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,2–1) 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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