Thermodynamics and Heat Engine

LESSON - 39 SATURATION CURVE AND MISSING QUANTITY AND NUMERICAL PROBLEM

Let us now consider any point M on the expansion curve and read off pressure and volume corresponding to the point M from the diagram.

Let, p_m is the absolute pressure of wet steam at M in bar and

V_m is volume of wet steam at point M in m³=  LM (say)

From the steam tables, obtain specific volume v_g of saturated steam corresponding to pressure p_m .

Then the volume of dry steam if wet steam at point M is fully dry = m v_g

Say this volume is represented by LN to the volume scale of the p V diagram.

This means that the point N represents the volume of dry steam if wet steam at point M is fully dry.

The dryness fraction of steam at point M =

Using the above procedure for other points on expansion curve, other points dry volume are obtained.  A curve passing through these points of dry volume is known as the saturation curve. In Fig 39.1 the saturation curve is shown in dotted line.

From the saturation curve the dryness fraction of the steam at all points of the expansion curve may be obtained. Say if dryness fraction of steam is required at the point ‘Q’ and ‘Y’, draw a horizontal line through ‘Q’ and ‘Y’ to cut vertical ordinate at point ‘P’ and ‘X’ and the saturation curve at ‘R’ and ‘Z’ respectively. Then the dryness fraction at Q and Y are given as Dryness fraction of steam at point Q = Dryness fraction of steam at point Y = It is observed from Fig. 39.1 that the steam is wet at the beginning of the expansion stroke and becomes dryer towards the end. This is so, because of the fact that the high pressure steam is hotter at the beginning of expansion stroke than the cylinder walls, this causes the steam to condense. During the expansion stroke, the steam pressure falls, and towards the end of the stroke, the walls will be hotter than the steam; this causes the re-evaporation of the condensed steam and the dryness fraction will consequently increase. If the walls of the cylinder were not steam jacketed, this increase in dryness fraction, at the end of expansion stroke, would not be obtained.

Fig. 39.1.  Saturation Curve

39.2. MISSING QUANTITY

The missing quantity is regarded as the difference between the actual mass of steam in the cylinder (m) and the indicated mass of dry saturated steam (m_i)

       Missing quantity =  m - m_i

As the actual mass of steam in cylinder (m) is the sum of mass of the steam admitted into the cylinder (m_a) and mass of cushion steam (m_c).

     Therefore, missing quantity= (m_a+m_c) – m_i

The mass of steam admitted into the cylinder (m_a) and mass of cushion steam (m_c) are determined as discussed in previous Lesson. Whereas, the indicated mass of steam (m_i) is determined from indicated diagram shown in Fig. 39.1. At point M, the indicated mass of steam is given by

m_i =

                 Where V_m is the volume at point M and

                              v_g is specific volume of saturated steam at p_m pressure

Sometimes the missing quantity is also expressed as the horizontal distance between the saturation line and the actual expansion curve is called missing quantity.

Thus in Fig 39.1, the missing quantity at point M i.e. at pressure P_m is

 MN = LN – LM = V_n– V_m

The missing quantity is mainly due to condensation of steam but a small amount is also due to leakage past the piston. The area between the saturation curve and the expansion curve represents the work lost per cycle due to this missing quantity.

The missing quantity due to steam condensation may be reduced by the following methods:

(a)    By supplying the superheated steam to the engine cylinder.

(b)   By jacketing the cylinder walls with steam supplied by the boiler.

(c)    By providing water separator in the steam pipe just before the steam enters the steam chest in steam engine.

(d)   By insulating steam pipe leading to the steam engine to prevent heat loss from the steam to surrounding.

(e)    By increasing the engine speed.

(f)    By compounding in which expansion takes place in more than one cylinder.

(g)   By a different arrangement of valves as in uniflow engine.

Problem 39.1: The following different events take place in a double acting, single cylinder steam engine having 55 cm cylinder diameter, 80 cm stroke length, clearance 5% of the stroke volume, steam consumption 70 kg/min and the speed 80 rpm.

Events	Fraction of stroke	Pressure
Cut-off	40% of forward stroke	9 bar
Release	90% of forward stroke	4.5 bar
Compression	80% of return stroke	2 bar

From the data given above, determine

 (a) mass of dry steam present in the cylinder at the points of cut-off, release and compression.

(b) dryness fraction of steam at cut-off and release.

(c) missing quantities at cut-off and release. If there is a difference in the missing quantities at cut-off and release, give reasons for that.

(d) percentage of re-evaporation during expansion.

Solution:   Given:   Steam engine type= Double acting and single cylinder Diameter of cylinder,d= 55 cm = 0.55 m; Stroke length, L = 80 cm = 0.8 m;               Steam consumption, M = 70 kg/min               Speed, N= 80 rpm              Clearance = 5% of the stroke volume,                i.e. Clearance volume, Vc = 0.05Vs                      = 0.05  × L       = 0.05 ×   = 0.0095 m3              Cut off =40% of the forward stroke     i.e.  (V1–Vc) = 0.4Vs

             Release = 90% of the forward stroke     i.e.  (V₂–V_c) = 0.9V_s

            Compression = 80% of return stroke     i.e.  (V₃–V_c) = (1 - 0.8V_s) = 0.2V_s

(a)   Determine mass of dry steam(indicated mass) present in the cylinder at the points of cut-off, release and compression:

        Formula: Mass of dry steam (indicated mass) at any point on indicated (p-v) diagram,

                          m_i = 

(i) Mass of dry steam (indicated mass) present in the cylinder at the points of cut-off i.e. point 1 at 9 bar, m_i,1:

                          Formula:    m_i,1 = =

 Finding unknown, V₁ ;

                                    As,   (V₁–V_c) = 0.4V_s

                                          

       V₁   = V_c + 0.4 V_s

        = V_c + 0.4 × L

        = 0.0095 + 0.4 x 0.8 = 0.0855 m³

                    Finding unknown,  ;

                            From steam table (dry saturated steam)

                            Specific volume of dry steam at cut off point 1 at 9 bar,  = 0.215 m³/kg

Answer:    Mass of dry steam (indicated mass) at cut-off, i.e. point 1 at 9 bar, m_i,1 = 

                      = =  0.398 kg/cycle

(ii) Mass of dry steam (indicated mass) present in the cylinder at the points of release i.e. point 2 at 4.5 bar, m_i,2:

Formula:      m_i,2 =

Finding unknown, V₂ ;

 As,   (V₂–V_c) = 0.9V_s

 V₂ = V_c + 0.9 V_s

      = V_c + 0.9

       = 0.0095 + 0.9 x  = 0.1805 m³

                Finding unknown,  ;

                       From steam table (dry saturated steam)  

                       Specific volume of dry steam at release point 2 at 4.5 bar,  = 0.414 m³/kg

Answer:    Mass of dry steam (indicated mass) at release, i.e. point 2 at 4.5 bar, m_i,2 =

                                   =  =  0.436 kg/cycle

(iii)  Mass of dry steam (indicated mass) present in the cylinder at the points of compression i.e. point 3 at 2 bar, m_i,3 :

Formula:     m_i,3  =   = 

Finding unknown, V₂ ;

 As,   (V₃–V_c) =  0.2V_s

 V₃ = V_c + 0.2 V_s

       = V_c + 0.2

       = 0.0095 + 0.2 x   = 0.0475 m³

Finding unknown, ;

From steam table (dry saturated steam)

Specific volume of dry steam at compression point 3 at 2 bar,  = 0.8854 m³/kg

Answer:    Mass of dry steam (indicated mass) at compression, i.e. point ‘3’ at 2 bar, m_i,3  = 

                             = = 0.0536 kg/cycle

(b)   Determine dryness fraction of steam at cut-off, x₁ and dryness fraction of steam at release, x₂:

Formula:  Dryness fraction at cut-off,  x₁  

=

Dryness fraction at release,  x₂

=

Finding unknown, m ;

Actual  mass of steam in cylinder between cut-off and release  (i.e. at cut off point ‘1’ and release point ‘2),

 m = mass of cushion steam per cycle between point ‘3’ and ‘4’ + mass of steam admitted in the engine per cycle from point ‘4’ to ‘1’.

= m_i,3 + m_a

Finding unknown, m_a;    

Mass of steam admitted per min. = M = 70 kg/min

Mass of steam admitted in to the engine cylinder per cycle from point ‘4’ to ‘1’,

m_a = = 0.4375 kg/cycle

 m = m_i,3 + m_a = 0.0536 + 0.4375 = 0.4911 kg/cycle

Answer:      Dryness fraction at cut-off, x₁ = = 0.8104 

 Dryness fraction at release,  x₂ =   = 0.8878

(c) Determine missing quantities at cut-off and release. If there is a difference in the missing quantities at cut-off and release, give reasons for that

Assumption: During actual cycle steam is dry and saturated at any point on the compression curve.

Formula: Missing quantity at cut-off per cycle

 = (Actual mass of steam in cylinder at cut-off point ‘1’) – (Indicated mass of steam present in the cylinder at the points of cut-off i.e. point ‘1’)

= m – m_i,1,             kg/cycle

 Missing quantity at cut-off per hour = (Missing quantity at cut-off per cycle) x 2 x N x 60,kg/h

Similarly, missing quantity at release per cycle

= (Actual mass of steam in cylinder at release point ‘2’) – (Indicated mass of steam present in the cylinder at the points of cut-off i.e. point ‘1’)

= m – m_i,2, kg/cycle

Missing quantity at cut-off per hour = (Missing quantity at cut-off per cycle) x 2 x N x 60,kg/h

Answer:    Missing quantity at cut-off per cycle= m – m_i,1= 0.4911 – 0.398 = 0.0931 kg/cycle

                    Missing quantity at cut-off per hour = 0.0931 x 2 x 80 x 60    = 939.76 kg/h

Answer:   Missing quantity at release per cycle = m – m_i,2= 0.4911 – 0.436 = 0.0551 kg/cycle

                     Missing quantity at release per hour = 0.0931 x 2 x 80 x 60    = 519.36 kg/h

The missing quantity at release is less than at cut-off. This is so because of the fact that the initial condensation is pronounced at cut-off than at the release. During the end of the expansion stroke heat flows from the wall to the steam and hence re-evaporation starts.

 (d)   Determine percentage of re-evaporation during expansion

   Formula:  The percentage of re-evaporation,

                        =

Answer:  The percentage of re-evaporation,

                          =

                          = × 100 = 44.73%

Problem 39.2: The dry and saturated steam is supplied to a non-condensing single cylinder, double acting steam engine which has a cylinder diameter 30 cm and stroke 50 cm. The clearance volume is 10% of the swept volume. The engine runs at 200 rpm. At a point on the expansion curve immediately after cut-off and at 40% of the stroke, the pressure taken from the indicator card is 4.5 bar. At a point on the compression curve at 90% of the return stroke, pressure is 1.6 bar.

Calculate the indicated and the actual steam consumption per hour if the missing quantity at the above mentioned point on the expansion curve after the cut-off is 0.007 kg per cycle.

Solution: Given: Steam engine type = Double acting and single cylinder             Diameter of cylinder,d= 30 cm = 0.30 m;             Stroke length, L = 50 cm = 0.5 m;              Speed, N= 200 rpm              Clearance = 10% of the stroke volume,  i.e. Clearance volume,   Vc = 0.1Vs              Point ‘5’ on the expansion curve immediately after cut off at 4.5 bar pressure                                                                                                 = 40% of the forward stroke               i.e.  (V5–Vc) = 0.4Vs               Point ‘6’ on the compression curve at 1.6 bar pressure                                                                                                  = 90% of return stroke              i.e.  (V6–Vc) = (1 - 0.9Vs) = 0.1Vs

(a)   Determine the indicated steam consumption per hour.

    Formula: Indicated steam consumption per stroke

                 = [(Indicated mass of steam present in the cylinder at the point ‘5’) - (Indicated mass of steam present in the cylinder at the point ‘6’)]

                 = (m_i,5 – m_i,6)       , kg/cycle

                 Indicated steam consumption per hour = (m_i,5 – m_i,6) x 2N x 60     , kg/hr

          Finding unknown, Indicated mass of steam present in the cylinder at the point ‘5’ at 4.5 bar, m_i,5:

                  m_i,5 = 

        Finding unknown, V₅ ;

        As,   (V₅–V_c) = 0.4V_s

     

         V₅ = V_c + 0.4 V_s

              = 0.1 V_s+ 0.4 V_s

              = 0.5 V_s = 0.5

              = 0.5 x   = 0.0177 m³

  Finding unknown,  ;

From steam table (dry saturated steam)

Specific volume of dry steam at cut off point 5  at 4.5 bar,    = 0.4123 m³/kg

              Indicated mass of steam at point ‘5’ at 4.5 bar, m_i,5 = ₌   = 0.0428 kg/cycle

         Finding unknown, Indicated mass of steam present in the cylinder at the point ‘6’ at 1.6 bar, m_i,6:

m_i,6 = 

 Finding unknown, V₆ ;

As,  (V₆–V_c) =  0.1V_s

 V₆ = V_c + 0.1 V_s

       = 0.1 V_s+ 0.1 V_s

      = 0.2V_s

      = 0.2

      = 0.2 x  = 0.00706 m³

Finding unknown,   ;

  From steam table (dry saturated steam)

Specific volume of dry steam at compression point 6 at 1.6 bar, = 1.111 m³/kg

Indicated massof steam at point 6 at 1.6 bar, m_i,6  =  = 0.00635 kg/cycle

Answer:    Indicated steam consumption per stroke = (m_i,5 – m_i,6)      

                                                                                          = (0.0428 - 0.00635) = 0.0364 kg/cycle

                Indicated steam consumption per hour = (m_i,5 – m_i,6) x 2N x 60    

                                                                                  =  0.03654 x 2 x 200 x 60 = 876.96 kg/h

(b)   Determine the actual steam consumption per hour if the missing quantity at the above mentioned point on the expansion curve after the cut-off is 0.007 kg per cycle.

     Given: Missing quantity point ‘5’ on the expansion curve after the cut-off = 0.007 kg per cycle.

Formula: Actual consumption of steam = [ Indicated steam consumption per hour + Missing quantityper hour],kg/h

Finding unknown, missing quantity;

            Missing quantity = 0.007 x 24000 = 168 kg/h

      Answer:  Actual consumption of steam

                                  = Indicated steam consumption per hour + Missing quantity per hour

                                  = 876.96 + (0.007 x 2N x 60)

 = 876.96 + (0.007 x 2 x 200 x 60)

 = 876.96 +168 = 1044.96 kg/h

Problem 39.3: Dry and saturated steam is generated at 13 bar from feed water at 50°C in a boiler. The steam passes at the rate of 1.5 kg/s through a steam pipe to the engine control valve. While passing through valve, the steam is throttled to 9 bar and steam is found to be 1% wet. The exhaust takes place at 0.3 bar. The i.p. developed by the engine is 625 kW. Calculate (a) the heat loss per minute in the steam pipe, (b) thermal efficiency of the engine only based on i.p.

Solution:

Given: Dry and saturated steam generated in boiler from 50°C feed water at pressure, p_A = 13 bar;

From Steam table (Dry saturated steam)

At pressure, p_A = 13 bar   →   h_g,A=  2785.7 kJ/kg, kJ/kgK 

Given: Steam condition after control valve i.e. at point ‘C’ = 9 bar and 1% wet;

From Steam table (Dry saturated steam)

At pressure, p_c = 9 bar   →   h_f,c= 742.64 kJ/kg,  h_fg,c=  2029.5kJ/kg

Given: Exhaust/ back pressure, p_b = 0.3 bar;

From Steam table (Dry saturated steam)

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

Given: Steam flow rate through steam pipe from point ‘A’ to ‘B’,  = 1.5 kg/s;

              Indicated power, i.p. = 625 kW.

(a)   Determine the heat loss per minute in the steam pipe;

           Formula: Assuming there is no pressure drop in the steam pipe,

                             The heat loss per kg in the pipe from point ‘A’ to ‘B’

= Heat in steam at point ‘A’ – heat in steam at point ‘B’

= Q_A– Q_B ,     kJ/kg

The heat loss per min in the pipe from point ‘A’ to ‘B’  

 = (Q_A– Q_B) x m    kJ/min.

 Finding unknown, Q_A;

                                Heat in steam at point ‘A’ , Q_A = h_A + W ;

                                             or                             Q_A = h_A

                             Finding unknown, h_A ;

                                From steam table,  h_A = hg,_A = 2785.7 kJ/kg;

                  Q_A = h_A= 2785.7 kJ/kg

Finding unknown, Q_B;

Heat in steam at point ‘B’ , Q_B = h_B + W ;

                                              or              Q_B = h_B

 Finding unknown, h_B;

                 In throttling process from ‘B’ to ‘C’ , enthalpy remains constant

                                h_B = h_{C, (9 bar, 1% wet)}

          Finding unknown, h_C;

         h_C = (h_f,c + x . h_fg,c)

              = 2751.48 kJ/kg

    h_B = h_C, = 2751.48 kJ/kg

  Q_B = h_B = 2751.48 kJ/kg

   Answer:  The heat loss in the pipe per kg = Q_A– Q_B

        = 2785.7 – 2751.48 = 34.22 kJ/kg

                   The heat loss per min in the pipe = (Q_A– Q_B) x    

     = 34.22 x 1.5 x 60 = 3079.8 kJ/min

(b) Determine the thermal efficiency of the engine only based on i.p. ;

        Formula:    Thermal efficiency of the engine only based on i.p.,

                           

     Answer:   η_th =   =  16.92%