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

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Solved Problems: Steam Power Cycle

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

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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