Chapter 2 Properties of Steam – Thermal Engineering

Chapter 2

Properties of Steam

2.1 ❐ PURE SUBSTANCE

A pure substance may be defined as a system that has the following characteristics:

  1. Homogeneous in composition: It means that the composition of every part of the system is same throughout. By composition, we mean the relative proportions of the chemical elements into which the sample of the system can be analysed.
  2. Homogeneous in chemical aggregation: It means that the chemical elements must be combined chemically in the same way in all parts of the system.
  3. Invariable in chemical aggregation: It means that the state of chemical combination of the system does not change with time and position.

Examples of pure substance are atmospheric air, mixture of steam-water, combustion products of a fuel, etc.

2.2 ❐ CONSTANT PRESSURE FORMATION OF STEAM

Consider 1 kg of ice under atmospheric pressure and at a temperature of −10°C contained in a vessel. If we heat the ice gradually keeping the pressure constant, the following changes will take place, as shown in Fig. 2.1:

Figure 2.1 Constant pressure temperature-heat added diagram for steam and water

  1. The temperature of ice will gradually increase till it just reaches the freezing temperature (represented by point ‘b’), which is 0°C at atmospheric pressure. This is shown by the straight line ab.
  2. On addition of more heat at point b, the ice starts melting, and with no rise in temperature till the whole ice has melted, up to point c, the ice is converted into water. The heat added during the process bc is called the latent heat of fusion of ice or simply latent heat of ice.
  3. Gradual addition of more heat beyond the point c leads to gradual increase in temperature till the water reaches the vapourisation or boiling point ‘d’. Under atmospheric pressure of 1.01325 bar, it is 100°C.
  4. On further heating beyond point d, the water will gradually turn into steam, with no rise in temperature. This steam contains some water particles and is called the wet steam. With further addition of heat, all the water particles in the steam disappear and this steam is called dry saturated steam or simply dry steam. This corresponds to point ‘e’. The heat added during de is called the latent heat of vapourisation of steam or simply latent heat of steam.
  5. If the dry saturated steam is further heated beyond point e, keeping the pressure constant, the process is called superheating. The steam so obtained is called the superheated steam. Let the steam be superheated to temperature ts°C, represented by point ‘f’. Then (tst)°C is called the degree of superheat. The heat added during process ef is called the heat of superheat.
2.3 ❐ PROPERTIES OF STEAM
  1. Enthalpy or total heat of water: It is defined as the quantity of heat required to raise the temperature of 1 kg of water from 0°C to its boiling point or saturation temperature corresponding to the pressure applied. It is also called enthalpy of saturated water or liquid heat and is represented by hf.
    hf = specific heat of water, cpw × rise in temperature
    = 4.187 × Δt, kJ/kg

    Note that enthalpy of saturated water can also be found from the steam table based on saturation temperature (Appendix A.1.1).

  2. Heat of steam or latent heat of vapourisation: Latent heat of steam at a particular pressure may be defined as the quantity of heat required to convert 1 kg of water at its boiling point into dry saturated steam at the same pressure. It is denoted by hfg.
  3. Dryness fraction or quality: Dryness fraction of a liquid−vapour mixture is defined as the ratio of mass of dry steam actually present to the mass of the total mixture. It is denoted by x.

    Let ms = mass of dry steam (saturated vapour), kg

    mw = mass of water vapour in steam (saturated liquid), kg

    x = dryness fraction of the sample

    For dry steam, mw = 0 and x = 1.

    The quality of steam is the dryness fraction expressed as a percentage.

    Wetness fraction = 1 − x

  4. Specific volume of wet steam: The specific volume of steam is the volume of 1 kg of dry steam.

    Let V be the total volume of the liquid-vapour mixture, Vf be the volume of the saturated liquid and Vg be the volume of the saturated vapour.

    Total volume of liquid and vapour can be expressed in terms of their specific volume as

    Vf = mwvf
    Vg = msvg

    where vf and vg are the specific volumes of saturated liquid and saturated vapour, respectively.

    Specific volume of the mixture is then

    = (1 − x)v f + xvg
    = v f + x(vgv f)

    Now vf << vg

  5. Enthalpy of wet steam: It may be defined as the quantity of heat required to convert 1 kg of water at 0°C, at constant pressure, into wet steam. Similar to the analysis of specific volume, one can write

    where hf is the specific enthalpy of saturated water and hfg is the change in specific enthalpy during vapourisation. When steam is dry saturated, then x = 1

  6. Total enthalpy of superheated steam:

    Let cps = specific heat of superheated steam

    ts = temperature of formation of steam, °C

    tsup = temperature of superheated steam, °C

    Then heat of superheat = cps (tsupts)

    Note that enthalpy of superheated steam can also be found from the steam table (Appendix A.1.3).

  7. Internal energy of steam: The internal energy of steam may be defined as the actual heat energy stored in steam above the freezing point of water. It is the difference between the total heat of steam and the external work of evaporation. Similar to the analysis of specific volume, one can write for 1 kg of wet steam,

    where uf is the specific internal energy of saturated water and ufg is the change in specific internal energy during evaporation. From the known value of enthalpy, internal energy can be found as follows:

     

    uwet = hpv1

    For superheated steam,

    hsup = hg + cps (tsupts)
    usup = hsuppvsup
  8. Entropy of water: Consider 1 kg of water to be heated from a temperature Ts to T at a constant pressure.

    Then, change of entropy =

    where cpw = specific heat of water

    The entropy of water is always reckoned above the freezing point of water.

    Therefore, Ts = saturation temperature of formation of steam

    T = absolute temperature of freezing water = 273 K

    Entropy of saturated water,

  9. Entropy of wet steam: Consider 1 kg of wet steam at an absolute pressure p bar.

    Let x = dryness fraction of steam

    Ts = temperature of formation of steam

    sf = the specific enthalpy of saturated water

    sfg = change in entropy during evaporation

    The entropy of wet steam can be calculated as

    When steam is dry saturated, then x = 1

  10. Entropy of superheated steam: Consider 1 kg of dry saturated steam at Ts temperature of formation, to be superheated to a temperature Tsup.

    Let cps = specific heat of steam at constant pressure during superheating

    Change of entropy per kg of steam during superheating

    Total entropy of superheated steam

    ssup = Entropy of dry saturated steam + Change of entropy during superheating

2.4 ❐ STEAM TABLES

In actual practice, it is quite cumbersome to calculate the relation between various quantities such as pressure, temperature, volume, enthalpy, latent heat and entropy, etc., of steam. Such quantities have been determined experimentally and recorded in the form of tables known as steam tables. Steam table is a complete data book that contains various properties, like volume, internal energy, enthalpy and entropy for 1 kg of water in different phases for a given pressure or temperature. Steam table is given in the Appendix A. The properties of steam are given in three different tables—saturated steam table based on temperature (Appendix A.1.1), saturated steam table based on pressure (Appendix A.1.2), and superheated steam table (Appendix A.1.3). When the temperature is selected as independent variable and the properties of steam are tabulated, the steam table is referred to as the saturated steam table based on temperature. When the pressure is chosen as independent variable, it is referred to saturated steam table based on pressure. For superheated steam, there is a separate table for different amount of degree of superheat.

Values for unknown pressure or temperature can be calculated by linear interpolation between two consecutive values. Saturated steam table based on pressure for a particular pressure (= 1 bar) is shown in Table 2.1.

 

Table 2.1 Saturated steam table based on pressure

2.5 ❐ TEMPERATURE−ENTROPY DIAGRAM FOR WATER AND STEAM

It is the plot of the saturation temperature of water and steam corresponding to various absolute pressure vs entropies at those saturation temperatures, as shown in Fig. 2.2. The curve on the left marked water line (saturated liquid line) shows the relationship between entropy and temperature before steam is formed. The dry steam line (saturated vapour line) shows the relationship between entropy and temperature when all the water has been converted into steam. The area in between two regions is called the wet region. The area to the right of the dry steam lines is called the region of superheat. The lines of constant dryness, constant volume, and constant total heat are also drawn on the diagram. The temperature at which the distinction between the entropy of water and that of dry steam disappears is called the critical temperature (374.15°C). The pressure corresponding to this temperature is called the critical pressure (221.2 bar).

Figure 2.2 Temperature−entropy diagram for water and steam

2.6 ❐ ENTHALPY−ENTROPY OR MOLLIER DIAGRAM OF STEAM

In Mollier diagram (Fig. 2.3), the vertical ordinate represents the enthalpy and the horizontal ordinate represents the entropy. The regions above and below the saturation line represent the superheated and wet conditions of steam, respectively. The lines of constant dryness fraction are shown in the wet steam region, whereas the lines of constant temperature and pressure in the region of both wet and superheat are shown.

From the thermodynamic property relationship, one can write,

 

Tds = dhvdp

For constant pressure process, we have

 

Tds = dh

From Eq. (2.13), it is clear that the slope of the constant pressure lines is equal to the corresponding saturation temperatures. It may be noted that as the pressure increases, the saturation temperature also increases. This is the reason for which constant pressure lines (isobar) are divergent on the hs diagram. The saturated liquid line and saturated vapour line meet at critical point.

Figure 2.3 Mollier diagram for steam

2.7 ❐ VARIOUS PROCESSES FOR STEAM

2.7.1 Constant Volume Process

The constant volume heating process is shown on the pv and Ts diagrams in Fig. 2.4. Consider 1 kg of wet steam at a certain pressure p1, and dryness fraction x1 is heated to state 2 in the wet region and then to state 3 in the superheated region.

Initial volume of steam v1 = vf 1 + x1(vg1vf1)

Final volume of steam in wet region v2 = vf2 + x2(vg2vf2)

Since the volume remains constant,

 

vf1 + x1(vg1vf1) = vf2 + x2(vg2vf2)

Figure 2.4 Constant volume heating: (a) p−v diagram, (b) T−s diagram

 

From steam table, internal energy of steam can be found as follows:

Initial internal energy of steam, u1 = uf1 + x1 ufg1

Final internal energy of steam in the wet region, u2 = uf2 + x2 ufg2

Since dv = 0, therefore δW = 0

For constant volume heating process, v2 = v3. After knowing the value of v3, the condition of steam can be found from superheated steam table. If pressure of state 3 is given (=p3), then from known values of p3 and v3, temperature of state 3 can be obtained. Similarly, if temperature at state 3 (=T3) is given, pressure can be found from known values of T3 and v3.

2.7.2 Constant Pressure Process

The constant pressure heating process is shown in Fig. 2.5. In this case, p1 = p2 = p (say).

For wet steam, during heating process 1–2,

Work done per kg of steam, w = pdv = p(v2v1)

where v1 = vf + x1(vgvf) and v2 = vf + x2(vgvf)

The specific enthalpy of steam, h1 = hf + x1 hfg and h2 = hf + x2 hfg

From first law of thermodynamics, heat transfer per kg of steam during constant pressure heating

 

q = h2h1
q = (hf + x2hfg) − (hf + x1hfg)

For superheated steam, specific volume and specific enthalpy of steam can be found from the superheated steam table. Then, during heating process 2-3,

Work done per kg of steam, w = pdv = p(v3 v2) = p(vsupv2)

and heat transfer per kg of steam

Figure 2.5 Constant pressure expansion

2.7.3 Isothermal Process

The isothermal heating process is shown in Fig. 2.6. The isothermal process for wet steam will be same as the constant pressure process. However, in the superheated region, the steam will behave as a gas and shall follow Boyle’s law, pv = const. Therefore, the process will be hyperbolic.

In the wet region,

 

q = h2h1 and w = p1 (v2v1)

In the superheated region,

 

q = h3h2 and w = p1 (v3v2)

Figure 2.6 Isothermal process

2.7.4 Hyperbolic Process

During a hyperbolic process, pv = const. If the superheated vapour behaves as an ideal gas, then the process during which pv = const. may be regarded isothermal, as shown in Fig. 2.7.

For 1 kg of steam in the wet region,

Initial volume of steam v1 = vf1 + x1(vg1vf1)

Final volume of steam v2 = vf2 + x2(vg2vf2)

As pv = const., therefore

or

Figure 2.7 Hyperbolic expansion

Now, h1 = hf 1 + x1hfg1 and h2 = hf 2 + x2hfg2 for wet steam

and hs2 = hg2 + cp (TsupT2) for superheated steam

Change is internal energy,

where v1 and v2 are the specific volumes of steam before and after heating.

Total heat supplied, q = w + (u2u1)

2.7.5 Reversible Adiabatic or Isentropic Process

The isentropic process of expansion (1-2) is shown in Fig. 2.8 on Ts diagram. The initial condition of steam may be wet, dry, or superheated. For initially wet steam as shown in Fig. 2.8(a), we have

 

s1 = s f1 + x1s fg1
s2 = s f2 + x2s fg2

Since s1 = s2, therefore

 

s f 1 + x1s fg1 = s f 2 + x2s fg2

For initially dry steam (Fig. 2.8(b)), s1 = sg1

Figure 2.8 Isentropic expansion: (a) Expanded steam wet, (b) Expanded steam dry saturated, (c) Expanded steam superheated

s2 = s f2 + x2s fg2

Then, sg1 = s f2 + x2s fg2

For initially superheated steam at temperature Tsup, s1 can be calculated from superheated steam table. After expansion, state 2 may be superheated (Fig. 2.8(c)), dry or wet. If the state after expansion is superheated, then the condition of steam can be calculated from the relationship s1 = s2. Note that if pressure is specified, then temperature can be calculated or otherwise if temperature is specified, then pressure can be calculated.

Now q = w + (u2u1)

where u1 = u f1 + x1u fg1 for wet steam

= ug1 for dry steam

= usup for superheated steam

Similarly, u2 can be found depending on the condition of steam after expansion.

But q = 0

w = u1u2

2.7.6 Polytropic Process

Let the steam expand from condition 1 to 2 according to the law pvn = constant. Then for wet steam,

Initial volume of steam, v1 = vf1 + x1(vg1vf1)

Final volume of steam v2 = vf2 + x2(vg2vf2 )

or

or

Also u1 = uf1 + x1ufg1

u2 = uf2 + x2ufg2

n = 1.13 for wet steam and 1.3 for superheated steam

For superheated steam on heating from wet steam, v2 can be calculated from superheated steam table for the known pressure and temperature conditions. Similarly, the condition of steam can also be found after knowing the value of v2 from Eq. (2.21).

2.7.7 Throttling Process

The throttling process occurs when steam is expanded through a small aperture as in the case of throat of a nozzle. During this process, no work is done and the enthalpy remains constant. The throttling process is shown on Ts and hs diagrams in Fig. 2.9.

 

h1 = hf1 + x1hfg1
h2 = hf2 + x2hfg2 for wet steam after expansion
= hf2 + hfg2 + cps (TsupTs), for superheated steam after expansion

Now h1 = h2

For wet steam after expansion

For superheated steam after expansion

Figure 2.9 Throttling process: (a) Ts diagram, (b) hs diagram

2.8 ❐ DETERMINATION OF DRYNESS FRACTION OF STEAM

The different methods commonly used for the measurement of dryness fraction of steam are as follows:

  1. Barrel Calorimeter
  2. Separating Calorimeter
  3. Throttling Calorimeter
  4. Combined Separating and Throttling Calorimeter.

The working principle of all the calorimeters is to bring the state of the substance from the two-phase region to the single phase region (either compressed liquid or superheated vapour region). This is because in the two-phase region, pressure and temperature are not independent variable, one is independent and the other is dependent. On the other hand, both in compressed liquid and superheated vapour region, both pressure and temperature are independent variables.

2.8.1 Barrel Calorimeter

The barrel calorimeter, shown in Fig. 2.10, consists of a copper barrel placed on wooden block and covered by a wooden cover. The barrel contains a known quantity of cold water and is surrounded by an outer vessel with air space in between, which acts as an insulator. The temperature is measured by a thermometer that passes through one of the holes in the wooden cover. The whole assembly is placed on the platform of a weighing bridge. The steam from the main steam pipe enters through the sampling tube via the control valve and flows into the cold water through fine exit holes of the ring provided at the end of the pipe. The condensation of steam takes place as it comes in contact with cold water, and as a result, the temperature of water rises. The quantity of steam condensed can be known from the difference of readings of the weighing bridge before and after the condensation of steam. The pressure of entering steam is indicated by the pressure gauge.

Figure 2.10 Barrel calorimeter

Let ms = mass of steam condensed

x = dryness fraction of steam

hfg = latent heat of steam

ts = temperature of formation of steam (corresponding to pressure indicated by the pressure gauge)

mw = mass of cold water in the calorimeter in the beginning

mc = water equivalent of calorimeter

t1 = initial temperature of cold water in the calorimeter

t2 = final temperature of the mixture after steam condensation

Heat lost by steam = ms[x × hfg + cps(tst2)]

Heat gained by cold water and calorimeter = (mw + mc) × cpw × (t2t1)

Heat lost by steam = Heat gained by cold water and calorimeter

This method gives only approximate results. The value of x calculated is always lower than the actual value as heat losses due to convection and radiation are not accounted for.

2.8.2 Separating Calorimeter

This calorimeter, shown in Fig. 2.11, consists of two concentric chambers that communicate with each other through an opening at the top. The steam to be tested flows from the main steam pipe through a sampling tube to the calorimeter pipe through the valve which must be fully open when testing. The metal basket has a large number of perforations through which the steam discharges. The water particles due to their heavier momentum get separated from the steam and get collected in the inner chamber. The quantity of water collected is indicated by the level in the gauge glass and the pointer that moves on a scale. The dry steam in the inner chamber moves up and then down again through the annular space between the two chambers. The gauge fitted has two scales: the inner one indicates the pressure, whereas the outer one indicates the rate of discharge of dry saturated steam during a predefined time interval. Then the steam then flows through the calibrated orifice to the bucket calorimeter which is placed on the platform balance by which the most of steam discharged can be further confirmed.

Figure 2.11 Separating calorimeter

Let ms = mass of dry steam condensed

mw = mass of suspended moisture collected

It is not possible to remove all the water particles from steam, and the dryness fraction calculated by this method is always greater than the actual. However, it is a quick method and is used for the measurement of dryness fraction for very wet steam.

2.8.3 Throttling Calorimeter

The throttling calorimeter, shown in Fig. 2.12, consists of the sampling tube through which the steam from the main steam pipe flows through the throttling valve and becomes superheated. The steam then flows into the inner chamber, flows down, and rises up again to enter the annular space. The loss of heat by radiation from the inner chamber is minimised by the hot steam around the outside of the inner chamber. The temperature of throttled steam is measured by the thermometer placed in the pocket filled with cylinder oil. To obtain good results, the steam issuing from the throttle valve must be superheated. To ensure this, a manometer is attached to measure the pressure of steam. Corresponding to this pressure, the saturation temperature of steam must be lower than the temperature indicated by the thermometer. The steam finally escapes through the exhaust. This calorimeter is suitable for steam having high dryness fraction.

Let p1 = initial pressure of wet steam before expansion

x1 = dryness fraction of wet steam

hfg1 = latent heat of vapourization of wet steam

tsup = temperature of superheated steam after expansion

hfg2 = latent heat of vapourization of steam after expansion

ts2 = saturation temperature of steam after expansion

cps = specific heat of superheated steam at constant pressure

Enthalpy of steam before throttling = hf 1 + x1 hfg1

Figure 2.12 Throttling calorimeter

Enthalpy of steam after throttling = hf 2 + hfg2 + cps (tsupts2)

Enthalpy before throttling = Enthalpy after throttling

 

hf 1 + x1 hfg1 = hf 2 + hfg2 + cps (tsupts2)

2.8.4 Combined Separating and Throttling Calorimeter

The combined separating and throttling calorimeter is shown in Fig. 2.13. The steam is first passed through the separating calorimeter where it losses most of its moisture and becomes comparatively drier. It is then passed through the throttling calorimeter where superheating takes place without change of enthalpy. The temperature and pressure of steam after throttling are measured by using a thermometer and pressure gauge, respectively.

Let p1 = pressure of wet steam entering the throttling calorimeter

x1 = dryness fraction of steam

hfg1 = latent heat of entering steam

mw = mass of suspended moisture collected

ms = mass of steam leaving the separating calorimeter and entering the throttling calorimeter

p2 = pressure of steam after throttling

hfg2 = latent heat of steam at pressure p2

Enthalpy of steam entering throttling calorimeter = Enthalpy of steam leaving throttling calorimeter

 

hf 1 + x1hfg1 = (hf 2 + hfg2) + cps (tsupts2)

Figure 2.13 Combined separating and throttling calorimeter

 

and x = dryness fraction of steam entering the separating calorimeter

Mass of dry steam entering the whole apparatus = ms x1

This calorimeter gives quite satisfactory results when the steam is considerably wet.

Example 2.1

Determine the condition of steam in the following cases:

  1. At a pressure of 10 bar and temperature 200°C.
  2. At a pressure of 10 bar and specific volume 0.175 m3/kg.

Solution

Given p = 10 bar; t = 200°C; v = 0.175 m3/kg

  1. Condition of steam at temperature of 200°C:

    From steam table (A.1.2 in the Appendix), corresponding to a pressure of 10 bar, we find the saturation temperature ts = 179.91°C

    Since the saturation temperature at 10 bar is (179.91°C) lower than the given temperature of the steam (200°C), therefore the given steam is superheated.

    The degree of superheat = 200 − 179.91= 20.09°C

  2. Condition of steam at a volume of 0.175 m3/kg:

    From steam table (A.1.2 in the Appendix), vf = 0.001127 m3/kg, vg = 0.19444 m3/kg.

    Since the volume of given steam (0.175 m3/kg) is less than the specific volume of the dry saturated steam (0.19444 m3/kg), therefore the given steam is wet. Therefore, the dryness fraction can be found to be

     

    0.175 = 0.001127 + x (0.19444 − 0.001127)

    or x = 0.8994

Example 2.2

Calculate the quantity of heat required to generate 1 kg of steam at a pressure of 8 bar from water at 30°C (a) when dryness fraction is 0.9, (b) when steam is just dry, and (c) when the temperature of steam is 300°C.

Solution

From steam tables (A.1.2 in the appendix) at 8 bar: hf = 721.1 kJ/kg, hfg = 2048.0 kJ/kg

And at 30°C (A.1.1 in the appendix), hf = 125.79 kJ/kg

  1. h = hf + x hfg = 721.1 + 0.9 × 2048.0 = 2564.3 kJ/kg

    Net heat required = hhf = 2564.3 − 125.79 = 2438.51 kJ/kg

  2. h = hf + hfg = hg = 2769.1 kJ/kg

    Net heat required = hhf = 2769.1 − 125.79 = 2643.31 kJ/kg

  3. ts = 170.4°C, tsup = 300°C

Therefore, the steam is in superheated condition. From superheated steam table (A.1.3 in the appendix), we have h = 3056.4 kJ/kg

Net heat required = hhf = 3056.4 − 125.77 = 2930.63 kJ/kg

Example 2.3

A pressure cooker contains 2 kg of steam at 5 bar and 0.9 dry. Find the quantity of heat that must be rejected so that the quality of steam becomes 0.5 dry.

Solution

The pressure cooker is a constant volume vessel. Therefore, W = 0

 

Q = ∆U = U2U1

From steam tables (A.1.2 in the appendix) at 5 bar: vf = 0.001093 m3/kg, vg= 0.3749 m3/kg, uf = 639.66 kJ/kg, ufg = 1921.6 kJ/kg,

 

u1 = uf + x1ufg = 639.66 + 0.9 × 1921.6 = 2369.1 kJ/kg
U1 = 2 × 2369.1 = 4738.2 kJ

Specific volume of the steam at 5 bar

 

v1 = vf + x1 (vgvf) = 0.001093 + 0.9 × (0.3749 − 0.001093) = 0.3375 m3/kg

From steam tables (A.1.2 in the appendix), the properties of steam at 2.5 and 2.75 bar are given below:

For constant volume of steam, we have

 

v1 = v2 = 0.3375 m3/kg

Specific volume of the steam at 2.5 bar

v = vf + x (vg vf ) = 0.001067 + 0.9 × (0.7187 − 0.001067) = 0.3599 m3/kg

Specific volume of the steam at 2.75 bar

v = vf + x(vg vf ) = 0.001070 + 0.9 × (0.6573 − 0.001070) = 0.3292 m3/kg

Using linear interpolation, the pressure corresponding to v2 = 0.3375 m3/kg is obtained as

or p2 = 2.68 bar

Using linear interpolation, we get

or u fg2 = 1994.83 kJ/kg

u2 = uf 2 + x2 ufg2 = 544.79 + 0.5 × 1994.83 = 1542.205 kJ/kg

U2 = m u2 = 2 × 1542.205 = 3084.41 kJ
Q = U2U1 = 3084.41 − 4738.2 = − 1653.79 kJ

Example 2.4

About 5400 m3/h of wet steam with dryness fraction 0.92 and at 10 bar is to be supplied by a boiler for a processing plant. Calculate (a) the mass of steam supplied per hour and (b) the quantity of coal of calorific value 15 MJ/kg to be burnt in the boiler if overall efficiency of the boiler is 30%.

Solution

  1. From steam tables (A.1.2 in the appendix) at 10 bar: vf = 0.001127 m3/kg, vg= 0.19444 m3/kg

    Specific volume of the steam v = vf + x (vg vf ) = 0.001127 + 0.92 × (0.19444 − 0.001127) = 0.179 m3/kg

  2. At p = 10 bar, hf = 762.79 kJ/kg, hfg = 2015.3 kJ/kg

    h = hf + x hfg = 762.79 + 0.92 × 2015.3 = 2616.866 kJ/kg

    H = ṁh = 30167.6 × 2616.866 = 78944566 kJ/h

    Heat supplied to boiler, Hb = = 263148553 kJ/h

    Quantity of coal burnt = = 17.54 tonnes/h

Example 2.5

Find the internal energy of 1 kg of superheated steam at a pressure of 10 bar and 300°C. If this steam is expanded to 1.5 bar and 0.9 dryness, then find the change in internal energy.

Solution

From steam table (Appendix A.1.3), at p = 10 bar and tsup = 300°C, for superheated steam,

 

u1 = 2793.2 kJ/kg

For dry saturated steam, at p = 1.5 bar, we have from table A.1.2 in the Appendix

 

uf = 466.92 kJ/kg, ufg = 2052.7 kJ/kg,
u2 = uf 2 + x2ufg2 = 466.92 + 0.9 × 2052.7 = 2314.35 kJ/kg
u = u2u1 = 2314.35 − 2793.2 = − 478.85 kJ/kg

Example 2.6

A rigid vessel is initially divided into two parts A and B by a thin partition. Part A contains 1 kg of steam at 4 bar, dry and saturated, and part B contains 2 kg of steam at 8 bar with dryness fraction of 0.90. The partition is removed and the pressure in the vessel after sometimes is found to be 6 bar. Find (a) the volume of the vessel and the state of steam and (b) the amount of heat transfer from steam to the vessel and the surroundings.

Solution

  1. At pA = 4 bar, from steam table A.1.2 in the Appendix, vgA = 0.4625 m3/kg, uA = 2553.6 kJ/kg
    VA = mAvgA = 1 × 0.4625 = 0.4625 m3

    At pB = 8 bar, from steam table A.1.2 in the Appendix, vfB = 0.001115 m3/kg, vgB = 0.2404 m3/kg

    VB = mB (vf B + xBvgB) = 2 × [0.001115 + 0.9 × (0.2404 − 0.001115)] = 0.433 m3

    Volume of vessel, V = VA + VB = 0.4625 + 0.433 = 0.8955 m3

    At p = 6 bar, from steam table A.1.2 in the Appendix, vf = 0.001101 m3/kg, vg = 0.3157 m3/kg,

    m = 1 + 2 = 3 kg

    Dryness fraction at 6 bar, x can be found to be

    or x = 0.945

  2. Now UA = mAuA = 1 × 2553.6 = 2553.6 kJ
    UB = mB (hfB + xBhfgBpBxBvgB × 102)
    = 2 (721.1 + 0.9 × 2048.0 − 8 × 0.9 × 0.240 × 102)
    = 4783.0 kJ

    At 6 bar, U = m[hf + x hfgpxvg × 102]

    = 3[670.6 + 0.944 × 2086.3 − 6 × 0.944 × 0.316 × 102)
    = 7383.25 kJ
    Q = U − (UA + UB)
    = 7383.25 − (2553.3 + 4783.0) = 46.95 kJ

Example 2.7

Calculate enthalpy and entropy for 5 kg of steam at 8 bar under the following conditions:

(a) dry and saturated, (b) wet steam having wetness of 40%, and (c) superheated steam at 250°C.

Solution

From steam table (Appendix A.1.2), at p = 8 bar, we have

hf = 721.1 kJ/kg, hfg = 2048.0 kJ/kg, hg = 2769.1 kJ/kg, sf = 2.0461 kJ/kgK, sfg = 4.6166 kJ/kgK, sg = 6.6627 kJ/kgK

At tsup = 250°C and p = 8 bar, we have from table A.1.3

h = 2950.0 kJ/kg, s = 7.0384 kJ/kgK
  1. H = m hg = 5 × 2769.1 = 13845.5 kJ

    S = m sg= 5× 6.6627 = 33.3135 kJ/K

  2. x = 1 − 0.4 = 0.6

    H = m (hf + x hfg) = 5(721.1 + 0.6 × 2048.0) = 9749.5 kJ

    S = m (sf + xsfg) = 5(2.0461 + 0.6 × 4.6166) = 24.08 kJ/K

  3. H = mh = 5 × 2950.0 = 14750.0 kJ

    S = ms = 5 × 7.0384 = 35.192 kJ/K

Example 2.8

About 1 kg of steam 0.85 dry at a pressure of 1 bar and dryness fraction 0.85 is compressed in a cylinder to a pressure of 2 bar according to pv1.25 = const. Determine the final condition of steam and the heat transfer through the cylinder walls.

Solution

From steam table (Appendix A.1.2), at p = 1 bar, vf1 = 0.001043 m3/kg, vg1 = 1.694 m3/kg

uf1 = 417.33 kJ/kg, ufg1 = 2088.7 kJ/kg

Specific volume of steam

v = vf + x (vgvf) = 0.001043 + 0.85 × (1.694 − 0.001043) = 1.44 m3/kg

Now p1v1n = p2v2n

At p2 = 2 bar, vf 2 = 0.001061 m3/kg, vg2 = 0.8857 m3/kg, uf 2 = 504.47 kJ/kg, ufg2 = 2025 kJ/kg

Now v2 = 0.8271 = vf 2+ x2 (vg2vf 2) = 0.001061 + x2 (0.8857− 0.001061)

or x2 = 0.934

Now u1 = uf 1 + x1ufg1 = 417.33 + 0.85 × 2088.7 = 2192.72 kJ/kg

u2 = uf 2 + x2ufg2 = 504.47 + 0.934 × 2025 = 2395.82 kJ/kg

u = u2u1 = 2395.82 − 2192.72 = 203.1 kJ/kg

Work done,

Heat transfer through the cylinder walls,

q = w + (u2u1) = − 85.68 + 203.1 = 117.42 kJ/kg

Example 2.9

Steam at a pressure of 10 bar and 0.9 dryness fraction expands to 1 bar hyperbolically. Find (a) work done and heat absorbed and (b) internal energy and change in enthalpy.

Take specific heat of steam at constant pressure = 2 kJ/kg.K.

Solution

From steam table (Appendix A.1.2), we have

v1 = vf 1+ x1 (vg1vf 1) = 0.001127 + 0.9 × (0.19444 − 0.001127) = 0.1751 m3/kg

Now, p1v1 = p2v2

Expansion ratio,

Work done, w = p1v1 ln r = 10 × 102 × 0.1751 ln 10 = 403.18 kJ/kg

v2 = 10v1 = 10 × 0.1751 = 1.751 m3/kg

Now, v2 > vg2. Therefore, the steam is superheated at p2 = 1 bar.

Using linear interpolation, we have

Tsup 2 = 111.47°C

h1 = hf 1 + x1hfg1 = 762.79 + 0.9 × 2015.3 = 2576.56 kJ/kg

Using linear interpolation, we get

h2 = 2699.19 kJ/kg

h = h2h1 = 2699.19 − 2576.56 = 122.63 kJ/kg

u = ∆h for hyperbolic process

q = w + ∆u = 403.18 + 122.63 = 525.81 kJ/kg

Example 2.10

Steam, which is initially dry saturated, expands isentropically from pressure of 15 bar to 0.15 bar. Find the index of isentropic expansion.

Solution

From steam table (Appendix A.1.2), we have

For isentropic expansion, s1 = s2

s1 = sf 1 + sfg1 = 2.315 + 4.130 = 6.445 kJ/kg.K
s2 = sf 2 + x2sfg2 = 0.7548 + x2 × 7.2536 = 6.445

or x2 = 0.7845

v1 = vg1 = 0.13177 m3/kg
v2 = vf 2+ x2 (vg2vf 2) = 0.001014 + 0.7845 × (10.022 − 0.001014) = 7.8625 m3/kg

Now, p1v1n = p2v2n

or n = 1.126

Example 2.11

Steam at a pressure of 8 bar and 0.95 dry is expanded to a pressure of 1.8 bar. Find the final condition of steam and the heat drop for each of the following methods of expansion:

  1. Adiabatic expansion
  2. Throttling expansion

Solution

Given that p1 = 8 bar, x1 = 0.95, p2 = 1.75 bar

From steam tables (Appendix A.1.2), at 8 bar, sf1 = 2.0461 kJ/kg.K, sfg1 = 4.6166 kJ/kg.K, hf1 = 721.1 kJ/kg, hfg1 = 2048 kJ/kg, and at 1.75 bar, sf2 = 1.4848 kJ/kg.K, sfg2 = 5.6868 kJ/kg.K, hf 2 =486.97 kJ/kg, hfg2 = 2213.6 kJ/kg

  1. During an adiabatic expansion, s1 = s2
    sf 1 + x1sfg1 = sf 2 + x2sfg2
    2.0461 + 0.95 × 4.6166 = 1.4848 + x2 × 5.6868
    x2 = 0.8699
  2. During throttling process, h1 = h2
    hf 1 + x1hfg1 = hf 2 + x2hfg2
    721.1 + 0.95 × 2048.0 = 486.97 + x2 × 2213.6
    x2 = 0.9847

Example 2.12

Steam at 10 bar flows into a barrel containing 150 kg of water at 10°C. The final temperature of water is 25°C and the increase in mass is 4 kg. Calculate the dryness fraction of steam. Take cpw = 4.187 kJ/kg.K.

Solution

Given that ms = 4 kg, p = 10 bar, mw = 150 kg, tw1 = 10°C, tw2 = 25°C, and cps = 4.187 kJ/kg.K

From steam table (Appendix A.1.2), at 10 bar, hfg = 2015.3 kJ/kg and ts = 179.91°C

Heat lost by steam = Heat gained by water

ms[x hfg + cpw(tstw2)] = mw (tw2tw1)cpw
4[x × 2015.3 + 4.187(179.91 − 25)] = 150 (25 − 10) × 4.187

Dryness fraction of steam, x = 0.847

Example 2.13

In a separating and throttling calorimeter, the steam main pressure is 8 bar absolute and the temperature after throttling is 130°C. The pressure in the throttling calorimeter is 0.001 bar gauge, and the barometer reading is 75 cm. If 0.15 kg of water is trapped in the separator and 1.8 kg of steam is passed through the throttling calorimeter, then determine the dryness fraction of steam in the steam main.

Solution

After throttling, p2 = 0.001 + 0.75 × 9.81 × 13600 × 10−5 = 1.01 bar

The saturation temperature for p2 = 1.01 bar, ts2 = 99.62°C

Degree of superheat = tsupts2 = 130 − 99.62 = 30.38°C

For superheated steam, from steam tables (Appendix A.1.3), for tsup = 130°C

p = 1.01 bar, we have by linear interpolation
hsup2 = 2736.32 kJ/kg

From steam tables (Appendix A.1.2),

At 8 bar hf1 = 721.1 kJ/kg, hfg1 = 2048 kJ/kg

h1 = hf 1 + x1hfg1

Now h1 = hsup2

721.1 + x1 + 2048.0 = 2736.32
x1 = 0.984

Dryness fraction, x = x1x2

= 0.984 × 0.923 = 0.908

Example 2.14

Calculate the enthalpy of 1 kg of steam at a pressure of 8 bar and dryness fraction of 0.8. How much heat would be required to raise 2 kg of this steam from water at 20°C?

Solution

Given that ms = 1 kg, p = 8 bar, x = 0.8, and tw = 20°C

Enthalpy of 1 kg of steam:

From steam table (Appendix A.1.2), corresponding to a pressure of 8 bar, we find that,

hf = 721.1 kJ/kg and hfg = 2048 kJ/kg

Enthalpy of 1 kg of wet steam,

h = hf + x hfg = 721.1 + 0.8 × 2048 = 2359.5 kJ

Heat required to raise 2 kg of this steam from water at 20°C:

We have calculated above the enthalpy or total heat required to raise 1 kg of steam from water at 0°C. Since the water, in this case, is already at 20°C, therefore,

Heat already in water = 4.2 × 20 = 84 kJ

Heat required per kg of steam = 2359.5 − 84 = 2275.5 kJ

and heat required for 2 kg of steam = 2 × 2275.5 = 4551 kJ

Example 2.15

Determine the quantity of heat required to produce 1 kg of steam at a pressure of 6 bar at a temperature of 25°C, under the following conditions:

  1. When the steam is wet having a dryness fraction 0.9
  2. When the steam is dry saturated
  3. When it is superheated at a constant pressure at 250°C, assuming the mean specific heat of superheated steam to be 2.3 kJ/kg.K.

Solution

Given that p = 6 bar, tw = 25°C, x = 0.9, tsup = 250°C, and cps = 2.3 kJ/kg.K

From steam table (Appendix A.1.2), corresponding to a pressure of 6 bar, we find that

hf = 670.4 kJ/kg, hfg = 2086.3 kJ/kg, and ts = 158.85°C
  1. When the steam is wet:

    Enthalpy or total heat of 1 kg of wet steam,

    h = hf + xhfg = 670.54 + 0.9 × 2086.3 = 2548.21 kJ

    Since the water is at a temperature of 25°C, therefore from steam tables (Appendix A.1.1)

    Enthalpy of water 104.87 kJ

    ∴ Heat actually required = 2548.21 − 104.87 = 2443.34 kJ

  2. When the steam is dry saturated:

    From steam tables (Appendix A.1.2), enthalpy or total heat of 1 kg of dry saturated steam,

    hg = 2756.8 kJ

    ∴ Heat actually required = 2756.8 − 104.87 = 2651.93 kJ

  3. When the steam is superheated:

    From steam tables (Appendix A.1.3), enthalpy or total heat of 1 kg of superheated steam,

    hsup = 2957.2 kJ

    ∴ Heat actually required = 2957.2 − 104.87 = 2852.33 kJ

Example 2.16

Steam enters an engine at a pressure of 12 bar with a 67°C of superheat. It is exhausted at a pressure of 0.15 bar and 0.95 dry. Find the drop in enthalpy of the steam. Assume specific heat of superheated steam cps = 2.1 kJ/kgK.

Solution

Given that p1 = 12 bar, tsupts = 67°C, p2 = 0.15 bar, and x = 0.95

From steam table (Appendix A.1.2), corresponding to a pressure of 12 bar, we find that

hf = 798.64 kJ/kg; hfg = 1986.2 kJ/kg ; hg = 2784.8 kJ/kg

Enthalpy or total heat of 1 kg of superheated steam,

hsup = hg + cps (tsupts)
= 2784.8 + 2 × 67 = 2925.5 kJ/kg

Similarly, from steam tables (Appendix A.1.2), corresponding to a pressure of 0.15 bar

hf = 225.91 kJ/kg; hfg = 2373.1 kJ/kg

Enthalpy or total heat of 1 kg of wet steam,

h = hf + xhfg = 225.91 + 0.95 × 2373.1 = 2480.36 kJ/kg

∴ Drop in enthalpy of the steam = hsuph = 2925.5 − 2480.36 = 445.14 kJ/kg

Example 2.17

A steam engine obtains steam from a boiler at a pressure of 15 bar and 0.98 dry. It was observed that the steam loses 21 kJ of heat per kg as it flows through the pipe line and pressure remaining constant. Calculate the dryness fraction of the steam at the engine end of the pipeline.

Solution

Given that p = 15 bar, x = 0.98, and heat loss = 21 kJ/kg

From steam table A.1.2 in the Appendix, corresponding to a pressure of 15 bar we find that

hf = 844.87 kJ/kg; hfg = 1947.3 kJ/kg

Enthalpy of wet steam at the boiler end.

h = hf + xhfg = 844.87 + 0.98 × 1947.3 = 2753.22 kJ/kg

Since the steam loses 21 kJ/kg of steam, therefore enthalpy of wet steam at the engine end,

h2 = 2753.22 − 21 = 27332.22 kJ

Let x2 = Dryness fraction of steam at the engine end.

Since the pressure remains constant, therefore hf and hfg is same. We know that

h2 = hf + x2hfg
2732.22 = 844.87 + x2 × 1947.3

or x2 = 0.9692

Example 2.18

A coal fired boiler plant consumes 400 kg of coal per hour. The boiler evaporates 3200 kg of water at 45°C into superheated steam at a pressure of 12 bar and 274.5°C. If the calorific value of the fuel is 32600 kJ/kg of coal, then determine the following:

  1. Equivalent evaporation
  2. Efficiency of boiler

Assume specific heat of superheated steam as 2.1 kJ/kg.K.

Solution

Given that mf = 400 kg/h, ms = 3200 kg/h, feed water temperature, t1 = 45°C, p = 12 bar,

Tsup = 274.5°C, C.V. = 32600 kJ/kg, cps = 2.1 kJ/kg.K

From steam table A.1.2 in the Appendix, at 12 bar, we have

Ts = 187.99°C, hg = 2784.2 kJ/kg

From steam table A.1.3 in the Appendix, using linear interpolation, enthalpy of superheated steam at exit of boiler (12 bar and 274.5°C) is found to be

hsup = 2989.29 kJ/kg

From steam table A.1.1 in the Appendix, enthalpy of water at 45°C,

hf1 = 188.42 kJ/kg

  1. Equivalent evaporation:
  2. Boiler efficiency:

Example 2.19

During a trial on an oil-fired smoke tube boiler for 1 h, the following data were recorded:

Steam pressure, p = 14 bar

Amount of water evaporated, m1= 5400 kg

Condition of steam, x = 0.92 dryness fraction

Amount of fuel burnt, mf = 540 kg

Calorific value of fuel used, C.V. = 42000 kJ/kg

Temperature of steam leaving the superheater, Tsup = 250°C

Temperature of feed water, t1 = 50°C

Determine the equivalent evaporation from and at 100°C with and without superheater, boiler efficiency, and the percentage of heat utilised in the superheater.

Solution

Given that p = 14 bar, m1 = 5400 kg, x = 0.92, mf = 540 kg, C.V. = 42000 kJ/kg, Tsup = 250°C, and t1 = 50°C

Amount of steam generated per kg of fuel,

Enthalpy of wet steam at 15 bar, 0.92 dry,

h = hf + x.hfg = 845.2 + (0.92 × 1947.2) = 2636.6 kJ/kg

From steam table A.1.2 in the Appendix, enthalpy of wet steam at 14 bar, 0.92 dry,

h = hf + x.hfg = 830.29 + (0.92 × 1959.7) = 2633.2 kJ/kg

From steam table A.1.3 in the Appendix, Enthalpy of superheated steam at 14 bar, 250°C,

hsup = 2927.2 kJ/kg

Equivalent evaporation without superheated from and at 100°C.

Equivalent evaporation with superheater from and at 100°C.

Boiler efficiency,

Percentage of heat utilized in the superheater,

Example 2.20

Following particulars refer to a steam power plant consisting of a boiler, superheater, and an economiser:

Steam pressure, p = 20 bar

Mass of steam generated, m1 = 10000 kg/h

Mass of coal used, mf = 1300 kg/h

Calorific value of coal, C.V. = 29000 kJ/kg

Temperature of feed water entering the economiser, t1 = 35°C

Temperature of feed water leaving the economiser, t2 = 105°C

Temperature of superheated steam leaving the superheater, tsup = 350°C

Dryness fraction of steam leaving the boiler, x = 0.98

Determine the following:

  1. Overall efficiency of the plant
  2. Equivalent evaporation from and at 100°C
  3. The percentage of heat utilised in boiler, superheater, and economiser

Solution

Given that p = 20 bar, m1 = 10000 kg/h, mf = 1300 kg/h, C.V. = 29000 kJ/kg, t1 = 35°C, t2 = 105°C, tsup = 350°C, x = 0.98

From steam table A.1.1 in the Appendix, enthalpy of feed water entering the economiser,

hf1 = 146.66 kJ/kg

Enthalpy of feed water leaving the economiser at 105°C,

hf2 = 440.13 kJ/kg

From steam table A.1.2 in the Appendix, enthalpy of steam leaving the boiler,

h = hf + x.hfg = 908.77 + 0.98 × 1890.7 (at 20 bar, 0.98 dry)
= 2761.66 kJ/kg

From steam table A.1.3 in the Appendix, enthalpy of steam leaving the superheater,

hsup = 3137.0 kJ/kg (at 20 bar, 350°C)

Mass of steam generated per kg of fuel,

  1. Overall efficiency of the plant:
  2. Equivalent evaporation from and at 100°C:
  3. Percentage of heat utilised in the boiler:

    Heat supplied in 1 kg of fuel = 29000 kJ

    Percentage of heat utilised in superheater:

    Percentage of heat utilised in economiser:

Note: Superheater is considered as a part of the boiler for calculation of equivalent evaporation.

Example 2.21

The following readings were recorded during a boiler trial of 6 h duration:

Mean steam pressure = 12 bar

Mass of steam generated = 40000 kg

Mean dryness fraction of steam = 0.85

Mean feed water temperature = 30°C

Coal used = 4000 kg

Calorific value of coal = 33400 kJ/kg

Calculate the following:

  1. Factor of equivalent evaporation
  2. Heat rate of boiler in kJ/hr
  3. Equivalent evaporation from and at 100°C
  4. Efficiency of the boiler

Solution

Given that p = 12 bar, m = 40000 kg, x = 0.85, t1 = 30°C, mf = 4000 kg, C.V. = 33400 kJ/kg, and duration of trial, T = 6 h

From steam table A.1.2 in the Appendix at 12 barhf = 798.64 kJ/kg, hfg = 1986.2 kJ/kg

∴ Enthalpy of steam, h = hf + x.hfg = 798.64 + 0.85 × 1986.2 = 2486.91 kJ/kg

From steam table A.1.1 in the Appendix, enthalpy of feed water at 30°C, hf 1 = 125.77 kJ/kg

  1. Factor of equivalent evaporation:
  2. Heat rate of boiler:
  3. Equivalent evaporation from and at 100°C:
  4. Boiler efficiency:

Example 2.22

During a trial for 8 h on a boiler, the following data were obtained:

Pressure of steam leaving the boiler = 14 bar

Condition of steam = 0.97 dryness fraction

Steam produced = 26700 kg

Temperature of feed water = 50°C

Mass of coal fired = 4260 kg

Calorific value of coal = 28900 kJ/kg

Air fuel ratio = 17

Temperature of flue gases leaving the boiler = 344°C

Boiler house temperature = 21°C

Specific heat of flue gases = 1.1 kJ/kg.K

Determine the following:

  1. Boiler efficiency
  2. Equivalent evaporation
  3. Heat lost to flue gases expressed as percentage of heat supplied

Solution

Given that p = 14 bar, x = 0.97, m = 26700 kg, t1 = 50°C, mf = 4260 kg, C.V. = 28900 kJ/kg, air−fuel ratio, = 17, tf = 344°C, t0 = 21°C, Cpg = 1.1 kJ/kg.K, and duration of boiler trial, T = 8 h

  1. From steam table A.1.2 in the Appendix, at 14 bar, hf = 830.29 kJ/kg and hfg = 1959.7 kJ/kg

    Enthalpy of steam, h = (hf + x.hfg) at 14 bar = 830.29 + 0.97 × 1959.7 = 2731.2 kJ/kg

    From steam table A.1.1 in the Appendix, enthalpy of water at 50°C, hf 1 = 209.31 kJ/kg

    ∴ Efficiency of boiler,

  2. Equivalent evaporation:
  3. Fuel burnt/hour, 532.5 kg/h

    Heat supplied in fuel = f × C.V. = 532.5 × 28900 = 15389250.0 kJ/h

    Mass of flue gases formed/hour,

    Heat carried away by flue gases,

    = g × cpg (tft0) = 9585 × 1.1 (344 − 21) = 3405550.5 kJ/h

    Heat loss of flue gases as per cent of heat supplied

Example 2.23

A gas-fired boiler operates at a pressure of 100 bar. The feed water temperature is 255°C. Steam is produced with a dryness fraction of 0.9, and in this condition, it enters a superheater. Superheated steam leaves the superheater at a temperature of 450°C. The boiler generates 1200 tonne of steam per hour with an efficiency of 92%. The gas used has a calorific value of 38 mJ/m3. Determine the following:

  1. The heat transfer per hour in producing wet steam in the boiler
  2. The heat transfer per hour in producing superheated steam in the superheater
  3. The gas used in m3/hour

Solution

Given that p = 100 bar, t1 = 256°C, x = 0.9, tsup = 450°C, m = 1200 tonne = 1200 × 103 kg

η = 0.92, C.V. = 38 MJ/m3 = 38 × 103 kJ/m3
  1. From table A.1.1 in the Appendix, enthalpy of feed water: hf 1 = 1109.72 kJ/kg

    From table A.1.2 in the Appendix, enthalpy of steam at 100 bar, 0.9 dry

    h = hf + x.hfg = 1407.53 + 0.9 × 1317.1 = 2592.92 kJ/kg

    From table A.1.3 in the Appendix, Enthalpy of superheated steam at 100 bar, 450°C

    hsup = 3240.8 kJ/kg

    ∴ Heat transfer/h producing wet steam in boiler

    = m(hhf1) = 1200 × 103 (2592.92 − 1109.72)
    = 17.7984 × 108 kJ/h
  2. Heat transfer/hour in producing superheated steam in superheater:
    = m(hsuph) = 1200 × 103 (3240.8 − 2592.92)
    = 77.7456 × 107 kJ/h
  3. Let the gas used in m3/hour be Vf.
    Vf = 73149.2 m3/h

Summary for Quick Revision

  1. A pure substance may be defined as a system which is homogeneous in composition and chemical aggregation and invariable in chemical aggregation.
  2. Properties of steam are as follows:
    1. Latent heat of steam, hfg = Quantity of heat required to convert 1 kg of liquid water at its boiling point into dry saturated steam at the same pressure.
    2. Dryness fraction, x =
      Quality of steam = x × 100
      Wetness fraction = 1 − x
    3. Enthalpy of wet steam, hwet = hf + x hfg
    4. Density of wet steam = kg/m3

      Density of dry saturated steam,

      Volume of wet steam, vwet = (1 − x)vf + xvg = vf + xvfg

    5. Internal energy of wet steam, uwet = u f + xu fg
    6. Entropy of wet steam, swet = s f + xs fg
  3. Steam processes are as follows:
    1. Constant volume process,
      v1 = v2; q = u2u1
    2. Constant pressure process,
      w = p (p2p1)

      q = (x2x1) hfg, for wet steam

      = hsuphg, for superheated steam.

    3. Isothermal process,

      In the wet region, q = h2h1 and w = p1 (v2v1)

    4. Hyperbolic process,
      p1v1 = p2v2
    5. Isentropic process,
      s1 = s2
    6. Polytropic process,

      n = 1.13 for wet steam

      = 1.3 for superheated steam

      p1(v1)n = p2(v2)n
    7. Throttling process,
      h1 = h2
      h1 = hf1 + x1 hfg1 for wet steam
      h2 = hf2 + x2 hfg2 for wet steam
      = hg + cps (TsupTs) for superheated steam.
  4. Determination of dryness fraction
    1. Barrel calorimeter,

      It gives approximate results which are always lower than the actual value.

    2. Separating calorimeter,

      The value of x calculated is always greater than the actual.

    3. Throttling calorimeter,

      It is suitable for steam with high dryness fraction.

    4. Combined separating and throttling calorimeter.

      It is suitable for considerably wet steam.

Multiple-choice Questions

  1. The dryness fraction of 1 kg of steam containing 0.8 kg of dry steam is
    1. 0.2
    2. 0.8
    3. 1.0
    4. 0.4
  2. If x1 and x2 be the dryness fractions of steam obtained in the separating and throttling calorimeters, respectively, then the dryness fraction is
    1. x1 + x2
    2. x1x2
    3. x1x2
  3. Superheating of steam is done at constant
    1. volume
    2. pressure
    3. temperature
    4. enthalpy
  4. With the increase of pressure, the enthalpy of evaporation of water
    1. decreases
    2. increases
    3. remains same
    4. changes randomly
  5. Only throttling calorimeter is used for measuring
    1. very low dryness fraction up to 0.7
    2. very high dryness fraction up to 0.98
    3. dryness fraction of only low pressure steam
    4. dryness fraction of only high pressure steam
  6. Constant pressure lines in the superheated region of the Mollier diagram have
    1. a positive slope
    2. a negative slope
    3. zero slope
    4. both positive and negative slope
  7. Specific volume of wet steam with dryness fraction x is
    1. xvf
    2. vf + x vfg
    3. x2vg
    4. x3vg
  8. Entropy of wet steam is given by
    1. sf + x sfg
    2. x sg
    3. sg + x sfg
    4. x sf
  9. The phase change at constant pressure (or constant temperature) from liquid to vapour is referred to as
    1. melting
    2. solidification
    3. sublimation
    4. vapourization
  10. The point that connects the saturated liquid line to the saturated vapour line is called the
    1. triple point
    2. critical point
    3. superheated point
    4. saturated point

Review Questions

  1. Define a pure substance.
  2. Define latent heat of fusion and latent heat of vapourisation.
  3. Define dryness fraction and wetness of steam.
  4. What is quality of steam?
  5. Write expressions for internal energy of steam when it is wet and when it is superheated.
  6. Write expressions for entropy of steam for wet and superheated.
  7. What is a Mollier diagram?
  8. When steam is heated at constant volume and its end condition is still wet, what is heat transfer?
  9. In the wet region, constant temperature process is also a constant pressure process. Say true or false.
  10. Hyperbolic process is also an isothermal process in the superheat region. Say true or false.
  11. What remains constant during isentropic process?
  12. What is throttling of steam?
  13. What are the drawbacks of barrel calorimeter?
  14. Which calorimeter is used for a very wet steam?

Exercises

2.1 A vessel of 1.35 m3 capacity is filled with steam at 13 bar absolute and 95% dry. The vessel and its contents cool until the pressure is 2 bar absolute. Calculate the mass of the contents in the vessel and the dryness fraction of steam after cooling. Neglect the volume of water present.

[Ans. 9.2 kg, 0.1625]

2.2 Steam at a pressure of 5 bar and temperature 200°C is expanded adiabatically to a pressure of 0.7 bar absolute. Determine the final condition of steam. For superheated steam, cps = 2.2 kJ/kg.K.

[Ans. 0.93]

2.3 Steam at a pressure of 20 bar absolute and dryness fraction 0.8 is throttled to a pressure of 0.5 bar. Determine the final condition of steam.

[Ans. 0.905]

2.4 About 2 kg of wet steam at 10 bar and 90% dry is expanded according to the law pv = constant to a pressure of 1 bar. Determine the final condition of steam and the change in internal energy.

[Ans. Superheated to 112°C, 257 kJ]

2.5 While conducting the dryness fraction test with a throttling calorimeter, it was found that the entering steam was at a pressure of 12 bar and a sample after being reduced to 1 bar in the calorimeter was at a temperature of 120°C. Estimate the dryness fraction of steam assuming cps = 2.1 kJ/kg.K.

[Ans. 0.965]

2.6 In a test with a separating and throttling calorimeter, the following observations were made:

Water separated = 2.04 kg

Steam discharged from throttling calorimeter = 20.6 kg at 150°C

Initial pressure of steam = 12 bar absolute

Final pressure of steam = 12.3 cm of mercury above atmospheric pressure

Barometer = 76 cm of mercury

Calculate the dryness fraction of steam entering the calorimeter.

[Ans. 0.905]

2.7 Determine the change in internal energy when 1 kg of steam expands from 10 bar and 300°C to 0.5 bar and 0.9 dry. Take cps = 2.1 kJ/kg.K.

[Ans. −669 kJ/kg]

2.8 One kg of steam at 10 bar exists at 200°C. Calculate the enthalpy, specific volume, density, internal energy, and entropy. Take cps = 2.1 kJ/kg.K.

[Ans. 2818 kJ/kg, 0.203 m3/kg, 4.93 kg/m3, 2615 kJ/kg, 6.673 kJ/kg.K]

2.9 A vessel contains one kg of steam which contains one-third liquid and two-third vapour by volume. The temperature of steam is 151.86°C. Calculate the quality, specific volume, and specific enthalpy of the mixture.

[Ans. 0.0576, 0.003245 m3/kg, 650.5 kJ/kg]

2.10 A cylinder fitted with a piston contains 0.5 kg of steam at 4 bar. The initial volume of steam is 0.1 m3. Heat is transferred to steam at constant pressure until the temperature becomes 300°C. Determine the heat transferred and work done during the process.

[Ans. 771 kJ, 91 kJ]

2.11 Steam at 10 bar and 0.9 dry initially occupies 0.35 m3. It is expanded according to the law pv1.25 = constant until the pressure falls to 2 bar. Determine the mass of steam used in the process, the work done, the change in internal energy, and the heat exchange between steam and surroundings.

[Ans. 2 kg, 192 kJ, −896 kJ, −704 kJ]

2.12 One kg of steam at 8.5 bar and 0.95 dry expands adiabatically to a pressure of 1.5 bar. The law of expansion is pv1.2 = constant. Determine the final dryness fraction of steam and the change in internal energy during the expansion.

[Ans. 0.792, −230 kJ/kg]

2.13 A throttling calorimeter is used to measure the dryness fraction of steam in the steam main where the steam is flowing at a pressure of 6 bar. The steam after passing through the calorimeter comes out at 100 kPa pressure and 120°C temperature. Calculate the dryness fraction of steam in the main. Assume cps = 2.09 kJ/kg.K.

[Ans. 0.985]

2.14 A separating and throttling calorimeter was used to determine the dryness fraction of steam flowing through a steam main at 900 kPa. The pressure and temperature after throttling were 105 kPa and 115°C, respectively. The mass of steam condensed after throttling was 1.8 kg and mass of water collected in the separating calorimeter was 0.16 kg. Determine the dryness fraction of steam flowing through the steam main. Take cps = 2.09 kJ/kg.K.

[Ans. 0.89]

ANSWERS TO MULTIPLE-CHOICE QUESTIONS
  1. b
  2. b
  3. b
  4. a
  5. b
  6. a
  7. b
  8. a
  9. a
  10. a