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

**1. A vessel of volume
0.04 m ^{3} 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 m^{3}

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 =} v_{1} = 0.001251 m^{3}/kg

v_{g =} v_{s}
= 0.050037 m^{3}/kg p = 39.776 bar

Total volume occupied by the liquid,

V_{1} = m_{1} × v_{1}

= 9
× 0.001251

= 0.0113
m^{3}.

Total volume of the vessel,

V
= Volume of liquid + Volume of steam

= V_{1} + V_{S}

0.4 =
0.0113 + V_{S}

V_{S}
= 0.0287 m^{3}.

Mass of steam, m_{s} = V_{S} / v_{s}

= 0.0287
/ 0.050037

= 0.574
kg.

Mass of mixture of liquid and steam, m = m_{1}
+ m_{s}

= 9
+ 0.574

= 9.574
kg.

Total specific volume of the mixture,

v =

= 0.04
/ 9.574

0.00418 m^{3} / 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
–39.776×10^{2} × 0.00418

= 1172
KJ / kg.

**Result:**

p = 39.776 bar

m = 9.574 kg

v = 0.00418 m^{3} / 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:**

p_{1} = 150 bar

T_{1} = 550°C

p_{2} = 40 bar

T_{3} = 550 °C

p_{3} = 0.1 bar

**To find:**

4. The
quality of steam at turbine exhaust, (x_{4})

5. cycle
efficiency and

6. The
steam rate.

**Solution:**

**1. The quality of steam at turbine
exhaust, (x _{4}):**

Properties of steam from steam tables at
150 bar & 550°C h_{1} = 3445.2 KJ/kg.

S_{1}=
6.5125 KJ/kg K

At 40 bar & 550°C

h_{3}
= 3558.9 KJ/kg.

S_{3}=
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

S_{1}
= S = 6.5125 KJ/kg K

S_{2}
= S_{g} at 40 bar

Therefore,

Exit of HP turbine is superheat

T_{sup}
= 332°C

h_{2}
= 3047.18 KJ/kg

S_{3}
= S_{g} at 0.1 bar

Steam is at wet condition.

S_{4}
= S_{3} = 7.2295 KJ/kg K

S4
= Sf4 + x4 Sfg4

7.2295
= 0.649 + x_{4} × 7.502

x_{4}
= 0.877

h4
= hf4 + x4 hfg4

=
191.8 + 0.877 × 2392.9

h_{4}
= 2290.37 KJ/kg K

**2) Cycle efficiency:**

D = (h_{1} –h_{2}) + (h_{3}
–h_{4}) / (h_{1} –h_{f4}) + (h_{3} –h_{2})

= (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 / (h_{1} –h_{2}) + (h_{3}
–h_{4})

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

^{ }

= 2.16
kg/Kw–hr.

**Result:**

The quality of steam at turbine exhaust, (x_{4})
= 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

p_{1} = p_{2} = 10 bar

T_{2} = 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:**

p_{1}
= 30 bar

p_{2}
= 0.05 bar

T_{1}
= 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 (x_{2})

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

h_{1}
= 2995.1 KJ/kg; S_{1}
= 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 m^{3}/kg.

1-2
= Isentropic expansion in the turbine

S_{1}
= S_{2} = 6.542 KJ/ kg K

S_{2}
= S_{f2} + x_{2} ×S_{fg2}

6.542
= 0.476 + x_{2} × 7.92

=
0.766

Therefore,
Quality of steam after expansion = 0.766 dry.

h_{2}
= h_{f2} + x_{2} ×h_{fg 2}

= 137.81
+ 0.766 × 2423.8

= 1994.43
KJ/kg.

h_{3}
= h_{fg2} = 137.8 KJ/kg.

Considering
the pump work, h_{4} –h_{3} = v_{f2} (p_{1} –p_{2})

h_{4}
= h_{3 +} v_{f2} (p_{1} –p_{2})

= 137.8
+ 0.001005 × (30 - 0.05) × 10^{2}

= 140.81
KJ/kg.

**Heat
supplied in the boiler:**

Q_{S}
= m × (h_{1} –h_{4})

= 2
× (2995.1 –140.81)

= 5708.58
KJ/s

= 20.55
× 10^{6} KJ /hr.

**Power
generated by the turbine:**

W_{T}
= m × (h_{1} –h_{2})

= 2
× (2995.1 –1994.43)

= 2001.34
KW.

**Rankine
efficiency by the plant:**

= (h_{1}
–h_{2}) - (h_{4} –h_{3}) / (h_{1} –h_{4})

= (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 × 10^{6} KJ /hr

Determine the quality of steam after
expansion (x_{2}) = 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%

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

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