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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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Mechanical : Engineering Thermodynamics : Properties of a Pure Substance and Steam Power Cycle : Solved Problems: Steam Power Cycle |