Introduction

Understanding the behaviour and properties of steam is critical to the successful operation of steam-related equipment and processes. The widespread use of steam is due to the availability of water and the relative safety of steam compared to other fluids. There are some disadvantages to using water and steam, but they are minor compared to the disadvantages of other, more thermodynamically superior fluids, such as ammonia, which are much more hazardous.

Steam Tables and Enthalpy

The 2nd Class student should already have a solid understanding of the Steam Tables from previous studies. Here is a brief review only.

Steam Tables were formulated to show the thermodynamic properties of water and steam at certain pressures and temperatures. A section of the various steam tables can be found in the PanGlobal Academic Supplement. Tables 1 and 2 cover the properties of saturated steam, which is steam at its saturation temperature. Table 1 is a “steam pressure table”, which gives the thermodynamic properties of the steam according to pressure. Table 2 is a “steam temperature table”, which gives the properties according to the saturation temperature of the steam. Table 3 has “properties of superheated steam”, which is steam that is at a higher temperature than the saturation temperature. It shows the properties at certain pressures, cross-referenced against steam temperatures. The properties of steam included in the Steam Tables are:

Interpolation:

Since the tables cannot show all values of pressure and temperature, a mathematical process of interpolation must be applied to obtain intermediate values. For example, the pressure table does not include values between 100 kPa and 125 kPa. Therefore, to find the liquid enthalpy, hf, for a pressure of 115 kPa (for example), the following interpolation is required:

125 kPa minus 100 kPa = 25 kPa

115 kPa minus 100 kPa = 15 kPa

So, 115 kPa is 15/25 of the way between 100 kPa and 125 kPa.

Also, the value of hf increases between 100 kPa and 125 kPa.

Therefore:

Enthalpy

A simple definition of “enthalpy” is “the heat transferred to or from a system”. In the steam tables, the stated enthalpy of water or steam at any given condition represents the amount of heat that must be added (transferred) to each kilogram of water at 0°C in order to convert it to that final condition. Since they represent the heat per kg, the enthalpies stated in the Steam Tables are referred to as specific enthalpy. It follows that total enthalpy is the total mass times the specific enthalpy, or:

Another, more technical, definition of enthalpy is “a thermodynamic quantity, equal to the internal energy of a system plus the product of its volume and pressure”. That is:

This can be demonstrated, using a couple of examples from the steam tables.

1. The specific enthalpy of water (hf) at 300 kPa absolute is given as 561.47 kJ/kg. Applying the formula should give the same value.

2. The specific enthalpy of dry, saturated steam (hg) at 300 kPa is given as 2725.3 kJ/kg. Applying the formula should give the same value.

Water to Steam

The amount of heat required to convert water into steam is determined by the pressure and temperature conditions. Figure 1 is a temperature-enthalpy diagram, which is intended to show the three stages required to convert one kg of water to one kg of superheated steam as heat is added and the enthalpy increases.

The first stage starts at point A, which is water at 0°C and an enthalpy of zero. From there, the water is heated and the process follows the liquid line. If the pressure is 101.3 kPa (atmospheric pressure), at point B on the liquid line there have been 419.04 kJ of heat added to the water and the temperature is now 100°C. The heat added is called sensible heat, since it causes an increase in the temperature of the water which can be “sensed” and measured. The term “sensible heat” has become less common, being superseded by the terms, “liquid enthalpy” or “enthalpy of water”.

In the next stage, from B to C on Figure 1, the pressure remains constant at 101.3 kPa while heat is added and the water is progressively converted to steam. This occurs at the saturation temperature of 100°C. The heat added is called latent heat (since it produces no change in temperature) or the more proper term, enthalpy of evaporation. As the figure shows, for water at atmospheric pressure, the enthalpy of evaporation (hfg) from B to C is 2257 kJ/kg.

At point C, enough heat has been added to convert 100% of the water into steam and, since there is no more liquid, the steam is now dry and saturated. It is called dry, saturated steam, where “saturated steam” means steam that is at the saturation temperature for the existing pressure. At any point between B and C, some of the water has not yet been converted to steam, so the steam is called wet steam. The proportion of steam in the mixture is called the dryness fraction, which increases from 0 (ie. 0% steam) at point B to 1 (ie. 100% steam) at point C.

At point C, further heat may be added to the steam. This heat, called superheat, increases the steam temperature to some value above saturation temperature. The steam may then be described as dry and superheated. For example, at point D in Figure 1, the 100°C, 101.3 kPa dry, saturated steam has been increased to 300°C superheated steam. It has 200°C of superheat.

Figure 1 also shows a comparison with steam at the higher pressure of 1553.8 kPa, which has a correspondingly higher saturation temperature of 200°C. The liquid enthalpy to reach saturation temperature is higher, at 852.45 kJ/kg. The enthalpy of evaporation to convert this water to dry, saturated steam is less, at 1940.7 kJ/kg.

As pressure increases, a point is reached where the densities of water and steam are virtually equal. At this pressure, called the critical pressure, there is no additional heat required to convert the water to steam. Liquid enthalpy (hf = 2099.3 kJ/kg) alone was sufficient to produce steam. Critical pressure is 22 090 kPa and the corresponding temperature is 374.136°C.

Specific Enthalpy

Specific enthalpies, in kJ/kg, are provided in the “Properties of Saturated Steam, Pressure Table” (Table 1 in the Academic Supplement) and in the “Properties of Saturated Steam, Temperature Table” (Table 2 in the Academic Supplement). Both tables have three columns for specific “Enthalpy”.

As long as the steam is indeed 100% dry, the relationship between these three specific enthalpies can be stated as:

Specific enthalpies are also provided in the “Table of Properties for Superheated Steam” (Table 3 in the Academic Supplement). Here, the symbol “h”, beside each of the listed pressures represents specific enthalpy. The corresponding row gives the specific enthalpies for the various columns of superheated steam temperatures at that pressure. Note that the number in brackets beneath each pressure is the saturation temperature at that pressure. Subtracting this value from the final superheat temperature will give the “degrees of superheat” that have been given to the steam. For example, from the superheat table, steam at 11 000 kPa and 700°C, has an enthalpy of 3864.5 kJ/kg. The saturation temperature is 318.15°C; therefore, the steam has 700 – 318.15 = 381.85°C of superheat.

Example 1

a) How much has the enthalpy changed when 50 kg of feedwater at 15°C is evaporated and leaves a 2000 kPa boiler as dry, saturated steam?

b) What is the total change in enthalpy if the superheater then increases the steam temperature to 800°C?

c) How much superheat has been added to the saturated steam?

Solution 1

a) From the steam (temperature) table, specific enthalpy of the feedwater at 15°C is:

From the steam (pressure) table, specific enthalpy of dry, saturated steam at 2000 kPa is:

The change in enthalpy per kg of water is, therefore:

Therefore, the change in enthalpy to convert 50 kg of water into 50 kg of dry, saturated steam:

b) From the superheated steam table, at 2000 kPa and 800°C, the specific enthalpy is:

Increase in specific enthalpy from feedwater to superheated steam:

Total enthalpy increase from feedwater to superheated steam:

c) Again, from the superheated steam table at 2000 kPa, the saturation temperature is 212.43. Therefore:

Degrees of superheat added = 800 – 212.43 = 587.57 (Ans.)

Another way to approach this type of question is to apply a simple equation,

Using this equation, the solution to Example 1(b), increase in enthalpy from feedwater to superheated steam, would be:

Dryness Fraction

Wet steam is steam (at saturation temperature) that contains a certain amount of free water. That is, some of the water has not yet evaporated, so the steam is not totally dry. The fraction of the steam that is dry is called the “dryness fraction”, often referred to as the “steam quality”. The letter, q, is used to represent the dryness fraction, which is properly stated as a decimal number between 0.0 and 1.0 (eg. 0.78 dry). It is also common to state the dryness as a percentage (eg. 78% dry).

For the same pressure and saturation temperature, the specific enthalpy of wet steam is less than the specific enthalpy of dry steam. This is because the water (at hf) has not received the total enthalpy of evaporation (hfg) required to convert it entirely into dry steam (at hg). Since the actual proportion of the enthalpy of evaporation received is dependant on the dryness fraction, the specific enthalpy (h) may be determined from:

Example 2

What is the total enthalpy of 100 kg of 2500 kPa steam, if it is 20% wet?

Solution 2

Since the steam is 20% wet, it must be 80% dry; so, q = 0.80

Specific enthalpy (hf) of saturated liquid at 2500 kPa = 962.11 kJ/kg

Specific enthalpy of evaporation (hfg) at 2500 kPa = 1841.0 kJ/kg

Specific enthalpy (h) of 80% dry steam at 2500 kPa is:

Example 3

100 kg of saturated steam at 290°C has a specific enthalpy of 2600 kJ/kg.

a) how dry is this steam and

b) how much total heat must be added to completely dry it.?

Solution 3

From the steam (temperature) table at 290°C:

a)

b) Heat required, Q, to totally dry the steam:

Specific Heat

By definition, the specific heat of water is the amount of heat required to increase the temperature of a unit mass of the water by 1°C. This can be expressed in J/g/°C or, more commonly, in kJ/kg/°C.

While the specific heat of water is often quoted as 4.2 kJ/kg/°C, this is the specific heat at atmospheric pressure. The specific heat actually varies with pressure and saturation temperature. This is because the average specific heat from 0°C to any given saturation temperature is equal to the liquid enthalpy divided by the saturation temperature. For example:

At saturation temperature of 100°C (pressure = 101.3 kPa), liquid enthalpy (hf) equals 419.04 kJ/kg. The average specific heat of the water between 0°C and 100°C is:

At saturation temperature of 200°C (pressure = 1553.8 kPa), liquid enthalpy (hf) equals 852.41 kJ/kg. The average specific heat of the water between 0°C and 200°C is:

And, from the steam tables, at a pressure of 7000 kPa, the saturation temperature is 285.88°C and hf = 1267.0 kJ/kg. The average specific heat of the water between 0°C and 285.88°C is:

The average specific heat of water over a certain range (not starting from 0°C), can be determined if the starting and ending enthalpies and temperatures are known. For example, if water at 280°C is heated to 300°C the average specific heat of the water during that process would be:

Specific heat is meaningless during a phase change (water to steam). However, the specific heat of superheated steam at a given pressure and through a given temperature range can be determined if the enthalpies and temperatures are known. For example, the specific heat of superheated steam at 6000 kPa between 400°C and 500°C, can be calculated as follows:

Specific Volume

The specific volume of any substance, including steam, is the volume occupied by a unit mass of the substance. In the steam tables, the units for specific volume, v, are “cm3/g”. However, alternate units are “× 10-3 m3/kg” or “litres/kg (L/kg)”.

The steam tables provide specific volumes for saturated liquid (vf) and saturated vapour (vg). We know that, at any given pressure, the specific volume of saturated steam is significantly higher than the volume of saturated water. Water expands significantly when converted to steam. This volume increase is much greater at lower pressures and gets progressively less as pressure increases. For example (refer to the steam tables):

Notice from the above that the specific volume of the steam at 100 kPa is about 1623 times that of the liquid. However, at the critical pressure of 22 090 kPa, the specific volumes of steam and water are identical.

Dryness fraction affects the specific volume of wet saturated steam. Since only a fraction of the steam has evaporated and expanded accordingly, the specific volume will be dependant on the dryness fraction. To calculate the specific volume for wet steam, the dryness fraction is used as follows:

Example 4

500 kg of saturated steam at 2500 kPa has a dryness fraction of 0.8? How many cubic meters does this steam occupy?

Solution 4

From the steam tables, at 2500 kPa:

Measuring the Dryness Fraction of Steam

Steam that is excessively wet causes corrosion and erosion in equipment. It also reduces plant thermal efficiencies and increases the risk of water hammer in piping or damage to equipment that uses the steam. In a steam plant that produces saturated steam, it may become necessary to monitor the dryness fraction of the steam at various locations in the system in order to maintain efficiency and avoid the difficulties associated with wet steam.

A calorimeter is a device that is used to measure the dryness fraction of wet steam and to determine the amount of heat in the steam. There are three types of calorimeter:

• Throttling

• Separating

• Electric

Throttling Calorimeters

Throttling calorimeters are used for steam with low moisture content. Their operation is based on the principle that enthalpy remains constant when steam pressure is reduced through an orifice or constriction. This is because there is no opportunity for heat to be removed from the steam. If the steam contains a moderate amount of moisture, this moisture will flash (ie. evaporate quickly) into superheated steam at the suddenly lower pressure.

The steam sample should be taken from a vertical pipe that is as close as possible to the steam user (eg. steam engine, heater, turbine) or to the steam source (boiler outlet). An example of a good sampling arrangement is shown in Figure 2.

A simple throttling calorimeter is illustrated in Figure 3. The steam is throttled through the orifice, O, into a lower pressure chamber, C. The temperature is measured with a thermometer, T, and the pressure is measured with a manometer, connected to valve, V2. The steam is allowed to escape to a vent through valve, V1. The body of the calorimeter and the piping to it are well insulated to prevent heat loss to the surroundings and an inadvertent increase in the moisture of the sample. The calorimeter must be warmed with steam for considerable time before the test, to ensure the body is at the same temperature as the steam. If the orifice is sized correctly, the steam will be at atmospheric pressure after throttling.

Throttling calorimeters are used for steam that ranges from about 0.965 dryness at 700 kPa to 0.93 dryness at 2700 kPa. If the moisture content is higher than this, a separating calorimeter is often required upstream of the throttling calorimeter to remove most of the moisture before the steam enters the throttling calorimeter. This is because if the steam is too wet entering the throttling calorimeter, it may still be wet after throttling rather than dry and containing superheat. It will also be at the same saturation temperature as the entering steam and, without measurable superheat, it won’t be possible to determine the enthalpy.

Separating Calorimeter

An example of a simple, separating calorimeter is shown in Figure 4. The steam enters at A and a change in direction as the steam passes through a tube with holes produces inertial forces that cause moisture to collect in the cavity, V. The amount of liquid collected is measured in the sight glass, G. The dryer steam exits at D and goes to the throttling calorimeter.

A separating calorimeter can be used on its own (no downstream throttling calorimeter) to calculate the dryness fraction. In this case, the steam leaving the calorimeter is subsequently condensed and the mass of this condensate plus the water in the sight glass are recorded for comparison.

A disadvantage to using a separating calorimeter without a throttling calorimeter is that the resulting calculation tends to underestimate the wetness of the steam because some of the moisture carries over with the steam.

Electric Calorimeters

Electric calorimeters are also used to measure the dryness fraction of wet steam. A measured amount of steam is heated with a known amount of electric heat until the steam becomes superheated. The enthalpy is computed from the resultant pressure and temperature from the Steam Tables and the enthalpy of the wet steam is calculated by subtracting the electric heat added. This requires an accurate measurement of the steam flow, which may be difficult to measure.

Measuring Dryness with a Throttling Calorimeter

A throttling calorimeter works on the basis that the throttled, superheated steam leaving the calorimeter has the same enthalpy as the wet steam entering the calorimeter. With this in mind the dryness fraction of the wet steam can be calculated by manipulation of the enthalpy equation for wet steam, as follows:

Enthalpy of wet steam entering calorimeter, h = hf + q hfg

Example 5

Wet steam at a pressure of 1000 kPa abs is sampled in a throttling calorimeter. If temperature is measured at 100°C and pressure is 100 kPa, what is the quality of the steam?

Solution 5

From the table for superheated steam, for steam at 100 kPa and 100°C:

From the steam (pressure) table, for saturated steam at 1000 kPa:

Measuring Dryness with a Separating Calorimeter

With a separating calorimeter, a rough estimate of the dryness can be made for steam that is too wet for a throttling calorimeter. In order to do so, the steam leaving the calorimeter must be condensed. Then the mass of the water collected in the calorimeter (m1) and the mass of the condensed steam (m2) can be entered into the following calculation for dryness fraction.

This calculation underestimates the wetness of the steam since some of the moisture usually carries out of the calorimeter with the steam.

Separating and Throttling Calorimeters in Series

A more accurate method of measuring dryness is to first pass the steam sample through a separating calorimeter and then pass the steam from that calorimeter through a throttling calorimeter to determine the remaining moisture in the steam.

Figure 5 illustrates the series arrangement of separating and throttling calorimeters.

If the steam entering the throttling calorimeter has a mass of m2 kg and a dryness fraction of q2, then the mass of the dry fraction of this steam will be q2 × m2 kg. The mass of water removed in the separating calorimeter is m1 kg, so the total mass of the original steam sample is m1 + m2 and the dryness fraction of the original sample is:

Now, if the steam entering the separating calorimeter has a dryness fraction of q1, we found earlier that this dryness fraction, using the separating calorimeter alone is:

We can now substitute q1 for $\frac{m_2}{m_1+m_2}$ in the previous equation, leaving:

Example 6

The dryness fraction of wet steam at 1400 kPa is to be determined by separating and throttling calorimeters in series. The mass of water obtained in the separating calorimeter is 0.4 kg. In the throttling calorimeter, temperature is 140°C and pressure is 100 kPa and 9 kg of condensate is collected. What is the dryness fraction of the steam?

Solution 6

First, find the dryness fraction using the separating calorimeter:

Now, we need to find q2, the dryness fraction entering the throttling calorimeter.

This will involve the equation, $\frac{q_2=h_{sup}–h_f}{h_{fg}}$.

From the saturated steam tables (pressure) at 1400 kPa, the liquid enthalpy (hf) is 830.3 kJ/kg and the enthalpy of evaporation (hfg) is 1959.7 kJ/kg. The enthalpy of the superheated steam (hsup) at 100 kPa and 140°C can be calculated from the superheated steam tables, using interpolation, as follows:

The dryness fraction, q2 , entering the throttling calorimeter is:

With q1 and q2 both known, the overall dryness fraction can be found:

Internal Energy

Internal energy follows the First Law of Thermodynamics, which states that work and heat are equivalent and interchangeable. The equation associated with this law states that the heat added to a system causes a change in both the internal energy and the work done on the system. That is:

In this equation, U represents internal energy. Q represents enthalpy, with a scale beginning at an enthalpy of zero at 0°C. In other words, Q = h. Also, the work done on the fluid causes changes in pressure and volume and the work, W, can be expressed as pv (pressure in kPa × specific volume in m3/kg).

By substituting these equivalent expressions, Q = ΔU + W may be rewritten as:

Let’s apply this to a boiler:

The feedwater enters the boiler at a pressure of p1 kPa and a specific volume of v1. The internal energy is u1. Thus the enthalpy, h1, of the feedwater is:

Heat is then added until the saturation temperature is reached. Most of the heat goes into increasing the internal energy through an increase in temperature. The pressure is constant and there is only a minor increase in the specific volume.

The enthalpy at the saturation temperature is:

An example from the steam table will verify this. At 2000 kPa:

During the next phase, evaporation, the pressure and temperature remain constant while the specific volume increases to vg. Part of the enthalpy change (hfg) is used to increase the internal energy (ug – uf) and the remaining enthalpy change increases the volume (vg – vf). The equation can be written for the evaporation phase as:

Again, using the example of 2000 kPa, this equation can be verified from the steam tables:

The internal energy of wet saturated steam is dependent on the dryness fraction. The internal energy, u, at any dryness between saturated liquid (uf) and dry saturated steam (ug) is:

The change in internal energy (ΔU) between any two conditions (U1 and U2) may be found using the simple formula:

Example 7

a) Using the basic equation, hf = uf + p1vf , determine the change in internal energy if 1 kg of feedwater at 40°C and 500 kPa is converted to 500 kPa steam with a dryness fraction of 0.97.

b) Apply the equation, ∆U = U2 – U1, and use the internal energy values from the Steam Tables. Compare this answer with (a).

Solution 7

a) First, find the internal energy, uf, of the feedwater at 40°C and 500 kPa.

The enthalpy of the feedwater at 40°C, from the Steam (temperature) Tables is 167.57 kJ/kg and the specific volume is 1.0078 cm3/kg. Note: even though these values are for a saturation pressure of 7.384 kPa, they can be applied at 500 kPa, since liquids are non-compressible so the specific volume doesn’t change. Therefore:

Now, to find the internal energy of the 500 kPa steam, we need to first find specific enthalpy and specific volume of the steam. From the Steam Tables, at 500 kPa, hf = 640.23 kJ/kg and hfg = 2108.5 kJ/kg. Therefore:

Also, at 500 kPa from the Steam Tables, vf = 1.0926 × 10-3 m3/kg and vg = 374.9 × 10-3 m3/kg. Therefore:

Now, the internal energy of the 500 kPa steam is:

Therefore, the change in internal energy is:

b) Applying the equation, ∆U = U2 – U1:

Note: The two answers are the same. Solution (b) would be the easiest to apply on an examination.

Entropy

Entropy has been defined and described in various ways and is not an easy concept to understand. Perhaps that’s because it defines the energy within a substance (at any given condition or state) that is NOT available to do work or to be transferred. It is much easier to understand something that IS available than something that is NOT available.

The unavailable energy (entropy) in a substance changes with temperature. As temperature increases, entropy also increases and as temperature decreases, entropy also decreases.

From the Second Law of Thermodynamics, heat naturally flows from a higher temperature source to a lower temperature destination (the heat sink). During the transfer of heat, the temperature of the source decreases, while the temperature of the sink increases. Consequently, the heat transferred into or out of each substance is transferred at some average temperature, which lies between the initial and final temperatures. The heat transfer process is dependent upon the temperatures of the working fluids.

During a thermodynamic process, an entropy change is a function of the heat transferred and the absolute temperature(s) at which it transfers. The entropy change for each of the transfer substances is simply the amount of heat transferred (kJ) divided by the mean (average) absolute temperature (K) during the transfer. Given that S = entropy, Q = heat transferred, and T = temperature, the change in entropy can be shown as:

Now, since ∆Q = change in enthalpy = H2 – H1, and Tav = (T2 + T1)/2, the equation may be written as:

If we attach the units, the equation becomes:

Usually, when considering thermodynamic processes, we use specific values (ie. per kilogram) for enthalpy and entropy, since these are available in thermodynamic tables, including steam table. In this case, specific entropy is denoted by a lower case ‘s’, and specific enthalpy by lower case ‘h’. The equation for a change in specific entropy can be written as: (note the units)

Example 8

One kilogram of water is heated from 0°C to 10°C. How much does the entropy of the water change?

Solution 8

T2 = 0°C = 273 K, T1 = 10°C = 283 K

From steam (temperature) table:

Entropy in a Process

The above example only considered the entropy change of the water, while ignoring the entropy change of the heat source. In reality, any heat transfer process requires the heat sink and the heat source to simultaneously undergo a change in entropy. However, keeping the above equation in mind, the heat source experiences less entropy change than the heat sink, since the heat source has a higher mean temperature than the heat sink. This is true even though the total heat given off by the source is equal to the total heat received by the sink.

We can explore this concept, using a typical heat transfer situation. Consider the simplified heat exchange arrangement shown in Figure 6. In this exchanger, saturated steam, at 120°C (393 K), is used to heat water, from 10°C to 40°C. Let’s assume that the steam condenses completely and the condensate temperature is also 120°C. In other words, the only heat transferred from the steam to the water is the latent heat (ie. the enthalpy of evaporation). From the steam tables, hfg at 120°C is 2202.6 kJ/kg. Therefore, in this scenario, each kg of steam transfers 2202.6 kJ of heat to the water.

To raise the water from 10°C (283 K) to 40°C (313 K), the heat that must be received is:

Having set the conditions, let’s calculate the entropy changes within the boundaries of this thermodynamic process. First, consider the 1 kg of steam:

Since heat is removed from the steam, the entropy change is negative and since the scenario involves only 1 kg of steam, the total entropy change is:

Now consider the water. Since heat is added to the water, the entropy change will be positive.

Since 1 kg of steam heats 17.54 kg of water, the total entropy change for the water is:

Now, the total entropy change for the system is the sum of the individual entropy changes. In this scenario:

Notice that there is a net entropy change and the change is positive, indicating an overall increase in entropy. This scenario demonstrates an important, universal principle about entropy. That is, within a closed thermodynamic system the net entropy always increases. This can be expanded to say that the potential for a quantity of energy to do work decreases whenever it is transferred or whenever it is converted from one form to another. Once converted, energy loses the ability on its own (ie. without work being done) to be converted back to its original, more energy intense state.

Temperature-Entropy Diagram

A useful diagram to help us understand entropy in relation to temperature is the temperature-entropy diagram (aka the T-S Diagram). Absolute temperature (K) is plotted on the y-axis and specific entropy (kJ/kgK) on the x-axis.

Figure 7 shows a T-S diagram for steam. The diagram illustrates the conversion of water to superheated steam (at some constant pressure).

• In stage A-B, water is heated from 273 K (0°C) to saturation temperature (Tsat). The entropy increase during this stage is shown as sf.

• In stage B-C, the water is evaporated to dry steam at constant temperature, Tsat. Entropy increase during this stage is shown as sfg.

• Entropy also increases during superheating, since the enthalpy and temperature of the steam increase.

As Figure 7 also demonstrates, the areas under the process line represent the change in enthalpy during each stage.

Let’s relate this T-S diagram to a particular condition from the steam tables and follow the entropy changes as water is converted from 0°C into steam. Consider superheated steam at 2000 kPa and 500°C. First, notice that the steam tables define the specific entropy of water as 0.000 kJ/kgK at 0°C (273K).

• As the water is heated at 2000 kPa to the corresponding saturation temperature of 212.42°C (485.42 K), the enthalpy increases and the specific entropy increases to sf, which equals 2.4474 kJ/kgK.

• During evaporation, heat is added at constant temperature and the entropy increases by sfg, which equals 3.8935 kJ/kgK. This results in a total entropy, sg, of 6.3409 kJ/kgK at dry, saturated condition.

• During superheating, heat is added at increasing temperature to 500°C (773 K). At this final condition, the superheated steam table shows the entropy to be 7.4317 kJ/kgK.

Entropy of Wet Steam

Since a change in entropy is dependent, much like enthalpy, on the transfer of heat, the entropy of wet steam is determined by how much of the latent heat of evaporation has been transferred. This is related directly to the dryness fraction of the steam and the resultant equation for the entropy of wet steam is very similar to that for enthalpy. Where ‘q’ represents the dryness fraction of the steam, the equation is:

Example 9

Calculate the specific entropy of wet steam at a pressure of 1400 kPa and a dryness of 0.96.

Solution 9

From the Steam Tables, the specific entropy of the saturated liquid at 1400 kPa is 2.2842 kJ/kgK and the entropy of evaporation is 4.1850 kJ/kgK. The entropy of the wet steam can be calculated as:

Example 10

One kilogram of dry saturated steam at 3000 kPa is throttled to a pressure of 1000 kPa. Calculate the change in entropy.

Solution 10

From the steam tables:

Now, realize that during a throttling process, enthalpy after throttling equals enthalpy before throttling (ie. throttling is a constant enthalpy process).

Therefore:

Since this is greater than the enthalpy (2778.1 kJ/K) of dry, saturated steam at 1000 kPa, the throttled steam must contain superheat. We now need to determine how much superheat (in °C) is in the steam. This requires interpolation, as follows:

We now need to determine the entropy at 1000 kPa, with 10.53°C superheat. This requires interpolation of entropy between saturation temp (179.91°C) and 200°C, as follows:

Therefore:

And:

Mollier Chart

A Mollier Chart, also called an Enthalpy-Entropy Chart (or H-S Chart), is very useful in analyzing thermodynamic cycles and performing related calculations. In this chart the y-axis represents specific enthalpy (in kJ/kg), while the x-axis represents specific entropy (in kJ/kgK).

Figure 8 shows a portion of the Mollier Chart. The main components of the chart are as follows: (Please Note: An actual Mollier Chart is very complex and is normally presented in a much larger size. The size of our page here restricts the clarity, so we have shown only a portion of the chart and have enlarged some of the labels).

Constant entropy lines: Vertical lines along the Entropy scale represent points on the chart that have the same entropy value. At any point over the length of a line, the value of entropy is the same. The points are “isentropic”.

Constant enthalpy lines: Horizontal lines along the Enthalpy scale represent points on the chart that have the same enthalpy value. At any point over the length of a line, the value of enthalpy is the same. The points are “isenthalpic”.

Saturation Line: This single, curved line represents all points (conditions) on the chart at which steam is 100% dry and saturated. All points below this line represent wet steam, while all points above the line represent superheated steam.

Moisture Lines: Curved lines, below the saturation line, represent points at which the moisture content of the steam is constant. For example, along the line marked “10”, the moisture content is 10% (ie. the steam is 10% wet). Realize that this also means the steam is 90% dry (ie has a dryness fraction of 0.9).

Pressure Lines: Vertically slanted and curved lines, represent points at which the absolute pressure (kPa) is constant. Every point on a line is at that pressure.

Superheat Lines: Horizontally slanted, upwardly curved lines, above the saturation line, represent the degrees of superheat in the steam. For example, each points along the line marked “200” has 200°C of superheat (ie. 200 degrees above the saturation temperature.

Temperature Lines: Downward curved lines, above the saturation line, represent the actual temperature (°C) of the superheated steam.

Following are three examples (11, 12, and 13) of problems that can be solved using the Mollier Chart. Each example will provide a step-by-step solution to the questions posed. For each example, the steps are marked and circled on the appropriate lines in the accompanying section of Mollier Chart.

Example 11

Steam at 500 kPa, with a dryness fraction of 0.96, is throttled down until it is dry and saturated. Using the Mollier Chart, what is the final pressure of the steam and what are the enthalpies and entropies before and after throttling?

Solution 11

First, you must realize that a throttling process for steam occurs at constant enthalpy. That is, the enthalpy before and after throttling will be the same.

Example 12

Superheated steam at a pressure of 10 000 kPa and a temperature of 700°C is expanded isentropically to a pressure of 100 kPa. Using the Mollier Chart, what is the change in enthalpy and the dryness fraction after expansion?

Solution 12

From the above chart information:

Change in enthalpy = 3870 – 2605 = 1265 kJ/kg

and Dryness fraction after expansion = 0.97 (Ans.)

Example 13

A boiler steam drum delivers steam that is 98% dry to a superheater, which adds 300 degrees of superheat to the steam. If the operating pressure is 2000 kPa and the pressure drop through the superheater is negligible, how much enthalpy change occurs in the superheater and what is the final steam temperature?

Solution 13

From the above chart information:

Change in enthalpy = 3495 – 2760 = 735 kJ/kg

and Final temperature of the steam = 513°C (Ans.)

Boiler Efficiency

It is important to know the thermal efficiency at which a boiler is operating. A high efficiency boiler is economical and contributes to a high efficiency for the overall plant. Many factors contribute to the thermal efficiency, including boiler design, type of fuel burned, variations in the boiler load factor, and maintenance practices. Therefore, the ability to calculate the efficiency of a boiler helps to ensure economical operation.

Boiler efficiency is the ratio of the heat transferred to the water and steam over the heat supplied to the boiler by the fuel. This can be written as:

The heat transferred to the steam is found by taking the difference between the specific enthalpy of the feedwater (hw) and the specific enthalpy of the steam produced (h1) and then multiplying this difference by the mass of steam produced per unit of time (ms). Therefore, the heat transferred to the water and steam is represented by ms (h1 – hw) kJ.

The heat supplied by the fuel is the mass of fuel used per unit of time ( mf), multiplied by the heating value of the fuel (HVf).

The equation for boiler efficiency becomes:

Example 14

A small boiler generates 14.9 kg of steam per kg of fuel oil burned. Fuel oil has a heating value of 44 200 kJ/kg. Feedwater is at 50°C and 1400 kPa. If steam is generated with a dryness of 0.941, what is the boiler efficiency?

Solution 14

Enthalpy of the feedwater, at 50°C, can be found in the Steam Temperature Tables as 209.33 kJ/kg.

Enthalpy of the steam is found as:

Now the boiler efficiency can be calculated, using the equation:

1. Describe the principle of a throttling calorimeter and explain why a separating calorimeter sometimes has to be added.

2. A combination separating and throttling calorimeter is connected to a main steam header. The following data was collected:

• Steam pressure – 2410 kPa

• Amount of water collected in the separating calorimeter – 0.87 kg

• Amount of condensate after the throttling separator – 12.5 kg

• Pressure of the steam in throttling calorimeter – 185 kPa

• Temperature of steam in throttling calorimeter – 135°C

Taking the specific heat of the throttled superheated steam as 2.25 kJ/kgK, calculate the main steam line dryness fraction.

3. One kilogram of steam at a pressure of 100 kPa and 40% dry, receives heat under constant volume raising the pressure to 250 kPa absolute.

Calculate the following:

a) volume of the steam

b) quality of the steam at the end of the process

c) work done

d) change in internal energy

4. Steam with a dryness of 0.97 is throttled from a pressure of 800 kPa. Using the Mollier Chart, what is the final pressure required for the steam to be dry and saturated? What are the enthalpies and entropies for these two points?

5. A boiler is fed with water at a temperature of 40°C and produces 2360 kg of dry steam per hour at a pressure of 1750 kPa and a temperature of 250°C. Heat losses are 40% of the heat supplied and the heating value of the fuel is 46 000 kJ/kg. What is the fuel flow required?

Self-Assessment

Question 1

p(v_g - v_f) is equal to the ____________________ when the water evaporates into steam in a boiler.

increase in entropy

change in specific volume

work done

increase in enthalpy

increase in volume

Question 2

On the enthalpy/entropy diagram, an isentropic process can be determined by following:

entropy lines running vertically

pressure lines running from lower left to upper right

enthalpy lines running horizontally

steam temperature lines running horizontally

quality lines running from left to right

Question 3

One of the following statements about throttling is not true.

throttling reduces steam temperature

throttling occurs when a vapour is passed through a small opening

throttling causes an increase in steam enthalpy

throttling is not reversible

throttling always causes an increase in steam quality

Question 4

When determining dryness fraction for average boiler pressures a separating calorimeter is generally used for steam when the moisture content is:

> 2 %

> 4 %

> 5 %

> 7 %

> 10 %

Question 5

When measuring dryness fraction using a throttling calorimeter it is important to be able to:

superheat the steam so that it is hot

preheat the steam prior to entry into the calorimeter

reduce the steam pressure to a manageable level

catch the condensate in a separator.

superheat the steam so that its temperature can be measured