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					                                                                                   PURE SUBSTANCES
DEFINITIONS:
 A pure substance is one which consists of only single species with characteristic feature of being of
   homogeneous and invariable composition irrespective of phase or phases in which it exists.
 A pure substance can exist in three different phases/states/conditions namely, solid, liquid and gas (vapour).
 Examples of pure substances are (1) a system composed of liquid and vapour phases of water (2) air
   (mixture of 79%-N and 21%-O by volume) if its composition does not change when undergoes a
   cooling/heating expansion/compression process (3) vapours of Fr-12, Fr-22 and NH3 used in refrigeration
   and air conditioning systems for cooling effect, etc.
 If some of the liquid is vaporized during a process, the system will be in chemically homogeneous but not in
   physical homogeneous because liquid density is more than that of vapour density.
 A phase is any homogeneous part of the system that is physically distinctive. It is essentially homogeneous,
   chemical and physical state of aggregation of molecules comprising the substance.
 Each phase of the system is separated from the other phases by the interfaces called phase boundaries. For
   example, considering a solid ice cube in liquid water, the two phases are separated by a phase boundary
   which is the surface of the cube.
 A change of phase is the transition from one phase to another phase of the substance. For example,
   melting/freezing is the change of phase between solid and liquid, vaporization/condensation is the change of
   phase between liquid and vapour and sublimation/deposition is the change of phase between solid and
   vapour.
 Whenever a change of phase occurs, all the mass of substance does not change its phase simultaneously.
 Melting is the process of changing the phase from solid to liquid.
 Freezing is the process of changing the phase from liquid to solid.
 Vaporization is the process of changing the phase from liquid to gas/vapour.
 Condensation is the process of changing the phase from gas/vapour to liquid.
 Sublimation is the process of changing the phase from solid to gas/vapour.
 Deposition is the process of changing the phase from gas/vapour to solid.
 Evaporation is the process of vapour generation from the surface of the liquid. It occurs at any temperature.
 Boiling is the process of formation of vapour from the whole mass of the liquid. When heat is added to a
   liquid, it starts boiling and vapours get generated both from the free surface as well as inside the liquid.
 Saturation temperature (ts) is the definite temperature at which all the liquid is transformed into vapour. At
   this constant temperature, the liquid is in equilibrium with its own vapour. For ex: ts for water is equal to
   100C.
 Saturation pressure (ps) is the pressure of the liquid/vapour phase corresponding to saturation temperature
   at which the liquid boils or is in thermal equilibrium with its own vapour. For ex: ps for water is equal to
   1.01325 bar corresponding to ts = 100C.
 Saturated liquid is the liquid which is in equilibrium with its own vapour at specified temperature/pressure
   at its boiling point.
 Saturated vapour is the vapour which is in equilibrium with its own liquid at specified
   temperature/pressure. Saturated vapour/steam may be wet or dry.
 Wet saturated steam is the saturated vapour which contains particles of liquid evenly distributed over the
   entire mass of the vapour.
 Dry saturated steam is the saturated vapour that contains no liquid particles.
 Saturation point (or) saturation temperature (or) boiling point is the temperature at which the molecules
   escaping from the liquid become equal to the molecules returning to it.
 Dryness fraction is the ratio of mass of dry saturated vapour to the total mass of the mixture. It is denoted
   by ‘x’.
   Let,        mf = mass of water vapour (saturated liquid) in suspension with saturated dry steam.
               mg = mass of dry saturated steam = mf + mg
               m = total mass of the mixture,
                                       mass of dry saturated vapour mg    mg
   then,       Dryness fraction, x                                   
                                          total mass of mixture      m m f  mg

 Wetness fraction is the ratio of mass of water or moisture in suspension with the total mass of the steam
   which contains it. It is denoted by ‘y’
                                          mass of water      m    mf            mg
               Wetness fraction, y                          f         1           1 x
                                       total mass of mixture m m f  mg      m f  mg

 Degree of superheat is the difference between the temperature of superheated steam (tsup) and temperature
   of saturation liquid (ts). Higher the degree of superheat, greater is the volume of steam occupied. At very
   high superheat, the steam behaves like an ideal gas.
STEAM GENERATION:
 The process of steam generation can be divided into three stages namely,
     1. Heating of water up to boiling point
     2. Evaporation of boiling water and its conversion into dry saturated steam, and
     3. Transformation of dry saturated steam into superheated steam.
 Considering an example of 1kg of ice at -10C to be heated to 200C generate steam.
 Consider the following T-Q, T-v and P-v diagrams.
             When heat is added to ice at -10C (point 1), its temperature increases to 0C without any
   Step-1    change in its phase. This quantity of heat used to increase the temperature without change in
             phase is called sensible heat of ice. (point 2, process 1-2)
             When more heat is added to ice at 0C, the ice is completely transformed into water without any
             change in its temperature. This quantity of heat used to change the phase at constant
   Step-2    temperature is called latent heat of fusion of ice. At atmospheric pressure of 1.01325 bar (1
             atm), its value is equal to 335 kJ/kg. There is a nominal decrease in volume for water and
             increase in volume for other substances. This is the peculiarity of water. (point 3, process 2-3)
             During supply of heat to water at 0C, it continues till boiling point is reached. This heat
             required to change from liquid state to vaporization is called sensible heat of water. This boiling
             point depends on pressure at which heat is supplied and is called saturation temperature. At
   Step-3
             normal atmospheric pressure, its value is equal to 100C. There is a decrease in volume of
             water up to 4C and thereafter volume increases up to the commencement of boiling (point 4,
             process 3-4).
             Evaporation of water starts at 100C and once again another change of phase from liquid to
             vapour occurs due to supply of latent heat of vaporization at this constant temperature of 100C
   Step-4
             whose value at constant pressure of 1 atm is equal to 2257 kJ/kg.. This process of vaporization
             which is both isothermal and isobaric, increases the volume of water. (point 4, process 4-5)
             At point 5, More heat is supplied to vapour at 100C to generate dry saturated steam at 200C.
             This quantity of heat is called sensible heat of vapour. Once the steam attains dry saturated
   Step-5
             state, its temperature and volume start increasing with further supply of heat. The steam so
             obtained is called supersaturated steam and it behaves like a perfect gas. (point 6, process 5-6)
 The horizontal distance between saturated liquid line and dry saturated vapour line becomes less and less
    with increase of pressure and saturation temperature. The point at which this distance becomes zero and
    represents highest temperature and pressure above which liquid no longer exists is called Critical point. For
    water, the temperature, pressure and volume corresponding to critical point are called critical temperature
    (tcritical = 374.14K), critical pressure (pcritical = 220.9 bar) and critical volume (vcritical = 0.005155 m3/kg)
SOLID-LIQUID-VAPOUR EQUILIBRIUM
 The figure shows the phase diagram drawn between pressure on y-axis
   and temperature on x-axis for water in which regions of solid, liquid and
   vapour are indicated.
 Single phase regions are separated from each other by saturation lines
 Equilibrium exists between (i)             Solid   &    vapour    phases along
   sublimation(1-T) or deposition(T-1) curves (ii) Solid and liquid phases
   along melting (2-T) or freezing (T-2) curves (iii) liquid and vapour phases
   along evaporation (T-C) or condensation (C-T) curves
 The slope of the melting/freezing curve is positive for most of the substances, but is negative for water.
 The evaporation line ends at critical point where the liquid and vapour phases are indistinguishable.
 The sublimation/deposition, melting/freezing and evaporation/condensation meet at a point called triple
   point at which all the three phases co-exist in equilibrium. This point exists at a definite temperature and
   pressure. For water, its values are ttp = 0.01C and ptp = 0.611 kPa
 The process where solid is directly transformed into vapour without touching liquid phase is called
   sublimation. The reverse of sublimation process is referred as deposition/ablimation.
ENTHALPY CHANGE DURING FORMATION OF ICE
 The quantity of heat required to change from ice at -10C to steam at 250C is called change in enthalpy.
 Change in enthalpy of ice from -10C to 0C is called sensible heat of ice which is equal to
    h12  c pidt where   c pi  specific heat of ice = 2.09 kJ/kg-K

                        h12  2.090   10  20.9kJ / kg  K
 Change in enthalpy during melting of ice at 0C is called latent heat of fusion of ice,
                        h23  335kJ / kg  K
 Change in enthalpy during heating of water from 0C to 100C is called sensible heat of ice which is equal
   to h34  c pwdt where c pw  specific heat of water = 4.186 kJ/kg-K

                        h34  4.186100  0  418.6kJ / kg  K
 Change in enthalpy during vaporization of water at 100C is called latent heat of vaporization of water,
                        h45  2256.9kJ / kg  K
 Change in enthalpy during vaporization from 100C to 250C is called sensible heat of vapour which is
   equal to h56  c psdt where c ps  specific heat of steam = 2.1 kJ/kg-K

                        h56  2.1250  100  315kJ / kg  K
 Total change in enthalpy during change of ice at 0C to steam at 250C is equal to
    h  h12  h23  h34  h45  h56  20.9  335  418.6  2256.9  315  3346.4kJ / kg  K
ENTROPY CHANGE DURING FORMATION OF ICE

 Change in entropy of ice from -10C to 0C is equal to s 1  c pi ln
                                                                                 t2
                                                                                 t1

 Change in entropy during melting of ice at 0C is equal to s 2 
                                                                               Qice Qice 335
                                                                                             1.23kJ / kg  K
                                                                                t2   t3   273

 Change in entropy during heating of water from 0C to 100C is equal to s 3  c pw ln
                                                                                                      t4
                                                                                                      t3

 Change in entropy during vaporization of water at 100C is equal to s 4 
                                                                                        Qwater Qwater 2257
                                                                                                          6.05kJ / kg  K
                                                                                         t4     t5     373

 Change in entropy during heating of water from 100C to 250C is equal to s 5  c ps ln
                                                                                                           t6
                                                                                                           t5

 Total change in entropy during change of ice at 0C to steam at 250C is equal to
                                                                          t2                 t                  t
           s   s 1  s 2  s 3  s 4  s 5  c pi ln       1.23  c pw ln 4  6.05  c ps ln 6
                                                                          t1                 t3                 t5
INDEPENDENT PROPERTIES OF PURE SUBSTANCE
Let us consider for a pure substance,
                                                                                                                   For liquid-vapour
   Property            Symbol                          For liquid                          For vapour
                                                                                                                        Mixture
    dryness                               When liquid starts boiling,           When vaporization completes
                           X                                                                                            0<x<1
    fraction                                 x = 0 (saturated water)             x = 1 (dry saturated vapour)
specific volume            V                         vl (or) vf (or) vw                           vg                       vfg
    pressure               P               psat = saturated pressure             psup = superheated pressure     pcr = critical pressure
  temperature              T              tsat = saturated temperature         tsup = superheated temperature   tcr = critical temperature
    specific
                           h                     hl (or) hf (or) hw                               hg                       hfg
   enthalpy
specific entropy           s                         sl (or) sf (or) sw                            sg                      sfg
     mass                  m                    ml (or) mf (or) mw                                mg                   m (or) mfg
specific internal
                           u                     ul (or) uf (or) uw                               ug                       ufg
    energy
 If the pure substance exists at saturation points (i.e. points on saturated liquid or saturated vapour curves),
   all the independent properties can be extracted from the tables at different saturation temperature and
   saturation pressure.
 If the pure substance exists in sub-cooled region (i.e. wet region) or superheated region (i.e. dry region), the
   properties are to be evaluated as discussed below.
 Wet region: The latent heat of ice (hfg) is to be added to saturated liquid, the enthalpy increases from hf to
   hg. Therefore,
                    hg  h f  h fg                                                                                                 (1)

 If x% of latent heat is supplied, then
                    hx  h f  xh fg                                                                                                (2)

                    s x  s f  xs fg                                                                                               (3)

                    u x  u f  xu fg                                                                                               (4)

                    vx  v f  xv fg                                                                                                (5)

 Superheated region: The enthalpy of a saturated vapour (hg) can directly be taken from tables. The enthalpy
   of superheated vapour is equal to the energy of the vapour when it is at superheated temperature (tsup). So
   the change in enthalpy at constant pressure process is given by,
                    hsup  hg  c p tsup  t sat                 (or)   hsup  hg  c p tsup  t sat                            (6)

                                             tsup                                               tsup
                    ssup  s g  c p ln                            (or)   ssup  s g  c p ln                                       (7)
                                             t sat                                              t sat

                    usup  u g  cv tsup  t sat                 (or)   usup  u g  cv tsup  t sat                            (8)

                    vsup       vs (or )v g
                                                                                                                                   (9)
                    tsup          t sat
 Sub-cooled region: In the sub-cooled region, the enthalpy is less than the enthalpy of a saturated liquid and
   adopting same procedure as that of superheated region, we have
                 hsub  h f  c pl t sat  t sub                                                       (10)

                                         t sat
                 ssub  s f  c pl ln                                                                    (11)
                                         t sub

                 usub  u f  cvl t sat  t sub                                                        (12)

                 vsub  vsat                                                                             (13)

EQUATION OF STATE FOR VAPOUR PHASE OF A SIMPLE COMPRESSIBLE SUBSTANCE
 Referring to the figure below, a-b-c-d represents the elementary Carnot
   cycle for dry saturated steam on p-v diagram.
 The point a lies on saturated liquid line and point b lies on dry saturated
   steam line.
 Let dtsup represents difference between two isothermal lines a-b and c-d.
 The efficiency of this elementary Carnot cycle is given by,
                                    t1  t 2 dt sat
                                                                              (1)
                                       t1     t sat
 The efficiency of this elementary Carnot cycle can also be given from the following relation,

                           
                                     work done     v  v f dp
                                                   g                            (2)
                                    heat supplied     h fg

    where dp is the difference in pressures of two isothermal lines.
 Equating the above two equations, we get
                                    dt sat vg  v f dp
                                        
                                    t sat      h fg

                 (or)      v   g    vf 
                                              h fg dt sat
                                                                                 (3)
                                              t sat dp
 This equation (3) is called Clapeyron’s equation which gives specific volume at any required pressure and
                              dt 
   temperature. The value of  sat  is obtained from slope of temperature-pressure curve which is plotted
                              dp 
                                  
   from experimental data.
PROBLEMS ON PURE SUBSTANCES
1. Calculate the dryness fraction of steam which has 1.25 kg of water in suspension with 40 kg of steam.
Solution:                  Mass of steam, = 40 kg,                        Mass of water, m f = 1.25 kg

        Mass of mixture, m  mg  m f = 40 + 1.25 = 41.25 kg

                                               mg         40
       Dryness fraction,                x                     0.9696
                                                 m       41.25
2. Using steam tables, determine the mean specific heat for superheated steam at 1 bar and between
   150C and 200C.
Solution:                  From steam tables of superheated steam and at 1 bar pressure, we have
                 At 150C:           hsup  2776.4kJ / kg

                 At 200C:           hsup  2875.3kJ / kg

                  2875.3 = 2776.4 + c ps (200 – 150)

                          2875.3  2776.4
                 c ps                     1.978kJ / kgK
                                50
3. A spherical shell of 30 cm in radius contains saturated steam and water at 300C. Calculate the mass
   of each if their volumes are equal.
                                                     4 3 4
Solution:        Total volume of the shell =           r   (0.3)3  0.113m3
                                                     3     3
                                                                      0.113
       Volume of vapour and that of water = V f  Vg                       0.0565m3
                                                                        2
       From steam tables, corresponding to saturation temperature of 300C, we have
                                            v f  0.001404m3 / kg ,             vg  0.0216m3 / kg

                                       Vg       0.0565
       Mass of vapour, mg                            2.616kg , and
                                       vg       0.0216

                                     Vf          0.0565
       Mass of water, m f                              40.242kg
                                      vf        0.001404

4. A tank contains 100 kg of liquid water and 5 kg of water vapour under saturation conditions at 20C.
   If the specific volume of saturated vapour at that temperature is taken as 57.78 m 3/kg, calculate the
   volume of tank and moisture content of the mixture.
Solution:        From steam tables, corresponding to saturation temperature of 20C,
   we have,                          v f  0.001002m3 / kg ,             vg  57.79m3 / kg      that gives

       Volume of water = V f  m f v f  100  0.001002  0.1002m3

       Volume of vapour = Vg  mg vg  5  57.79  288.95m3

       Total volume, V  V f  Vg  0.1002  288.95  289m3

       Moisture content of the mixture refers to wetness fraction,
                mass of water         mf     100     100
       y                                              0.952
             total mass of mixture m f  mg 100  5 105

5. A closed container holds 1 kg of water at 153.3C temperature with the composition by volume 1/3
   liquid and 2/3 vapour. Determine the pressure, quality, volume and enthalpy of the mixture.
Solution:        Since a mixture of water and vapour is considered, its pressure would be the same as the
saturation pressure corresponding to 153.3C. Then from steam tables,                    psat  5.2bar

                           Vf       1/ 3 1                             mf vf        1
       Given,                                       (or)                      
                           Vg       2/3 2                              mg v g       2

            mf           0.361                m
                                 165  mg  f
            mg       2  0.001094             165
                                    mg              m f / 165           1
       Dryness fraction, x                                               0.00602
                               m f  mg        m f  m f / 165       166

       For wet steam with dryness fraction, x  0.00602 ,
                v  1  x v f  xv g  1  0.00602  0.001094  0.00602  0.361  3.2606  103 m3 / kg
                h  h f  xh fg  646.7  0.00602  2103.8  659.363.2606 103 kJ / kg

6. A vessel of 0.3 m3 capacity contains 1.5 kg of water and steam in equilibrium at a pressure of 5 bar.
   Calculate the volume and mass of liquid and water vapour.
                                                                         0.3
Solution:      The specific volume of the given mixture, v                   0.2m3 / kg
                                                                         1.5
       From steam tables, the specific volumes of saturated liquid and saturated vapour at 5 bar pressure are
       noted as

                       v f  0.001093m3 / kg                and         v g  0.375m3 / kg

                       v  xv g  1  xv f

                       0.2  x  0.375  1  x0.001093  x  0.532
                                     mg
       Now by definition, x             m g  x m  m g 
                                                          
                                 m  mg           f       
                                  f

        Mass of vapour, mg  x m  mg   0.532 1.5  0.798kg
                                       
                                f      

            Mass of water, m f  1.5  0.798  0.702kg

        Volume of vapour, V g  mg v g  0.798  0.375  0.2992m3

            Volume of water, V f  m f v f  0.702  0.001093  0.00135m3

7. A vessel of volume 0.04 m3 contains a mixture of saturated water and saturated steam at a
   temperature of 250C. The mass of the liquid present is 9 kg. Find the pressure, the mass, the specific
   volume, the enthalpy, the entropy and the internal energy.
Solution:      From steam tables, corresponding to saturation temperature of 20C,
                psat  39.73bar ,         v f  0.0012512m3 / kg               vg  0.05013m3 / kg

                                          h f  1085.36kJ / kg                 h fg  1716.2kJ / kg

                                          s f  2.7927kJ / kgK                 s fg  3.2802kJ / kgK

               Volume of liquid,         V f  m f v f  9  0.0012512  0.01126m3

               Volume of vapour,         Vg  V  V f  0.04  0.01126  0.02874m3

                                          Vg       0.02874
               Mass of vapour, mg                         0.575kg
                                          vg       0.05013

               Total mass of mixture, m  m f  mg  9  0.575  9.575kg
                                         mg        0.575
               Quality of mixture, x                    0.06 (or) 6%
                                          m        9.575
               v  v f  xv fg  0.0012512  0.06(0.05013  0.0012512)  0.00418m3 / kg

               h  h f  xh fg  1085.36  0.06 1716.2  1188.32kJ / kg

               s  s f  xs fg  2.7927  0.063.2802  2.9895kJ / kg  K

               u  h  pv  1188.32  3.973  103  0.00418  1171.72kJ / kg

       Also at 250C,                  u f  1080.39kJ / kg               u g  1522kJ / kg

               u  u f  xu fg  1080.39  0.06 1522  1071.71kJ / kg

8. Calculate the enthalpy and entropy of steam at 10 bar when it is (a) saturated (b) 0.95 dry (c)
   superheated to 260C. Take cp of steam = 2.09 kJ/kg-K.
Solution:
   (a) When steam is saturated and is at 10 bar, the properties can be obtained directly from the saturated
       steam tables: at 10 bar,         hg  2776.2kJ / kg          and   s g  6.5828kJ / kg  K

   (b) When steam is 0.95 dry x  0.95 and is at 10 bar,

                        hx  h f  x hg  h   h0.95  762.6  0.95  2776.2  762.2  2676kJ / kg
                                    
                                           f 
                                              
                        s x  s f  x s g  s   s0.95  2.1382  0.95  6.5828  2.1382  6.36kJ / kg  K
                                     
                                             f
                                               

                                                   
   (c) When steam is superheated at 260C tsup  260C ,        
       From superheated steam tables, t sat  179.9C

        hsup  hg  c pg  tsup  t sat   2776.2  2.09  260  179.9  2943.6kJ / kg
                                        
                                        

                                 tsup                        260  273
            ssup  s g  c pg ln        6.5828  2.09  ln              6.923kJ / kg  K
                                 t sat                      179.9  273

9. Calculate the amount of heat supplied to a boiler to generate 10 kg/s of steam at a pressure of 10 bar
   and 0.95 dry if the water is supplied to it at 30C and at the same pressure.
Solution:
   From first law of thermodynamics,               dQ  dh  dW
                                                   dQ  dh dW  0 in constant pressure process
                                                          
                                                          
                                                   dQ  m h2  h1   
       Where, h2 is enthalpy of steam at 10 bar and 0.95 dry and is given by,

                    h2  h f  0.95   hg  h f     762.6  0.95  2776  2762.2  2676kJ / kg
                                                  
                                                  

       Where h1 is enthalpy of steam at 30C and is equal to h1  125.8kJ / kg

        Heat supplied, dQ  10  2676 125.8  25502kJ / s  25.5MW
10. One kg of steam at 20 bar exits in the following conditions (a) 0.9 dry (b) dry and saturated (c) at a
   temperature of 250C. Find its enthalpy, specific volume, density, internal energy and entropy in each
   case. Assume cp of steam = 2.01 kJ/kg-K.
Solution:
   (a) When steam is 0.9 dry,

              hx  h f  x hg  h   h0.9  908.6  0.9  2797.2  908.6  2608.3kJ / kg
                          
                                 f
                                   

              s x  s f  x s g  s   s0.9  2.446  0.9  6.336  2.446  5.947kJ / kg  K
                           
                                   f
                                     
              v0.9  xv g  0.9  0.0995  0.08955m3 / kg

                        1              1
              0.9                         11.166m3 / kg
                       v0.9         0.08955

              u0.9  h  pv0.9  2608.3  2000  0.08955  2428.6kJ / kg

   (b) When steam is dry saturated,

       hg  2797.2kJ / kg ,         s g  6.3366kJ / kg  K ,         v g  0.0995m3 / kg ,     10.05kg / m3

              u  h  pv  2797.2  2000  0.0995  2598.2kJ / kg
   (c) When steam is superheated, tsup  250C ,                      t sat  212.4C ,

              hsup  hg  c pg  tsup  t sat   2797.2  2.01 250  212.4  2872.7kJ / kg
                                             
                                             

                                   tsup
              ssup  s g  c pg ln        6.4865kJ / kg  K
                                   t sat

                       v
                                                             250  273  0.1072m3 / kg ,
                           g                     0.0995
              v               t         
               sup t
                       sat
                                    sup       212.4  273
                           1            1
              sup                         9.328kg / m3
                       vsup          0.1072

              usup  h  pv sup  2872.7  2000  0.1072  2658.3kJ / kg

11. Ten kg of water at 45C is heated at constant pressure of 10 bar until it becomes superheated vapour
   at 300C. Find the changes in volume, enthalpy, internal energy and entropy.
                                    (Ans: 2.569 m3, 28627.5 kJ, 26047.6kJ, 64.842 kJ/K)
12. A rigid vessel of o.86 m3 capacity contains 1 kg of steam at 2 bar. Evaluate the specific volume,
   temperature, dryness fraction, internal energy, enthalpy and entropy of steam.
                  (Ans: 0.86 m3/kg, 120.2C, 0.97, 2468.5 kJ/kg, 2640.5 kJ/kg, 6.959 kJ/kg-K)
13. A mass of wet steam at 165C is expanded at constant quality of 80% to a pressure of 3 bar. Use
   steam tables to determine the initial pressure of steam and changes in enthalpy and entropy during
   the expansion process.
                                                 (Ans: 7 bar, 59 kJ/kg, 0.163kJ/kg-K)
14. Steam initially at 60 bar and 450C is allowed to cool to 200C at constant volume. Find the final
   condition of syeam and heat transfer if the mass of hte steam is 0.75 kg.
                                             (Ans: 0.4038, -1423.35 kJ)
15. Two kg of dry saturated steam at 11 bar is to be generated from water available at 45C. Determine
   the quantity of heat to be supplied for raising the steam. If subsequently the steam is to be
   superheated by 50C, what additional heat input is required. Take cp for superheated steam = 2.25
   kJ/kg-K and cp for water = 4.18 kJ/kg-K.
                                          (Ans: 25714.8kJ/kg, 112.5kJ/kg)
QUESTIONS ON PURE SUBSTANCES
1. Define pure substance, phase of a substance, critical point, triple point, dryness fraction.
2. Differentiate between evaporation, boiling and vaporization of steam.
3. Differentiate between saturation liquid and saturation vapour for a pure substance.
4. Expalin phase equilibrium diagram for water with neat diagram.
5. Explain the process of steam generation at constant pressure process. Show the stages on p-v, t-v and t-s
   diagrams.
6. Differentiate between sub-cooled region, wet region and superheated region for a pure substance.
7. Express various independent properties of a pure substance for a dryness fraction ‘x’ .
8. Explain equation of state for vapour phase of a simple compressible pure substance.
9. Explain Mollier diagram and its features with a neat sketch.
10. List out the advantages of using superheated steam.

				
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Description: PURE SUBSTANCES,VAPOUR POWER CYCLES