Question

Give 2 examples of how to determine a thermodynamic properties (H, S, G, etc) of a substance by using the thermodynamic diagrams (PH Diagram, TS Diagram and Mollier Diagram).

Give 2 examples of how to determine a thermodynamic properties (H, S, G, etc) of a substance by using the tables of thermodynamic properties.

Each example must be different from each other

Answer 1

MOLLIER CHART

EXAMPLE 1 :

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

Solution:

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

EXAMPLE 2 : Determine if steam at 450°C and 1 bar is saturated or superheated. Find the enthalpy and entropy of this steam.

SOLUTION :

See the Mollier diagram in Figure below

Identify the point of intersection of the 450°C line (or 450°C isotherm) and the constant pressure line (or isobar) of 1 bar. This point of intersection of the two lines is labeled A. Mollier diagram is the superheated steam region.

Therefore, the steam at 450°C and 1 bar is superheated.

Enthalpy Determination: To determine the enthalpy at point A, draw a straight horizontal line from point A to the left till it intersects with the diagonal enthalpy line. This horizontal line intersects the enthalpy line at an enthalpy value of, approximately, 3380 kJ/kg.

Therefore, h(A), or enthalpy at point A, is 3380 kJ/kg.

Entropy Determination: To determine the entropy at point A, draw a straight vertical line from point A to the bottom, until it intersects with the entropy line. The vertical line intersects the entropy line at, approximately, 8.7kJ/kg.°K.

Therefore, s(A), or entropy at point A, is 8.7kJ/kg.°K.

STEAM TABLES :

EXAMPLE 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

EXAMPLE 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