Entropy - Mollier Diagram

Hlock 2 Steam Engineering Principles and Heat Transfer Entropy - A Basic Understanding Module 2.15

Enthalpy

E n t r o p y - A B a s i c U n d e r s t a n d i n g

What is entropy? In some ways, it is easier to say what it is not! It is not a physical property of steam like pressure or temperature or mass. A sensor cannot detect it, and it does not show on a gauge. Rather, it must be calculated from things that can be measured. Entropy values can then be listed and used in calculations; in particular, calculations to do with steam flow, and the production of power using turbines or reciprocating engines.

It is, in some ways, a measure of the lack of quality or availability of energy, and of how energy tends always to spread out from a high temperature source to a wider area at a lower temperature level. This compulsion to spread out has led some observers to label entropy as 'time's arrow'. If the entropy of a system is calculated at two different conditions, then the condition at which the entropy is greater occurs at a later time. The increase of entropy in the overall system always takes place in the same direction as time flows.

That may be of some philosophical interest, but does not help very much in the calculation of actual values. A more practical approach is to define entropy as energy added to or removed from a system, divided by the mean absolute temperature over which the change takes place.

To see how this works, perhaps it is best to start off with a diagram showing how the enthalpy content of a kilogram of water increases as it is heated to different pressures and evaporated into steam.

Since the temperature and pressure at which water boils are in a fixed relationship to each other, Figure 2.15.1 could equally be drawn to show enthalpy against temperature, and then turned so that temperature became the vertical ordinates against a base of enthalpy, as in Figure 2.15.2.

Enthalpy of saturated steam

Pressure

Fig. 2.15.1 The enthalpy/pressure diagram

Enthalpy of evaporation

Enthalpy of water

2.15.2 The Steam and Condensate Loop

Tem

pera

ture

t in

°C

200

300

100

0

Saturated water line

Evaporation lines

Dryness fraction lines

Superheated steam region

Dry saturated steam line

Lines of constant pressure

Block 2 Steam Engineering Principles and Heat Transfer Entropy - A Basic Understanding Module 2.15

Critical 400 point

0 500 1000 1500 2000 2500 3000

Enthalpy h in kJ/kg

Fig. 2.15.2 The temperature/enthalpy diagram

Lines of constant pressure originate on the saturated water line. The horizontal distance between the saturated water line and the dry saturated steam line represents the amount of latent heat or enthalpy of evaporation, and is called the evaporation line; (enthalpy of evaporation decreases with rising pressure). The area to the right of the dry saturated steam line is the superheated steam region, and lines of constant pressure now curve upwards as soon as they cross the dry saturated steam line.

A variation of the diagram in Figure 2.15.2, that can be extremely useful, is one in which the horizontal axis is not enthalpy but instead is enthalpy divided by the mean temperature at which the enthalpy is added or removed. To produce such a diagram, the entropy values can be calculated. By starting at the origin of the graph at a temperature of 0°C at atmospheric pressure, and by adding enthalpy in small amounts, the graph can be built. As entropy is measured in terms of absolute temperature, the origin temperature of 0°C is taken as 273.15 K. The specific heat of saturated water at this temperature is 4.228 kJ/kg K. For the purpose of constructing the diagram in Figure 2.15.3 the base temperature is taken as 273 K not 273.15 K.

The Steam and Condensate Loop 2.15.3


275 K

Tem

pera

ture

274 K

4.228/274.5

273 K

4.228/273.5

Change in enthalpy Mean temperature

∅ H

T(mean) =

By assuming a kilogram of water at atmospheric pressure, and by adding 4.228 kJ of energy, the water temperature would rise by 1 K from 273 K to 274 K. The mean temperature during this operation is 273.5 K, see Figure 2.15.3.

Fig. 2.15.3 The cumulative addition of 4.228 W of energy to water from 0°C

The width of the element representing the added enthalpy = 4.228 273.5 = 0.015459 kJ kg K.

This value represents the change in enthalpy per degree of temperature rise for one kilogram of water and is termed the change in specific entropy. The metric units for specific entropy are therefore kJ/kg K.

This process can be continued by adding another 4.228 kJ of energy to produce a series of these points on a state point line. In the next increment, the temperature would rise from 274 K to 275 K, and the mean temperature is 274.5 K.

4.228 The width of this element representing the added enthalpy = _______ = 0.015 403 kJ kg K.

274.5

It can be seen from these simple calculations that, as the temperature increases, the change in entropy for each equal increment of enthalpy reduces slightly. If this incremental process were continuously repeated by adding more heat, it would be noticed that the change in entropy would continue to decrease. This is due to each additional increment of heat raising the temperature and so reducing the width of the elemental strip representing it. As more heat is added, so the state point line, in this case the saturated water line, curves gently upwards.

At 373.14 K (99.990C), the boiling point of water is reached at atmospheric pressure, and further additions of heat begin to boil off some of the water at this constant temperature. At this position, the state point starts to move horizontally across the diagram to the right, and is represented on Figure 2.15.4 by the horizontal evaporation line stretching from the saturated water line to the dry saturated steam line. Because this is an evaporation process, this added heat is referred to as enthalpy of evaporation,

At atmospheric pressure, steam tables state that the amount of heat added to evaporate 1 kg of water into steam is 2256.71 kJ. As this takes place at a constant temperature of 373.14 K, the mean temperature of the evaporation line is also 373.14 K.

The change in specific entropy from the water saturation line to the steam saturation line is therefore:

2 256.71 = 6.047891 kJ kg K 373.14



Tem

pera

ture

(T)

473 K

373 K

273 K

1

3

2

4

6.0 6.5 7.0 7.5 8.0 8.5 9.0

1 bar

0.5 bar

0.2 bar

0.1 bar

0.04 bar

0.01 bar

650°C 600°C

550°C 500°C

450°C 400°C

350°C 300°C

250°C 200°C

150°C 100°C

50°C

Ξ = 0.95

Ξ = 0.90 Ξ = 0.85

Ξ = 0.80 Ξ = 0.75

Ξ = 0.70

3800

3600

3400

Spec

ific e

ntha

lpy

(kJ/

kg) 3200

3000

2800

2600

2400

2200

2000

1800

The diagram produced showing temperature against entropy would look something like that in Figure 2.15.4, where: 0 1 is the saturated water line. 0 2 is the dry saturated steam line. 0 3 are constant dryness fraction lines in the wet steam region. 0

4 are constant pressure lines in the superheat region.

Entropy (S)

Fig. 2.15.4 The temperature - entropy diagram

What use is the temperature - entropy diagram (or T - S diagram)? One potential use of the T - S diagram is to follow changes in the steam condition during processes occurring with no change in entropy between the initial and final state of the process. Such processes are termed Isentropic (constant entropy). Unfortunately, the constant total heat lines shown in a T - S diagram are curved, which makes it difficult to follow changes in such free and unrestricted expansions as those when steam is allowed to flow through and expand after a control valve. In the case of a control valve, where the velocities in the connecting upstream and downstream pipes are near enough the same, the overall process occurs with constant enthalpy (isenthalpic). In the case of a nozzle, where the final velocity remains high, the overall process occurs with constant entropy. To follow these different types of processes, a new diagram can be drawn complete with pressures and temperatures, showing entropy on the horizontal axis, and enthalpy on the vertical axis, and is called an enthalpy - entropy diagram, or H - S diagram, Figure 2.15.5.

400 bar 200 bar 100 bar 50 bar 20 bar 10 bar 5 bar 2 bar

Specific entropy (kJ/kg K)

Fig. 2.15.5 The H - S diagram



Specific enthalpy (kJ/kg)

Specific enthalpy (kJ/kg)

3800

3600

3400

3200

3000

2800

2600

2400

2200

2000

1800

1 bar

0.5 bar

0.2 bar

0.1 bar

0.04 bar

0.01 bar

3800

3600

3400

3200

3000

2800

2600

2400

2200

2000

1800

1 bar

0.5 bar

0.2 bar

0.1 bar

0.04 bar

0.01 bar

The H - S diagram is also called the Mollier diagram or Mollier chart, named after Dr. Richard Mollier of Dresden who first devised the idea of such a diagram in 1904.

Now, the isenthalpic expansion of steam through a control valve is simply represented by a straight horizontal line from the initial state to the final lower pressure to the right of the graph, see Figure 2.15.6; and the isentropic expansion of steam through a nozzle is simply a line from the initial state falling vertically to the lower final pressure, see Figure 2.15.7.


200 250

400350°C

550°

C 500°C

650600°C

°C

450°C °C

300°C °C

°C

50°C 100°C

150°C

= 0.90

X = 0.95

X

X = 0.80

X = 0.85

X = 0.75 X = 0.70

6.0 6.5 7.0 7.5 8.0 8.5 9.0


Fig. 2.15.6 Isenthalpic expansion, as through a control valve

400 bar 200 bar 100 bar 50 bar 20 bar 10 bar 5 bar 2 bar 650 °C

300°C °C

400 350°C

450°C °C

550500°C

600°C °C

200 250

°C 150°C

50°C

100°C

X = 0.95

X = 0.90 X = 0.85 X = 0.80 X = 0.75 X = 0.70

6.0 6.5 7.0 7.5 8.0 8.5 9.0


Fig. 2.15.7 Isentropic expansion, as through a nozzle



650600°C

°C

200 250

°C

300°C °C

400350°C

550500°C

°C

450°C °C

50°C 100°C

150°C

= 0.95 Ξ

Ξ = 0.80

Ξ = 0.85 Ξ = 0.90

Ξ = 0.75 Ξ = 0.70

1 bar

0.5 bar

0.2 bar

0.1 bar

0.04 bar

0.01 bar

3800

3600

3400

Spec

ific e

ntha

lpy

(kJ/

kg) 3200

3000

2800

2600

2400

2200

2000

1800

An isentropic expansion of steam is always accompanied by a decrease in enthalpy, and this is referred to as the 'heat drop' (H) between the initial and final condition. The H values can be simply read at the initial and final points on the Mollier chart, and the difference gives the heat drop. The accuracy of the chart is sufficient for most practical purposes.

As a point of interest, as the expansion through a control valve orifice is an isenthalpic process, it is assumed that the state point moves directly to the right; as depicted in Figure 2.15.6. In fact, it does not do so directly. For the steam to squeeze through the narrow restriction it has to accelerate to a higher speed. It does so by borrowing energy from its enthalpy and converting it to kinetic energy. This incurs a heat drop. This part of the process is isentropic; the state point moves vertically down to the lower pressure.

Having passed through the narrow restriction, the steam expands into the lower pressure region in the valve outlet, and eventually decelerates as the volume of the valve body increases to connect to the downstream pipe. This fall in velocity requires a reduction in kinetic energy which is mostly re-converted back into heat and re-absorbed by the steam. The heat drop that caused the initial increase in kinetic energy is reclaimed (except for a small portion lost due to the effects of friction), and on the H - S chart, the state point moves up the constant pressure line until it arrives at the same enthalpy value as the initial condition.

The path of the state point is to be seen in Figure 2.15.8, where pressure is reduced from 5 bar at saturation temperature to 1 bar via, for example, a pressure reducing valve. Steam's enthalpy at the upstream condition of 5 bar is 2748 kJ/kg.


6.0 6.5 7.0 7.5 8.0 8.5 9.0


Fig. 2.15.8 The actual path of the state point in a control valve expansion

It is interesting to note that, in the example dicussed above and shown in Figure 2.15.8, the final condition of the steam is above the saturation line and is therefore superheated. Whenever such a process (commonly called a throttling process) takes place, the final condition of the steam will, in most cases, be drier than its initial condition. This will either produce drier saturated steam or superheated steam, depending on the respective positions of the initial and final state points. The horizontal distance between the initial and final state points represents the change in entropy. In this example, although there was no overall change in enthalpy (ignoring the small effects of friction), the entropy increased from about 6.8 kJ/kg K to about 7.6 kJ/kg K.



Entropy always increases in a closed system In any closed system, the overall change in entropy is always positive, that is, it will always increase. It is worth considering this in more detail, as it is fundamental to the concept of entropy. Whereas energy is always conserved (the first law of thermodynamics states that energy cannot be created or destroyed), the same cannot be said about entropy. The second law of thermodynamics says that whenever energy is exchanged or converted from one form to another, the potential for energy to do work gets less. This really is what entropy is all about. It is a measure of the lack of potential or quality of energy; and once that energy has been exchanged or converted, it cannot revert back to a higher state. The ultimate truth of this is that it is nature's duty for all processes in the Universe to end up at the same temperature, so the entropy of the Universe is always increasing. Example 2.15.1 Consider a teapot on a kitchen table that has just been filled with a certain quantity of water containing 200 kJ of heat energy at 100°C (373 K) from an electric kettle. Consider next that the temperature of the air surrounding the mug is at 20°C, and that the amount of heat in the teapot water would be 40 kJ at the end of the process. The second law of thermodynamics also states that heat will always flow from a hot body to a colder body, and in this example, it is certain that, if left for sufficient time, the teapot will cool to the same temperature as the air that surrounds it. What are the changes in the entropy values for the overall process? For the teapot:

Initial enthalpy in the teapot = 200 kJ Initial teapot temperature = 373 K ( 100 °C)

Final teapot temperature (the air temperature) = 293 K ( 20

°C) Mean teapot temperature T(mean) = 373 K + 293 K2

Mean teapot temperatu ( ) re T m e a n = 333 K ( 60°C)

Final enthalpy in the teapot = 40 kJ

Enthalpy delivered by the teapot to its surroundings = 200 - 40 kJ

Enthalpy delivered by the teapot to its surroundings = 160 k

Entropy delivered by the teapot to its surroundings = Enthalpy change ( m e a n )

160 kJ Entropy delivered by the teapot to its surroundings =

333 K Entropy delivered by the teapot to its surroundings = - 0.48 kJ K

Because the heat is lost from the teapot, convention states that its change in entropy is negative.

For the air: Initial air temperature = 293 K (20°C)

At the end of the process, the water in the teapot would have lost 160 kJ and the air would have gained 160 kJ; however, the air temperature would not have changed because of its large volume, therefore:

T(mean) for the air = 293 K (20°C) 160 kJ

Entropy received by the air = 293 K

Entropy received by the air = + 0.546 kJ K Because the heat is received by the air, convention states that its change in enthalpy is positive.

Therefore:

The overall change in entropy of the teapot and surroundings = - 0.48 + 0.546 kJ/K The

overall change in entropy of the teapot and surroundings = + 0.066 kJ/K



650 °C

600°C °C

250 °C

300°C °C

400350°C

450°C °C

550500°C

200

50°C 100°C

150°C Ξ = 0.95

Ξ = 0.90

Ξ = 0.80

Ξ = 0.85

Ξ = 0.75 Ξ = 0 . 7 0

1 bar

0.5 bar

0.2 bar

0.1 bar

0.04 bar

0.01 bar

3800

3600

3400

Spec

ific e

ntha

lpy

(kJ/

kg) 3200

3000

2800

2600

2400

2200

2000

1800

Practical applications - Heat exchangers In a heat exchanger using saturated steam in the primary side to heat water from 20°C to 60°C in the secondary side, the steam will condense as it gives up its heat. This is depicted on the Mollier chart by the state point moving to the left of its initial position. For steady state conditions, dry saturated steam condenses at constant pressure, and the steam state point moves down the constant pressure line as shown in Figure 2.15.9.

Example 2.15.2 This example considers steam condensing from saturation at 2 bar at 120°C with an entropy of 7.13 kJ/kg K, and an enthalpy of about 2 700 kJ/kg. It can be seen thatthe state point moves from right to left, not horizontally, but by following the constant 2 bar pressure line. The chart is not big enough to show the whole condensing process but, if it were, it would show that the steam's final state point would rest with an entropy of 1.53 kJ/kg K and an enthalpy of 504.8 kJ/kg, at 2 bar and 120°C on the saturated water line.


6.0 6.5 7.0 7.5 8.0 8.5 9.0


Fig. 2.15.9 The initial path of the state point for condensing steam

It can be seen from Figure 2.15.9 that, when steam condenses, the state point moves down the evaporation line and the entropy is lowered. However, in any overall system, the entropy must increase, otherwise the second law of thermodynamics is violated; so how can this decrease in entropy be explained?

As for the teapot in the Example 2.15.1, this decrease in entropy only reflects what is happening in one part of the system. It must be remembered that any total system includes its surroundings, in Example 2.15.2, the water, which receives the heat imparted by the steam.

In Example 2.15.2, the water receives exactly the same amount of heat as the steam imparts (it is assumed there are no heat losses), but does so at a lower temperature than the steam; so, as entropy is given by enthalpy/temperature, dividing the same quantity of heat by a lower temperature means a greater gain in entropy by the water than is lost by the steam. There is therefore an overall gain in the system entropy, and an overall spreading out of energy.



Table 2.15.1 Relative densities/specific heat capacities of various solids

Material Relative density

Specific heat capacity kJ/kg °C

Aluminium 2.55 - 2.80 0.92 Andalusite 0.71 Antimony 0.20 Apatite 0.83 Asbestos 2.10 - 2.80 0.83 Augite 0.79 Bakelite, wood filler 1.38 Bakelite, asbestos filler 1.59 Barite 4.50 0.46 Barium 3.50 2.93 Basalt rock 2.70 - 3.20 0.83 Beryl 0.83 Bismuth 9.80 0.12 Borax 1.70 - 1.80 1.00 Boron 2.32 1.29 Cadmium 8.65 0.25 Calcite 0 - 37°C 0.79 Calcite 0 - 100°C 0.83 Calcium 4.58 0.62 Carbon 1.80 - 2.100 0.71 Carborundum 0.66 Cassiterite 0.37 Cement, dry 1.54 Cement, powder 0.83 Charcoal 1.00 Chalcopyrite 0.54 Chromium 7.10 0.50 Clay 1.80 - 2.60 0.92 Coal 0.64 - 0.93 1.08 - 1.54 Cobalt 8.90 0.46 Concrete, stone 0.79 Concrete, cinder 0.75 Copper 8.80 - 8.95 0.37



Material

Relative density

Specific heat capacity kJ/kg °C

Corundum 0.41 Diamond 3.51 0.62 Dolomite rock 2.90 0.92 Fluorite 0.92 Fluorspar 0.87 Galena 0.20 Proxylin plastics 1.42 - 1.59 Quartz, 12.8 - 100°C 2.50 - 2.80 0.79 Quartz, 0°C 0.71 Rock salt 0.92 Rubber 2.00 Sandstone 2.00 - 2.60 0.92 Serpentine 2.70 - 2.80 1.08 Silk 1.38 Silver 10.40 - 10.60 0.25 Sodium 0.97 1.25 Steel 7.80 0.50 Stone 0.83 Stoneware 0.79 Talc 2.60 - 2.80 0.87 Tar 1.20 1.46 Tellurium 6.00 - 6.24 0.20 Tin 7.20 - 7.50 0.20 Tile, hollow 0.62 Titanium 4.50 0.58 Topaz 0.87 Tungsten 19.22 0.16 Vanadium 5.96 0.50 Vulcanite 1.38 Wood 0.35 - 0.99 1.33 - 2.00 Wool 1.32 1.38 Zinc blend 3.90 - 4.20 0.46 Zinc 6.90 - 7.20 0.37



Table 2.15.2 Relative densities/specific heat capacities of various liquids

Liquid Relative density

Specific heat capacitiy kJ/kg °C

Acetone 0.7900 2.13 Alcohol, ethyl, 0°C 0.7890 2.30 Alcohol, ethyl, 40°C 0.7890 2.72 Alcohol, methyl, 4 - 10°C 0.7960 2.46 Alcohol, methyl, 15 - 21°C 0.7960 2.51 Ammonia 0°C 0.6200 4.60 Ammonia 40°C 4.85 Ammonia 80°C 5.39 Ammonia 100°C 6.19 Ammonia 114°C 6.73 Anilin 1.0200 2.17 Benzol 1.75 Calcium chloride 1.2000 3.05 Castor oil 1.79 Citron oil 1.84 Diphenylamine 1.1600 1.92 Ethyl ether 2.21 Ethylene glycol 2.21 Fuel oil 0.9600 1.67 Fuel oil 0.9100 1.84 Fuel oil 0.8600 1.88 Fuel oil 0.8100 2.09 Gasoline 2.21 Glycerine 1.2600 2.42 Kerosene 2.00 Mercury 19.6000 1.38 Naphthalene 1.1400 1.71 Nitrobenzole 1.50 Olive oil 0.91 - 0.9400 1.96 Petroleum 2.13 Potassium hydrate 1.2400 3.68 Sea water 1.0235 3.93 Sesame oil 1.63 Sodium chloride 1.1900 3.30 Sodium hydrate 1.2700 3.93 Soybean oil 1.96 Toluol 0.8660 1.50 Turpentine 0.8700 1.71 Water 1.0000 4.18 Xylene 0.861 - 0.8810 1.71



Table 2.15.3 Specific heat capacities of gases and vapours

Gas or vapour Specific heat capacity

kJ/kg °C (constant pressure)

Acetone 1.31 Air, dry, 0°C 1.00 Air, dry, 100°C 1.01 Air, dry, 200°C 1.03 Air, dry, 300°C 1.05 Air, dry, 400°C 1.07 Air, dry, 500°C 1.09 Alcohol, C2 H5 OH 1.66 Alcohol, CH3 OH 1.53 Ammonia 1.76 Argon 0.30 Benzene, C6 H6 0.98 Bromine 0.19 Carbon dioxide 0.62 Carbon monoxide 0.71 Carbon disulphide 0.55 Chlorine 3.43 Chloroform 0.54 Ether 1.95 Hydrochloric acid 0.56 Hydrogen 10.00 Hydrogen sulphide 0.79 Methane 1.86 Nitrogen 0.71 Nitric oxide 0.69 Nitrogen tetroxide 4.59 Nitrous oxide 0.69 Oxygen 0.65 Steam, 0.5 bar a saturated 1.99 Steam, 2 bar a saturated 2.13 Steam, 10 bar a saturated 2.56 Steam, 0.5 bar a 150°C 1.95 Steam, 2 bar a 200°C 2.01 Steam, 10 bar a 250°C 2.21 Sulphur dioxide 0.49



Questions

1. What is true of entropy? a It is not a physical property of steam �

b It reflects the quality of energy during a process �

c It is energy change divided by the mean temperature of the change �

d It is all of the above �

2. From the Mollier diagram in Figure 2.15.8, if the initial state point was saturated steam at 2 bar, the final state point was at 0.01 bar, and the expansion was isentropic, what is the approximate heat drop in one kilogram of steam?

a 2 000 kJ �

b 2 700 kJ �

c 700 kJ �

d 1 000 kJ �

3. What always accompanies an isentropic expansion of steam? a An increase in entropy �

b An increase in enthalpy �

c A decrease in entropy �

d A decrease in enthalpy �

4. What always accompanies an isenthalpic expansion of steam? a An increase in entropy �

b An increase in enthalpy �

c A decrease in entropy �

d A decrease in enthalpy �

5. What is true about steam as it condenses? a It does so at constant entropy and temperature �

b It does so at constant enthalpy and reducing temperature �

c Both enthalpy and entropy reduce and temperature remains constant �

d Both enthalpy and entropy increase �

6. From the Mollier diagram in Figure 2.15.8, if the initial state point was saturated steam at 2 bar, the final state point was at 0.01 bar, and the expansion was isenthalpic, what is the approximate heat drop in one kilogram of steam?

a None �

b 1.5 kJ �

c 700 kJ �

d 2 000 kJ �

Answer


Block 2 Steam Engineering Principles and Heat Transfer Entropy- Its Practical Use Module 2.16

Module 2.16

Entropy - Its Practical Use


Block 2 Steam Engineering Principles and Heat Transfer Entropy - Its Practical Use Module 2.16

Entropy - Its

Practical Use Practical use of entropy It can be seen from Module 2.15 that entropy can be calculated. This would be laborious in practice, consequently steam tables usually carry entropy values, based on such calculations. Specific entropy is designated the letter's' and usually appears in columns signifying specific values for saturated liquid, evaporation, and saturated steam, sf, sfg and sg respectively. These values may equally be found in charts, and both Temperature - Entropy (T - S) and Enthalpy - Entropy (H - S) charts are to be found, as mentioned in Module 2.15. Each chart has particular use in specific circumstances.

The T - S chart is often used to determine the properties of steam during its expansion through a nozzle or an orifice. The seat of a control valve would be a typical example.

To understand how a T - S chart is applied, it is worth sketching such a chart and plotting the steam properties at the start condition, reading these from the steam tables.

Example 2.16.1 Steam is expanded from 10 bar a and a dryness fraction of 0.9 to 6 bar a through a nozzle, and no heat is removed or supplied during this expansion process. Calculate the final condition of the steam at the nozzle outlet? Specific entropy values quoted are in units of kJ1k g °C. At 10 bar a, steam tables state that for dry saturated steam:

Specific entropy of saturated water (sf) = 2.1389

Specific entropy of evaporation of dry saturated steam (sfs) = 4.4471

At the inlet condition, as the dryness fraction is 0.9:

Specific entropy of evaporation present = 0.9 x 4.4471

= 4.0024

Specific entropy of the inlet steam = 2.1389 + 4.0024 Specific entropy of the inlet steam = 6.1413

As no heat is added or removed duringthe expansion, the process is described as being adiabatic and isentropic, that is, the entropy does not change. It must still be 6.1413 kJ1k g K at the very moment it passes the throat of the nozzle. At the outlet condition of 6 bar a, steam tables state that:

Specific entropy of saturated water (sf) = 1.9316

Specific entropy of evaporation of dry saturated steam (sfg) = 4.8285

ut, in this example, since the total entropy is fixed at 6.1413 kJ/kg K:

Specific entropy of evaporation present = 6.1413 - 1.9316

= 4.209 7 4.2097

Therefore: Dryness fraction = _______ 4.828 5

Dryness fraction = 0.871 8



T (°C)

10 bar a 180°C

6 bar a 159°C

2.1389 2.1389 + 4.0024 = 6.1413

0.9 dry

1.9316 + 4.2097 = 6.1413

2.1389 + 4.4471 = 6.586

1.9316 + 4.8285 = 6.76 1.9316

0.8718 dry

6.1413 S (kJ/kg °C)

By knowing that this process is isentropic, it has been possible to calculate the dryness fraction at the outlet condition. It is now possible to consider the outlet condition in terms of specific enthalpy (units are in kJ1kg). Φροµ στεαµ ταβλεσ, ατ τηε ινλετ πρεσσυρε οφ 10 βαρ α:

Specific enthalpy of saturated water (hf) = 762.9 Specific enthalpy of evaporation of dry saturated steam (hfg) = 2014.83

As the dryness fraction is 0.9 at the inlet condition: Specific enthalpy of evaporation present = 0.9 x 2014.83

= 1813.35

Specific enthalpy of the inlet steam = 762.9 + 1813.35 Σπεχιφιχ εντηαλπψ οφ τηε ινλετ στεαµ = 2576.25

Φροµ στεαµ ταβλεσ, ατ τηε ουτλετ χονδιτιον οφ 6 βαρ α:

Specific enthalpy of saturated water (hf) = 670.74 Specific enthalpy of evaporation of dry saturated steam (hfg) = 2085.98

Βυτ ασ τηε δρψνεσσ φραχτιον ισ 0.8718 ατ τηε ουτλετ χονδιτιον:

Specific enthalpy of evaporation present = 0.8718 x 2085.98

= 1818.56

Total specific enthalpy of the outlet steam = 670.74 + 1818.56

Τοταλ σπεχιφιχ εντηαλπψ οφ τηε ουτλετ στεαµ = 2489.30

It can be seen that the specific enthalpy of the steam has dropped in passing through the nozzle from 2 576.25 to 2489.30 kJ1kg, that is, a heat drop of 86.95 kJ1kg.

This seems to contradict the adiabatic principle, which stipulates that no energy is removed from the process. But, as seen in Module 2.15, the explanation is that the steam at 6 bar a has just passed through the nozzle throat at high velocity, consequently it has gained kinetic energy. As energy cannot be created or destroyed, the gain in kinetic energy in the steam is at the expense of its own heat drop. The above entropy values in Example 2.16.1 can be plotted on a T - S diagram, see Figure 2.16.1.

Fig. 2.16.1 The T - S diagram for Example 2.16.1



m u Kinetic energy (E)= ___

2 g J Equation 2.16.1

Equation 2.16.2 u= 2 2EgJ

m

u = 2 0 0 0 x E u = 4 4 . 7 2 E

u = 44.72 h Equation 2.16.3

Further investigation of kinetic energy in steam What is the significance of being able to calculate the kinetic energy of steam? By knowing this value, it is possible to predict the steam velocity and therefore the mass flow of steam through control valves and nozzles.

Kinetic energy is proportional to mass and the square of the velocity.

It can be further shown that, when incorporating Joule's mechanical equivalent of heat, kinetic energy can be written as Equation 2.16.1:

Where: E = Kinetic energy (kJ) m = Mass of the fluid (k g) u = Velocity of the fluid (m1s) g = Acceleration due to gravity (9.806 65 m1s2) J = Joule's mechanical equivalent of heat (101.972 m k g1kJ)

By transposing Equation 2.16.1 it is possible to find velocity as shown by Equation 2.16.2:

For each kilogram of steam, and by using Equation 2.16.2

u2=2EgJ u = 2 2 x E x 9.806 65 x 101.972 u = 2 000

2 x E

As the gain in kinetic energy equals the heat drop, the equation can be written as shown by Equation 2.16.3:

Where: h = Heat drop in kJ1kg

By calculating the adiabatic heat drop from the initial to the final condition, the velocity of steam can be calculated at various points along its path; especially at the throat or point of minimum pass area between the plug and seat in a control valve.

This could be used to calculate the orifice area required to pass a given amount of steam through a control valve. The pass area will be greatest when the valve is fully open. Likewise, given the valve orifice area, the maximum flowrate through the valve can be determined at the stipulated pressure drop. See Examples 2.16.2 and 2.16.3 for more details.

Example 2.16.2 Consider the steam conditions in Example 2.16.1 with steam passing through a control valve with an orifice area of 1 cm2. Calculate the maximum flow of steam under these conditions.

The downstream steam is at 6 bar a, with a dryness fraction of 0.8718. Specific

volume of dry saturated steam at 6 bar a (sg) equals 0.315 6 m31k g.

Specific volume of saturated steam at 6 bar a and a dryness fraction of 0.8718 equals 0.3156 m31k g x 0.8718 which equates to 0.2751 m31k g.



Equation 2.16.3 u = 44.72 h

Velocity m/s x Orifice area m2 The mass flow = ____________________________ kg s Specific volume m kg 3

Equation 2.16.4

Point of interest

Thermodynamic textbooks will usually quote Equation 2.16.3 in a slightly different way as shown in Equation 2.16.5:

u = 2 h Equation 2.16.5

Where: u = Velocity of the fluid in m1s h = Heat drop in J1kg 2 = Constant of proportionality incorporating the gravitational constant 'g'.

Considering the conditions in Example 2.16.3:

u = 2 h u = 2 x 86950

u = 417 m s

This velocity is exactly the same as that calculated from Equation 2.16.3, and the user is free to practise either equation according to preference.

Heat drop ( h) = 86.95 kJ kg Heat drop ( h) = 86 950 J kg

The heat drop in Example 2.16.1 was 86.95 kJ1k g, consequently the velocity can be calculated using Equation 2.16.3:

u = 44.72 86.95 u = 44.72 x 9.32 u = 417 m s

The mass flow is calculated using Equation 2.16.4:

An orifice area of 1 cm2 equals 0.0001 m2 417 x 0.000 1

The mass flow = __________ kg s 0.2751

The mass flow = 0.152 kg s ( 547 kg h)

The above calculations in Example 2.16.2 could be carried out for a whole series of reduced pressures, and, if done, would reveal that the flow of saturated steam through a fixed opening increases quite quickly at first as the downstream pressure is lowered.

The increases in flow become progressively smaller with equal increments of pressure drops and, with saturated steam, these increases actually become zero when the downstream pressure is 58% of the absolute upstream pressure. (If the steam is initially superheated, CPD will occur at just below 55% of the absolute upstream pressure).

This is known as the 'critical flow' condition and the pressure drop at this point is referred to as critical pressure drop (CPD). After this point has been reached, any further reduction of downstream pressure will not give any further increase in mass flow through the opening.



1.0 0.8 0.58 P/P1

Veloc

ity

P, P u

S S - u

In fact if, for saturated steam, the curves of steam velocity (u) and sonic velocity (s) were drawn for a convergent nozzle (Figure 2.16.2), it would be found that the curves intersect at the critical pressure. P, is the upstream pressure, and P is the pressure at the

throat.

Fig. 2.16.2 Steam and acoustic velocities through a nozzle

The explanation of this, first put forward by Professor Osborne Reynolds (1842 - 1912) of Owens College, Manchester, UK, is as follows:

Consider steam flowing through a tube or nozzle with a velocity u, and let s be the speed of sound (sonic velocity) in the steam at any given point, s being a function of the pressure and density of the steam. Then the velocity with which a disturbance such as, for example, a sudden change of pressure P, will be transmitted back through the flowing steam will be s - u.

Referring to Figure 2.16.2, let the final pressure P at the nozzle outlet be 0.8 of its inlet pressure P1. Here, as the sonic velocity s is greater than the steam velocity u, s - u is clearly positive. Any change in the pressure P would produce a change in the rate of mass flow.

When the pressure P has been reduced to the critical value of 0.58 P1, s - u becomes zero, and any further reduction of pressure after the throat has no effect on the pressure at the throat or the rate of mass flow.

When the pressure drop across the valve seat is greater than critical pressure drop, the critical velocity at the throat can be calculated from the heat drop in the steam from the upstream condition to the critical pressure drop condition, using Equation 2.16.5.

Control valves The relationship between velocity and mass flow through a restriction such as the orifice in a control valve is sometimes misunderstood.

Pressure drop greater than critical pressure drop It is worth reiterating that, if the pressure drop across the valve is equal to or greater than critical pressure drop, the mass flow through the throat of the restriction is a maximum and the steam will travel at the speed of sound (sonic velocity) in the throat. In other words, the critical velocity is equal to the local sonic velocity, as described above.

For any control valve operating under critical pressure drop conditions, at any reduction in throat area caused by the valve moving closer to its seat, this constant velocity will mean that the mass flow is simultaneously reduced in direct proportion to the size of the valve orifice.

Pressure drop less than critical pressure drop For a control valve operating such that the downstream pressure is greater than the critical pressure (critical pressure drop is not reached), the velocity through the valve opening will depend on the application.



velocity m/s x orifice area m2 The mass flow = ___________________________ kg s Specific volume m kg 3

Εθυατιον 2.16.4

Pressure reducing valves If the valve is a pressure reducing valve, (its function is to achieve a constant downstream pressure for varying mass flowrates) then, the heat drop remains constant whatever the steam load. This means that the velocity through the valve opening remains constant whatever the steam load and valve opening. Constant upstream steam conditions are assumed.

It can be seen from Equation 2.16.4 that, under these conditions, if velocity and specific volume are constant, the mass flowrate through the orifice is directly proportional to the orifice area.

Temperature control valves In the case of a control valve supplying steam to a heat exchanger, the valve is required to reduce the mass flow as the heat load falls. The downstream steam pressure will then fall with the heat load, consequently the pressure drop and heat drop across the valve will increase. Thus, the velocity through the valve must increase as the valve closes.

In this case, Equation 2.16.4 shows that, as the valve closes, a reduction in mass flow is not directly proportional to the valve orifice, but is also modified by the steam velocity and its specific volume.

Example 2.16.3 Find the critical velocity of the steam at the throat of the control valve for Example 2.16.2, where the initial condition of the steam is 10 bar a and 90% dry, and assuming the downstream pressure is lowered to 3 bar a.

Specific enthalpy at 10 bar a, 0.9 dryness fraction = 2 576.26 kJ1kg

Specific entropy at 10 bar a, 0.9 dryness fraction = 6.14129 kJ1kg K

For wet steam, critical pressure can be taken as 58% of the absolute upstream pressure, therefore:

Pressure of steam at the throat = 0.58 x 10 bar a

= 0.58 bar a

At the throat condition of 5.8 bar a, and from steam tables: Specific entropy of saturated water (sf) = 1.91836

Specific entropy of evaporation of dry saturated steam (sfg) = 4.8538

But, in this example, since the total entropy is fixed at 6.14129 kJ1k g K:

Specific entropy of evaporation present = 6.14129 - 1.918 36

= 4.22293 4.222 93

Therefore, the dryness fraction at the throat at the throat = __________ 4.85 38

Δρψνεσσ φραχτιον = 0.8701

From steam tables, at the throat condition of 5.8 bar a: Specific enthalpy of saturated water (hf) = 665.008

Specific enthalpy of evaporation of dry saturated steam (hfg) = 2090.23



Equation 2.16.5

But as the dryness fraction is 0.8701 at the throat condition:

Specific enthalpy of evaporation present = 0.8701 x 2 090.23 = 1818.71

Total specific enthalpy of the outlet steam = 665.008 + 1818.71

= 2 483.72

Therefore, the heat drop at critical pressure drop = 2 576.72 - 2483.72 Heat drop at critical pressure drop = 92.54 kJ/kg (92 540 J/kg)

The velocity of the steam through the throat of the valve can be calculated using Equation 2.16.5:

u = 2 x 9 2 5 4 0 u = 4 3 0 m s

The critical velocity occurs at the speed of sound, consequently 430 m1s is the sonic velocity for the Example 2.16.3.

Noise in control valves If the pressure in the outlet of the valve body is lower than the critical pressure, the heat drop at a point immediately after the throat will be greater than at the throat. As velocity is directly related to heat drop, the steam velocity will increase after the steam passes the throat of the restriction, and supersonic velocities can occur in this region.

In a control valve, steam, after exiting the throat, is suddenly confronted with a huge increase in space in the valve outlet, and the steam expands suddenly. The kinetic energy gained by the steam in passing through the throat is converted back into heat; the velocity falls to a value similar to that on the upstream side of the valve, and the pressure stabilises in the valve outlet and connecting pipework.

For the reasons mentioned above, valves operating at and greater than critical pressure drop will incur sonic and supersonic velocities, which will tend to produce noise. As noise is a form of vibration, high levels of noise will not only cause environmental problems, but may actually cause the valve to fail. This can sometimes have an important bearing when selecting valves that are expected to operate under critical flow conditions.

It can be seen from previous text that the velocity of steam through control valve orifices will depend on the application of the valve and the pressure drop across it at any one time.

Reducing noise in control valves There are some practical ways to deal with the effects of noise in control valves.

Perhaps the simplest way to overcome this problem is to reduce the working pressure across the valve. For instance, where there is a need to reduce pressure, by reducing pressure with two valves instead of one, both valves can share the total heat drop, and the potential for noise in the pressure reducing station can be reduced considerably.

Another way to reduce the potential for noise is by increasing the size of the valve body (but retaining the correct orifice size) to help ensure that the supersonic velocity will have dissipated by the time the flow impinges upon the valve body wall.

In cases where the potential for noise is extreme, valves fitted with a noise attenuator trim may need to be used.

Steam velocities in control valve orifices will reach, typically, 500 m1s. Water droplets in the steam will travel at some slightly lower speed through a valve orifice, but, being incompressible, these droplets will tend to erode the valve and its seat as they squeeze between the two.

It is always sensible to ensure that steam valves are protected from wet steam by fitting separators or by providing adequate line drainage upstream of them.



T (°C)

2.1389 2.1389 + 4.0024 = 6.1413

1.9316

1.9316 + 4.2097 = 6.1413

2.1389 + 4.4471 = 6.586

1.9316 + 4.8285 = 6.76 0.9 dry

0.8718 dry

6.1413 S (kJ/kg °C)

10 bar a 180°C

6 bar a 159°C

Summing up of Modules 2.15 and 2.16 The T - S diagram, shown in Figure 2.16.1, and reproduced below in Figure 2.16.3, shows clearly that the steam becomes wetter during an isentropic expansion (0.9 at 10 bar a to 0.8718 at 6 bar a)

in Example 2.16.1.

Fig. 2.16.3 A T-S diagram showing wetter steam from an isentropic expansion

At first, this seems strange to those who are used to steam getting drier or becoming superheated during an expansion, as happens when steam passes through, for example, a pressure reducing valve.

The point is that, during an adiabatic expansion, the steam is accelerating up to high speed in passing through a restriction, and gaining kinetic energy. To provide this energy, a little of the steam condenses (if saturated steam), (if superheated, drops in temperature and may condense) providing heat for conversion into kinetic energy.

If the steam is flowing through a control valve, or a pressure reducing valve, then somewhere downstream of the valve's seat, the steam is slowed down to something near its initial velocity. The kinetic energy is destroyed, and must reappear as heat energy that dries out or superheats the steam depending on the conditions.

The T - S diagram is not at all convenient for showing this effect, but the Mollier diagram (the H - S diagram) can do so quite clearly.

The Mollier diagram can depict both an isenthalpic expansion as experienced by a control valve, (see Figure 2.15.6) by moving horizontally across the graph to a lower pressure; and an isentropic expansion as experienced by steam passing through a nozzle, (see Figure 2.15.7) by moving horizontally down to a lower pressure. In the former, the steam is usually either dried or superheated, in the latter, the steam gets wetter.

This perhaps begs the question, 'How does the steam know if it is to behave in an isenthalpic or isentropic fashion?' Clearly, as the steam accelerates and rushes through the narrowest part of the restriction (the throat of a nozzle, or the adjustable gap between the valve and seat in a control valve) it must behave the same in either case.

The difference is that the steam issuing from a nozzle will next meet a turbine wheel and gladly give up its kinetic energy to turn the turbine. In fact, a nozzle could be thought of as a device to convert heat energy into kinetic energy for this very purpose.

In a control valve, instead of doing such work, the steam simply slows down in the valve outlet passages and its connecting pipework, when the kinetic energy appears as heat energy, and unwittingly goes on its way to give up this heat at a lower pressure.

It can be seen that both the T - S diagram and H - S diagram have their uses, but neither would have been possible had the concept of entropy not been utilised.



Questions

1. From the T - S diagram shown in Figure 2.16.1, had the initial state point been 100% dry saturated steam at 10 bar, what would have been its specific entropy?

α 6.586 κϑ1κ γ Κ �

β 2.1389 κϑ 1κ γ Κ �

χ 6.1413 κϑ 1κ γ Κ �

δ 6.76 κϑ1κγ Κ �

2. From the T - S diagram shown in Figure 2.16.1, had the initial state point been 100% dry saturated steam at 10 bar, and the final pressure 6 bar, in which region would the final state point have been?

α Τηε συπερηεατεδ ρεγιον �

β Ον τηε σατυρατεδ στεαµ λινε �

χ Τηε ωετ στεαµ ρεγιον �

δ Ον τηε σατυρατεδ ωατερ λινε �

3. In a steam control valve, the heat drop from the initial condition to that at the valve throat is calculated to be 50 ki/kg. What is the velocity of steam passing through the valve orifice?

α 416.65 µ1σ �

β 316.23 µ1σ �

χ Σονιχ ϖελοχιτψ �

δ Συπερσονιχ ϖελοχιτψ �

4. In Question 3, the orifice area is known to be 50 mm2, and the specific volume of steam at the downstream pressure is 0.3 m3/kg. What is the mass flowrate?

α Χριτιχαλ φλοω �

β 200.01 κ γ1η �

χ 189.74 κ γ1η �

δ 40 κ γ1η �

5. A pressure control valve is set to reduce and maintain pressure from 10 bar g to 7 bar g. The velocity through the valve orifice at full-load is 400 m/s. What is the velocity through the orifice at half-load?

α 200 µ1σ �

β 800 µ1σ �

χ 282.8 µ1σ �

δ 400 µ1σ �

6. What can be done to reduce noise in valves operating under critical conditions? α Υσε τωο ϖαλϖεσ ιν σεριεσ ινστεαδ οφ ονε �

β Υσε α ϖαλϖε ωιτη τηε σαµε σιζε σεατ βυτ ηαϖινγ α λαργερ βοδψ �

χ Υσε α ϖαλϖε ωιτη α νοισε αττενυατιον τριµ �

δ Ανψ οφ τηε αβοϖε �

Answers

z.16.1D The Steam and Condensate Loop

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Entropy - Mollier Diagram