Thermodynamics 5 Problem 5. (35 points) Consider a steam turbine in which steam

Question

Thermodynamics 5

Problem 5. (35 points) Consider a steam turbine in which steam enters at 10.45 MPa and 780 K with a flow rate of 38.739 kg/s. As shown in the sketch, a portion of the steam is extracted from the turbine after partial expansion at three different locations. The extracted steam is then led to various heat exchangers. The mass flow rates and the temperature and pressure at each extraction point are given in the following table: Extraction Location m (kg/s) P (MPa) T (K) 4343 3.054 620 2 482 0.332 4.345 3 2.871 0.136 EX-0.949 The remaining wet steam exits the turbine at 11.5 kPa with a quality of 0.88. Determine the power output from the turbine. Assume that the turbine is well-insulated (adiabatic). Also neglect changes kinetic and potential energy Use the Problem Solving Method for this problem Include a T-v diagram with all states marked and constant pressure curves drawn at each pressure. List all properties used in the Properties section and list the reference where you found each property. You can use the textbook tables or the online NIST tables. Inlet Y V 1 2 3 Exit

Explanation / Answer

since turbine well insulated (adiabatic),change in potential and kinetic energy is neglected

power output = M'in(hin)-M'1(h1)-M'2(h2)-M'3(h3)-M'ex(hex)

here hin,h1,h2,h3,hex are specific enthaphy of steam states at inlet,1,2,3,exit;hf&hg values at saturated state

u can obtain these values from : http://www.peacesoftware.de/einigewerte/wasser_dampf_e.html

as hin(10.45Mpa,780K)=3387.295 Kj/Kg

h1(3.054Mpa,620K)=3107.415 Kj/Kg

h2(0.332Mpa,482K)=2882.642 Kj/Kg

h3 is saturated state at 0.136MPa since quality is given as x=0.949

h3=hf+x(hg-hf); here hf=specific enthalphy of saturatedwater,hg= specific enthalphy of saturatedsteam

hf=454.739 Kj/kg,hg=2688.681 Kj/Kg

h3(sat 0.136MPa)=454.739+0.949(2688.681-454.739)=2574.749 Kj/Kg

hex(sat 11.5KPa)= hf+x(hg-hf) given x=0.88

hf=203.351 Kj/Kg , hg=2588.779 Kj/Kg

hex=203.351+0.88(2588.779-203.351)=2302.527Kj/Kg

and M'ex is obtained by conservation of mass (i.e M'in=M'1+M'2+M'3+M'ex)

38.739=4.343+4.345+2.871+M'ex

M'ex=27.18Kg/s

so substituting all these values

power output = 38.739(3387.295)-4.343(3107.415)-4.345(2882.642)-2.871(2574.749)-27.18(2302.527)

power output=35225.052KW

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