LESSON - 20 GENERATION OF STEAM AND INTERNAL ENERGY AND ENTROPY (EXTENSIVE PROPERTIES) OF STEAM

 

20.1. GENERATION OF STEAM

During generation of steam, the water in practice has to be pumped to the boiler at constant pressure and after transformation at constant pressure to dry or superheated steam; it has to be used in the primer mover (Turbine). During this steam generation process, the water may acquire compressed (subcooled) liquid state, saturated liquid state, wet steam state, saturated or dry vapor state, and super-heated vapor state. These states are discussed here in brief. 

The Subscripts used to express different states of water from liquid to vapor phase change process are as follows:

0      :    for zero °C

sub   :   for subcooled liquid

f        :   for saturated liquid

g       :   for saturated vapour

fg    :     for evaporation indicates difference in the property of saturated vapor and saturated liquid.

sup   :   for superheated vapor

 

                         Fig. 20.1.: Saturated, Wet, Superheat and Subcooling states

 20.1.1.  Compressed or Subcooled Liquid:

If the temperature of a liquid is less than the saturated temperature at the given pressure (P) the liquid is said to be compressed (subcooled) liquid. The difference between the saturation temperature t_s and actual subcooled liquid temperature t_sub i.e. (t_s-t_sub) is termed as the degree of subcooling (Fig. 20.1). Subcooled and compressed liquid states are referred with respect to either saturation temperature or saturation pressure. Say for example, let us consider liquid water at room temperature, say 30°C at atmospheric pressure. For 1 atmospheric pressure the corresponding saturation temperature is 100°C. However, the water exists at a temperature less than the saturation temperature corresponding to the atmospheric pressure. As we refer the temperature as the reference here, it is appropriate to say the water is in the subcooled state. Now let us talk about the same water with respect to its pressure. At 30°C, the corresponding saturation pressure will be definitely less than 1 atmospheric pressure, however, the water exists at 1 atmospheric pressure and hence we can state now that the water exists at a saturation pressure higher than that corresponding to its temperature 30°C, hence it is in compressed liquid state.

20.1.1.1.    Extensive properties in the Subcooled liquid region

In the subcooled liquid region, the properties can be calculated by the following relations:

Specific volume (v_sub),             

Let specific volume occupied by 1 kg of subcooled liquid at temperature t_Sub = v_sub (can be measured)

Specific Enthalpy (h_sub)

h_sub -  h₀ = C_P,sub (t_sub – t₀); 

h_sub =  h₀ + C_P,sub (t_sub – t₀);    

where  C_P,sub  is  specific heat of subcooled liquid at constant pressure.

             h₀ = 0  because 0°C is arbitrarily assigned as datum

                        (the value of zero enthalpy)

Specific Internal Energy (u_sub)

u_sub  = h_sub – p. v_sub

Specific Entropy (s_sub)

s_sub - s₀ = C_p,sub . ln 

s_sub = s₀ + C_p,sub . ln              

            s₀ = 0  because 0°C is arbitrarily assigned as datum (the value of zero entropy)

20.1.2.  Saturated liquid

The liquid at saturation temperature (T_f =T_s) at the given pressure (P) is termed as saturated liquid (Fig. 20.1).

20.1.2.1.   Extensive properties at saturated liquid

       At saturated liquid point, the properties can be calculated by the following relations:

 Specific volume (v_f),        

Let specific volume occupied by 1 kg of saturated liquid at temperature t_s = v_sub    (measured)

 Specific Enthalpy (h_f)

h_f -  h_o = C_P,f  (t_f – t_o);     

h_f =  h_o + C_P,f  (t_f – t_o); 

where  C_P,f   is  specific heat of liquid at constant pressure.

h_o = 0  because 0°C is arbitrarily assigned as datum (the value of zero enthalpy)

 Specific Internal Energy (u_f)

u_f  = h_f – p. v_f

Specific Entropy (s_f)

s_f – s_o = C_p,f . ln      

s_f  = s_o + C_p,f . ln 

       s_o = 0  because 0°C is arbitrarily assigned as datum  the value of zero entropy)

20.1.3.  Wet steam (Two phase mixture)

Referring Figure 20.1, when the substance lies between the saturated liquid and saturated vapor point, at the given pressure (P) it is termed as wet steam. The wet steam is the mixture of dry steam and suspended water particles. The steam ordinarily generated in any boiling vessel is wet steam. By applying heat to wet steam, it becomes dry and saturated as all the suspended water particles evaporate.

The relative amount of each phase present in wet steam (i.e. in two phase mixture) determines the quality or dryness fraction of steam (x).

The dryness / quality of wet steam (an equilibrium liquid-vapor mixture), x , is defined as the mass fraction of vapor in the mixture.

                  x =  

  if     m_g = mass of vapor per kg of mixture,

          m_f = mass of suspended water particles per kg of mixture,

 Then the   dryness / quality of wet steam,  x = 

The above equation shows that the dryness fraction of saturated liquid is zero or 0% because vapor is absent in saturated liquid state, whereas, the dryness fraction of saturated vapor is unity or 100% because suspended water particles are absent in saturated vapor state. This shows that the dryness fraction between saturated liquid and saturated vapor state varies between 0 and 1. Say if the dryness fraction of steam is 0.9, it means that in one kg of wet steam 0.9 kg is the dry steam and 0.1 kg is the suspended water particles.

The quality of steam ‘x’ is regarded as the intensive property of steam.

20.1.3.1.   Curve of constant quality In Fig. 20.2, Say with reference to some constant pressure p1 and saturation temperature t1:    Point ‘G’ on saturated vapor line denotes saturated vapor or 100 per cent vapor and 0 per cent liquid and  Point ‘F’ on saturated liquid line denotes 0 per cent vapor and 100 per cent liquid.  Point ‘Q’ between ‘F’ and ‘G’, represent some quality x1 of the wet steam.

Fig. 20.2. Curves of constant quality

Now say on some other constant pressure p₂ and saturation temperature t₂:

• Point ‘g’ on saturated vapor line denotes saturated vapor or 100 per cent vapor and 0 per cent liquid and

• Point ‘f’ on saturated liquid line denotes 0 per cent vapor and 100 per cent liquid at the same temperature.

• Find point ‘q’ between ‘f’ and ‘g’ where the quality of wet steam is same, x_1, as it was for point ‘Q’ with respect to points ‘G’ and ‘F’.

Similarly, find same quality points (x₁) on other pressures. Then the line joining ‘Q’, ‘q’ and other same quality points is called the curve of constant quality corresponding to x₁.

Similarly we can draw other curves of constant quality (say x_2, x₃, x₄ and so on) on p-V diagram.

20.1.3.2.  Extensive properties of wet steam i.e. in the liquid+vapor region

In the liquid+vapor region, the properties can be calculated by the following relations:

Specific volume (v),

V = V_g +V_f

      where,  V is total volume of the chamber having wet steam (mixture of steam and water) and    V_g and V_f  are volume of steam and water in the chamber respectively.

m. v = m_{g .}v_g + m_{f .}v_f

  where v is specific volume of chamber having wet steam (mixture of steam and water)

                        v_g and v_f are specific volume of steam and water in the chamber respectively.

         where     v_g =  = V  m³/kg                 (can be measured)

                         m is total mass of wet steam (mixture of steam and water)

                        m_g  and  m_f  are mass of steam and water in the chamber respectively.

or        

or               

or        

or        

or         

or                           

or        

or          v =  v_f + x .v_fg                                      

                     where    v_fg = v_g – v_f 

For substances such as water, at pressures far below the critical point, the above specific-volume equations may often be simplified to

v = x.v_g

because v_f is very small in comparison to v_g. This is not of course permissible when x is very small.

Specific Enthalpy (h)

    h = x.h_g + (1− x).h_f  = h_f + x. h_fg           

                where    h_fg (Latent heat of evaporation) = (h_g – h_f)

Specific Internal Energy (u)

    u  = h – p. v

Specific Entropy (s)

    s = x.s_g + (1 − x).s_f   = s_f + x .s_fg              

                           where    s_fg = s_g – s_f =            (as q = Tds)

20.1.4.  Dry or Saturated vapor

When whole mass of liquid is converted into vapor at saturation temperature (T_s) at the given pressure (p), it is termed as saturated vapor (Fig. 20.1).

 20.1.4.1.     Extensive properties at saturated vapor

At saturated vapor point, the properties can be calculated by the following relations:

Specific volume (v_g)

v_g = v_f  + s_fg

Specific Enthalpy (h_g)

h_g = h_f  + h_fg

Specific Internal Energy (u_g)

           u_g = u_f  + u_fg

or        u_g  = h_g – p. v_g

Specific Entropy (s_g)

s_g = s_f  + s_fg

20.1.5.  Super-heated vapour state

 If the temperature of the vapor is greater than the saturation temperature (T_s) corresponding to the given pressure (p), the vapor is said to be in superheated state (Fig. 20.1). The difference between the temperature of superheat vapor t_sup and the saturation temperature ( t_s) at that pressure is called the degree of superheat.

Thus, degree of superheat = (t_sup − t_s)

20.1.5.1.    Extensive properties in the vapour region

In the superheated region, the properties can be calculated by the following relations:

Specific volume (v_sup),

Approximate volumes of superheated vapor may be determined assuming that it behaves like a perfect gas from the dry saturation point.

  

Specific Enthalpy (h_sup)

h_sup -  h_g = C_P,g  (t_sup – t_g);      or      h_sup -  h_g = C_P,g  (t_sup – t_s); 

h_sup =  h_g + C_P,g  (t_sup – t_s); 

where  C_P,g  is specific heat of vapor at constant pressure

Specific Internal Energy (u_sup)

u_sup  = h_sup – p. v_sup

Specific Entropy (s_sup)

s_sup – s_g = C_p,g . ln             or        s_sup – s_g = C_p,g. ln 

s_sup= s_g + C_p,g. ln 

20.1.5.2.   Curves of constant superheating.

  Refer Fig. 20.3.

   Say with reference to some constant pressure p₁ and its saturation temperature t_sat,1: 

• Point ‘G’ on saturated vapor line.

• Say the saturated vapor at point ‘G’ is superheated at its constant pressure ‘p₁’ to point ‘S’ where superheating temperature is ‘t_sup,S’. For point ‘S’ say degrees of superheating is (t_sup,S – t_sat,1) = t₁°C.

Now say on some other constant pressure p₂ and its saturation temperature t_sat,2: 

• Say Point ‘g’ is on saturated vapor line.

• Now heat the saturated vapor at point ‘g’ at its constant pressure ‘p₂’ until point ‘s’ ,so that the degree of superheating (t_sup,s– t_sat,2) at point ‘s’ is same as it was for point ‘S’ for pressure ‘p₁’ i.e. t₁°C.        

Similarly, on other pressures, find same t₁°C degree of superheating points. Then the line joining ‘S’, ‘s’ and other same degree of superheating points is called the curve of constant superheating corresponding to t₁°C.

Similarly we can draw other constant superheating curves (say t_2, t₃, t₄ and so on) for different degree of superheating on p-V diagram.

Fig. 20.3. Curves of constant quality and curves of constant superheat.

Problem 20.1: Find the dryness fraction of steam if 2 kg of water is in suspension with 90 kg of dry and saturated steam.

Solution:

Given: Mass of dry and saturated steam = m_g = 90 kg

              Mass of water in suspension = m_f = 2 kg

 Determine dryness fraction, x:

            Formula: dryness fraction 

Answer:      x =  = 0.9782