LESSON - 35 RANKINE CYCLE AND MODIFIED RANKINE CYCLE APPLIED TO STEAM ENGINE, INDICATOR DIAGRAMS, MEAN EFFECTIVE PRESSURE, AND DIAGRAM FACTOR

 35.1. MODIFIED RANKINE CYCLE APPLIED TO STEAM ENGINE

The ideal cycle on which the steam engine works is known as the Rankine cycle. The Rankine cycle is represented by 1-2-3-4-1 cycle on p-v and T-s diagrams in Fig. 35.1.

 Fig. 35.1. Rankine cycle and Modified Rankine cycle of steam  engine on p-v and T-s diagrams

Process  4-1: Admission of high pressure steam into the engine cylinder

  1-2: Isentropic expansion of steam in the engine cylinder

  2-3: Exhaust of steam in the condenser or atmosphere.

Thus the work done during Rankine cycle by engine is represented by the area 1-2-3-4-1

In practice it is not economical to expand the steam to the extreme toe of the pV diagram, which is represented by point 2 in above Figure. Because the work obtain during this final portion of the expansion stroke is extremely small and, in fact, it is not sufficient to overcome the friction of the moving parts of the engine. Therefore, the expansion stroke of the piston in Rankine cycle is terminated at 2'. The Rankine cycle with incomplete expansion of steam is called the incomplete expansion cycle or modified Rankine cycle. 

The modified Rankine cycle is represented by 1-2'-3'-3-4-1 cycle on p-v and T-s diagram in Fig. 35.1. In this cycle the expansion is stopped at a pressure called release pressure which is above the condenser (back) pressure and then it falls to the condenser (back) pressure at constant volume. 

Heat energy supplied by the steam in the cylinder

₃q₁ = q_A = Area under process 3-4-1 on T-s diagram = (h_g,1 - h_f,3)          

Work done during cycle by steam engine = ₁w_2' +  _2'w_3' 

                                                    = (h_g,1 – h_2')  +     

                                                                        = (h_g,1 – h_2')  + v_2' (p₂' – p₃) 

Therefore, Thermal efficiency   

35.2. INDICATOR DIAGRAMS OF STEAM ENGINE

Indicator diagram: A diagram which shows the variation of volume against the pressure of a steam inside the cylinder for one complete cycle of operation is called the pressure-volume (p-V) diagram. It is produced with the help of an instrument known as indicator. Since it is produced by indicator, this pressure-volume diagram is also called the indicator diagram.

Theoretical or Hypothetical Indicator diagram: In a steam engine, if it is assumed that the opening and closing of steam ports occur instantaneous and there is no pressure drop due to condensation or heat loss, no wire drawing due to restricted valve opening and no compression, the indicator diagram thus obtained will be known as theoretical or hypothetical Indicator diagram. The theoretical Indicator diagram ‘acrdkf’ for single acting steam engine is shown in Fig. 35.2 by firm lines.

Actual indicator diagram: If all losses and imperfections in steam engine are considered then the indicator diagram will be known as actual Indicator diagram. The actual indicator diagram ‘ACRK’ for single acting steam engine superimposed on the theoretical Indicator diagram ‘acrdkf’ is shown in Fig. 35.2 by dotted lines.

Deviation of actual indicator diagram to theoretical indicator diagrams:

The deviation of actual indicator diagram to that of theoretical indicator diagrams of steam engine is explained with the help of different operations of actual steam engine in Fig. 35.2.

1. In actual diagram, for steam admission, the steam port is opened slightly before the piston has reached the dead centre position and the steam is admitted at the point ‘A’ This is done to have a maximum steam pressure when the piston commences its working stroke and to open the steam port sufficiently wide so that the maximum amount of steam may then enter into the cylinder.

2. The actual admission line ‘AC’ is shown below the theoretical line ‘ac’; this is because of pressure loss due to condensation in main steam pipe and throttling in the admission valve.

3. There is rounding-off of the diagram at cut-off as the cut-off steam does not take place instantaneously due to the time factor in the closing of the admission valve.

4. As the actual admission pressure is lower than the theoretical admission pressure, so the actual expansion line ‘CR’ is lower than the theoretical line ‘cr’. The actual expansion line ‘CR’ is not a true hyperbola as it is assumed in theoretical diagram; this is owing to the varying interchange of heat between the cylinder walls and steam in cylinder. At the start of expansion stroke the steam in cylinder is hotter than the walls, thus causing condensation of steam due to heat transfer from steam to cylinder walls. At the end of expansion stroke, owing to its low pressure, the steam in cylinder is colder than the cylinder walls, thus causing re-evaporation of condensed steam due to heat transfer from cylinder walls to steam.

5. The release of steam is done at the point ‘R’ slightly before the piston reaches dead center as instantaneous drop in pressure is not possible.

6. There is rounding-off of the diagram at the point of release as the opening of exhaust valve is not instantaneous.

7. The actual exhaust line ‘RK’ is above the theoretical line ‘dk’ as the steam is being forced out of the cylinder overcoming the resistance.

8. The discharge of steam is stopped at ‘K’ before the piston is reached at the end of stroke. The entrapped steam called cushion steam is now compressed during remaining stroke. The pressure of the cushion steam is raised and thus a smooth change from exhaust to admission is made. It is denoted by ‘KA’.

9. The rounding off the toe

Fig. 35.2. Indicator diagram

The area of the actual diagram is smaller than the theoretical diagram. So the work done by the actual steam engine is less than its theoretical work.

35.3. MEAN EFFECTIVE PRESSURE (p_m) OF STEAM ENGINE

In steam engine, the pressure which is responsible for driving the piston is the difference between the steam pressure acting on one side and the opposite pressure on the other side of the piston. So,  effective pressure = steam pressure on one side  - steam pressure on the other side. In Fig. 35.3, it can be seen from the indicator diagram of steam engine (cycle ‘ABCD’), the effective pressure does not remain constant during the whole stroke of the piston. So it is necessary to determine the average or mean effective pressure (mep) from the indicator diagram to calculate the work done per cycle.  Mean effective pressure can be regarded as imaginary constant effective pressure which acted on a piston during one stroke, would do the same amount of work as is done by varying effective pressure during one cycle ‘ABCD’. In Fig 35.3., the Mean effective pressure is represented by line ‘ab’.  Net work done per cycle = mep x swept volume

Fig. 35.3. Mean effective pressure on indicator diagram

                                           = p_m x  V_s                                                                   ……….........(35.1)

From equation (35.1),

           p_m  =                                  ……….……………. (35.2)                                                       

Therefore, Mean effective pressure can be defined as a measure of work output per cycle per unit cylinder size.

Method of determining the mean effective pressure (mep) from indicator diagram

The mean effective pressure, p_m is calculated from indicator diagram as follows:

             Mean effective pressure, p_m  (bar) = h (cm) x pressure scale (bar/cm)

where, pressure scale (bar/cm) is pressure per unit height of diagrams

             h (cm) is mean height of indicator diagram  = 

                                 The net area of indicator diagram is measured either by counting squares on graph paper or by use of planimeter.

35.3.1.  Actual mean effective pressure (amep), p_am

If the mean effective pressure is determined from the actual indicator diagram, it is known as actual mean effective pressure (amep), p_am.

or         p_am  =                                            …………….(35.2a)

35.3.2. Theoretical mean effective pressure (tmep), p_tm

If it is calculated from the area of a theoretical indicator diagram, it will be the theoretical mean effective pressure (tmep), p_tm.     

or   p_tm  =                                     …………….(35.2b)

35.3.3. Diagram Factor

The area of the actual indicator diagram is less than the area of the theoretical indicator diagram as shown in Fig. 35.4. The ratio of the areas of these two diagrams is called the diagram factor or card factor denoted k.

 

Fig. 35.4. Diagram factor

So   k =     

Let L be the length of the diagrams

Therefore,  k =    

                       =    

Let p be the pressure scale i.e pressure per unit height of diagrams

Then,  k =  

           where, p_im and p_tm are actual and theoretical mean effective pressure, respectively.                  

So,          p_am = k  p_tm                       

The diagram factor varies from 0.6 to 0.9

35.4. THEORETICAL WORK DONE AND MEAN EFFECTIVE PRESSURE OF STEAM ENGINE

The theoretical work done and mean effective pressure of steam Engine can be computed with the help of a theoretical indicator diagram shown in Fig. 35.5. Say r’, x’ and c’ are certain fractions.          pa = admission pressure in absolute          pb = back pressure in absolute          Vs = F’D’ = swept volume in one stroke               or    clearance volume Va = c’Vs = Vf                ……………...(35.3)

Fig.35.5. Theoretical indicator diagram

or      V_c -V_a = r’V_s                                         ……………………….(35.4)

or      V_c = r’V_s + V_a  =   r’V_s + c’V_s   = (c’+ r’) V_s         …………..(35.5)

 

or      V_k –V_f = x’V_s                                          ……………………….(35.6)

or      V_k = x’V_s + V_f  =   x’V_s + c’V_s   = (c’+ x’) V_s              ………..(35.7)

or      V_r = V_d = V_s + V_a  =   V_s + c’V_s   = (c’+ 1) V_s           ………..(35.8)

or      V_d – V_k = V_s + c’V_s - c’V_s – x’V_s  = (1- x’) V_s           ………..(35.9)

The area of the hypothetical indicator diagram represents the theoretical work done per cycle, so

W = area  acrdkf   = area acCF + area crDC - area dkKD – area kfFK

                                = W_ac +W_cr – W_dk – W_kf                              ……………….. …………..(35.10)

(a)  Work done by the steam at constant pressure

           W_ac = area acCF = aF x ac = p_a (V_c –V_a) = p_a r’ V_s                     ………….. …………..(35.11)

(b)  Work done by the steam hyperbolically

           W_cr = area under the curve cr = area crDC

                 = 

                =       from equation (35.5)  and (35.8)                 …………(35.12)

(c)    Work done upon the steam at constant pressure

            W_dr = area dkKD = kK x kd = p_b (V_d –V_k) = p_b (1 – x’) V_s                                                  …………..(35.13)

(d)   Work done upon the steam hyperbolically

           W_kf = area under the curve kf = area kfFK

                = 

                =     from equation (35.3)  and (35.7)                                        …………(35.14)

From eq. (35.10), (35.11), (35.12), (35.13) and (35.14) we get, the theoretical work done per cycle

          W = W_ac +W_cr – W_dk – W_kf          

              =      

              =       

               =                                ....……(35.15)     

Cases:-

(i)  If there is compression and clearance,

From eqt. (35.2b)

The theoretical mep,    p_tm =      

  p_tm  =                                                 .....…..(35.16)

(ii) If there is no compression i.e. x’ = 0,  the area of work done is acrdf  Fig. 35.6 and from eqt. (35.16), we have                      ptm  =	Fig. 35.6. p-V diagram without compression.
(iii) If there is no compression and clearance i.e. x’ = 0 and c’ = 0, the area of work done is acrdf  Fig. 35.7 and from eqt. (35.16), we have                  ptm  =           From equation (35.5)  and (35.8)          Ratio of expansion,                or                So    ptm  =	Fig. 35.7. p-V diagram without compression and clearance.

W = area of work done = area acrdf    Fig. 35.6

             = area under ac + area under cr – area under df

             = 

Problem 35.1: Find the theoretical mean effective pressure with the help of the following data relating to a steam engine: Admission pressure 5 bar gauge, back pressure 0.2 bar gauge, cut off at 0.32 of the stroke.

Solution: Given: Admission pressure,  pa = 5 bar gauge                                               = 5 + 1.01325                                               = 6.01325 bar      Back pressure, pb = 0.2 bar gauge                              = 0.2 + 1.01325                              = 1.21325 bar Cut off = 0.32 of the stroke     i.e.  = r′ = 0.32

Determine the theoretical mean effective pressure of a steam engine, p_tm :

Formula:     p_tm = 

Answer:    p_tm  =   

                             = 

                            = 2.9 x 10⁵   N/m²

Problem 35.2: Steam at a pressure of 9 bar is admitted into a steam engine and then expands hyperbolically up to a pressure of 1.5 bar.  The exhaust pressure is 1.1 bar. Find the actual mean effective pressure neglecting clearance and compression and assuming diagram factor as 0.85.

Solution:

 Given: Admission pressure,  p_a = 9 bar

Exhaust/Back pressure, pb = 1.1 bar             Release pressure,  pr = 1.5 bar             Expansion = Hyperbolically             Diagram factor = 0.85.             Clearance and compression = Neglected Find the actual mean effective pressure, Pam ; Formula:     Pam = k  ptm                                      Finding unknown, ptm:                                         ptm =                                                         where, r is ratio of expansion

                                             Finding unknown, r:

As the expansion ‘cr’ is hyperbolic, so

         p_cV_c= p_rV_r 

 or     9 x 10⁵ V_c = 1.5 x 10⁵  V_r   

 or                 = 6

 or                  r    = 6

                                       p_tm   = 

                                                                                   = 3.0876 x 10⁵   N/m²  

Answer:  P_am = k  p_tm  = 0.85 x 3.0876 x 10⁵ 

         = 2.62  x 10⁵  N/m²    or  2.62 bar