Thermodynamics and Heat Engine4 (3+1)

LESSON - 38 MASS OF STEAM IN THE CYLINDER

Mass of steam admitted into the cylinder per power cycle (m_a) can be calculated from

                                                                                               ……………..(38.1)

where,  M is mass of steam admitted per min (kg/min) also called actual steam consumption in steam engine.

       n is number of power cycle per minute  

       For double acting engine, n = 2N  and for single acting,  n =  N    

       Where, N = number of crankshaft revolutions per minute (rpm)

The actual steam consumption (M) can be determined experimentally by one of following methods:

i)    Recording the quantity of feed water delivered to the boiler

ii)   The quantity of steam condensed in the condenser

iii)  Measuring the amount of steam supplied to the engine with the help of an orifice-meter.

Mass of cushion steam (m_c ) can be calculated from any point on the compression curve ‘K-A’ of the indicated diagram. To do this it is first necessary to calibrate the pressure line and volume line of the indicator diagram so that the absolute pressure and volume of any point may be read off.

Calibration of indicator diagram:

Pressure line calibration:

Let ‘de’ represents the atmospheric pressure line on indicator diagram.

Let S be pressure in bar per cm of vertical ordinate.

Pressure scale =   of vertical ordinate

The distance ‘od’ represents the atmospheric pressure to this pressure scale;

The line ‘ab’ is drawn at distance ‘od’ from atmospheric pressure line to represent the absolute zero pressure.

 where,   od =   cm

The pressure may now be marked on the vertical ordinate to the scale of 1 cm = S bar absolute pressure, commencing with its zero at ‘o’.

Volume line calibration: The length ‘ab’ of the indicator diagram represents the stroke volume, hence        ‘ab’ =  Where,  D is the diameter of the piston in meter              L is the stroke in meter. Then volume scale  = , m3 per cm of horizontal ordinate The distance ‘oa’ represents the clearance volume to this volume scale; The line ‘od’ is drawn at distance ‘oa’ from ‘aA’ to represent the absolute zero volume.

Fig. 38.1. Mass of steam in cylinder

     where   oa =         , cm

 The volume may now be marked on the horizontal line ‘ob’ to the volume scale, commencing with its zero at ‘o’.

 The absolute pressure and volume at any point may now be determined from the calibrated p-V diagram.

Mass of cushion steam (m_c) calculation from calibrated indicator diagram

The cushion steam is trapped in the cylinder during the compression stroke therefore the mass of cushion steam is calculated from the process ‘KA’ on the indicator diagram.

Select a point ‘f’ on the compression curve, towards the end of the stroke where the steam may be assumed to be just dry because compression tends to dry the steam. Read off the pressure and volume of the steam at point ‘f’ from the calibrated p-v diagram.

Let p_f = absolute pressure of steam at ‘f’   , bar

V_f = volume of steam at ‘f’   , m³

v_f = specific volume of dry steam corresponding to pressure p_f,  m³/kg  (from steam table)

  

Mass of steam in cylinder at any point between cutoff and release is

              (kg)

38.2. SPECIFIC STEAM CONSUMPTION (SSC):

The amount of steam consumed per unit power output is called the specific steam consumption. It is usually expressed in kg per kW hr. It is also called steam rate. 

The specific steam consumption may be expressed on the basis of either i.p. or  b.p. If specific steam consumption is based on i.p., it is known as indicated specific steam consumption. Similarly, if b.p. is taken as basis, it is known as brake specific steam consumption.

The mass of steam consumed is determined from eqn (38.1),

                     M = m_a . n                     (kg/min)

                         = m_a . n . 60                (kg/hr)

                         = m_a . 2N . 60          (for double acting)

                         = m_a . N . 60          (for single acting)

Problem 38.1: Estimate the specific Steam consumption in kg per i.p. per hour of a double acting engine from the following data:-

Cylinder diameter 60 cm, stroke 90 cm, rpm 88, admission pressure 8 bar, back pressure 1.8 bar, cut-off occurs at 20% of the stroke for both ends. The specific volume of steam at the point of cut off is 0.24 m³/kg. The diagram factor may be assumed equal to 0.8.  

Solution:    Given:  Steam engine = double acting engine             Diameter of cylinder, d= 60 cm = 0.6 m;             Stroke length, L = 90 cm = 0.9 m;             Speed, N = 88 rpm;              Admission pressure,  pa = 8 bar;              Exhaust/Back pressure, pb = 1.8 bar;      Cut off = 20% of the stroke  or  Volume at cut-off = Vc = 0.2V                              i.e.,   = r′ = 0.20

            Therefore, ratio of expansion  = 5

            Specific volume at the point of cut-off = v_c   = 0.24 m³/kg

            Diagram factor, k = 0.8

Determine the specific Steam consumption in kg per i.p. per hour of a double acting engine

Formula:   Specific steam consumption =         , kg/kWh

        Finding unknown, i.p. and M ;

              Finding unknown, i.p.;

             Indicated power of double acting steam engine, i.p. =             , kW

                                                                                                         = 2 × 

              Finding unknown,  p_am ;

                          The actual mean effective pressure is given by

                           p_am (actual) = k  p_tm (theoretical)

                                       

Finding unknown,  p_tm ;

                                     The theoretical mean effective pressure is given by

                                     p_tm =

                                             = 2.375 bar

                       p_am (actual) = k  p_tm (theoretical) = 0.8 x 2.375 = 1.9    bar

          i.p. = 2 ×    

               = 2 ×                                 

               = 141.82 kW

            Finding unknown,  M;

                We have,   Volume at cut-off = V_c = 0.2V =

                                                                        = 0.2 x (π/4) (0.6)² x 0.9 = 0.05089 m³

                       Mass of steam admitted per stroke, m_a   = = = 0.212 kg/cycle

  Steam consumption per hour, M = ma . 2N . 60

                                                                 = 0.212 x 2 x 88 x 60 = 2238.72 kg/h

Answer:       Specific steam consumption =  = 15.78 kg/kWh

Problem 38.2: Dry and saturated steam at 15 bar is admitted into a double acting steam engine. The steam exhausts at 0.4 bar. The original cut-off was 40% of the stroke. In order to effect a saving in economy it was decided that cut-off should occur earlier at 20% of the stroke. The power output was kept constant by increasing the speed. If diagram factor in both cases remains to be the same calculate (a) the ratio of new original speed and (b) saving in steam consumption.

Solution:

Given:   Steam engine = double acting engine

               Admission pressure,  p_a = 15 bar;

               Exhaust/Back pressure, p_b = 0.4 bar;

Original cut off = 40% of the stroke

         i.e. Volume at 40 cut off, V_C,40% = 0.4Vs

               New cut off = 20% of the stroke

        i.e. Volume at 20 cut off, V_C,20% = 0.2Vs

              Diagram factor for 40% (original) cut off = Diagram factor for 20% (new)  cut off  = k

             Assume,  Compression and Clearance = Neglected

(a) Determine the ratio of new and original speed,  :

Formula:  Since power developed remains the same for 40% (original) cut off and 20% (new) cut off,

                        i.e.     i.p. _40%  = i.p. _20% 

                    Hence,  2 × = 2 ×

                  or            

                          Finding unknown, p_am,40%:

                              The actual mean effective pressure is given by

                               p_am,40% (actual) = (k.p_tm)_40%  = k x 11.09

                                         Finding unknown, p_tm,40% :

                                         The theoretical mean effective pressure is given by     

          

                                            Finding unknown, r_40%:

                                             When cut-off occurs at 40% of the stroke,

                                     i.e.,   =   = 0.40

                                     Therefore, ratio of expansion  r_40%  = = 2.5

                                             Therefore,  p_{tm, 40%}    = = 11.09 bar

                      p_am,40% (actual) = (k.p_tm)_40%  = k x 11.09

                      Finding unknown, p_am,20%:

                      The actual mean effective pressure is given by

                        p_am,20% (actual) = k.(p_tm)_20%                           

                                         Finding unknown, p_tm,20% :

                                            The theoretical mean effective pressure is given by

                                             p_{tm, 20%} =

                                                Finding unknown, r_20% :

                                                    When cut-off occurs at 20% of the stroke,

                                                    i.e.,    = r_20%′ = 0.20

                                                   Therefore, ratio of expansion   r_20% = = 5

                                            Therefore, p_{tm, 20%} = = 7.428 bar

                  p_am,20% (actual) = (k.p_tm)_20%  = k x 7.428

 Answer:   = 1.492

(b) Saving in steam consumption, S%:

          Formula:  Percentage saving in steam consumption is

  

       Where, M_40% is mass of steam supplied per hour when cut- takes place at 40% of the stroke is

                                                        

                                                        

                and  M_20% is mass of steam supplied per hour when cut- takes place at 20% of the stroke is

    

   

Answer:    Percentage saving in steam consumption is

× 100

 = × 100 = 25.4%