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Unformatted text preview: 1. Steam Cycle Problem A steam power plant consists of a boiler (with superheater), turbine, generator and
low— and high pressure feedwater preheaters. The condensates from the
preheaters are lead to the condenser. a) Sketch a picture of the connections in the power plant b) Calculate the extraction massflows from the turbine 0) Calculate the electrical power output and the total efficiency The following is given: Feed water flow 100 kg/s
Pressure after feed water pump 60 bar
Temperature after superheater 500°C
Pressure in condenser 0.1 bar
Pressure after condensate pump 5 bar
Temperature after HP preaheater 150°C
Temperature after LP preheater 100°C
lsentropic efﬁciency of the turbine 0.8
Boiler efficiency 0.83
Mechanical efﬁciency 0.99
Generator efficiency 0.98 For the preheaters and the condenser it is given that the outcoming feedwater
temperature is the same as the saturation temperature of the extracted
steam. However, observe that the steam extracted is NOT saturated. It is also
given that the extractionflow after the feedwater preheater is saturated liquid
water. SOLUTION a) Sketch of the plant In =100 kgi's
p =60 bar
t = 500 deg C equals the saturation ' — —_— _ —
pressure at I px' 150 C I equals the
9" saturation I pressure at b) Extraction flows The extraction massflows can be found through heat balances over the feedwater
preheaters. All enthalpies must then be found for incoming and outgoing flows for
each preheater according to zmm mm = z mour*hour The pressure in line Y equals the saturation pressure at 150°C (similar in line X): by = 4,8 bar
px = 1 bar The steam extractions are thus performed at 4,8 bar respective 1 bar. With which
enthalpies are the extraction flows entering each feedwater preheater? These
enthalpies can be found in the expansionline of the turbine. The expansion line is determined by the inlet and outlet of the turbine, i.e. by the admission data (60 bar and 500°C) for inlet and by the isentropic efficiency and
condenser pressure for the outlet. The enthalpy after the expansion is (by the isentropic efﬁciency definition from the
thermodynamics) h2 = h1_ rlls * (h1  h2is) From hs diagram: h1 = 3425 kJ/kg and h2i5 = 2180 kJ/kg Thus the enthalpy in the turbine outlet is h2 = 3425— 0.8 * ( 3425 — 2180) kJ/kg = 2429 lekg Return to the h—s diagram with the calculated enthalpy and the condenser pressure
= 0,1 bar as indata to mark the turbine outlet condition and thereafter draw the expansion line. In the expansion line read the extraction enthalpies at respectively 4,8 bar and 1
bar. 9 hy = 2940 kJikg and hx = 2710 kJ/kg Heat Balance High Pressure Feedwater Preheater (4,8 bar): W = OUT my*hy+m*h5=my*hyv+m*h7 m = main flow = 100 kgis he = liquid water = CF, * t = 4,18 * 100 = 418 lekg (Cp for liquid water is constant (= 4,18 kJ/kgK) independently of the water pressure
and temperature as liquid water is regarded as incompressible. The reference state
for enthalpies is 0°C thus the enthalpy in point 6 is h5 = Cp*(100 — 0)....) h? = liquid water = CF, * t = 4,18 * 150 = 627 lekg hy' = (sat. liq water at 4,8 bar) = 533 kJ/kg (or h; = by as both points have 150°C and liquid water) With the values of enthalpies and main mass flow the steam extraction ﬂow
becomes: my = m * (h? — he, )l (m — th ) = 100*(527 — 418 )/(2940 — 633) = 9,06 kgls Heat Balance Low Pressure Feedwater Preheater (1 bar): W = OUT mx*hx+m*h4=mx*hx+m*h5 The enthalpy change over a pump can be neglected thus:
h4 = h3 = (sat. liq. water at 0,1 bar) = 190,5 kJ/kg and h5 = he = 418 kJ/kg me = (sat. liq. water at 1 bar) = 418 lekg (= h5 as both points have the same
temp. and are in liquid state) mx = m * (hf, — h4 )l (hX — hx)=100*(418 — 190,5 M2710 —418)= 9,93 kgls Answer b) my = 9,1 kgls and mx = 9,9 kgls c) Power output and efficiency The steam turbine power output can be expressed in 2 ways, depending on which
viewpoint is taken: Alternative 1: Following the ﬂow through the turbine
PST = m* “‘1  hv) + (m — mY)*(hY  hx) + (m — mY  mx)*(hx  “2) Alternative 2: Regard the turbine as a black box with energy in = energy out In = Out (Obs.i Do not forget to count the power output as energy out) m*h1=my*hy+mx*hx+(m—my—mx)*h2+P3T Both expressions will lead to PST = 92180 les = 92,18 MW Taking into account the mechanical and generator efficiencies:
Pel = 92,18*0,99*0,98 = 89, 4 MW The efficiency of the steam plant is calculated by ﬁrst estimating the fuel input to
the boiler (= heat recovered to the steam cycle divided by boiler efﬁciency): PFUEL = m * (ht  h?) / nBOILER PFUEL = 100*(3425 — 627)/ 0,83 kW = 337,1 MW ncycle = Pal / PFUEL = 89,4 / 337,1 = 0,265 Answer c) Pe. = 89,4 MW and ncycle = 26,5% ...
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This note was uploaded on 10/22/2011 for the course EML 4450 taught by Professor Greska during the Fall '06 term at FSU.
 Fall '06
 Greska

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