8-A.) The ideal (internally reversible, frictionless) components of a steam powe

Question

8-A.)

The ideal (internally reversible, frictionless) components of a steam power-plant are shown. Superheated vapor at 4000 kPa (40 bars) & 500 degree C flows steadily into an adiabatic turbine at (1), which is connected to a condenser at (2). Saturated liquid exits from the condenser at 20 kPa, entering the adiabatic pump at (3). High-pressure liquid leaves the pump at 4000 kPa (40 bars), flowing into the boiler at (4). Neglect changes in kinetic and potential energy. The steam mass flow rate is 50 kg/s. Assume source TH = TMAX.CYCLE and cold sink Tc =TMIN.CYCIE. For the write-up clearly sketch the components, carefully draw the T-s diagram and organize a state table. Find the turbine & pump power, W t.out. W p.in (MW) Find the heat transfer rates, Q iN.hot> Q out.cold (MJ/s), & entropy creation rates, (MJ/K-s) Find Rankine cycle efficiency vs. the reversible limit, = W cycle/Q inHot =

Explanation / Answer

Process 1 to 2 is Isentropic (s1 = s2)

Process 2 to 3 is Constant Pressure(P2 = P3 = 20 kPa)

Process 3 to 4 is Isentropic(s3 = s4)

Process 4 to 1 is Constant Pressure(P1 = P4 = 4000 kPa)

From Steam Tables

At P = 4000 kPa and T = 500 Deg

h1 = 3445.21 KJ/kg

s1 = 7.09 KJ/kg.K

P2 = 20 kPa and s2 = 7.09

At P2 = 20 kPa (from steam tables)

hf = 251.38 KJ/kg

hg = 2609.7 KJ/kg

T2 = 60.06 Deg

sf = 0.8319 KJ/kg.K

sg = 7.9085 KJ/kg.K

s2 = (1-x)*sf + x*sg

7.09 = (1-x)*0.8319 + x*7.9085

x = 0.884

h2 = (1-x)*hf + x*hg

h2 = (1-0.884)*251.38 + 0.884*2609.7

h2 = 2336.14 KJ/kg

P3 = 20 kPa

h3 = hf = 251.38 KJ/kg

s3 = sf = 0.8319

T3 = 60.06 Deg

P4 = 4000 kPa and s4 = s3 = 0.8319

h4 = 255.58

Wturbine,OUT = massflow*(h1-h2) = 50*(3445.21 - 2336.14) KW = 55,453.5 KW

Wpump,IN = massflow*(h4-h3) = 50*(255.58-251.38) = 210 KW

QIN,Hot = m*(h1-h4) = 50*3445.21-255.58) = 159,481.5 KW

Qout,Cold = m*(h2-h3) = 50*(2336.14 - 251.38) = 104,238 KW

Entropy creation rates

S(2 to 3) = m*(s3-s2) = 50*(0.8319-7.09) = -312.905

S(4 to 1) = m*(s1-s4) = 312.905

Th = 500 deg = 773 K

Tc = 60.06 = 333.06 K

Rankine cycle eff = (Wnet/Qin)*100

Wnet = Wout - Win = 55,453.5 - 210 = 55,243.5 KW

Rankinecycle Eff = (55,243.5/159,481.5)*100 = 34.64%

Revesibble Eff = {1-(Tc/Th)}*100 = {1-(333.06/773)}*100 = 56.91%

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