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ترمودینامیک

مکانیک مهندسیدانشکده

1

The Simple Ideal Rankine Cycle9-1

© The McGraw-Hill Companies, Inc.,1998

2

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Rankine Cycle: Actual Vapor Power Deviation and Pump and Turbine Irreversibilities

9-2

• (Fig. 9-4)

(a) Deviation of actual vapor power cycle from the ideal Rankine cycle.(b) The effect of pump and turbine irreversibilities on the ideal Rankine cycle.

3

Effect of Lowering Condenser Pressure on the Ideal Rankine cycle

9-3

4

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Effect of Increasing Boiler Pressure on the Ideal Rankine cycle

9-4

5

The Ideal Reheat Rankine Cycle9-5

6

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Ideal Regenerative Rankine Cycle with Open Feedwater Heater

9-6

7

Ideal Regenerative Rankine Cycle with Closed Feedwater Heater

9-7

8

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A Steam Power Plant With One Open and Three Closed Feedwater Heaters

9-8

9

An Ideal Cogeneration Plant9-9

10

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Schematic and T-s Diagram9-10

(Fig. 9-23)

11

12

A supercritical Rankine cycle.

Today many modern steam power plants operate at supercritical pressures (P > 22.06 MPa) and have thermal efficiencies of about 40% for fossil-fuel plants and 34% for nuclear plants.

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Mercury-Water Binary Vapor Cycle

9-11

(Fig. 9-24)

13

Combined Gas-Steam Power Plant9-12

14

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15

Brayton CycleOur study of gas power cycles will involve the study of those heat engines in which the working fluid remains in the gaseous state throughout the cycle. We often study the ideal cycle in which internal irreversibilities and complexities (the actual intake of air and fuel, the actual combustion process, and the exhaust of products of combustion among others) are removed.

We will be concerned with how the major parameters of the cycle affect the performance of heat engines. The performance is often measured in terms of the cycle efficiency.

thnet

in

W

Q

16

Air-Standard Assumptions

In our study of gas power cycles, we assume that the working fluid is air, and the air undergoes a thermodynamic cycle even though the working fluid in the actual power system does not undergo a cycle.

To simplify the analysis, we approximate the cycles with the following assumptions:

•The air continuously circulates in a closed loop and always behaves as an ideal gas.

•All the processes that make up the cycle are internally reversible.

•The combustion process is replaced by a heat-addition process from an external source.

•A heat rejection process that restores the working fluid to its initial state replaces the exhaust process.

•The cold-air-standard assumptions apply when the working fluid is air and has constant specific heat evaluated at room temperature (25oC or 77oF).

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17

The closed cycle gas-turbine engine

18

The T-s and P-v diagrams for the Closed Brayton Cycle

Process Description 1-2 Isentropic compression (in a

compressor)2-3 Constant pressure heat addition 3-4 Isentropic expansion (in a turbine)4-1 Constant pressure heat rejection

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Recall processes 1-2 and 3-4 are isentropic, so

Since P3 = P2 and P4 = P1, we see that

3 32 4

1 4 1 2

orT TT T

T T T T

The Brayton cycle efficiency becomes

th Brayton

T

T, 1 1

2

Is this the same as the Carnot cycle efficiency?

Since process 1-2 is isentropic,

20

where the pressure ratio is rp = P2/P1 and

th Brayton

p

k kr

, ( ) /

1

11

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21

Example

The ideal air-standard Brayton cycle operates with air entering the compressor at 95 kPa, 22oC. The pressure ratio rp is 6:1 and the air leaves the heat addition process at 1100 K. Determine the compressor work and the turbine work per unit mass flow, the cycle efficiency, the back work ratio, and compare the compressor exit temperature to the turbine exit temperature. Assume constant properties.

Apply the conservation of energy for steady-flow and neglect changes in kinetic and potential energies to process 1-2 for the compressor. Note that the compressor is isentropic.

E E

m h W m h

in out

comp

1 1 2 2

The conservation of mass gives

m m

m m min out

1 2

22

For constant specific heats, the compressor work per unit mass flow is

( )

( )

( )

W m h h

W mC T T

wW

mC T T

comp

comp p

comp

comp

p

2 1

2 1

2 1

Since the compressor is isentropic

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23

w C T T

kJ

kg KK

kJ

kg

comp p

( )

. ( . )

.

2 1

1005 492 5 295

19815

The conservation of energy for the turbine, process 3-4, yields for constant specific heats (let’s take a minute for you to get the following result)

( )

( )

( )

W m h h

W mC T T

wW

mC T T

turb

turb p

turbturb

p

3 4

3 4

3 4

Since process 3-4 is isentropic

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Since P3 = P2 and P4 = P1, we see that( 1) /

4

3

( 1) / (1.4 1) /1.4

4 3

1

1 11100 659.1

6

k k

p

k k

p

T

T r

T T K Kr

w C T TkJ

kg KK

kJ

kg

turb p

( ) . ( . )

.

3 4 1005 1100 659 1

442 5

We have already shown the heat supplied to the cycle per unit mass flow in process 2-3 is

( ) . ( . )

.

m m m

m h Q m h

qQ

mh h

C T TkJ

kg KK

kJ

kg

in

inin

p

2 3

2 2 3 3

3 2

3 2 1005 1100 492 5

609 6

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25

The net work done by the cycle is

w w w

kJ

kg

kJ

kg

net turb comp

( . . )

.

442 5 19815

244 3

The cycle efficiency becomes

th Braytonnet

in

w

q

kJ

kgkJ

kg

or

,

.

..

244 3

609 60 40 40%

26

The back work ratio is defined as

BWRw

w

w

w

kJ

kgkJ

kg

in

out

comp

turb

19815

442 50 448

.

..

Note that T4 = 659.1 K > T2 = 492.5 K, or the turbine outlet temperature is greater than the compressor exit temperature. Can this result be used to improve the cycle efficiency?

What happens to th, win /wout, and wnet as the pressure ratio rp is increased? Consider the T-s diagram for the cycle and note that the area enclosed by the cycle is the net heat added to the cycle. By the first law applied to the cycle, the net heat added to the cycle is equal to the net work done by the cycle. Thus, the area enclosed by the cycle on the T-s diagram also represents the net work done by the cycle.

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Let's take a closer look at the effect of the pressure ratio on the net work done.

w w w

C T T C T T

C T T T C T T T

C Tr

C T r

net turb comp

p p

p p

p

p

k k p p

k k

( ) ( )

( / ) ( / )

( ) ( )( )/

( )/

3 4 2 1

3 4 3 1 2 1

3 1 1

1

1 1

11

1

28

Note that the net work is zero when/( 1)

3

1

1

k k

p p

Tr and r

T

For fixed T3 and T1, the pressure ratio that makes the work a maximum is obtained from:

dw

drnet

p

0

This is easier to do if we let X = rp(k-1)/k

w C TX

C T Xnet p p 3 111

1( ) ( )

dw

dXC T X C Tnet

p p 3

210 1 1 0 0[ ( ) ] [ ]

Solving for X

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Then, the rp that makes the work a maximum for the constant property case and fixed T3 and T1 is

For the ideal Brayton cycle, show that the following results are true.•When rp = rp, max work, T4 = T2

•When rp < rp, max work, T4 > T2

•When rp > rp, max work, T4 < T2

The following is a plot of net work per unit mass and the efficiency for the above example as a function of the pressure ratio.

0 2 4 6 8 10 12 14 16 18 20 22120

140

160

180

200

220

240

260

280

0.15

0.20

0.25

0.30

0.35

0.40

0.45

0.50

0.55

0.60

Pratio

wn

et

kJ

/kg

th

,Bra

yto

n

T1 = 22C

P1 = 95 kPa

T3 = 1100 K

t = c = 100%

rp,max

30

Regenerative Brayton Cycle

For the Brayton cycle, the turbine exhaust temperature is greater than the compressor exit temperature. Therefore, a heat exchanger can be placed between the hot gases leaving the turbine and the cooler gases leaving the compressor. This heat exchanger is called a regenerator or recuperator. The sketch of the regenerative Brayton cycle is shown below.

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We define the regenerator effectiveness regen as the ratio of the heat transferred to the compressor gases in the regenerator to the maximum possible heat transfer to the compressor gases.

q h h

q h h h h

q

q

h h

h h

regen act

regen

regen

regen act

regen

,

, max '

,

, max

5 2

5 2 4 2

5 2

4 2

32

For ideal gases using the cold-air-standard assumption with constant specific heats, the regenerator effectiveness becomes

5 2

4 2

regen

T T

T T

Using the closed cycle analysis and treating the heat addition and heat rejection as steady-flow processes, the regenerative cycle thermal efficiency is

th regenout

in

q

q

h h

h h

,

1

1 6 1

3 5

Notice that the heat transfer occurring within the regenerator is not included in the efficiency calculation because this energy is not heat transferred across the cycle boundary.

Assuming an ideal regenerator regen = 1 and constant specific heats, the thermal efficiency becomes (take the time to show this on your own)

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When does the efficiency of the air-standard Brayton cycle equal the efficiency of the air-standard regenerative Brayton cycle? If we set th,Brayton = th,regen then

Recall that this is the pressure ratio that maximizes the net work for the simple Brayton cycle and makes T4 = T2. What happens if the regenerative Brayton cycle operates at a pressure ratio larger than this value?

34

For fixed T3 and T1, pressure ratios greater than this value cause T4 to be less than T2, and the regenerator is not effective.

What happens to the net work when a regenerator is added?

What happens to the heat supplied when a regenerator is added?

The following shows a plot of the regenerative Brayton cycle efficiency as a function of the pressure ratio and minimum to maximum temperature ratio, T1/T3.

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Example: Regenerative Brayton Cycle

Air enters the compressor of a regenerative gas-turbine engine at 100 kPa and 300 K and is compressed to 800 kPa. The regenerator has an effectiveness of 65 percent, and the air enters the turbine at 1200 K. For a compressor efficiency of 75 percent and a turbine efficiency of 86 percent, determine

(a) The heat transfer in the regenerator.(b) The back work ratio.(c) The cycle thermal efficiency.

Compare the results for the above cycle with the ones listed below that have the same common data as required. The actual cycles are those for which the turbine and compressor isentropic efficiencies are less than one.

(a) The actual cycle with no regeneration, = 0.(b) The actual cycle with ideal regeneration, = 1.0.(c) The ideal cycle with regeneration, = 0.65.(d) The ideal cycle with no regeneration, = 0.(e) The ideal cycle with ideal regeneration, = 1.0.

We assume air is an ideal gas with constant specific heats, that is, we use the cold-air-standard assumption.

36

The cycle schematic is the same as above and the T-s diagram showing the effects of compressor and turbine efficiencies is below.

s

T 100 kPa

800 kPa

T-s Diagram for Gas Turbine with Regeneration

1

2s

2a5

3

4s

4a

6

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Summary of Results

Cycle type Actual Actual Actual Ideal Ideal Ideal

regen 0.00 0.65 1.00 0.00 0.65 1.00

comp 0.75 0.75 0.75 1.00 1.00 1.00

turb 0.86 0.86 0.86 1.00 1.00 1.00

qin kJ/kg 578.3 504.4 464.6 659.9 582.2 540.2

wcomp kJ/kg 326.2 326.2 326.2 244.6 244.6 244.6

wturb kJ/kg 464.6 464.6 464.6 540.2 540.2 540.2

wcomp/wturb 0.70 0.70 0.70 0.453 0.453 0.453

th 24.0% 27.5% 29.8% 44.8% 50.8% 54.7%

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Compressor analysis

The isentropic temperature at compressor exit is( 1) /

2 2

1 1

( 1) /

(1.4 1) /1.422 1

1

800300 ( ) 543.4

100

k k

s

k k

s

T P

T P

P kPaT T K K

P kPa

To find the actual temperature at compressor exit, T2a, we apply the compressor efficiency

comp

isen comp

act comp

s

a

s

a

a

comp

s

w

w

h h

h h

T T

T T

T T T T

K K

K

,

,

( )

.(543. )

.

2 1

2 1

2 1

2 1

2 1 2 1

1

3001

0 754 300

624 6

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Since the compressor is adiabatic and has steady-flow

2 1 2 1( )

1.005 (624.6 300) 326.2

comp a p aw h h C T T

kJ kJK

kg K kg

Turbine analysis

The conservation of energy for the turbine, process 3-4, yields for constant specific heats (let’s take a minute for you to get the following result)

( )

( )

( )

W m h h

W mC T T

wW

mC T T

turb a

turb p a

turbturb

p a

3 4

3 4

3 4

40

Since P3 = P2 and P4 = P1, we can find the isentropic temperature at the turbine exit. ( 1) /

4 4

3 3

( 1) /

(1.4 1) /1.444 3

3

1001200 ( ) 662.5

800

k k

s

k k

s

T P

T P

P kPaT T K K

P kPa

To find the actual temperature at turbine exit, T4a, we apply the turbine efficiency.

turb

act turb

isen turb

a

s

a

s

a turb s

a

w

w

h h

h h

T T

T T

T T T T

K K

K T

,

,

( )

. ( . )

.

3 4

3 4

3 4

3 4

4 3 3 4

2

1200 086 1200 662 5

737 7

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The turbine work becomes

w h h C T T

kJ

kg KK

kJ

kg

turb a p a

3 4 3 4

1005 1200 737 7

464 6

( )

. ( . )

.

The back work ratio is defined as

BWRw

w

w

w

kJ

kgkJ

kg

in

out

comp

turb

326 2

464 60 70

.

..

42

Regenerator analysis

To find T5, we apply the regenerator effectiveness.

regena

a a

a regen a a

T T

T T

T T T T

K K

K

5 2

4 2

5 2 4 2

624 6 0 65 737 7 624 6

6981

( )

. . ( . . )

.

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43

( )

. ( . . )

.

m h Q m h

m m m

qQ

mh h

C T T

kJ

kg KK

kJ

kg

a a regen

a

regen

regen

a

p a

2 2 5 5

2 5

5 2

5 2

1005 6981 624 6

739

To find the heat transferred from the turbine exhaust gas to the compressor exit gas, apply the steady-flow conservation of energy to the compressor gas side of the regenerator.

44

4 4 6 6

4 6

4 6 4 6

6 4

( )

73.9

737.7

1.005

664.2

a a regen

a

regen

regen a p a

regen

a

p

m h Q m h

m m m

Qq h h C T T

m

kJq kg

T T KkJC

kg K

K

&& &

& & &

&

&

Using qregen, we can determine the turbine exhaust gas temperature at the regenerator exit.

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Heat supplied to cycle

Apply the steady-flow conservation of energy to the heat exchanger for process 5-3. We obtain a result similar to that for the simple Brayton cycle.

q h h C T T

kJ

kg KK

kJ

kg

in p

3 5 3 5

1005 1200 6981

504 4

( )

. ( . )

.

Cycle thermal efficiency

The net work done by the cycle is

(464.6 326.2) 138.4

net turb compw w w

kJ kJ

kg kg

46

The cycle efficiency becomes

th Braytonnet

in

w

q

kJ

kgkJ

kg

or

,

.

.. .

138 4

504 40 274 27 4%

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Other Ways to Improve Brayton Cycle Performance

Intercooling and reheating are two important ways to improve the performance of the Brayton cycle with regeneration.

48

The T-s diagram for this cycle is shown below.

Sketch the P-v diagram

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Intercooling

When using multistage compression, cooling the working fluid between the stages will reduce the amount of compressor work required. The compressor work is reduced because cooling the working fluid reduces the average specific volume of the fluid and thus reduces the amount of work on the fluid to achieve the given pressure rise.

To determine the intermediate pressure at which intercooling should take place to minimize the compressor work.

For the adiabatic, steady-flow compression process, the work input to the compressor per unit mass is

4 32 4

1 31 2

= = compw v dP v dP v dP vdP 0

50

For the isentropic compression process

2 2 1 1 4 4 3 3

2 1 4 3

1 2 1 3 4 3

( 1) /( 1) /

2 41 3

1 3

= ( ) ( )-1 -1

( ) ( )-1 -1

( / 1) ( / 1)-1

1 1-1

comp

k kk k

k kw P v Pv P v Pv

k k

k kRR T T T T

k k

kR T T T T T T

k

k P PR T T

k P P

Notice that the fraction kR/(k-1) = Cp.

Can you obtain this relation another way? Hint: apply the first law to processes 1-4.

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For two-stage compression, let’s assume that intercooling takes place at constant pressure and the gases can be cooled to the inlet temperature for the compressor, such that P3 = P2 and T3 = T1.

The total work supplied to the compressor becomes

To find the unknown pressure P2 that gives the minimum work input for fixed compressor inlet conditions T1, P1, and exit pressure P4, we set

dw P

dP

comp ( )2

2

0

52

This yields

P PP2 1 4

or, the pressure ratios across the two compressors are equal.

P

P

P

P

P

P2

1

4

2

4

3

Intercooling is almost always used with regeneration. During intercooling the compressor final exit temperature is reduced; therefore, more energy must be supplied in the heat addition process to achieve the maximum temperature of the cycle. Regeneration can make up part of the required heat transfer.

To supply only compressed air, using intercooling requires less work input. The next time you go to a home supply store where air compressors are sold, check the larger air compressors to see if intercooling is used. For the larger air compressors, the compressors are made of two piston-cylinder chambers. The intercooling heat exchanger is often a pipe with a attached fins that connects the large piston-cylinder chamber with the smaller piston-cylinder chamber. Often the fly wheel used to drive the compressor has spokes shaped like fan blades that are used to increase air flow across the compressor and heat exchanger pipe to improve the intercooling effect.

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Reheating

When using multistage expansion through two or more turbines, reheating between stages will increase the net work done (it also increases therequired heat input). The regenerative Brayton cycle with reheating was shown above.

The optimum intermediate pressure for reheating is the one that maximizes the turbine work. Following the development given above for intercooling and assuming reheating to the high-pressure turbine inlet temperature in a constant pressure steady-flow process, we can show the optimum reheat pressure to be

P P P7 6 9or the pressure ratios across the two turbines are equal.

P

P

P

P

P

P6

7

7

9

8

9

54

Terminology for Reciprocating Devices

The following is some terminology we need to understand for reciprocating engines—typically piston-cylinder devices. Let’s look at the following figures for the definitions of top dead center (TDC), bottom dead center (BDC), stroke, bore, intake valve, exhaust valve, clearance volume, displacement volume, compression ratio, and mean effective pressure.

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The compression ratio r of an engine is the ratio of the maximum volume to the minimum volume formed in the cylinder.

rV

V

V

VBDC

TDC

max

min

The mean effective pressure (MEP) is a fictitious pressure that, if it operated on the piston during the entire power stroke, would produce the same amount of net work as that produced during the actual cycle.

MEPW

V V

w

v vnet net

max min max min

56

Internal Combustion Engine

Below is an eight-cylinder, four stroke internal combustion engine.

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Otto Cycle: The Ideal Cycle for Spark-Ignition Engines

Consider the automotive spark-ignition power cycle.

ProcessesIntake stroke Compression stroke Power (expansion) stroke Exhaust stroke

Often the ignition and combustion process begins before the completion of the compression stroke. The number of crank angle degrees before the piston reaches TDC on the number one piston at which the spark occurs is called the engine timing. What are the compression ratio and timing of your engine in your car, truck, or motorcycle?

58

The air-standard Otto cycle is the ideal cycle that approximates the spark-ignition combustion engine.

Process Description 1-2 Isentropic compression 2-3 Constant volume heat addition 3-4 Isentropic expansion 4-1 Constant volume heat rejection

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59

The P-v and T-sdiagrams are

60

Thermal Efficiency of the Otto cycle:

thnet

in

net

in

in out

in

out

in

W

Q

Q

Q

Q Q

Q

Q

Q

1

Now to find Qin and Qout.

Apply first law closed system to process 2-3, V = constant.

Thus, for constant specific heats,

Q U

Q Q mC T T

net

net in v

,

, ( )

23 23

23 3 2

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61

Apply first law closed system to process 4-1, V = constant.

Thus, for constant specific heats,

Q U

Q Q mC T T

Q mC T T mC T T

net

net out v

out v v

,

, ( )

( ) ( )

41 41

41 1 4

1 4 4 1

The thermal efficiency becomes

th Ottoout

in

v

v

Q

Q

mC T T

mC T T

,

( )

( )

1

1 4 1

3 2

62

th Otto

T T

T T

T T T

T T T

,

( )

( )

( / )

( / )

1

11

1

4 1

3 2

1 4 1

2 3 2Recall processes 1-2 and 3-4 are isentropic, so

Since V3 = V2 and V4 = V1, we see that

T

T

T

T

or

T

T

T

T

2

1

3

4

4

1

3

2

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63

The Otto cycle efficiency becomes

th Otto

T

T, 1 1

2Is this the same as the Carnot cycle efficiency?

Since process 1-2 is isentropic,

where the compression ratio is r = V1/V2 and

th Otto kr,

1

11

64

We see that increasing the compression ratio increases the thermal efficiency. However, there is a limit on r depending upon the fuel. Fuels under high temperature resulting from high compression ratios will prematurely ignite, causing knock.

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65

Example

An Otto cycle having a compression ratio of 9:1 uses air as the working fluid. Initially P1 = 95 kPa, T1 = 17oC, and V1 = 3.8 liters. During the heat addition process, 7.5 kJ of heat are added. Determine all T's, P's, th, the back work ratio, and the mean effective pressure.

Process Diagrams: Review the P-v and T-s diagrams given above for the Otto cycle.

Assume constant specific heats with Cv = 0.718 kJ/kg K, k = 1.4.

Process 1-2 is isentropic; therefore, recalling that r = V1/V2 = 9,

66

The first law closed system for process 2-3 was shown to reduce to (your homework solutions must be complete; that is, develop your equations from the application of the first law for each process as we did in obtaining the Otto cycle efficiency equation)

Q mC T Tin v ( )3 2

Let qin = Qin / m and m = V1/v1

vRT

P

kJ

kg KK

kPa

m kPa

kJ

m

kg

11

1

3

3

0 287 290

95

0875

. ( )

.

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67

qQ

mQ

v

V

kJ

m

kg

m

kJ

kg

inin

in

1

1

3

3 37 5

0875

38 10

1727

.

.

.

Then,T T

q

C

K

kJ

kgkJ

kg K

K

in

v

3 2

698 4

1727

0 718

3103 7

..

.

68

Using the combined gas law (V3 = V2)

P PT

TMPa3 2

3

2

9 15 .

Process 3-4 is isentropic; therefore,

1 1 1.4 1

34 3 3

4

1 1(3103.7)

9

1288.8

k kV

T T T KV r

K

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69

Process 4-1 is constant volume. So the first law for the closed system gives, on a mass basis,

Q mC T T

qQ

mC T T

kJ

kg KK

kJ

kg

out v

outout

v

( )

( )

. ( . )

.

4 1

4 1

0 718 1288 8 290

717 1

The first law applied to the cycle gives (Recall ucycle = 0)

w q q q

kJ

kg

kJ

kg

net net in out

( . )

.

1727 717 4

1009 6

70

The thermal efficiency is

th Ottonet

in

w

q

kJ

kgkJ

kg

or

,

.

. .

1009 6

1727

0585 58 5%

The mean effective pressure is

max min max min

1 2 1 2 1 1

3

3

(1 / ) (1 1/ )

1009.6

12981

0.875 (1 )9

net net

net net net

W wMEP

V V v v

w w w

v v v v v v r

kJ

m kPakgkPa

m kJ

kg

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71

The back work ratio is (can you show that this is true?)

2 112 2 1

exp 34 3 4 3 4

( ) ( )

( ) ( )

0.225 22.5%

comp v

v

w C T Tu T TBWR

w u C T T T T

or

Air-Standard Diesel Cycle

The air-standard Diesel cycle is the ideal cycle that approximates the Diesel combustion engine

Process Description 1-2 Isentropic compression 2-3 Constant pressure heat addition 3-4 Isentropic expansion 4-1 Constant volume heat rejection

The P-v and T-s diagrams are

72

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73

Thermal efficiency of the Diesel cycle

th Dieselnet

in

out

in

W

Q

Q

Q, 1

Now to find Qin and Qout.

Apply the first law closed system to process 2-3, P = constant.

Thus, for constant specific heats

Q U P V V

Q Q mC T T mR T T

Q mC T T

net

net in v

in p

,

,

( )

( ) ( )

( )

23 23 2 3 2

23 3 2 3 2

3 2

74

Apply the first law closed system to process 4-1, V = constant (just as we did for the Otto cycle)

Thus, for constant specific heats

Q U

Q Q mC T T

Q mC T T mC T T

net

net out v

out v v

,

, ( )

( ) ( )

41 41

41 1 4

1 4 4 1

The thermal efficiency becomes

th Dieselout

in

v

p

Q

Q

mC T T

mC T T

,

( )

( )

1

1 4 1

3 2

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75

th Dieselv

p

C T T

C T T

k

T T T

T T T

,

( )

( )

( / )

( / )

1

11 1

1

4 1

3 2

1 4 1

2 3 2What is T3/T2 ?

PV

T

PV

TP P

T

T

V

Vrc

3 3

3

2 2

2

3 2

3

2

3

2

where

where rc is called the cutoff ratio, defined as V3 /V2, and is a measure of the duration of the heat addition at constant pressure. Since the fuel is injected directly into the cylinder, the cutoff ratio can be related to the number of degrees that the crank rotated during the fuel injection into the cylinder.

76

What is T4/T1 ?PV

T

PV

TV V

T

T

P

P

4 4

4

1 1

1

4 1

4

1

4

1

where

Recall processes 1-2 and 3-4 are isentropic, so

PV PV PV PVk k k k1 1 2 2 4 4 3 3 and

Since V4 = V1 and P3 = P2, we divide the second equation by the first equation and obtain

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77

th Diesel

ck

c

kck

c

k

T T T

T T T

k

T

T

r

r

r

r

k r

,

( / )

( / )

( )

( )

11 1

1

11 1

1

11 1

1

1 4 1

2 3 2

1

2

1

Therefore,

When rc > 1 for a fixed r, .

But, since , .

78

th Diesel th Otto, ,

r rDiesel Otto th Diesel th Otto, ,

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Refrigeration Cycles

79

80

The vapor compression refrigeration cycle is a common method for transferring heat from a low temperature to a high temperature.

The above figure shows the objectives of refrigerators and heat pumps. The purpose of a refrigerator is the removal of heat, called the cooling load, from a low-temperature medium. The purpose of a heat pump is the transfer of heat to a high-temperature medium, called the heating load. When we are interested in the heat energy removed from a low-temperature space, the device is called a refrigerator. When we are interested in the heat energy supplied to the high-temperature space, the device is called a heat pump. In general, the term heat pump is used to describe the cycle as heat energy is removed from the low-temperature space and rejected to the high-temperature space.

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81

The performance of refrigerators and heat pumps is expressed in terms of coefficient of performance (COP), defined as

COPQ

W

COPQ

W

RL

net in

HPH

net in

Desired output

Required input

Cooling effect

Work input

Desired output

Required input

Heating effect

Work input

,

,

Both COPR and COPHP can be larger than 1. Under the same operating conditions, the COPs are related by

COP COPHP R 1

Can you show this to be true?

Refrigerators, air conditioners, and heat pumps are rated with a SEER number or seasonal adjusted energy efficiency ratio. The SEER is defined as the Btu/hr of heat transferred per watt of work energy input. The Btu is the British thermal unit and is equivalent to 778 ft-lbf of work (1 W = 3.4122 Btu/hr

Refrigeration systems are also rated in terms of tons of refrigeration. One ton of refrigeration is equivalent to 12,000 Btu/hr or 211 kJ/min. How did the term “ton of cooling” originate?

82

Reversed Carnot Refrigerator and Heat Pump

Shown below are the cyclic refrigeration device operating between two constanttemperature reservoirs and the T-s diagram for the working fluid when the reversedCarnot cycle is used. Recall that in the Carnot cycle heat transfers take place atconstant temperature. If our interest is the cooling load, the cycle is called the Carnotrefrigerator. If our interest is the heat load, the cycle is called the Carnot heat pump.

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83

The standard of comparison for refrigeration cycles is the reversed Carnot cycle. A refrigerator or heat pump that operates on the reversed Carnot cycle is called a Carnot refrigerator or a Carnot heat pump, and their COPs are

COPT T

T

T T

COPT T

T

T T

R Carnot

H L

L

H L

HP Carnot

L H

H

H L

,

,

/

/

1

1

1

1

Notice that a turbine is used for the expansion process between the high and low-temperatures. While the work interactions for the cycle are not indicated on the figure, the work produced by the turbine helps supply some of the work required by the compressor from external sources.

Why not use the reversed Carnot refrigeration cycle?•Easier to compress vapor only and not liquid-vapor mixture.•Cheaper to have irreversible expansion through an expansion valve.

84

The Vapor-Compression Refrigeration Cycle

The vapor-compression refrigeration cycle has four components: evaporator, compressor, condenser, and expansion (or throttle) valve. The most widely used refrigeration cycle is the vapor-compression refrigeration cycle. In an ideal vapor-compression refrigeration cycle, the refrigerant enters the compressor as a saturated vapor and is cooled to the saturated liquid state in the condenser. It is then throttled to the evaporator pressure and vaporizes as it absorbs heat from the refrigerated space.

The ideal vapor-compression cycle consists of four processes.

Ideal Vapor-Compression Refrigeration Cycle Process Description 1-2 Isentropic compression 2-3 Constant pressure heat rejection in the condenser3-4 Throttling in an expansion valve4-1 Constant pressure heat addition in the evaporator

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85

The P-h diagram is another convenient diagram often used to illustrate the refrigeration cycle.

86

The ordinary household refrigerator is a good example of the application of this cycle.

COPQ

W

h h

h h

COPQ

W

h h

h h

RL

net in

HPH

net in

,

,

1 4

2 1

2 3

2 1

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87

Example

Refrigerant-134a is the working fluid in an ideal compression refrigeration cycle. The refrigerant leaves the evaporator at -20oC and has a condenser pressure of 0.9 MPa. The mass flow rate is 3 kg/min. Find COPR and COPR, Carnot for the same Tmax and Tmin , and the tons of refrigeration.

Using the Refrigerant-134a Tables, we have

12

2 2

11 2

2 11

3

3

21

238.41278.23

90020

0.9456 43.790.94561.0

3

900

0

s

soo

ss

StateState kJ

kJh Compressor exithCompressor inlet kg

kgP P kPakJT C

s T CkJkg K s sx

kg K

State

Condenser exit

P kPa

x

3 4

44 13

4 3

4101.61 0.358

0.4053200.3738

.0

o

StatekJh x

Throttle exitkgkJ

skJ T T Cs kg K

kg K h h

88

1 4 1 4

, 2 1 2 1

( )

( )

(238.41 101.61)

(278.23 238.41)

3.44

LR

net in

Q m h h h hCOP

W m h h h h

kJ

kgkJ

kg

The tons of refrigeration, often called the cooling load or refrigeration effect, are

1 4( )

13 (238.41 101.61)

min 211min

1.94

LQ m h h

kg kJ Ton

kJkg

Ton

,

( 20 273)

(43.79 ( 20))

3.97

LR Carnot

H L

TCOP

T T

K

K

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89

Another measure of the effectiveness of the refrigeration cycle is how much input power to the compressor, in horsepower, is required for each ton of cooling.

The unit conversion is 4.715 hp per ton of cooling.

, 4.715

4.715

3.44

1.37

net in

RL

W

COPQ

hp

Ton

hp

Ton

90

Actual Vapor-Compression Refrigeration Cycle

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91

Heat Pump Systems

92

Other Refrigeration Cycles

Cascade refrigeration systems

Very low temperatures can be achieved by operating two or more vapor-compressionsystems in series, called cascading. The COP of a refrigeration system alsoincreases as a result of cascading.

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93

Multistage compression refrigeration systems

94

Multipurpose refrigeration systems

A refrigerator with a single compressor can provide refrigeration at several temperatures by throttling the refrigerant in stages.

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95

Liquefaction of gases

Another way of improving the performance of a vapor-compression refrigeration system is by using multistage compression with regenerative cooling. The vapor-compression refrigeration cycle can also be used to liquefy gases after some modifications.

96

Gas Refrigeration Systems

The power cycles can be used as refrigeration cycles by simply reversing them. Of these, the reversed Brayton cycle, which is also known as the gas refrigeration cycle, is used to cool aircraft and to obtain very low (cryogenic) temperatures after it is modified with regeneration. The work output of the turbine can be used to reduce the work input requirements to the compressor. Thus, the COP of a gas refrigeration cycle is

COPq

w

q

w wR

L

net in

L

comp in turb out

, , ,

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97

Absorption Refrigeration Systems

Another form of refrigeration that becomes economically attractive when there is a source of inexpensive heat energy at a temperature of 100 to 200oC is absorption refrigeration, where the refrigerant is absorbed by a transport medium and compressed in liquid form. The most widely used absorption refrigeration system is the ammonia-water system, where ammonia serves as the refrigerant and water as the transport medium. The work input to the pump is usually very small, and the COP of absorption refrigeration systems is defined as

COPQ

Q W

Q

QR

L

gen pump in

L

gen

Desired output

Required input

Cooling effect

Work input ,

98

Thermoelectric Refrigeration Systems

A refrigeration effect can also be achieved without using any moving parts by simply passing a small current through a closed circuit made up of two dissimilar materials. This effect is called the Peltier effect, and a refrigerator that works on this principle is called a thermoelectric refrigerator.

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liquefaction Process

99

زمینه تولید در فعال شرکتهايLNG در مقیاس کوچک تا متوسط

در زمینه ایستگاه هاي مقیاس کوچک مختلفهاي چرخه

از چند ایستگاه مقیاس کوچک هایینمونه

ماژولهاسازي استاندارد

مطرح در زمینه طراحی کدهاياستانداردها و

تجهیزات چرخه مایع سازي

برآورد هزینه ها از چند مرجع براي ایستگاه هاي مقیاس کوچک

شکست پروژه و زمان بندي براي راه اندازي ایستگاه نمونه

100

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51

ردمبمختلف بر اساس ترمودینامیکیچرخه هاي

چرخه

ترمودینامیکی

Single Mixed Refrigerant

(SMR)

With propane precooling (C3/MR)

Without precooling

Dual Mixed refrigerant

(DMR)

With propane precooling (C3/MR)

Without precooling

Mixed fluid cascade (MFC)

N2

Single cycle Dual cycle

Niche

101

مقیاس LNGشرکتهاي فعال و چرخه هاي مختلف در زمینه ایستگاه هاي تولید کوچک

102

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LNGمقایسه فن آوري هاي مختلف تولید

103

Single Mixed Refrigerant (SMR)

مبرد مخلوط تک مبدله رایج ترین و ارزانترین فرایند تولیدLNG است .مخلوط گازهاي مبرد شامل متان، اتان، پروپان، پنتان، و نیتروژن است .هر چه این . دترکیبات این مخلوط باید به گونه اي باشد که منحنی مشخصه آن با منحنی مشخصه گاز طبیعی منطبق شو

. انطباق بیشتر شود بازده سیکل تبرید بیشتر خواهد شد

104

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Extensions of Single Mixed Refrigerant

زیادي ایجاد می کند که منجر به مصرف توان باالي کمپرسور خواهد شد سرمایشاین فرایند نرخ.

اهش دهدالبته دو مرحله اي کردن فرایند تراکم و استفاده از سرد کردن میانی می تواند کار مصرفی جهت تراکم را ک.

نسبت تراکم در چرخه نسبت به چرخهNiche LNG کمتر خواهد بود.

ول به منظور افزایش بازدهی چرخه می توان از یک چرخه پیش سرد کن اولیه هم استفاده نمود که این امري متدا

سانتیگراد قبل از چرخه -35است تا طی آن دماي گاز به حدود پروپانعموما سردکنچرخه پیش مبرد. است

.اصلی می رسد سرمایش

105

Dual Nitrogen Refrigerant

106

Statoil BHP

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Dual nitrogen refrigerant

:نیترون سرمایشبرخی محاسن چرخه

اجزاي چرخه ساده است و تعداد تجهیزات مورد نیاز، زیاد نیست

راه اندازي و نگهداري کم هزینه تجهیزات

فقط در فاز گازي خواهد بود –تکفاز بودن سیکل.

نیتروژنایمنی عملکرد ایمن در فشار باال و قابلیت تراکم دو مرحله اي و غیرقابل اشتعال بودن

107

Niche LNG

چرخه دو در نیتروژن و متان گاز تبرید، چرخه در مبرد ارايد که است باز دیگري و بسته سیکل یک .مجزاست

سردسازي فرآیند در ولی متفاوتند فرایند و فشار .هستند هماهنگ

دیگر و شیرها و ها لوله اندازه باال فشار چرخه در که مواردي در سیکل این و است کوچکتر تجهیزات

.دارد مزیت دارد وجود فضا محدودیت

و مانند می فاز تک چرخه طی در عامل سیال دو هر ازف به مربوط تجهیزات به نیاز و فاز دو جریان مسایل

.ندارد وجود مایع

108

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The NicheLNG process

109

شبیه سازي هاي عددي چرخه ترمودینامیکی

110

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Hysysشبیه سازي با نرم افزار

111

Dual N2و NicheLNGمقایسه چرخه

112

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113

SMR

DMR

Hysysشبیه سازي با نرم افزار

114

APCI

MATLABو بهینه سازي در Hysysشبیه سازي با نرم افزار

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معیارهاي انتخاب چرخه مناسب

115

معیارهاي انتخاب چرخه مناسب

وزن و حجم کم تجهیزات با توجه به محدودیت هاي فضا و وزن

انعطاف پذیري نسبت به ترکیبات گاز تغذیه ورودي مختلف

قابلیت راه اندازي سریع و توقف در مواقع ضروري

عملکرد ساده و در عین حال بهینه

هزینه به صرفه راه اندازي و تعمیر و نگهداري

هزینه کم تبدیلLNG به گاز

116

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چرخه بر حسب ظرفیت معیارهاي انتخاب

تولید

117

مقایسه تطبیقی چرخه هاي مختلف

118

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مبردهاو چرخه مخلوط تراکمیمقایسه چرخه هاي تبرید

119

Compact

Efficiency

Operation

Safety

فشار .دباش می نیاز نیتروژن چرخه به نسبت بیشتري فضاي مبردها مخلوط چرخه در ذیرپ اشتعال ،مبردها مخلوط چرخه خالف بر همچنین و است باالتر نیتروژن چرخه در

.نیست

مبردها مخلوط چرخه در ترمودینامیکی باالي بازدهی

Start همچنین ،مبردها مخلوط چرخه مبرد در پیچیدگی و تجهیزات باالي تعداد up flare به نیاز و کُند

.ندارد احتراق قابلیت مبرد که باشد می نیتروژن چرخه مبرد نوع اساس بر چرخه ایمن

Liquefaction processes Specifications

120

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61

هاي میزان سرمایه گذاري مورد نیاز براي بخش

مختلف

121

ه اطالعات کلی راجع به چرخه ها در ایستگاMINI LNGهاي

122

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62

تن در 50با ظرفیت LNGمقایسه چرخه هاي مختلف مایع سازي روز

LNGتن 50روزانه : ظرفیت تولید

:د ازدر این خصوص روشهاي مختلفی پیشنهاد شده است ولی شاخصترین آنها عبارتن

SMR

C3/MR

N2 Expander

123

124

TOTALمدل شرکت N2محاسبه توان مصرفی در چرخه

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125

APCIمدل مبردهامحاسبه توان مصرفی در چرخه مخلوط

126

بر حسب کاهش دما در دو چرخه مختلف سرمایشنمودار میزان

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64

127

128

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میزان توان مصرفی در روش مقایسه بین

N2 وSMR

129

130

ت کممقایسه بین سرمایه گذاري در چرخه هاي مختلف براي ایستگاه هاي ظرفی

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دمبرمشخصات گاز ورودي در ایران و انتخاب

مناسب

•

131

از، مقدار چاه گ جغرافیاییشده و بسته به محل تشکیلاساساً از متان طبیعیگازگاز طبیعی ایران تقریباً .(درصد متغیر است 98تا 80 بینمتان موجود در آن

ان مشابه با گاز مت بسیارگاز اینخواص بنابراین) درصد متان است 5/96داراي . است

:د ازعبارتن اهمیت ترتیبنیز در آن وجود دارد که به دیگرياز متان، مواد غیربه )درصد 8تا 1 بین( اتان)1)درصد 2تا ( پروپان) 2)درصد یکاز کمتر( پنتانو بوتان) 3

نیتروژنموادي مانند همچنین (N2) ،دي اکسیدکربن (CO2) ،هیدروژن سولفید)H2S ( ر شده د بیان درصدهاي. شود یافت طبیعیاست در گاز ممکن نیزو آب

ودي مختلف، این مقادیر تا حد چاههايمی باشند و براي صحیح کلیباال به طور .تغییر می کند

ملی تشرکاست که براي استفاده در موتور خودروها، براساس استاندارد ذکر شایان

.درصد متان است، استفاده می شود 80از بیششامل که طبیعیگاز از گاز

احتمالی مبردهايمشخصات ترکیب

132

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67

LNGنحوه ذخیره سازي

:منابع ذخیره براي این کار عبارتند از

مخازن ذخیره با سطوح فوالدي داخل و خارج و عایق پرلیت منبسط شده بعنوان عایق و کف -1

کامال عایق شده

مخازن با دیواره داخلی فلزي خارجی بتونی -2

شود می بینی پیش اول نوع ذخیره مخازن )مکعب متر 200-100( نظر مورد کم حجم به توجه با

160- حدود دماي با مایع به توجه با C 9 فوالد از مخزن کف و داخلی الیه در% Ni استاندارد با

ASTM A553 گریدهاي از یکی نیز جوشکاري الکترود .میشود استفاده آن معادل یا

INCONEL دماي رد تست این پایین، بسیار دماي در ضربه به مقاومت به نیاز به توجه با .است -

196C باشد می انجام ابلق ایران در.

133

134

استاندارد تست مخازن ذخیره

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نمایی از یک مخزن ذخیره

135

استانداردهاي مرتبط

مربوط به المانهاي پروسه می باشند و بعضی LNGاستانداردهاي موجود در رابطه با فرایند •:از آنها عبارتند از

API 625 وBS EN14620 part 1,2 مخازن ذخیرهLNG BS EN 1473 پلنتهاينصب و راه اندازي LNG در

خشکی

EN 1160 مشخصات کلیLNG EN 12066 عایقکاري و تستهاي آب بندي EN 13480 طراحیPIPING ASTM C549 پرلیت عایقکاري مخازن EN 10045 وEN875 تستهاي ضربه در دماي بسیار کم

136

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رو و چالشهاي پیش تجهیزات اصلی مورد نیاز

137

Screw Compressor

Cryogenic Heat Exchanger

Special PumpsKind of Mixture

Refrigerant Components

MaterialsStorage TanksSpecial Welding

138

مبدل حرارتی دما پایین

10/28/2018

70

139

مبدل حرارتی دما پایین

140

10/28/2018

71

تگاه بدون پیش سرمایش مناسب براي ایسروش تبرید مخلوط مبردها چرخه هاي مختلف مقیاس کوچکدر LNGتولید

141

ه چند ایستگاه نمونبه همراه چرخه و

ظرفیت تولید

142

10/28/2018

72

Small Scale Liquefiers,

Operating and Planned

143

144

10/28/2018

73

Small Scale Liquefiers,

Operating and Planned

145

Lindeشرکت

محدوده تولید ایستگاه مقیاس کوچک این شرکت

در روز LNGتن 1600تا 100

146

10/28/2018

74

Lindeدر شرکت ماژولهااستاندارد سازي

147

Lindeشرکت

148

10/28/2018

75

Lindeشرکت

149

150

10/28/2018

76

محدوده تولید ایستگاه مقیاس کوچک این شرکت

در روز LNGتن 400تا 5

چرخه تبرید نیتروژنی

چرخه مخلوط مبردها

151

152

10/28/2018

77

Hamaworthyشرکت

153

154