AT2251

APPLIED THERMODYNAMICS AND

HEAT TRANSFER

UNIT-I GAS POWER CYCLES 10+3

Air standard cycles-Otto-Diesel-Dual-Work output,Efficiency and MEP calculations comparison of the cycles for same compression ratio and heat addition,same compression ratio and heat rejection,same peak pressure, peak temperature and heat rejection, same peak pressure and heat input,same peak pressure and work output , Brayton cycle with intercooling, reheating and regeneration.

UNIT-II:RECIPROCATING AIR COMPRESSORS &REFRIGERATION CYCLES 10+3

Single acting and double acting air compressors, work required, effect of clearance volume, volumetric efficiency, isothermal efficiency, free air delivery, multistage compression, condition for minimum work. Fundamentals of refrigeration, C.O.P., reversed carnot cycle, simple vapour compression refrigeration system, T-S, P-H diagrams, simple vapour absorption refrigeration system, desirable properties of an ideal refrigerant. UNIT-III: CONDUCTION 10+2 Basic Concepts Mechanism of Heat Transfer Conduction, Convection and Radiation General Differential equation of Heat Conduction Fourier Law of Conduction Cartesian and Cylindrical Coordinates One Dimensional Steady State Heat Conduction Conduction through Plane Wall, Cylinders and Spherical systems Composite Systems Conduction with Internal Heat Generation Extended Surfaces Unsteady Heat Conduction Lumped Analysis Use of Heislers Chart.

UNIT-IV: CONVECTION 10+2

Basic Concepts Convective Heat Transfer Coefficients Boundary Layer Concept Types of Convection Forced Convection Dimensional Analysis External Flow Flow over Plates, Cylinders and Spheres Internal Flow Laminar and Turbulent Flow Combined Laminar and Turbulent Flow over Bank of tubes Free Convection Dimensional Analysis Flow over Vertical Plate, Horizontal Plate, Inclined Plate, Cylinders and Spheres.

UNIT-V: RADIATION 8+2

Basic Concepts, Laws of Radiation Stefan Boltzman Law, Kirchoff Law Black Body Radiation Grey body radiation Shape Factor Algebra Electrical Analogy Radiation Shields Introduction to Gas Radiation. TOTAL-60

Chapter - 1

GAS POWER CYCLES

AIR STANDARD CYCLES Theoretical Analysis The accurate analysis of the various processes taking place in an internal combustion engine is a very complex problem. If these processes were to be analyzed experimentally, the analysis would be very realistic no doubt. It would also be quite accurate if the tests are carried out correctly and systematically, but it would be time consuming. If a detailed analysis has to be carried out involving changes in operating parameters, the cost of such an analysis would be quite high, even prohibitive. An obvious solution would be to look for a quicker and less expensive way of studying the engine performance characteristics. A theoretical analysis is the obvious answer. A theoretical analysis, as the name suggests, involves analyzing the engine performance without actually building and physically testing an engine. It involves simulating an engine operation with the help of thermodynamics so as to formulate mathematical expressions which can then be solved in order to obtain the relevant information. The method of solution will depend upon the complexity of the formulation of the mathematical expressions which in turn will depend upon the assumptions that have been introduced in order to analyze the processes in the engine. The more the assumptions, the simpler will be the mathematical expressions and the easier the calculations, but the lesser will be the accuracy of the final results. The simplest theoretical analysis involves the use of the air standard cycle, which has the largest number of simplifying assumptions. A Thermodynamic Cycle In some practical applications, notably steam power and refrigeration, a thermodynamic cycle can be identified. A thermodynamic cycle occurs when the working fluid of a system experiences a number of processes that eventually return the fluid to its initial state. In steam power plants, water is pumped (for which work WP is required) into a boiler and evaporated into steam while heat QA is supplied at a high temperature. The steam flows through a turbine doing work WT and then passes into a condenser where it is condensed into water with consequent rejection of heat QR to the atmosphere. Since the water is returned to its initial state, the net change in energy is zero, assuming no loss of water through leakage or evaporation.

An energy equation pertaining only to the system can be derived. Considering a system with one entering and one leaving flow stream for the time period t1 to t2

)1(systemff EEEWQ outin =+ Q is the heat transfer across the boundary, +ve for heat added to the system and ve for heat taken from the system. W is the work transfer across the boundary, +ve for work done by the system and -ve for work added to the system

infE is the energy of all forms carried by the fluid across the boundary into the system

outf

E is the energy of all forms carried by the fluid across the boundary out of system Esystem is the energy of all forms stored within the system, +ve for energy increase -ve for energy decrease In the case of the steam power system described above

+===+ )2(PTRA WWWQQQ

All thermodynamic cycles have a heat rejection process as an invariable characteristic and the net work done is always less than the heat supplied, although, as shown in Eq. 2, it is equal to the sum of heat added and the heat rejected (QR is a negative number). The thermal efficiency of a cycle, th, is defined as the fraction of heat supplied to a thermodynamic cycle that is converted to work, that is

)3(A

RA

Ath

QQQ

QW

+=

=

This efficiency is sometimes confused with the enthalpy efficiency, e, or the fuel conversion efficiency, f

)4(cf

e QmW=

This definition applies to combustion engines which have as a source of energy the chemical energy residing in a fuel used in the engine. Any device that operated in a thermodynamic cycle, absorbs thermal energy from a source, rejects a part of it to a sink and presents the difference between the energy absorbed and energy rejected as work to the surroundings is called a heat engine. A heat engine is, thus, a device that produces work. In order to achieve this purpose, the heat engine uses a certain working medium which undergoes the following processes:

1. A compression process where the working medium absorbs energy as work. 2. A heat addition process where the working medium absorbs energy as heat from a source. 3 An expansion process where the working medium transfers energy as work to the

surroundings. 4. A heat rejection process where the working medium rejects energy as heat to a

sink.

If the working medium does not undergo any change of phase during its passage through the cycle, the heat engine is said to operate in a non-phase change cycle. A phase change cycle is one in which the working medium undergoes changes of phase. The air standard cycles, using air as the working medium are examples of non-phase change cycles while the steam and vapor compression refrigeration cycles are examples of phase change cycles. Air Standard Cycles The air standard cycle is a cycle followed by a heat engine which uses air as the working medium. Since the air standard analysis is the simplest and most idealistic, such cycles are also called ideal cycles and the engine running on such cycles are called ideal engines. In order that the analysis is made as simple as possible, certain assumptions have to be made. These assumptions result in an analysis that is far from correct for most actual combustion engine processes, but the analysis is of considerable value for indicating the upper limit of performance. The analysis is also a simple means for indicating the relative effects of principal variables of the cycle and the relative size of the apparatus. Assumptions 1. The working medium is a perfect gas with constant specific heats and molecular

weight corresponding to values at room temperature. 2. No chemical reactions occur during the cycle. The heat addition and heat rejection

processes are merely heat transfer processes. 3. The processes are reversible.

4. Losses by heat transfer from the apparatus to the atmosphere are assumed to be zero in this analysis.

5. The working medium at the end of the process (cycle) is unchanged and is at the same condition as at the beginning of the process (cycle).

Selecting an idealized process one is always faced with the fact that the simpler the assumptions, the easier the analysis, but the farther the result from reality. The air cycle has the advantage of being based on a few simple assumptions and of lending itself to rapid and easy mathematical handling without recourse to thermodynamic charts or tables or complicated calculations. On the other hand, there is always the danger of losing sight of its limitations and of trying to employ it beyond its real usefulness. Equivalent Air Cycle A particular air cycle is usually taken to represent an approximation of some real set of processes which the user has in mind. Generally speaking, the air cycle representing a given real cycle is called an equivalent air cycle. The equivalent cycle has, in general, the following characteristics in common with the real cycle which it approximates: 1. A similar sequence of processes. 2. Same ratio of maximum to minimum volume for reciprocating engines or

maximum to minimum pressure for gas turbine engines. 3. The same pressure and temperature at a given reference point. 4. An appropriate value of heat addition per unit mass of air. The Carnot Cycle This cycle was proposed by Sadi Carnot in 1824 and has the highest possible efficiency for any cycle. Figures 1 and 2 show the P-V and T-s diagrams of the cycle. Assuming that the charge is introduced into the engine at point 1, it undergoes isentropic compression from 1 to 2. The temperature of the charge rises from Tmin to Tmax. At point 2, heat is added isothermally. This causes the air to expand, forcing the piston forward, thus doing work on the piston. At point 3, the source of heat is removed and the air now expands isentropically to point 4, reducing the temperature to Tmin in the process. At point 4, a cold body is applied to the end of the cylinder and the piston reverses, thus compressing the air isothermally; heat is rejected to the cold body. At point 1, the cold body is removed and the charge is compressed isentropically till it reaches a temperature Tmax once again. Thus, the heat addition and rejection processes are isothermal while the compression and expansion processes are isentropic. From thermodynamics, per unit mass of charge

Heat supplied from point 1 to 2 )5(ln2

322 v

vvp=

Heat rejected from point 4 to 1 )6(ln4

144 v

vvp=

Now p2v2 = RTmax (7)

And p4v4 = RTmin (8)

Since Work done, per unit mass of charge, W = heat supplied heat rejected

4

1min

2

3max lnln v

vRTvv

RTW =

( )( ) )9(ln minmax TTrR =

We have assumed that the compression and expansion ratios are equal, that is

)10(4

1

2

3

vv

vv

=

Heat supplied Qs = R Tmax ln (r) (10)

Hence, the thermal efficiency of the cycle is given by

( )( )( )

)11(

lnln

max

minmax

max

minmax

TTT

TrRTTrR

th

=

=

From Eq. 11 it is seen that the thermal efficiency of the Carnot cycle is only a function of the maximum and minimum temperatures of the cycle. The efficiency will increase if the minimum temperature (or the temperature at which the heat is rejected) is as low as possible. According to this equation, the efficiency will be equal to 1 if the minimum temperature is zero, which happens to be the absolute zero temperature in the thermodynamic scale. This equation also indicates that for optimum (Carnot) efficiency, the cycle (and hence the heat engine) must operate between the limits of the highest and lowest possible temperatures. In other words, the engine should take in all the heat at as high a temperature as possible and should reject the heat at as low a temperature as possible. For the first condition to be achieved, combustion (as applicable for a real engine using fuel to provide heat) should begin at the highest possible temperature, for then the irreversibility of the chemical reaction would be reduced. Moreover, in the cycle, the expansion should proceed to the lowest possible temperature in order to obtain the

maximum amount of work. These conditions are the aims of all designers of modern heat engines. The conditions of heat rejection are governed, in practice, by the temperature of the atmosphere. It is impossible to construct an engine which will work on the Carnot cycle. In such an engine, it would be necessary for the piston to move very slowly during the first part of the forward stroke so that it can follow an isothermal process. During the remainder of the forward stroke, the piston would need to move very quickly as it has to follow an isentropic process. This variation in the speed of the piston cannot be achieved in practice. Also, a very long piston stroke would produce only a small amount of work most of which would be absorbed by the friction of the moving parts of the engine. Since the efficiency of the cycle, as given by Eq. 11, is dependent only on the maximum and minimum temperatures, it does not depend on the working medium. It is thus independent of the properties of the working medium. Piston Engine Air Standard Cycles The cycles described here are air standard cycles applicable to piston engines. Engines bases on these cycles have been built and many of the engines are still in use. The Otto Cycle The Otto cycle, which was first proposed by a Frenchman, Beau de Rochas in 1862, was first used on an engine built by a German, Nicholas A. Otto, in 1876. The cycle is also called a constant volume or explosion cycle. This is the equivalent air cycle for reciprocating piston engines using spark ignition. Figures 5 and 6 show the P-V and T-s diagrams respectively. At the start of the cycle, the cylinder contains a mass M of air at the pressure and volume indicated at point 1. The piston is at its lowest position. It moves upward and the gas is compressed isentropically to point 2. At this point, heat is added at constant volume which raises the pressure to point 3. The high pressure charge now expands isentropically, pushing the piston down on its expansion stroke to point 4 where the charge rejects heat at constant volume to the initial state, point 1. The isothermal heat addition and rejection of the Carnot cycle are replaced by the constant volume processes which are, theoretically more plausible, although in practice, even these processes are not practicable. The heat supplied, Qs, per unit mass of charge, is given by

cv(T3 T2) the heat rejected, Qr per unit mass of charge is given by

cv(T4 T1) and the thermal efficiency is given by

( )( )

)19(1

11

1

2

3

1

4

2

1

23

14

=

=

TTTT

TT

TTTT

th

Now 3

4

1

4

3

1

1

2

2

1

TT

VV

VV

TT

=

=

=

And since 2

3

1

4

3

4

2

1

TT

TThavewe

TT

TT

==

Hence, substituting in Eq. 19, we get, assuming that r is the compression ratio V1/V2

)20(11

1

1

1

1

1

2

2

1

=

=

=

r

VV

TT

th

In a true thermodynamic cycle, the term expansion ratio and compression ratio are synonymous. However, in a real engine, these two ratios need not be equal because of the valve timing and therefore the term expansion ratio is preferred sometimes. Equation 20 shows that the thermal efficiency of the theoretical Otto cycle increases with increase in compression ratio and specific heat ratio but is independent of the heat added (independent of load) and initial conditions of pressure, volume and temperature.

Figure 7 shows a plot of thermal efficiency versus compression ratio for an Otto cycle. It is seen that the increase in efficiency is significant at lower compression ratios. This is also seen in Table 1 given below.

From the table it is seen that if: CR is increased from 2 to 4, efficiency increase is 76% CR is increased from 4 to 8, efficiency increase is only 32.6% CR is increased from 8 to 16, efficiency increase is only 18.6% Mean effective pressure: It is seen that the air standard efficiency of the Otto cycle depends only on the compression ratio. However, the pressures and temperatures at the various points in the cycle and the net work done, all depend upon the initial pressure and temperature and the heat input from point 2 to point 3, besides the compression ratio.

A quantity of special interest in reciprocating engine analysis is the mean effective pressure. Mathematically, it is the net work done on the piston, W, divided by the piston displacement volume, V1 V2. This quantity has the units of pressure. Physically, it is that constant pressure which, if exerted on the piston for the whole outward stroke, would yield work equal to the work of the cycle. It is given by

)21(21

32

21

VVQ

VVWmep

=

=

where Q2-3 is the heat added from points 2 to 3. Now

R 1 0 2 0.242 3 0.356 4 0.426 5 0.475 6 0.512 7 0.541 8 0.565 9 0.585 10 0.602 16 0.67 20 0.698 50 0.791

)22(11

1

1

1

2121

=

=

rV

VVVVV

Here r is the compression ratio, V1/V2 From the equation of state:

)23(1

101 p

TmR

MV =

R0 is the universal gas constant Substituting for V1 from Eq. 3 in Eq. 2 and then substituting for V1 V2 in Eq. 1 we get

)24(11

10

132

A

r

TMRmp

Qmep

=

The quantity Q2-3/M is the heat added between points 2 and 3 per unit mass of air (M is the mass of air and m is the molecular weight of air); and is denoted by Q, thus

)24(11

10

1

B

r

TRmpQ

mep

=

We can non-dimensionalize the mep by dividing it by p1 so that we can obtain the following equation

)25(11

1

101

=

TRmQ

rp

mep

Since ( )10 = vcmR

, we can substitute it in Eq. 25 to get

[ ])26(

111

1

11

=

rTc

Qp

mep

v

The dimensionless quantity mep/p1 is a function of the heat added, initial temperature, compression ratio and the properties of air, namely, cv and . We see that the mean effective pressure is directly proportional to the heat added and inversely proportional to the initial (or ambient) temperature. We can substitute the value of from Eq. 20 in Eq. 26 and obtain the value of mep/p1 for the Otto cycle in terms of the compression ratio and heat added. In terms of the pressure ratio, p3/p2 denoted by rp we could obtain the value of mep/p1 as follows:

( )( )( )( ) )27(11

11 1

1

=

rrrr

pmep p

We can obtain a value of rp in terms of Q as follows:

)28(111

+

= rTcQr

vp

Another parameter, which is of importance, is the quantity mep/p3. This can be obtained from the following expression:

)29(1

11

11

13 +

=

rTcQrp

mepp

mep

v

Choice of Q We have said that

)30(32M

QQ =

M is the mass of charge (air) per cycle, kg. Now, in an actual engine

)31(/

32

cyclekJinQFM

QMQ

ca

cf

=

=

Mf is the mass of fuel supplied per cycle, kg

Qc is the heating value of the fuel, kJ/kg Ma is the mass of air taken in per cycle F is the fuel air ratio = Mf/Ma Substituting for Eq. (B) in Eq. (A) we get

)32(M

QFMQ ca=

)33(111

21

1

21

rVVV

And

VVV

MM

Now a

=

So, substituting for Ma/M from Eq. (33) in Eq. (32) we get

)34(11

=

rFQQ c

For isooctane, FQc at stoichiometric conditions is equal to 2975 kJ/kg, thus

Q = 2975(r 1)/r (35)

At an ambient temperature, T1 of 300K and cv for air is assumed to be 0.718 kJ/kgK, we get a value of Q/cvT1 = 13.8(r 1)/r. Under fuel rich conditions, = 1.2, Q/ cvT1 = 16.6(r 1)/r. Under fuel lean conditions, = 0.8, Q/ cvT1 = 11.1(r 1)/r

The Diesel Cycle This cycle, proposed by a German engineer, Dr. Rudolph Diesel to describe the processes of his engine, is also called the constant pressure cycle. This is believed to be the equivalent air cycle for the reciprocating slow speed compression ignition engine. The P-V and T-s diagrams are shown in Figs 8 and 9 respectively. The cycle has processes which are the same as that of the Otto cycle except that the heat is added at constant pressure. The heat supplied, Qs is given by

cp(T3 T2)

Whereas the heat rejected, Qr is given by

cv(T4 T1) and the thermal efficiency is given by

( )( )

)36(1

111

1

2

32

1

41

23

14

=

=

TT

T

TTT

TTcTTc

p

vth

From the T-s diagram, Fig. 9, the difference in enthalpy between points 2 and 3 is the same as that between 4 and 1, thus

1432 = ss

=

2

3

1

4 lnlnTT

cTT

c pv

=

2

3

1

4 lnlnTT

TT

=

2

3

1

4

TT

TT and 1

1

1

2

2

1 1

=

=

rVV

TT

Substituting in eq. 36, we get

)37(1

1111

2

3

2

31

=

TT

TT

rth

Now ratiooffcutrVV

TT

c ===2

3

2

3

( ) )38(1111 1

= c

c

rr

r

When Eq. 38 is compared with Eq. 20, it is seen that the expressions are similar except for the term in the parentheses for the Diesel cycle. It can be shown that this term is always greater than unity.

Now e

c rr

VV

VV

VV

r ===1

2

4

3

2

3 where r is the compression ratio and re is the expansion ratio

Thus, the thermal efficiency of the Diesel cycle can be written as

)39(1

111 1

=

e

e

rr

rr

r

Let re = r since r is greater than re. Here, is a small quantity. We therefore have

1

11

=

=

=r

rr

rr

rrr

e

We can expand the last term binomially so that

L+

+

+

+=

3

3

2

21

11rrrr

Also ( )

=

=

=

r

rr

rr

rrr

e

11

We can expand the last term binomially so that

( ) ( )( )L+

+++

++

+=

3

3

2

2

!321

!2111

rrrr

Substituting in Eq. 39, we get

( ) ( )( )

)40(!321

!21

11

3

3

2

2

3

3

2

2

1

+

+

+

+++

++

+

= L

L

rrr

rrrr

Since the coefficients of 332

,,rrr r , etc are greater than unity, the quantity in the

brackets in Eq. 40 will be greater than unity. Hence, for the Diesel cycle, we subtract

1

1r

times a quantity greater than unity from one, hence for the same r, the Otto cycle

efficiency is greater than that for a Diesel cycle.

If r is small, the square, cube, etc of this quantity becomes progressively

smaller, so the thermal efficiency of the Diesel cycle will tend towards that of the Otto cycle. From the foregoing we can see the importance of cutting off the fuel supply early in the forward stroke, a condition which, because of the short time available and the high pressures involved, introduces practical difficulties with high speed engines and necessitates very rigid fuel injection gear. In practice, the diesel engine shows a better efficiency than the Otto cycle engine because the compression of air alone in the former allows a greater compression ratio to be employed. With a mixture of fuel and air, as in practical Otto cycle engines, the maximum temperature developed by compression must not exceed the self ignition temperature of the mixture; hence a definite limit is imposed on the maximum value of the compression ratio. Thus Otto cycle engines have compression ratios in the range of 7 to 12 while diesel cycle engines have compression ratios in the range of 16 to 22.

We can obtain a value of rc for a Diesel cycle in terms of Q as follows:

)41(111

+

= rTcQr

pc

We can substitute the value of from Eq. 38 in Eq. 26, reproduced below and obtain the value of mep/p1 for the Diesel cycle.

[ ])26(

111

1

11

=

rTc

Qp

mep

v

In terms of the cut-off ratio, we can obtain another expression for mep/p1 as follows:

( ) ( )( )( ) )42(11

11

1

=

rrrrr

pmep cc

For the Diesel cycle, the expression for mep/p3 is as follows:

)43(1

13

= rp

mepp

mep

Modern high speed diesel engines do not follow the Diesel cycle. The process of heat addition is partly at constant volume and partly at constant pressure. This brings us to the dual cycle. The Dual Cycle An important characteristic of real cycles is the ratio of the mean effective pressure to the maximum pressure, since the mean effective pressure represents the useful (average) pressure acting on the piston while the maximum pressure represents the pressure which chiefly affects the strength required of the engine structure. In the constant-volume cycle, shown in Fig. 10, it is seen that the quantity mep/p3 falls off rapidly as the compression ratio increases, which means that for a given mean effective pressure the maximum pressure rises rapidly as the compression ratio increases. For example, for a mean effective pressure of 7 bar and Q/cvT1 of 12, the maximum pressure at a compression ratio of 5 is 28 bar whereas at a compression ratio of 10, it rises to about 52 bar. Real cycles follow the same trend and it becomes a practical necessity to limit the maximum pressure when high compression ratios are used, as in diesel engines. This also indicates that diesel engines will have to be stronger (and hence heavier) because it has to withstand higher peak pressures. Constant pressure heat addition achieves rather low peak pressures unless the compression ratio is quite high. In a real diesel engine, in order that combustion takes place at constant pressure, fuel has to be injected very late in the compression stroke (practically at the top dead center). But in order to increase the efficiency of the cycle, the fuel supply must be cut off early in the expansion stroke, both to give sufficient time for the fuel to burn and thereby increase combustion efficiency and reduce after burning but also reduce emissions. Such situations can be achieved if the engine was a slow speed type so that the piston would move sufficiently slowly for combustion to take place despite the late injection of the fuel. For modern high speed compression ignition engines it is not possible to achieve constant pressure combustion. Fuel is injected somewhat earlier in the compression stroke and has to go through the various stages of combustion.

Thus it is seen that combustion is nearly at constant volume (like in a spark ignition engine). But the peak pressure is limited because of strength considerations so the rest of the heat addition is believed to take place at constant pressure in a cycle. This has led to the formulation of the dual combustion cycle. In this cycle, for high compression ratios, the peak pressure is not allowed to increase beyond a certain limit and to account for the total addition, the rest of the heat is assumed to be added at constant pressure. Hence the name limited pressure cycle. The cycle is the equivalent air cycle for reciprocating high speed compression ignition engines. The P-V and T-s diagrams are shown in Figs.11 and 12. In the cycle, compression and expansion processes are isentropic; heat addition is partly at constant volume and partly at constant pressure while heat rejection is at constant volume as in the case of the Otto and Diesel cycles. The heat supplied, Qs per unit mass of charge is given by

cv(T3 T2) + cp(T3 T2)

Whereas the heat rejected, Qr per unit mass of charge is given by

cv(T4 T1) and the thermal efficiency is given by

( )( ) ( )

)44(11

11

)44(11

11

)44(1

3

3

1

2

2

3

2

3

1

2

1

4

3

33

2

32

1

41

2323

14

C

TT

TT

TT

TT

TT

TT

B

TT

TTT

T

TTT

ATTcTTc

TTc

pv

vth

+

=

+

=

+

=

From thermodynamics

)45(2

3

2

3prp

pTT

== the explosion or pressure ratio and

)46(3

3

3

3crV

VTT

== the cut-off ratio.

Now, 1

2

2

3

3

3

3

4

1

4

1

4

pp

pp

pp

pp

pp

TT

==

Also

=

=

=

rr

VV

VV

VV

pp

c1

4

3

3

3

4

3

3

4

And rpp

=1

2

Thus cprrTT

=1

4

Also 12

1

1

2 =

=

rVV

TT

Therefore, the thermal efficiency of the dual cycle is

( ) ( ) )46(11111 1

+

=

cpp

cp

rrrrr

r

We can substitute the value of from Eq. 46 in Eq. 26 and obtain the value of mep/p1 for the dual cycle. In terms of the cut-off ratio and pressure ratio, we can obtain another expression for mep/p1 as follows:

( ) ( ) ( )( )( ) )47(11

111

1

+=

rrrrrrrrr

pmep cppcp

For the dual cycle, the expression for mep/p3 is as follows:

)48(3

1

13

=

pp

pmep

pmep

Since the dual cycle is also called the limited pressure cycle, the peak pressure, p3, is usually specified. Since the initial pressure, p1, is known, the ratio p3/p1 is known. We can correlate rp with this ratio as follows:

)49(1

1

3

= rp

prp

We can obtain an expression for rc in terms of Q and rp and other known quantities as follows:

( ) )50(111 11

+

= pvc rrTc

Qr

We can also obtain an expression for rp in terms of Q and rc and other known quantities as follows:

)51(1

111

+

+

=

c

vp r

rTcQ

r

Figure 13 shows a constant volume and a constant pressure cycle, compared with a limited pressure cycle. In a series of air cycles with varying pressure ratio at a given compression ratio and the same Q, the constant volume cycle has the highest efficiency and the constant pressure cycle the lowest efficiency. Figure 14 compares the efficiencies of the three cycles for the same value of

1r

rQ for the same initial conditions and three values of p3/p1 for the dual cycle. It is

interesting to note that the air standard efficiency is little affected by compression ratio above a compression ratio of 8 for the limited pressure cycle. The curves of mep/p3 versus compression ratio for the same three cycles as above are given in Fig. 10. It is seen that a considerable increase in this ratio is obtained for a limited pressure cycle as compared to the constant volume or constant pressure cycles.

BBRRAAYYTTOONN CCYYCCLLEE

Brayton Cycle is the ideal cycle for gas turbine engines. Electric power

generation and aircraft propulsion are major applications for gas-turbine engines.

4 PROCESSES: 1-2 Isentropic compression 2-3 Constant pressure heat addition 3-4 Isentropic expansion 4-1 Constant pressure heat rejection Performing energy balance, we get:

( )( )

3 2

out 4 1

net outth 1

m p

p

m m

q C T T

q C T T

w qq q

=

=

= =

&

& &

( )( )

41

4 1 1th

33 22

2

11 1

1

TTT T TTT T T T

= =

(1) Note: P2 = P3 P1 = P4 Also:

1 1

3 32 2

1 1 4 4

k kk kP TT P

T P P T

= = =

341 2

TTT T

=

Substitute in (1), we get:

1th22

1

11 1T TTT

= =

(2) or

th 111 k k

pr =

where the pressure ratio, 21.p

Pr P=

DEVELOPMENT OF GAS TURBINES Early (1950) gas turbines had simple-cycle efficiencies of about 17% because of

low compressor and turbine efficiencies and low inlet temperature of the turbine. Effort to improve cycle efficiency concentrated in three areas:

(1) Increasing the turbine inlet temperature increased from about 500C to 1425C presently.

(2) Increasing the efficiencies of compressors and turbines. Efficiencies

improved due to designing the components aerodynamically with minimum losses.

(3) Adding modifications to the basic cycle, such as regeneration, inter-

cooling, and reheating. ADVANTAGES OF PRESENT GAS TURBINES

1. High efficiencies 2. Lower capital cost 3. Shorter installation time 4. Better emission characteristics 5. Being used for base-load as well as peak load

Capacities and Efficiencies Range.

Inlet

TemperaturePressure

Ratio Capacity

(MW) Efficiency

540 6.5 2 26

1425 135.5 282 39.5 ACTUAL GAS TURBINE CYCLES Actual gas turbine cycles differ from Brayton cycle. In the actual cases: (1) Pressure drops during heat addition and heat rejection processes. (2) The expansion and compression processes in the turbine and compressor,

respectively, are not isentropic.

The actual processes in the turbine and compressor can be accounted for by the isentropic efficiencies:

sca

ww

=

(3)

ats

ww

=

(4) BRAYTON CYCLE WITH REGENERATION Heating the high-pressure air leaving the compressor by the hot exhaust gases in a

counterflow heat exchanger is known as regeneration (see Figure 8.38). The thermal efficiency of the Brayton cycle increases as a result of decrease in the

heat input (thermo-fuel) for the same net power output. Regeneration is used only when the compressor exit temperature is less than the

turbine exit temperature. Referring to the T-s diagram, the regenerator effectiveness is given as:

reg, act 5 2reg, max 4 2

q h hq h h

= =

(5) Considering the cold air-standard assumptions, equation (5) reduces to:

5 24 2

T TT T

=

(6) Assuming cold air-standard assumptions show that the thermal efficiency of an

ideal Brayton cycle with regeneration is given as:

( )1

1th

3

1k

kp

T rT

=

(7) Comment on the effect of temperature and pressure ratios on the thermal

efficiencies.

See Example 8.7. BRAYTON CYCLE WITH INTERCOOLING, REHEATING, AND REGENERATION Using multistage compression with intercooling reduces the total work of the

compressor operating between two pressures. Similarly using multistage expansion with reheating increases the workout of a

turbine operating between two pressures. Even-though intercooling and reheating improves the back work ratio of a gas

turbine cycle, but it does guarantee an improvement in the thermal efficiency (why?).

Intercooling and reheating have to be used in conjunction with regeneration for

the thermal efficiency to improve. The best performance is achieved when equal pressure ratios are maintained

across each stage. For example (considering Figure 8.44) when

6 82 41 3 7 9

and .P PP PP P P P

= =

PROBLEM A gas turbine at Dammam Electrical Power Station takes in 108,000 kg/h of filtered outside air at 27C and compresses it to 6.516 atmospheres. The combustion of gas adds 30 MW of heat to the air. If the turbine exhausts to atmospheric pressure and both the compressor and turbine are 75% efficient: (a) Draw the T-s diagram taking the inlet to the compressor as State 1. (b) Determine the net power output. SOLUTION a) T-s diagram.

T

s

b) Determination of the net power output: Given: 1 27 273 300 K, 6pT r= + = =

net

2 1c 2 1

11 1

1

300.19 kJ kg300 K at

Pr 1.386

T c

sa

c

Wh h h h

hT T

=

= =

== =

2 2

1 1

2 1

2Hence

2

2 1

Pr 6.516Pr

Pr Pr * 6.516 1.386 6.516 9.031510 K

513.32 kJ kg513.32 300.19 284.17 kJ kg

0.75300.19 284.17 584.36 kJ kg

s

s

c

a c

PP

Th

h h

= =

= = ==

=

= =

= + = + =

3

in 3 2 3 in 2 2

3

3 3

4 44 3

3 3

4

30 10 584.3630.50

1000 584.36 1584.36 kJ kg

1457.3 K, and Pr 533Pr 1Pr PrPr 6.516

533Pr 81.86.516

a a aQq h h h q h hm

h

TPP

= = + = + = +

= + =

= =

= =

= =

&

&

Hence, h4s = 954.47 kJ/kg ( )3 4t t sh h =

0.75(1584.36 954.47) 472.42 kJ kg

472.42 284.17 188.25 kJ kgt

net t c

= =

= = =

Chapter 2

RECIPROCATING AIR COMPRESSORS &

REFRIGEREATION CYCLES

Reciprocating Air Compressors

Reciprocating air compressors are positive displacement machines, meaning that they increase the pressure of the air by reducing its volume. This means they are taking in successive volumes of air which is confined within a closed space and elevating this air to a higher pressure. The reciprocating air compressor accomplishes this by a piston within a cylinder as the compressing and displacing element.

Single-stage and two-stage reciprocating compressors are commercially available.

Single-stage compressors are generally used for pressures in the range of 70 psig to 100 psig.

Two-stage compressors are generally used for higher pressures in the range of 100 psig to 250 psig.

Note that

1 HP ~ 4 CFM at 100 psi

and that 1 to 50 HP are typically for reciprocating units. Compressors 100 hp and above are typically Rotary Screw or Centrifugal Compressors.

The reciprocating air compressor is single acting when the compressing is accomplished using only one side of the piston. A compressor using both sides of the piston is considered double acting.

Load reduction is achieved by unloading individual cylinders. Typically this is accomplished by throttling the suction pressure to the cylinder or bypassing air either within or outside the compressor. Capacity control is achieved by varying speed in engine-driven units through fuel flow control.

Reciprocating air compressors are available either as air-cooled or water-cooled in lubricated and non-lubricated configurations and provide a wide range of pressure and capacity selections.

Refrigeration

Refrigeration is the process of removing heat from an enclosed space, or from a substance, and moving it to a place where it is unobjectionable. The primary purpose of refrigeration is lowering the temperature of the enclosed space or substance and then maintaining that lower temperature. The term cooling refers generally to any natural or artificial process by which heat is dissipated. The process of artificially producing extreme cold temperatures is referred to as cryogenics.

Cold is the absence of heat, hence in order to decrease a temperature, one "removes heat", rather than "adding cold." In order to satisfy the Second Law of Thermodynamics, some form of work must be performed to accomplish this. This work is traditionally done by mechanical work but can also be done by magnetism, laser or other means.

COP

Where

is the change in heat at the heat reservoir of interest, and is the work consumed by the heat pump.

(Note: COP has no units, therefore in this equation, heat and work must be expressed in the same units.)

The COP for heating and cooling are thus different, because the heat reservoir of interest is different. When one is interested in how well a machine cools, the COP is the ratio of the heat removed from the cold reservoir to input work. However, for heating, the COP is the ratio of the heat removed from the cold reservoir plus the heat added to the hot reservoir by the input work to input work:

Where

is the heat moved from the cold reservoir (to the hot reservoir).

Derivation

According to the first law of thermodynamics, in a reversible system we can show that Qhot = Qcold + W and W = Qhot Qcold, where Qhot is the heat given off by the hot heat reservoir and Qcold is the heat taken in by the cold heat reservoir. Therefore, by substituting for W,

For a heat pump operating at maximum theoretical efficiency (i.e. Carnot

efficiency), it can be shown that and , where Thot and Tcold are the absolute temperatures of the hot and cold heat reservoirs respectively.

Hence, at maximum theoretical efficiency,

Similarly,

It can also be shown that COPcooling = COPheating 1. Note that these equations must use the absolute temperature, such as the Kelvin scale.

COPheating applies to heat pumps and COPcooling applies to air conditioners or refrigerators. For heat engines, see Efficiency. Values for actual systems will always be less than these theoretical maximums.

Example

A geothermal heat pump operating at COPheating 3.5 provides 3.5 units of heat for each unit of energy consumed (e.g. 1 kWh consumed would provide 3.5 kWh of output heat). The output heat comes from both the heat source and 1 kWh of input energy, so the heat-source is cooled by 2.5 kWh, not 3.5 kWh.

A heat pump of COPheating 3.5, such as in the example above, could be less expensive to use than even the most efficient gas furnace.

A heat pump cooler operating at COPcooling 2.0 removes 2 units of heat for each unit of energy consumed (e.g. such an air conditioner consuming 1 kWh would remove heat from a building's air at a rate of 2 kWh).

The COP of heat pumps (300%-350% efficient) make them much more efficient than high-efficiency gas-burning furnaces (90-99% efficient), and electric heating (100%). However, this does not always mean they are less expensive to operate. The 2008 US average price per therm (100,000 BTU) of electricity was $3.33 while the average price per therm of natural gas was $1.33. Using these prices, a heat pump with a COP of 3.5 would cost $0.95 to provide one therm of heat, while a high efficiency gas furnace with 95% efficiency would cost $1.40 to provide one therm of heat. With these average prices, the heat pump costs 32% less to provide the same amount of heat. The savings (if any) will depend on the actual cost of electricity and natural gas, which can both vary widely.

Conditions of use

While the COP is partly a measure of the efficiency of a heat pump, it is also a measure of the conditions under which it is operating: the COP of a given heat pump will rise as the input temperature increases or the output temperature decreases because it is linked to a warm temperature distribution system like under floor heating.

Reversed Carnot Cycle: Reversed Carnot cycle is shown in Fig.6.1. It consists of the following

processes. Process a-b: Absorption of heat by the working fluid from refrigerator at constant low temperature T2 during isothermal expansion. Process b-c: Isentropic compression of the working fluid with the aid of external work. The temperature of the fluid rises from T2 to T1. Process c-d: Isothermal compression of the working fluid during which heat is rejected at constant high temperature T1. Process d-a: Isentropic expansion of the working fluid. The temperature of the working fluid falls from T1 to T2.

Vapor-compression cycle

The vapor-compression cycle is used in most household refrigerators as well as in many large commercial and industrial refrigeration systems. Figure 1 provides a schematic diagram of the components of a typical vapor-compression refrigeration system.

The thermodynamics of the cycle can be analyzed on a diagram as shown in Figure. In this cycle, a circulating refrigerant such as Freon enters the compressor as a vapor. From point 1 to point 2, the vapor is compressed at constant entropy and exits the compressor superheated. From point 2 to point 3 and on to point 4, the superheated vapor travels through the condenser which first cools and removes the superheat and then

condenses the vapor into a liquid by removing additional heat at constant pressure and temperature. Between points 4 and 5, the liquid refrigerant goes through the expansion valve (also called a throttle valve) where its pressure abruptly decreases, causing flash evaporation and auto-refrigeration of, typically, less than half of the liquid.

That results in a mixture of liquid and vapor at a lower temperature and pressure as shown at point 5. The cold liquid-vapor mixture then travels through the evaporator coil or tubes and is completely vaporized by cooling the warm air (from the space being refrigerated) being blown by a fan across the evaporator coil or tubes. The resulting refrigerant vapor returns to the compressor inlet at point 1 to complete the thermodynamic cycle.

The above discussion is based on the ideal vapor-compression refrigeration cycle, and does not take into account real-world effects like frictional pressure drop in the system, slight thermodynamic irreversibility during the compression of the refrigerant vapor, or non-ideal gas behavior (if any).

More information about the design and performance of vapor-compression refrigeration systems is available in the classic "Perry's Chemical Engineers' Handbook".

Vapor absorption cycle

In the early years of the twentieth century, the vapor absorption cycle using water-ammonia systems was popular and widely used. After the development of the vapor compression cycle, the vapor absorption cycle lost much of its importance because of its low coefficient of performance (about one fifth of that of the vapor compression cycle). Today, the vapor absorption cycle is used mainly where fuel for heating is available but

electricity is not, such as in recreational vehicles that carry LP gas. It's also used in industrial environments where plentiful waste heat overcomes its inefficiency.

The absorption cycle is similar to the compression cycle, except for the method of raising the pressure of the refrigerant vapor. In the absorption system, the compressor is replaced by an absorber which dissolves the refrigerant in a suitable liquid, a liquid pump which raises the pressure and a generator which, on heat addition, drives off the refrigerant vapor from the high-pressure liquid. Some work is required by the liquid pump but, for a given quantity of refrigerant, it is much smaller than needed by the compressor in the vapor compression cycle. In an absorption refrigerator, a suitable combination of refrigerant and absorbent is used. The most common combinations are ammonia (refrigerant) and water (absorbent), and water (refrigerant) and lithium bromide (absorbent).

Required Properties of Ideal Refrigerant:

1) The refrigerant should have low boiling point and low freezing point.

2) It must have low specific heat and high latent heat. Because high specific heat decreases the refrigerating effect per kg of refrigerant and high latent heat at low temperature increases the refrigerating effect per kg of refrigerant.

3) The pressures required to be maintained in the evaporator and condenser should be low enough to reduce the material cost and must be positive to avoid leakage of air into the system.

4) It must have high critical pressure and temperature to avoid large power requirements.

5) It should have low specific volume to reduce the size of the compressor.

6) It must have high thermal conductivity to reduce the area of heat transfer in evaporator and condenser.

7) It should be non-flammable, non-explosive, non-toxic and non-corrosive.

8) It should not have any bad effects on the stored material or food, when any leak develops in the system.

9) It must have high miscibility with lubricating oil and it should not have reacting properly with lubricating oil in the temperature range of the system.

10) It should give high COP in the working temperature range. This is necessary to reduce the running cost of the system.

11) It must be readily available and it must be cheap also.

Important Refrigerants:

Properties at -150C

(1) Ammonia (NH3)(R-717)

Latent heat = 1312.75 kJ/Kg

Specific volume = 0.509 m3/kg

(2) DichloroDifluoro methane (Freon12) (R-12) [C Cl2 F2]

Latent heat = 162 kJ/Kg

Specific volume = 0.093 m3/kg

(3) Difluoro monochloro methane or Freon-22 (R-22) [CH Cl F2]

Latent heat = 131 kJ/Kg

Specific Volume = 0.15 m3/kg.

Chapter 3

CONDUCTION Conduction will take place if there exist a temperature gradient in a solid (or

stationary fluid) medium.

Energy is transferred from more energetic to less energetic molecules when neighboring molecules collide. Conductive heat flow occurs in direction of the decreasing temperature since higher temperature is associated with higher molecular energy. Fourier's Law expresses conductive heat transfer as

q = k A dT / s (1)

Where,

q = heat transferred per unit time (W, Btu/hr)

A = heat transfer area (m2, ft2)

k = thermal conductivity of the material (W/m.K or W/m oC, Btu/(hr oF ft2/ft))

dT = temperature difference across the material (K or oC, oF)

s = material thickness (m, ft)

Example - Heat Transfer by Conduction

A plane wall constructed of solid iron with thermal conductivity 70 W/moC, thickness 50 mm and with surface area 1 m by 1 m, temperature 150 oC on one side and 80oC on the other.

Conductive heat transfer can be calculated as:

q = (70 W/moC) (1 m) (1 m) ((150 oC) - (80 oC)) / (0.05 m)

= 98,000 W

= 98 kW

STEADY STATE CONDUCTION - ONE DIMENSION

Examine Fouriers law of heat conduction in simple one dimensional system. Several

physical shapes may fall in the category of one-dimensional systems: cylindrical and

spherical systems are one dimensional when the temperature in the body is a function

only of radial distance and is independent of azimuth angle or axial distance.

Consider the plane wall

Application of Fouriers law gives (By integration):

( )q k Ax

T T= . . 2 1

The thermal conductivity is considered as constant. If k varies with temperature to some

linear relation :

k = k0(1+T), then the equation is:

( )q k Ax

T T T T= + 0 2 1 22

12

2. .[ ( )]

If more than one layer is present then:

( ) ( ) ( )q k A T Tx

k AT T

xk A

T TxA A

B

B

C

C

=

=

=

. . . . . .2 1 3 2 4 3

Solving these equations simultaneously the heat flow is written as:

qT T

xk A

xk A

xk A

A

A

B

B

C

C

=

+ +

1 4

..

..

..

The heat transfer may be considered as a flow, and the combination of thermal

conductivity, thickness of material and area as a resistance to this flow. The temperature

is the potential, or driving, function for the heat flow, and the Fourier equation may be

written

Heat flowthermal potential difference

thermal resis ce=

tan

a relation quite like Ohms law in electric-circuit theory.

IVR

=

Thermal resistance = x / kA of R = xi / kiA

Thermal Conductivity and Common Units

Btu/(h ft2 oF/ft)

Btu/(h ft2 oF/in)

Btu/(s ft2 oF/ft)

MW/(m2 K/m)

kW/(m2 K/m)

W/(m2 K/m)

W/(m2 K/cm)

W/(cm2 oC/cm)

W/(in2 oF/in)

kJ/(h m2 K/m)

J/(s m2 oC/m)

kcal/(h m2 oC/m)

cal/(s cm2 oC/cm)

Chapter 4

CONVECTION Heat energy transferred between a surface and a moving fluid at different

temperatures is known as convection.

In reality this is a combination of diffusion and bulk motion of molecules. Near the surface the fluid velocity is low, and diffusion dominates. Away from the surface, bulk motion increases the influence and dominates.

Convective heat transfer may take the form of either

forced or assisted convection natural or free convection

Forced or Assisted Convection

Forced convection occurs when a fluid flow is induced by an external force, such as a pump, fan or a mixer.

Natural or Free Convection

Natural convection is caused by buoyancy forces due to density differences caused by temperature variations in the fluid. At heating the density change in the boundary layer will cause the fluid to rise and be replaced by cooler fluid that also will heat and rise. This continues phenomena is called free or natural convection. Boiling or condensing processes are also referred as a convective heat transfer processes.

The heat transfer per unit surface through convection was first described by Newton and the relation is known as the Newton's Law of Cooling.

The equation for convection can be expressed as:

q = k A dT (1)

Where,

q = heat transferred per unit time (W)

A = heat transfer area of the surface (mo)

k = convective heat transfer coefficient of the process (W/m2K or W/m2oC)

dT = temperature difference between the surface and the bulk fluid (K or oC)

Heat Transfer Coefficients - Units

1 W/m2K = 0.85984 kcal/h m2 oC = 0.1761 Btu/ ft2 h oF 1 Btu/ft2 h oF = 5.678 W/m2 K = 4.882 kcal/h m2 oC 1 kcal/h m2 oC = 1.163 W/m2K = 0.205 Btu/ ft2 h oF

Convective Heat Transfer Coefficients

The convection heat transfer coefficient - k - is dependent on the type of media, gas or liquid, the flow properties such as velocity, viscosity and other flow and temperature dependent properties.

In general the convective heat transfer coefficient for some common fluids is within the ranges:

Air : 10 - 100 (W/m2K) Water : 500 - 10,000 (W/m2K)

Example - Convective Heat Transfer

A fluid flows over a plane surface 1 m by 1 m with a bulk temperature of 50oC. The temperature of the surface is 20oC. The convective heat transfer coefficient is 2,000 W/m2oC.

q = (2,000 W/m2oC) ((1 m) (1 m)) ((50oC) - (20oC))

= 60,000 (W)

= 60 (kW)

Chapter 5

RADIATION Heat transfer through radiation takes place in form of electromagnetic waves

mainly in the infrared region. Radiation emitted by a body is a consequence of thermal agitation of its composing molecules. Radiation heat transfer can be described by a reference to the so-called 'black body'.

The Black Body

A black body is defined as a body that absorbs all radiation that falls on its surface. Actual black bodies don't exist in nature - though its characteristics are approximated by a hole in a box filled with highly absorptive material. The emission spectrum of such a black body was first fully described by Max Planck.

A black body is a hypothetic body that completely absorbs all wavelengths of thermal radiation incident on it. Such bodies do not reflect light, and therefore appear black if their temperatures are low enough so as not to be self-luminous. All blackbodies heated to a given temperature emit thermal radiation.

The radiation energy per unit time from a blackbody is proportional to the fourth power of the absolute temperature and can be expressed with Stefan-Boltzmann Law as

q = T4 A (1)

Where,

q = heat transfer per unit time (W)

= 5.6703 10-8 (W/m2K4) - The Stefan-Boltzmann Constant

T = absolute temperature Kelvin (K)

A = area of the emitting body (m2)

The Stefan-Boltzmann Constant in Imperial Units

= 5.6703 10-8 (W/m2K4)

= 0.1714 10-8 (Btu/(h ft2 oR4) )

= 0.119 10-10 (Btu/ (h in2 oR4))

Gray Bodies and Emissivity Coefficients

For objects other than ideal blackbodies ('gray bodies') the Stefan-Boltzmann Law can be expressed as

q = T4 A (2)

Where,

= emissivity of the object (one for a black body)

For the gray body the incident radiation (also called irradiation) is partly reflected, absorbed or transmitted.

The emissivity coefficient lies in the range 0 < < 1 depending on the type of material and the temperature of the surface. The emissivity of some common materials

oxidized Iron at 390 oF (199 oC) > = 0.64 polished Copper at 100 oF (38 oC) > = 0.03 emissivity coefficients for some common materials

The Net Radiation Loss Rate

If a hot object is radiating energy to its cooler surroundings the net radiation heat loss rate can be expressed as

q = (Th4 - Tc4) Ac (3)

Where,

Th = hot body absolute temperature (K)

Tc = cold surroundings absolute temperature (K)

Ac = area of the object (m2)

Radiation Constants for some common Building Materials

Example - Radiation from the surface of the Sun

If the surface temperature of the sun is 5800 K and if we assume that the sun can be regarded as a black body the radiation energy per unit time can be expressed by modifying (1) like

q / A = T4

= (5.6703 10-8 W/m2K4) (5800 K) 4

= 6.42 107 W/m2