Thermodynamics and Heat Engine4 (3+1)

LESSON - 40 GOVERNING OF SIMPLE STEAM ENGINE AND NUMERICAL PROBLEM

40.1.1. By throttling the steam:

In this, governing is achieved by throttling the steam with the help of centrifugal governor which is mounted on steam engine and gets drive from steam engine shaft through belt and pulley arrangement (Fig. 40.1).

Suppose the engine is running at its rated speed and load and, corresponding to this the area ACRKF on the p-v diagram is shown in Fig. 40.2. Now suppose the load on the engine decreases, then the power developed (area ACRKF on the p-v diagram)in the engine will be more than the demand and, as a consequence of this the engine speed will increase. As thefly-balls revolve with a spindle, which is driven by the engine through bevel gears, the increase in engine speed will result in the increase of centrifugal force on the fly-balls and hence fly-balls moves outwards and upwards.  The sleeve which is connected to the fly-balls through links and can only slide up and down on spindle also rises upwards with fly-balls. Finally, the upward movement of the sleeve partly closes the throttling valveat the other end of the bell crank lever and. This throttles the steam from pressure p_ato a suitable lower pressure p_a',as shown in p-v diagram in Fig. 40.2,so that the heat drops and load for doing work is reduced from (h₁ –h₂) to (h₁' –h₂'),as shown in h-s diagram,and consequently, the work developed in the engine is reduced from ACRKF to A'C'R'KF which is just equal to the demanded load. Under this condition the speed of the engine drops to the normal value required and the speed is maintained constant.If the load is still further reduced, the steam is throttled to the required, lower pressure p_a'', so that, the power developed is equal the demand, and the speed is maintained constant. With increase in load reverse action takes place.

Fig. 40.1. Centrifugal Governor.

Fig. 40.2. p-v and T-s diagram of throttle governing

During throttling process:

(i)  Steam cut-off point ‘C’ remains constant,

(ii)  there is no change in condenser vacuum ‘p_b’,

(iii) expansion‘C-R’ follows the hyperbolic law,

(iv) engine runs at constant speed and

(v) the enthalpy of steam  1-1′ or  1-1′′ remains constant.

This type of governing is simple and low in cost but it has the disadvantage of low thermal efficiency at part load. It is suitable for small engines where engine efficiency is of no importance.

In Fig. 40.3, the steam consumption has been plotted against the indicated power of the engine; a straight line graph is obtained. This line is known as Willan’s line. Thus showing the directly proportionallybetween the steam consumption and indicated power.

It is expressed as

M = A + B . (i.p.)                                                                              ……………. (40.1)

Where M is steam consumption in kg/h.   A and B are constants of proportionality.   The equation (40.1) may be derived in the following manner We know that the mean effective pressure neglecting clearance and compression is given by         pim  = where, p is admission pressure at which steam is admitted into cylinder and pb is back pressure.

Fig. 40.3. Willian’s Line

 and   i.p. for double acting steam engine is given by    i.p. =

                                                                                                 

or       i.p.  = k₁ p -  k₂                                  ……………………………………..  (40.2)

where k₁ = = constant     and     k₂  =  = constant

Due to throttling, dry and saturated steam becomes superheated. Therefore, gas laws can be applied. This fact may be verified using steam tables. Thus at cut-off point

       pv = constant            or     v =

Where p and v are pressure and specific volume of admitted steam respectively.

and    V= volume of steam admitted per stroke

  Mass of steam admitted per stroke,         

For throttling governing, at all admission pressures (p_a, p_a' and p_a'') volume of steam admitted per stroke is also constant (i.e. V_c = V_c’  = V_c’’) and hence we may write.

m_a = p x constant

Now, the mass of steam consumed per hour for a double acting steam engine will be

 M = m_a x 2N x 60    , kg/h         =  2N x  p x constant  =   k₃ p

  or       where,  k₃   is another constant

 From eqn (40.2),     

 From above two eqn. , we get          

 or   M =

 or   M = A + B (i.p.)

Where, A and B are constants.

40.1.2. Cut-off Governing:

In the cut-off governing, the governing is achieved by varying the amount of steam supplied to the engine i.e. by varying the steam cut-off point.

Suppose an engine is working at its rated load and speed and the power developed is given by the indicator diagram area ACRKF with steam cut-off the point at ‘C’ as shown in Fig. 40.4.Now suppose the load on the engine is reduced, then the power developed (area ACRKF on the p-v diagram) in the engine will be more than the demand and, as a consequence of this the engine speed will increase. To maintain constant speed, the cut-off point of steam is varied to a suitable point C', so that, the steam consumption is reduced causing the power developed shown by area AC'R'KF is just equal the load demanded. Under this condition the speed will decrease to the normal value, and thus, by this method the speed is maintained constant. If the load is further reduced, then the cut-off is varied to another suitable point C'', so that, the power developed is just equal to the load demanded. If the load increases, the reverse action takes place.

In cut-off governing, the admission pressure remains constant. The variation of the rate of steam consumption against the i.p. developed is found as follows: Neglecting clearance and compression effects, the mean effective pressure, pim  = where p is admission pressure at which steam is admitted into cylinder and pb is back pressure.    In the cut-off governing, p and pb remain constant. Hence pim  =

Fig. 40.4. Cut-off Governing

Again,  i.p. for double acting steam engine = 

But   2 LAN is a constant for an engine.

    

                                                                                              ………………………………..(40.3)

The mass of steam consumed per hour for a double acting steam engine will be

 M = m_a x 2N x 60           kg/h        

where  Mass of steam admitted per stroke,     

where V is volume of steam admitted per stroke and not constant for cut-of governing (i.e. V_c ≠ V_c’  ≠ V_c’’).

    v = specific volume of admitted steam and a constant for cut-of governing (i.e. v_c = v_c’ = v_c’’ ).

                                                                     ………………………………..(40.4)

 Note that the volume of steam admitted per stroke not constant in Cut- off governing but varies with  

But the swept volume V_r is constant hence

Putting the value of  V in equation (40.4), we have

                  ( as   N = constant)

or                                                                                                          ………………………………..(40.5)

From equations (40.3) and (40.5), we get

or     

or                                                                                      ………………………………..(40.6)

where,    The equations (40.3) and (40.4) show the variation of steam rate consumption (M) with expansion ratio (r) and indicated power (i.p.).This variation is also plotted in Fig. 40.5. A smooth curve is obtained for steam consumption versus indicated power plot. A straight line can be obtained if the ratio of expansion is constant; this condition is not fulfilled in a cut off governed engine as it is fulfilled by throttled governed engine.

Fig. 40.5. Variation of M with i.p. and r

In cut-off governing, the part load performance is better, but to meet very light loads, the cut-off has to be extremely small and this promotes condensation. From experiment, it has been found that if the cut-off is reduced below 10%,(as shown by the dotted line C′′′R′′′ in Fig. 40.4).then towards the end of expansion R′′′D, the piston has to do work on the steam.

If the cut-off is 100% (as shown by the dotted line DC′′′′ in Fig. (40.4), then there is no time left for expansion.

40.2. PERFORMANCE COMPARISON OF NON-EXPANSION ENGINES AND EXPANSION ENGINES

Fig. 40.6 shows the indicated diagram of non expansion and expansion steam engines.

The area ACRDF is the hypothetical indicated work when the cut-off occur at 50% of the stroke (expansion engines) and area AC’RDF is the hypothetical indicated work thewhen cut-off is at 100% i.e., the steam is admitted throughout the stroke length (non expansion engines). In case of non expansion engines, the increase in work-done is given by the area CC’R as compare to expansion engines whereas the steam consumption of non expansion engines is twice the expansion engines i.e. in case of non expansion engines, the increase in work-done is not proportional to the increase in steam consumption. So, non expansion engines are not at all economical as work done per kg of steam consumption is more as compare to expansion engines. However, non-expansive engines are used in rolling mill where large and intermittent power is needed.

 The effect of cut-off on power, steam consumption  and work-done per kg of steam consumption is also shown in Fig. 40.7.

Fig. 40.6. Indicated diagram of steam engine.

Fig. 40.7. Performance of steam Engine

Problem 40.1: A throttle governed steam engine consumes 350 kg of steam per hour when delivering i.p. 20 kW. The steam consumption at no load condition is 60 kg/h. Calculate the specific steam consumption per i.p. when the engine develops i.p. 12 kW.

If the steam is supplied to the engine is at 11 bar dry and saturated and the back pressure is 0.3 bar, calculate the indicated and brake thermal efficiencies of the engine assuming 90% mechanical efficiency.

Solution: 

Given: Steam consumption per hour, M₁ = 350 kg/h when delivering Indicated power, i.p.₁ = 20 kW;

            Steam consumption per hour, M₂ = 60 kg/h  when delivering Indicated power, i.p.₂ = 0 kW;

     Indicated power, i.p. = 12 kW;

     Admission pressure, p_a = 11 bar;

     Back pressure, p_b = 0.3 bar;

     η_mech = 90%

(a)   Determine the specific steam consumption per i.p. when the engine develops i.p. 12 kW;

       Formula:  Specific steam consumption per i.p.

                  s.s.c. =

                             =

                  s.s.c. when the engine develops i.p. 12 kW  =

       Finding unknown, M_{(12) ;}

                 We have from Willian’s line,    M_(i.p.) = A + B (i.p.)                                                               ………. (40.7)

                  Where, M_(i.p.)  is the steam consumption per hour when delivering i.p. indicated power and

                   A and B are constants.

Finding constants ‘A’ and ‘B’

Steam consumption per hour, M₁ = 350 kg/h when delivering Indicated power, i.p.₁ = 20 kW, then from Willian’s line, we have

                               350 = A + 20 x B

Steam consumption per hour, M₂ = 60 kg/h when delivering Indicated power, i.p.₁ = 0 kW, then from Willian’s line, we have

                                60 = A + 0 x B

By solving above two equations, we have,   

             A = 60   and     B =  = 14.5                    

  By substituting constants A and B in equation (40.7), we have             

    M_(i.p.) = 60 + 14.5 (i.p.)

    M(₁₂₎ =  60 + 14.5 (i.p.)

              = 60 + x 12 = 234 kg/h

Answer:   Specific steam consumption per i.p. when the engine develops i.p. 12 kW  

                    s.s.c. =    = 19.5 kg/kw.h

 (b) Determine the indicated thermal efficiencies of the engine η_i,th ;

       Formula: Indicated thermal efficency (η_ith) =

Finding unknown,

From steam tables (saturated steam) , we have

At 11 bar   →   h_g,11bar = 2797.7 kJ/kg

At 0.3 bar   →   h_f,0.3bar = 289.3 kJ/kg

Answer:     Indicated thermal efficiency, η_i,th = 

                                                                              =  = 7.35 %

 (c) Determine the brake thermal efficiencies of the engine assuming 90% mechanical efficiency η_b,th .

Formula:    Brake thermal efficiency, η_b,th =  η_i,th x η_mech

      Answer:   Brake thermal efficiency, η_b,th =  η_i,th x η_mech  

                                                                              = 7.35 x 0.9 = 6.62%

40.3. HEAT BALANCE SHEET 

Refer Fig. 40.8. In order to prepare a heat balance sheet for steam engine cylinder, the engine should be tested over a period of time under the conditions of constant load and steam supply. The following heat quantities should he included in the heat balance:

 

 

Fig. 40.8. Heat balancing of steam engine

Heat in steam supplied to engine, Q_steam

Heat supplied to engine =  h_steam           , kW                            ………………..  (40.1)

    where   = mass of steam supplied to the cylinder obtained from the steam condensed by the condenser, kg/s

                 = mass of steam supplied to steam jacket measured from the jacket drains, kg/s

                 h_steam = total heat of steam supplied, kJ/kg  (from steam table with respect to pressure of steam and condition (dry, superheated or wet) of steam entering the cylinder)  

Heat energy converted into indicated power of the engine, Q_i.p.

The heat energy converted into indicated power should be obtained from the indicator diagram by the following equation given in lesson 36.

 For

• Single acting engine,

                       Q_i.p. = I.P. =  , kW                       

• Double acting engine having mean effective pressure and effective piston area different on head and crank end sides of the piston,

                       Q_i.p. = I.P. = , kW

• Double acting engine having mean effective pressure and effective piston area same on head and crank end sides of the piston

                        Q_i.p. = I.P. =  , kW                                     ………………..  (40.2)

Heat energy rejected in exhaust steam, Q_exhaust

This is obtained from condenser and hot well

Heat rejected to exhaust steam,

                  Q_exhaust =  Heat given to condensing water + Heat to hot well

                                 = m_c .C_p (t₂ - t₁) + m_cyl h_well          , kW                                                     .....................................………(40.3)

Where   m_c = mass of circulating cooling water in the condenser, kg/s

C_p = specific heat of cooling water, kJ/kg K

(t₂ - t₁)  = temperature rise of cooling water in condenser; K

                              h_well  = total heat of condensed steam in hot well,  kJ/kg (from steam table with respect to condenser pressure and saturated water condition of steam)  

Heat energy rejected in steam jacket drain, Q_jacket

Heat drained from steam jacket,

            Q_jacket  =  m_jacket  x h_jacket                                        ,   kW                                             ................. .................... (40.4)

    where m_jacket  = mass of steam in jacket, kg/s

 h_jacket = total heat of water drained from the steam jackets at the temperature of the steam in the steam jacket

Heat unaccounted, Q^._unaccounted

 Heat lost by radiation and other unaccounted for

       Q^._unaccounted = Q_steam – [Q_i.p. + Q_exhaust + Q_jacket]        ; kW 

Heat balance sheet

Heat balance sheet may be put in tabular form as : —

40.4. PERFORMANCE OF STEAM ENGINES  Fig. 40.8 shows the performance of steam engine. The diagram shows that specific steam consumption decreases with increasing load to the point of maximum thermal efficiency. In the lower range of load the mechanical efficiency increases faster but in the middle portion of the load i.e. (40-90% load) the increase is slower and at about 90 to 100% load, the curve is more or less flat. The same trend found in the case of thermal efficiency.

Fig. 40.8. Performance of steam engine.

Problem: 40.2: The following data were obtained during a test on a single cylinder double acting steam engine:-

Engine stroke = 30 cm Piston diameter = 22 cm Piston rod diameter = 5 cm	Coal consumption = 0.4 kg/minute Calorific value of coal (CV)= 35170 kJ/kg of coal
Mean speed, N = 124 rpm	Steam consumption = 4 kg/min
Indicator area of cover end = 7 cm2 Indicator area of crank end = 8 cm2	Steam pressure = 3 bar (dry and saturated)
Length of indicator card = 7 cm Strength of indicator spring = 2 bar/cm of compression.	Condition of steam at the exit of steam engine,  x2 = 0.939 Condenser pressure = 1 bar
Load on brake = 1250 N Spring balance reading = 70 N	Cooling water flow rate = 51 kg/minute Rise in temperature of cooling water (t2-t1) = 20°C
Radius of brake wheel = 50 cm Radius of rope = 2 cm

Calculate the i.p., b.p., mechanical, thermal, Rankine, relative and overall efficiencies. Also draw up a heat balance sheet for engine cylinder.

Solution:

Given:    Engine type = Single cylinder and double acting

               Engine stroke, L = 30 cm = 0.3m

Piston diameter, D = 22 cm= 0.22

Piston rod diameter, d = 5 cm= 0.05m

Mean speed, N = 124 rpm

Indicator area of cover end = 7 cm²

Indicator area of crank end = 8 cm²

Length of indicator card = 7 cm

Strength of indicator spring = 2 bar/cm of compression.

Load on brake, W = 1250 N

Spring balance reading, S = 70 N

Radius of brake wheel, R = 50 cm = 0.5m

Radius of rope, r = 2 cm = 0.02m

 Coal consumption, m_f = 0.4 kg/minute

Calorific value of coal (C.V.) = 35170 kJ/kg of coal

Steam consumption, m_cyl = 4 kg/min

Cooling water flow rate, m_c = 51 kg/minute

Rise in temperature of cooling water, (t₂-t₁) = 20°C

Steam pressure, p₁ = 3 bar (dry and saturated)

From Steam table (Dry saturated steam)

At pressure, p₁ = 3 bar   →   h_g =  2724.7 kJ/kg,

Condition of steam at the exit of steam engine, x₂ = 0.939

Condenser pressure, p₂ = p₃ = 1 bar

From Steam table (Dry saturated steam)

At pressure, p₂ = p₃ = 1 bar   →   h_g =  2675.4 kJ/kg;    h_f= 417.15 kJ/kg ; h_fg = 2257.9 kJ/kg

 

  (a)   Determine i.p.:

Formula:    Total (i.p.) indicated power developed by the engine

                          = +  ,     kW

Finding unknown, P_am,1, P_am,2  , A₁ and A₂:

        Actual mean effective pressure for the cover end is given by

                   p_am,1 (cover end) =  = = 2 bar

       Actual mean effective pressure for the crank end is

                   p_am,2 (crank end) =  =  = 2.285 bar

         Effective area on which pressure acts on the cover end side is given by

                   A₁ (cover end) = =  0.03803 m²

         Effective area on which pressure acts on the crank end side is

                    A₂ (crank end) ==  0.03606 m²

  Answer:    Total indicated power developed by the engine,

                    i.p. =

                           =

                            = 4.71 + 5.107 = 9.817 kW

(b)   Determine b. p.:

Formula: b.p. developed by the engine;    

                        Brake power (bp) =   (kW or kJ/S)    

Finding unknown, T;

    T = (W-s) x R_eff

        = (W - S) x (R + r)

                     = (1250 - 70) x (0.5 + 0.002) = 613.6   N-m

    Answer: Therefore, Brake power (bp) =  =   7.967   (kW or kJ/S)

(c)    Determine mechanical efficiency, η_m:

Formula:   η_m =

Answer: Mechanical efficiency,  η_m =  × 100 = 81.15%

(d)   Determine indicated thermal efficiency:

Formula: The indicated thermal efficiency is    =

      Answer:    Indicated thermal efficiency,  η_i,th  =

                                                                                    =  = 6.38%

(e)    Determine brake thermal efficiency:

Formula: The brake thermal efficiency is  η_b,th =                     

       Answer:  The brake thermal efficiency, η_b,th =

                                                                                     = = 5.17%

(f)     Determine Rankine thermal efficiency:

Formula: The Rankine thermal efficiency is  =                                             

Finding unknown,  ;    

                        After expansion in the steam engine, the condition of steam at the exit of steam engine is  x₂ = 0.939 :  

    For wet steam, we have     = h_{f,1 bar} + x₂ h_{fg, 1bar}

                                                                   = 417.15 + 0.939 (2257.9)

       = 2537.3 kJ/kg

      Answer:     Rankine efficiency, η_th  = 

 =

                                                                     = 9.21%

(g)   Determine Relative efficiency:

Formula:   Relative efficiency =  

      Answer:  Relative efficiency =  = =  9.27%

(h)   Determine overall efficiency

Formula:  Overall efficiency (η_o) =  

       Answer:  Overall efficiency (η_o) =  =   = 3.39%       

(i)   Heat balance sheet for engine cylinder.

Heat supplied in steam, Q_steam = m_cyl x h_g,3bar  =  = 181.64 kW

Heat equivalent of i.p., Q_ip= 9.817 kW

Heat rejected to cooling water = m_c c_p (t₂-t₁) =   = 71.06 kW

Heat to hot-well = m_cyl x h_f,1bar  =  = 27.83 kW

Heat rejected in exhaust steam, Q_exhaust = heat to cooling water + heat to hot well 

                                                                       = 71.6 + 27.83 = 98.8 kW

Heat unaccounted, Q^._unaccounted = Q_steam – [Q_ip  + Q_exhaust]

                                                         = 181.64 – [9.817+98.8] = 72.93

Heat balance sheet may be put in tabular form as : —