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 = dh − vdp

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 h−s diagram. The saturated liquid line and saturated vapour line meet at critical point.

Figure 2.3 Mollier diagram for steam

VARIOUS PROCESSES FOR STEAM

1 Constant Volume Process

The constant volume heating process is shown on the p−v and T−s 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 = v_f 1 + x₁(v_g1 − v_f1)

Final volume of steam in wet region v₂ = v_f2 + x₂(v_g2 − v_f2)

Since the volume remains constant,

 

v_f1 + x₁(v_g1 − v_f1) = v_f2 + x₂(v_g2 − v_f2)

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 = u_f1 + x₁ u_fg1

Final internal energy of steam in the wet region, u₂ = u_f2 + x₂ u_fg2

Since dv = 0, therefore δW = 0

For constant volume heating process, v₂ = v_3. 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 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(v₂ − v₁)

where v₁ = v_f + x₁(v_g − v_f) and v₂ = v_f + x₂(v_g − v_f)

The specific enthalpy of steam, h₁ = h_f + x₁ h_fg and h₂ = h_f + x₂ h_fg

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

 

q = h₂ − h₁

q = (h_f + x₂h_fg) − (h_f + x₁h_fg)

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(v_sup − v₂)

and heat transfer per kg of steam

Figure 2.5 Constant pressure expansion

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 = h₂ − h₁ and w = p₁ (v₂ − v₁)

In the superheated region,

 

q = h₃ − h₂ and w = p₁ (v₃ − v₂)

Figure 2.6 Isothermal process

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 = v_f1 + x₁(v_g1 − v_f1)

Final volume of steam v₂ = v_f2 + x₂(v_g2 − v_f2)

As pv = const., therefore

or 

Figure 2.7 Hyperbolic expansion

Now, h₁ = h_f 1 + x₁h_fg1 and h₂ = h_f 2 + x₂h_fg2 for wet steam

and h_s2 = h_g2 + c_p (T_sup − T₂) 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 + (u₂ − u₁)

5 Reversible Adiabatic or Isentropic Process

The isentropic process of expansion (1-2) is shown in Fig. 2.8 on T−s 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 = s_g1

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

s₂ = s _f2 + x₂s _fg2

Then, s_g1 = s _f2 + x₂s _fg2

For initially superheated steam at temperature T_sup, 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 + (u₂ − u₁)

where u₁ = u _f1 + x₁u _fg1 for wet steam

= u_g1 for dry steam

= u_sup for superheated steam

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

But q = 0

w = u₁ − u₂

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 = v_f1 + x₁(v_g1 − v_f1)

Final volume of steam v₂ = v_f2 + x₂(v_g2 − v_f2 )

or

or 

Also u₁ = u_f1 + x₁u_fg1

u₂ = u_f2 + x₂u_fg2

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).

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 T−s and h−s diagrams in Fig. 2.9.

 

h₁ = h_f1 + x₁h_fg1

h₂ = h_f2 + x₂h_fg2 for wet steam after expansion

= h_f2 + h_fg2 + c_ps (T_sup − T_s), for superheated steam after expansion

Now h₁ = h₂

For wet steam after expansion

For superheated steam after expansion

Figure 2.9 Throttling process: (a) T−s diagram, (b) h−s diagram