Steam at 50 bar and 400°C enters the steam turbine to expand adiabatically to bar pressure, with an isentropic efficiency of 85%. The surrounding is at 100K 15 degrees C . If the steam flow rate is 100 kg/min, calculate the power generated. Also calculate the exergy flow rate at the inlet and outlet, and second law efficiency.
SOLUTION:
Given,
P1=50bar=5000kpa
T1=400°C=673K
P2=0.5bar=50kpa
η=85%=0.85
Po=100kpa
To=15°C=288K
mrate=100/60 kg/s =1.667 kg/s
To find,
P=?
B1 rate=?
B2 rate=?
ηII =?
For state1,
p1=50 bar
T1=400°C
(From the steam table),
v1 =0.05779 m^ 3 /kg,
h1 =3198.3 kJ/kg,
u1 =2909.4 kJ/kg,
s1 =6.651 kJ/kgK
For state2,
P2=0.5bar=50kpa
s2=s1
At this state, s 2 =7.595 kJ/kg and S2 > Sg2 So, it is a wet steam condition.
(From the steam table),
vf2'=0.001030 m^3/kg v g2'=3.24010 m^ 3 /kg
hf2'=340.6 kJ/kg h fg2'=2305.4 kJ/kg
uf2'=340.5kJ/kg
sf2'=1.091kJ/kgK
u fg2'=2143.4 kJ/kg
s fg2'=6.504kJ/kgK
h g2'=2646.0 kJ/kg
Here, s 2'=s 1
s f2' + x 2 s fg2' = s1 h2'=hf2' + x2' hfg2'
1.091 + x 2' (6.504) =6.651 h2'=340.6 + (0.8548)(2305.4)= 2311.25kJ/kg
x 2' = 0.8548
Already given,
η=0.85
η=(h1-h2)/(h1-h2')
0.85=(3198.3-h2)/(3298.3-2311.25)
h2=2444.30 kJ/kg
h2< hg2, (wet steam condition),
hf2 + x2 hfg2 = h2 s2=sf2 +x2 sfg2
340.6+ x2 (2305.4)= 2444.30 s2=1.091 + (0.9125)(6.504)
x2= 0.9125 s2=7.025 kJ/kg.K
Turbine actual work,
wta =(h1-h2)= 3198.3-2444.3= 754 kJ/kg
Power,
p= mrate . wta = 1.667(754) = 1256.67kW
At 15°C (From the steam table),
ho=hf=62.9 kJ/kg so=sf=0.224kJ/kg .K
Inlet exergy flow rate,
B1 rate= m rate . [(h1-ho)-To(s1-so)] = 1.667[(3198.3-62.9)-288(6.651-0.224)]=2141.3kW
Outlet exergy flow rate,
B2 rate= m rate . [(h2-ho)-To(s2-so)] = 1.667[(2444.3-62.90-288(7.025-0.224)]=704.66kW
Second law efficiency,
ηII = P/( B1 rate - B2 rate) = 1256.67 / (2141.3 - 704.66) = 87.48%
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USE CALCULATOR
MECHPEDIA CALCULATOR
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