gas specific heats and dissociation .

(c ) Exhaust losses due to the exhaust valve opening before the end of the expansion stroke . This promotes gas exchange but reduces the expansion work . Taylor (1966) estimates these losses as 2 per cent .

Since the fuel is injected towards the end of the compression stroke in compression ignition engines (unlike the spark ignition engine where it is pre-mixed with the air) the compression process will be closer to ideal in the compression ignition engine than in the spark ignition engine . This is another reason for the better fuel economy of the compression ignition engine . : The actual cycle

Comparing between actual cycles and air standard cycles

: 1- .

2- .The Actual Cycle

The eff of the actual cycle is low than the air standard cycle .

1- The fluid used is a mixture of air and fuel and the exhaust gases .

2- Energy loss is caused by time when the valves are opened or closed .

3- The specific heat varies with temp.

4- Due to the chemical dissociation there will energy loss .

5- The combustion isn't complete

6- The engine loses heat directly .

7- There are heat losses with exhaust gases .

8- The engine loses heat directly .

Air Standard Cycle1- The efficiency is high .

2- The fluid used in the cycle is air follows the ideal gas PV= m R T

3- The used gas has a constant mass of air in the closed system or moves with constant flourate in closed cycle .

4- The value of specific heat is constant .

5- Chemical reactions don't occur .

6- The compression and expansion occurs with constant entropy .

7- The received heat and rejected heat occurs reversibly .

8- There is no change I the kinetic & the potential energies that is why they are neglected .

9- The engine works without friction . : Gas Turbine

:

Gas Turbine

Gas turbine . types which burn fuels such as oil and natural gas . Instead of using heat to produce steam , as in steam turbines , the gas turbine hot gas is used directly . Used for the operation of gas turbine generators , ships , race cars , as used in jet engines .

Most of the gas turbine systems of the three main parts

1- the air compressor , 2- the combustion chamber , 3- turbine

The socalled air compressor with the combustion chamber , usually , the gas generator . In most systems , gas turbine , the air compressor and turbine boats on both sides of a horizontal axis , located between the combustion chamber . And is part of the power turbine air compressor .

Absorbs quantity of air compressor and air ldguetha , so pressure is increasing . In the combustion chamber , compressed air mixes with fuel and burning the mixture . The more air pressure , improved combustion of fuel mixed with air . The burning gases expand rapidly and to flow into the turbine , leading to the rotation of the wheels of the turbine . And moving hot gases through the various stages in the same manner as the gas turbine flows through the steam turbine vapor . The Downs fixed mobile gas resat member to change the rotor speed .

And benefit most from the gas turbine systems , the hot gases emerging from the turbine . In some systems are some of these gases , and go to a so called renewed . And these gases are used to heat the compressed air after leaving the air compressor . Before entering the combustion chamber reduces heat the compressed air in this way the amount of fuel used for combustion process . In jet engines , most of the gas used to produce momentum . See : Jet propulsion .

Gas turbines operate at a temperature higher than the steam turbines . And increase the efficiency of the turbine the more the temperature of operation ; a standard operating temperature of most gas turbines is 8750C or more .

Gas turbine (Gas Turbine) for this type of turbine has many uses , it is used in aircraft and jet propelled means of transport and maritime and land in the area of oil as well for use in power plants , especially in peak hours exceeded . Advantages of speed of operation (unlike the vapor turbine , which needs to prepare a preliminary arrangements) .

Contents

[Hide]

1 gas turbine components

0 1.1 ancillary equipment

2 fuel gas turbine

3 disadvantages of gas turbine

4 external links

[Changed] the components of gas turbine Gas turbine consists of the following main parts :

air compressor (The Air Compressor) : take the air from the atmosphere and the pressure to raise tens of air pressure .

the combustion chamber (The Combustion Chamber) : the mixed compressed air from the pressurized air with fuel and burn together by means of special Balachtal , and the combustion products of various gases and high temperatures up to 1000 degrees Fahrenheit , the pressure is high .

Turbine (The Turbine) : It would be centered with the horizontal axis attached directly to the air pressure on the one hand and from other mechanical contraceptives to be recycled (such as when a generator for example) and through the gearbox (Gear Box) to reduce the speed as the speed of rotation of the turbine are very high .

Interference resulting from the combustion gases in the turbine Vtstdm quill of the many and then to the chimney.

Ancillary Equipment

Gas turbines need to operate safely and the safety of certain machinery and equipment assistance (Auxiliaries) , as follows :

* Filters the air before entering the air pressure (air in take filters).

* initial operation of any assistant (the first operation of Starter) , which is either electric motor or diesel engine or a steam turbine (starting steam turbine) .

* the means or system setting .

* cooling system .

* system control and measurement equipment , temperature and pressure at each stage of the work and an integrated operating system , such as (mark 4, mark 5, mark 6) contains a processor , or more .

Fuel Gas turbine

Gas turbine operating on many types of fuel , it operates on natural gas (Natural Gas) and on diesel , gasoline and even on the crude oil (with some additions , chemical and arrangements) .

References

1- Thermodynamic , M.M ABBott .

2- Engineering thermodynamics , D.B. SPALDING and E. H. Cole .

3- Thermodynamics

ViRGil MORING FAIRES

4- Thermodynamics

KENNETH WARK , JR .

5- Gas turbine theory

H , COHEN

G.F.C. ROGERS

6- Engineering thermodynamics

G. P. Gupta and R. Praksh

7- Internal Combustion Engines ;

Harper and Row Inc. boundary

cylinder

Q=0

N=0

Q in Q out

No heat

boundary

Isolate

System

piston

1

1`

in

out

Qout

Qin

System

1

wout

win

boundary

2

Pg = Pabs Patm

Patm = atmospheric pressure

h = height of the liquid

EMBED Equation.3

body

F

x1

x2

EMBED Equation.3

V2v

WF = P . V

EMBED Equation.3

EMBED Equation.3

1

2

Q(+)

Q(-)

N1

Entry

W

P1 u1

V1 C1

Exit

Z2

Q

C1

P1V1 , u1

EMBED Equation.3

EMBED Equation.3

Q = m (h2 h1)

steam

water

1

1

2

water

W = m (h1 h2)

input

2

output

W = m (h1 h2)

2

1

EMBED Equation.3

Q1-2 = m EMBED Equation.3

Q1-2 = m cp EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

Cp - Cv

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

R0 = R * M

PV = C

P1 V1 = P2 V2 = C

(1)

(2)

(pressure)

(P)

(V) volume

PV = C

EMBED Equation.3

P

P1 = P2

(1)

(2)

V

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

P

V1

V2

V

1

2

WD

P

P2

P1

V

V1 = V2

1

2

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

Source

Heat engine

Sin k

T hot

T cold

Qout

Qin

W0

Q2

Q1

W

Heat engine

Boiler

condenser

pump

T hot

Qout

Win

T cold

Qin

Expansion valve

EMBED Equation.3

P

V2

V1

P0

EMBED Equation.3

PAGE 90

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