Solved Problems: Pure Substance and Steam Power Cycle

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:

2)  p, 2) m, 3) v, 4) h, 5) S,6) ΔU

Solution:

From the Steam tables corresponding to 250°C, v_{f =} v₁ = 0.001251 m³/kg

v_{g =} v_s = 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 + V_S

0.4    = 0.0113 + V_S

V_S = 0.0287 m³.

Mass of steam, m_s = V_S / v_s

= 0.0287 / 0.050037

= 0.574 kg.

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

= 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 = v_f + x v_fg

0.00418 = 0.001251 + x (0.050037 –0.001251)

x = 0.06

From Steam table corresponding to 250 °C,

h_f = 1085.8 KJ / kg

h_fg = 1714.6 KJ / kg

s_f = 2.794 KJ / kg K

s_fg = 3.277 KJ / kg K.

Enthalpy of mixture,

h = h_f + x h_fg

= 1085.8 + 0.06 × 1714.6

= 1188.67 KJ / kg Entropy of mixture,

s  = s_f + x s_fg

=  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:

p = 39.776 bar

m = 9.574 kg

v = 0.00418 m³ / kg

h = 1188.67 KJ / kg

S = 2.99 KJ /kg K

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:

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

5.     cycle efficiency and

6.     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.        

S_f= 0.649 KJ/kg K   S_fg= 7.502 KJ/kg K

1-2 = isentropic

S₁ = S = 6.5125 KJ/kg K

S₂ = S_g at 40 bar

Therefore,

Exit of HP turbine is superheat

T_sup = 332°C

h₂ = 3047.18 KJ/kg

S₃ = S_g 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 –h_f4) + (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:

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

cycle efficiency = 44.26%

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:

Change in volume,

Change in Enthalpy,

Change in Internal energy,

Change in Entropy.

Result:

Change in volume, ΔV= 2.5699 m3.

Change in Enthalpy,=h28637 KJ.

Change in Internal energy, ΔU   =   26067.1   KJ.

Change in Entropy, ΔS   =   64.87   KJ/K.

9)    A steam boiler generates steam at 30 bar, 300 °C at the rate of 2 kg/s. This steam is expanded isentropically in a turbine to a condenser pressure of 0.05 bar, condensed at constant pressure and pumped back to boiler.

e)     Find the heat supplied in the boiler per hour.

f)      Determine the quality of steam after expansion.

g)    What is the power generated by the turbine?

h)    Estimate the Ranking efficiency considering pump work.

Given Data:

p₁ = 30 bar

p₂ = 0.05 bar

T₁ = 300°C

m = 2 kg / s

To find:

Find the heat supplied in the boiler per hour (Q_S)

Determine the quality of steam after expansion (x₂)

What is the power generated by the turbine (W_T)

Estimate the Ranking efficiency considering pump work ( Ŋ)

Solution:

2.  Heat supplied in the boiler per hour (Q_S):

Properties of steam from the steam table

At 30 bar & 300°C

h₁ = 2995.1 KJ/kg;                          S₁ = 6.542 KJ/kg K;

At 0.05 bar

h_f2 = 137.8 KJ/kg;                          h_fg2 = 2423.8 KJ/kg;

S_f2 = 0.476 KJ/kg K;                       S_fg2 = 7.920 KJ/kg K;

V_f2 = 0.001005 m³/kg.

1-2 = Isentropic expansion in the turbine

S₁ = S₂ = 6.542 KJ/ kg K

S₂ = S_f2 + x₂ ×S_fg2

6.542 = 0.476 + x₂ × 7.92

= 0.766

Therefore, Quality of steam after expansion = 0.766 dry.

h₂ = h_f2 + x₂ ×h_{fg 2}

= 137.81 + 0.766 × 2423.8

= 1994.43 KJ/kg.

h₃ = h_fg2 = 137.8 KJ/kg.

Considering the pump work, h₄ –h₃ = v_f2 (p₁ –p₂)

h₄ = h_{3 +} v_f2 (p₁ –p₂)

= 137.8 + 0.001005 × (30 - 0.05) × 10²

= 140.81 KJ/kg.

Heat supplied in the boiler:

Q_S = m × (h₁ –h₄)

= 2 × (2995.1 –140.81)

= 5708.58 KJ/s

= 20.55 × 10⁶ KJ /hr.

Power generated by the turbine:

W_T = m × (h₁ –h₂)

= 2 × (2995.1 –1994.43)

= 2001.34 KW.

Rankine efficiency by the plant:

 = (h₁ –h₂) - (h₄ –h₃) / (h₁ –h₄)

= (2995.1 –1994.43) - (140.81 –137.8) / (2995.1 –140.81)

= 35 %

Result:

Find the heat supplied in the boiler per hour (Q_S) = 20.55 × 10⁶ KJ /hr

Determine the quality of steam after expansion (x₂) = 0.766 dry

What is the power generated by the turbine (W_T) = 2001.34 KW.

Estimate the Ranking efficiency considering pump work ( Ŋ) = 35%