LESSON - 21 USE OF STEAM TABLES AND MOLLIER CHART AND NUMERICAL PROBLEMS

 21.1. STEAM TABLES AND MOLLIER CHART

It is advisable to use Steam-tables or Mollier chart directly for easy and accurate estimation of the properties of steam, such as pressure, temperature, specific volume, enthalpy, entropy etc., which we have calculated by different relations for water/steam in different states under steam  generation section. The properties in Steam tables have been determined experimentally.

In the following discussion, the Steam tables and Mollier chart are used to demonstrate the use of steam tables and Mollier chart to determine the properties of steam in different states.

21.1.1. Steam Tables

21.1.1.1. Extensive properties at saturated liquid and saturated vapor state

In steam tables, extensive properties at saturated liquid and at saturated vapor for 1 kg of liquid/vapor are given as shown in Table 21.1(a) and Table 21.1(b). In Table 21.1(a) these properties are listed with reference to saturation temperature and in Table 21.1(b) with reference to saturation pressure. Therefore, it is more convenient to use Table 21.1(a) when temperature is given and Table 21.1(b) when pressure is given. In both the tables, the values of specific volume (v_f), enthalpy (h_f), and entropy (s_f) of water in saturated liquid state and values of specific volume (v_g), enthalpy (h_g), and entropy (s_g) of steam in saturated vapor state are directly noted down. The values of internal energy (u_f) of water in saturated liquid state and values of internal energy (u_g) of steam in saturated vapor state are calculated by using following relations.

u_f = h_f - pv_f       and    u_g = h_g – pv_g      

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

For wet steam, the values of specific volume (v), internal energy (u), enthalpy (h), and entropy (s) are calculated with the following relations.

      v = x.v_g + (1 -x).v_f  

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 specific-volume (v) 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.

h = x.h_g + (1 -x).h_f  = h_f + x. h_fg          where    h_fg (Latent heat of evaporation) = (h_g – h_f)

u  = h – p. v

s = x.s_g + (1 -x).s_f   = s_f + x .s_fg                    where    s_fg = s_g – s_f

In the above equations, all the properties of water in saturated liquid state and steam in saturated vapor state are found as discussed in the previous section of “Extensive properties at saturated liquid and saturated vapor state”.

21.1.1.3. Extensive properties in superheated vapor state (vapor region)

The properties of superheated steam are given as shown in Table 21.2 (a, b, c) separately. They depend not only on pressure but also on the superheating temperature T_sup. The values of specific volume of superheated steam (v_sup), enthalpy of superheated steam (h_sup), and entropy of superheated steam (s_sup) are directly noted down from Table 21.2 (a), Table 21.2 (b) and Table 21.2 (c), respectively. The values of internal energy of superheated steam (u_sup) is calculated by using the following relation.

     u_sup = h_sup – pv_sup      

Table 21.1 (a). Steam tables for saturated water and steam (temperature)

 Table 21.1 (b). Steam tables for saturated water and steam (pressure)

 Table 21.2 (a). Steam tables for specific volume of superheated steam 

 Table 21.2 (b). Steam tables for enthalpy of superheated steam  

 Table 21.2 (c). Steam tables for entropy of superheated steam  

21.1.2. Mollier or Enthalpy-Entropy (h-s) diagram

The Mollier diagram is a is plot of enthalpy (h) versus entropy (s) as shown in Fig. 21.1. It is also known as the h-s diagram. This diagram has a series of constant temperature lines, constant pressure lines, constant quality lines, and constant volume lines. The Mollier diagram is used only when quality is greater than 50% and for superheated steam. For any state, at least two properties should be known to determine the other unknown properties of steam at that state.

The commercially available Mollier diagram is truncated from a point beyond the critical point i.e. it shows only a portion of this diagram which is drawn in colors. In such truncated diagram property of liquid cannot be read.

                Fig. 21.1. Mollier or Enthalpy-Entropy (h-s) diagram

 Problem 21.1: Find the specific volume, enthalpy and internal energy of wet steam at 18 bar with dryness fraction (x) = 0.85, by using Steam Tables and Mollier chart.

Solution:

Given: Pressure of steam, p= 18 bar;            Dryness fraction, x= 0.85

(a)   BY USING STEAM TABLES

By using steam tables (for dry saturated steam):

            From steam tables for dry saturated steam at 18 bar pressure, we have:

             t_s = 207.11°C,   h_f = 884.5 kJ/kg, h_g = 2794.8 kJ/kg, h_fg =1910.3 kJ/kg,  v_g=0.110 m³/kg.

(i) Determine specific volume of wet steam, v :

Formula: Specific volume of wet steam,v = x.v_g 

Answer:    v = x.v_g  =0.85 x 0.110 = 0.0935 m³/kg.

(ii)  Specific enthalpy of wet steam, h :

Formula: Specific enthalpy of wet steam, h = h_f + xh_fg

Answer:         h = h_f + xh_fg= 884.6 + 0.85 x 1910.3 = 2508.35 kJ/kg.

(iii)  Specific Internal energy of wet steam,

Formula: Specific internal energy of wet steam, u = h – pv

Answer:    u = h – pv= 2508.35 – 18 x 10² (0.0935) = 2340.75 kJ/kg

(b)   BY USING MOLLIER CHART

Locate point ‘1’ at an intersection of 18 bar pressure line and 0.85 dryness fraction line.

Read the value of enthalpy (h) and specific volume (v) from Mollier diagram corresponding to point ‘1’.

(i)    Specific enthalpy of wet steam,  h = 2508 KJ/kg

(ii)   Specific volume of wet steam,    v = 0.0935 m³/kg

(iii)  Specific Internal energy of wet steam, u

                 u  = h – pv

                      = 2508 – 18 x 10² (0.0935) = 2340  kJ/kg

Problem 21.2: Find the dryness fraction, specific volume and internal energy of steam at 7 bar and enthalpy of 2550 kJ/kg.

Solution:

Given: Pressure of steam, p=7 bar;      Enthalpy of steam, h = 2550 kJ

By using steam table (for dry saturated steam):

         From steam tables for dry saturated steam at 7 bar pressure, we have:

         t_s = 164.96°C, h_f = 697.1 kJ/kg,  h_g = 2762.0 kJ/kg, h_fg = 2064.9 kJ/kg,  v_g = 0.273 m³/kg.

(i)   Determine dryness fraction, x ;

Formula: For wet steam we have equation,

h = h_f + x h_fg

or   

       Answer:   Dryness fraction,    x =  = 0.897

 (ii) Determine specific volume of wet steam, v :

Formula:   v = x.v_g

      Answer:    v = x.v_g = 0.897 x (0.273) = 0.2449   m³/kg.

(iii) Determine specific internal energy of wet steam, u:

Formula:  u = h – pv

Answer:   u = h – p v= 2550 – 7 x 10² (0.2449) = 2379.67 kJ/kg

Note:- Also try above problem by using Mollier chart.

Problem 21.3:  Find the internal energy of 1 kg of steam at 20 bar when

(i)  It is superheated, its temperature being 400°C.

(ii) It is wet, its dryness being 0.9

Assume superheated steam to behave as a perfect gas from the commencement of superheat and thus obeys Charle’s law. Specific heat for steam = 2 kJ/kg K.      

Solution:

Given:    Mass of steam, m = 1 kg;        Pressure of steam, p = 20 bar

By using steam table (for dry saturated steam):

     From steam tables for dry saturated steam at 20 bar, we have:

     t_s = 212.4 °C; h_f = 908.6 kJ/kg;  h_g = 2797.2 kJ/kg;  h_fg=1888.6 kJ/kg;   v_g = 0.0995 m³/kg

(i)  Steam is superheated, its temperature being 400°C.

Given: Temperature of superheated steam, t_sup = 400°C or T_sup=  400+273 = 673 K

                  Specific heat of superheated steam, C_ps = 2.3 kJ/kg K.

Determine internal energy of 1 kg of superheated steam;

Formula: Internal energy of superheated steam, u_sup = h_sup – p.v_sup

Finding unknown, h_sup ;

h_sup = h_f + h_fg +C_ps (t_sup – t_s)

       = 908.6 +1888.6 +2.3 (400 – 212.4)   = 3228.68 kJ/kg

Finding unknown,v_sup ;

v_sup can be found out by Charle’s law

                      or  v_sup =    = 0.1379 m³/kg

Answer:    Internal energy, u_sup = h_sup- p .v_sup

                                        = 3228.68 – 20 x 10⁵ x 0.1379 x 10^-3 = 2952.88 KJ/kg

 (ii) Steam is wet, its dryness (x) being 0.9

Given: Dryness fraction, x = 0.9 ;     Pressure of steam, p = 20 bar

By using steam table (for dry saturated steam):

     From steam tables for dry saturated steam at 20 bar, we have:

     t_s = 212.4 °C; h_f = 908.6 kJ/kg;  h_g = 2797.2 kJ/kg;  h_fg=1888.6 kJ/kg;    v_g = 0.0995 m³/kg

 

Determine internal energy of 1 kg of wet steam

Formula: Internal energy of wet steam, u = h – p.v

                         Finding unknown, h  and  v;

                        For wet steam, we have

 h = h_f+ x. h_fg  = 908.6 + 0.9 x 1888.6 = 2608.34 KJ/kg

 v = x v_g= 0.9 x 0.0995 = 0.8955m³/kg

        Answer:    Internal energy, u = h - p .v

                                    = 2608.34– 20 x 10⁵ x 0.8955 x 10^-3 = 2429.24 KJ/kg

Note:- Also try this problem by using Mollier chart.

Problem 21.4: Steam at 120 bar has a specific volume of 0.01721 m³/kg, find the temperature, enthalpy and the internal energy.

Solution:

Given: Pressure of steam, p = 120 bar;       Specific volume, v = 0.01721 m³/kg

             By using steam table (for dry saturated steam):

                 From steam tables for dry saturated steam at 120 bar, we have:

                 t_s = 324.65°C  ; h_f = 1491.77 kJ/kg;  h_g = 2689.2 kJ/kg;  h_fg= 1197.4 kJ/kg;    v_g = 0.0143 m³/kg

Determine temperature, enthalpy and the internal energy of steam.

First it is necessary to find whether the steam is wet, dry saturated or superheated.

Since the actual specific volume of steam (0.01721 m³/kg) is more than the value of v_g (0.0143 m³/kg), the steam is superheated.

Therefore, v_sup = v = 0.01721 m³/kg.

By using steam table (for superheated steam):

       From steam tables for superheated steam at 120 bar and v_sup = 0.01721 m³/kg, we have:

        t_sup = 350°C; h_sup = 2847.7   kJ/kg

Answer:   Superheated steam temperature, t_sup = 350°C

                                                      Enthalpy,       h_sup = 2847.7   kJ/kg

                                                  Internal energy, u_sup = h_sup – p.v_sup

                                                                                    = 2847.7 – 120 x 10⁵ x10^-3 x 0.01721  = 2641.18  kJ/kg.

Note:- Also try this problem by using Mollier chart.