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Chapter: Mechanical - Engineering Thermodynamics - Properties of a Pure Substance and Steam Power Cycle

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Formation of Steam and Properties

Formation of Steam and Properties
Imagine unit mass of ice below the freezing point, enclosed in a cylinder by a piston under a constant load of 1 atmosphere (1 atm. = 1.01325 bar = 101.325 kPa).


FORMATION OF STEAM AND PROPERTIES

 

Imagine unit mass of ice below the freezing point, enclosed in a cylinder by a piston under a constant load of 1 atmosphere (1 atm. = 1.01325 bar = 101.325 kPa). If heat is added to the cylinder while keeping the pressure constant, the temperature rises and ice expands until a temperature of 273.15 K (00C) is reached (AB) as shown in Fig. Further heating does not raise the temperature of ice but causes a change to the liquid phase (BC). The change of phase occurs at a constant temperature and by reduction of specific volume. The heat required for this process is known as latent heat of fusion. Further heating results in a rise of temperature of liquid and a further contraction in volume until the temperature is about 40C and subsequent expansion until a temperature of 373.15 K (1000C) is reached (point D). At this point a second phase change occurs at constant temperature with a large increase in volume until the liquid has been vaporised (point E). The heat required in this case is called the latent heat of vaporisation. When vaporisation is complete, the temperature rises again on heating (line EF). The heat transferred to a substance while the temperature changes is sometimes referred to as sensible heat. This constant pressure lines are called isobars.



If the pressure is reduced, there is a slight rise in the melting point and also there is a marked drop in the boiling point of water and a marked increase in the change in volume, which accompanies evaporation. When the pressure is reduced to 0.006113 bar (0.6113 kPa), the melting point and boiling point temperatures become equal and change of phase, ice-water-steam, are represented by a single line. The temperature at which this occurs has been accepted internationally as a fixed point for the absolute temperature scale and is by definition 273.16 K. Only at this temperature and pressure of 0.6112 kPa, can ice, water and steam coexists in thermodynamic equilibrium in a closed vessel and is known as triple point. If the pressure is reduced further, the ice, instead of melting, sublimes directly into steam.




p-v, p-T, T-v, T-s, h-s DIAGRAMS

 

Consider now the behaviour at pressure above atmospheric. The shape of the curve is similar to that of the atmospheric isobar, but there is a marked reduction in the change in volume accompanying evaporation. At a sufficiently high pressure, this change in volume falls to zero and the horizontal portion of the curve reduces to a point of inflexion. This is referred to critical point. The values pressure and temperature of water at which critical point reached are

 

pc = 221.2 bar = 22.12 MPa ; Tc = 647.3 K ; vc = 0.00317 m3/kg.




The pressure at which liquid vaporises or condenses is called saturation pressure corresponding to a given temperature. Alternately, the temperature at which this phenomena occur is called saturation temperature corresponding to the given pressure. A vapour in a state lying along the saturated vapour line is also called dry saturated vapour and the vapour lying right of this line is called superheated vapour.



DEFINITION AND APPLICATIONS

 

Saturation temperature: Temperature at which a pure substance changes phase at a

 

given pressure.

 

Saturation pressure: Pressure at which a pure substance changes phase at a given

 

temperature.

 

Latent heat: The amount of energy absorbed or released during a phase-change. Melting/freezing: Latent heat of fusion.

Evaporation/condensation: Latent heat of vaporization.

 

Temperature at which water starts boiling depends on the pressure => if the pressure is fixed, so is the boiling temperature.




SLOVED PROBLEMS

 

1. A vessel of volume 0.04 m3 contains a mixture of saturated water and steam at a temperature of 250°C. The mass of the liquid present is 9 kg. Find the pressure, mass, specific volume, enthalpy, entropy and internal energy. [April/May 2012,2015]

 

Given Data:

 

Volume, V = 0.04 m3

 

Temperature, T = 250°C

 

Mass, m = 9 kg

 

To find:

 

1)  p, 2) m, 3) v, 4) h, 5) S,6) ΔU

 

Solution:

 

From the Steam tables corresponding to 250°C, vf = v1 = 0.001251 m3/kg

 

vg = vs = 0.050037 m3/kg p = 39.776 bar

Total volume occupied by the liquid,

 

V1 = m1 × v1

 

=  9 × 0.001251

 

= 0.0113 m3.

 

Total volume of the vessel,

 

V = Volume of liquid + Volume of steam

 

= V1 + VS

 

0.4    = 0.0113 + VS

 

VS = 0.0287 m3.

 

Mass of steam, ms = VS / vs

 

= 0.0287 / 0.050037

 

= 0.574 kg.

 

Mass of mixture of liquid and steam, m = m1 + ms

 

= 9 + 0.574

 

= 9.574 kg.

 

Total specific volume of the mixture,

 

v =

 

= 0.04 / 9.574

 

= 0.00418 m3 / kg.

 

We know that,

 

v = vf + x vfg

 

0.00418 = 0.001251 + x (0.050037 –0.001251) x = 0.06

 

From Steam table corresponding to 250 °C, hf = 1085.8 KJ / kg

 

hfg = 1714.6 KJ / kg

 

sf = 2.794 KJ / kg K

 

sfg = 3.277 KJ / kg K.

 

Enthalpy of mixture,

 

h = hf + x hfg

 

= 1085.8 + 0.06 × 1714.6

 

=             1188.67 KJ / kg Entropy of mixture,

s  = sf + x sfg

 

=  2.794 + 0.06 × 3.277

 

=               2.99 kJ / kg K. Internal energy, u = h –p v

 

= 1188.67 –39.776×102 × 0.00418

 

= 1172 KJ / kg.

 

Result:

 

1)    p = 39.776 bar

 

2)    m = 9.574 kg

 

3)    v = 0.00418 m3 / kg

4)    h = 1188.67 KJ / kg

 

5)    S = 2.99 KJ /kg K

 

6)    ΔU= 1172 KJ / kg.

 

 

 

 

2). A steam power plant uses steam at boiler pressure of 150 bar and temperature of 550°C with reheat at 40 bar and 550 °C at condenser pressure of 0.1 bar. Find the quality of steam at turbine exhaust, cycle efficiency and the steam rate. [May/June 2014]

 

Given Data:

 

p1 = 150 bar

 

T1 = 550°C

p2 = 40 bar

 

T3 = 550 °C

 

p3 = 0.1 bar

 

To find:

 

1.     The quality of steam at turbine exhaust, (x4)

 

2.     cycle efficiency and

 

3.     The steam rate.

 

Solution:

 

1. The quality of steam at turbine exhaust, (x4):

 

Properties of steam from steam tables at 150 bar & 550°C h1 = 3445.2 KJ/kg.

 

S1= 6.5125 KJ/kg K At 40 bar & 550°C

h3 = 3558.9 KJ/kg.

 

S3= 7.2295 KJ/kg K

 

At 40 bar

 

Tsat = 250.3°C = 523.3 K

hf =1087.4 KJ/kg. hfg = 1712.9 KJ/kg.

Sf= 2.797 KJ/kg K         Sfg= 3.272   KJ/kg K

At 0.1 bar            

hf =191.8 KJ/kg.   hfg = 2392.9          KJ/kg.

Sf= 0.649 KJ/kg K         Sfg= 7.502 KJ/kg K

1-2 = isentropic

S1 = S = 6.5125 KJ/kg K

 

S2 = Sg at 40 bar

 

Therefore,

Exit of HP turbine is superheat

 

Tsup = 332°C

 

h2 = 3047.18 KJ/kg

 

S3 = Sg at 0.1 bar

 

Steam is at wet condition.

 

S4 = S3 = 7.2295 KJ/kg K

 

S4 = Sf4 + x4  Sfg4

 

7.2295 = 0.649 + x4 × 7.502

 

x4 = 0.877

 

h4 = hf4 + x4  hfg4

 

= 191.8 + 0.877 × 2392.9

 

h4 = 2290.37 KJ/kg K

 

2) Cycle efficiency:

 

D = (h1 –h2) + (h3 –h4) /  (h1 –hf4) + (h3 –h2)

 

= (3445.2 –3047.15) + (3558.9 –2290.37) / (3445.2 –191.8) + (3558.9 –3047.18)

= 0.4426 × 100

= 44.26%

 

3) Steam rate:

= 3600 / (h1 –h2) + (h3 –h4)

= 3600 / (3445.2 –3047.15) + (3558.9 –2290.37)

= 2.16 kg/Kw–hr.

 

Result:

 

1.     The quality of steam at turbine exhaust, (x4) = 0.877

 

2.     cycle efficiency = 44.26%

 

3.     The steam rate = 2.16 kg/Kw–hr.

 

 

 

 

3). Ten kg of water 45 °C is heated at a constant pressure of 10 bar until it becomes superheated vapour at 300°C. Find the change in volume, enthalpy, internal energy and entropy.

 

Given Data:

 

m= 10 kg

 

p1 = p2 = 10 bar

 

T2 = 300°C

 

To find:

 

1)    Change in volume,

2)    Change in Enthalpy,

3)    Change in Internal energy,

 

4)    Change in Entropy.

 

Solution:

 

From steam tables, corresponding to 45°C,

v1=vf1=0.001010 m3 / kg;          h1 = hf1 = 188.4 KJ/kg;

s1 = sf1 = 0.638 KJ/kg K                 

From steam tables, corresponding to 10 bar and 300°C,       

h2      = 3052.1 KJ/kg;   s2 =   7.125 KJ/kg K;

v2      = 0.258 m3 /kg;            

Change   in   Volume,2–v1)     V   =   m   (v

= 10 (0.258 –0.001010)

= 2.5699 m3.

Change          in   Enthalpy,21)                     h   =   m   (h

 

= 10 (3052.1 –188.4)

 

= 28637 KJ.

 

 

Change          in   Entropy,2–s1)                 S   =   m   (s

 

= 10 (7.125 –0.638)

 

= 64.87 KJ/K.

 

 

Change          in   Internal2–u1)   energy,                              U   =   m   (u

 

= m [(h2 - h1) –(p2v2 –p1v1)]

 

= m [(h2 - h1) –p1 (v2 –v1)

 

= 10 [(3052.1 –188.4) –1000 (0.258 –0.001010)]

 

= 26067.1 KJ.

 

Result:

1)                               Change in volume, ΔV= 2.5699 m3.

2)                               Change in Enthalpy,=h28637 KJ.

3)                               Change in Internal energy, ΔU   =   26067.1   KJ.

4)                               Change in Entropy, ΔS   =   64.87   KJ/K.

 

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