Water enters the steam generator of the power plant shown in the figure at 7MPa,

Question

Water enters the steam generator of the power plant shown in the figure at 7MPa, 50 degrees C, and exits at 7MPa, 600 degrees C. The steam is then fed into a well-insulated turbine, developing a power output of 31.6 MW before exiting the turbine as a two-phase, liquid vapor mixture at 10kPa with a quality of 90%. Combustion gases enter the steam generator at 730 degrees C and are cooled to 170 degrees C. For steady operation, determine: A) the mass flow rate of combustion gases in kg/s.   B) the rate of energy transfer between the gass and vapor streams in the steam generator in kJ/s.   The combustion gases can be modeled as air as an ideal gas with constant specific heats at 600 K. Ignore any heat losses from the components to the surroundings and neglect all kinetic and potential energy effects.

Explanation / Answer

for state 1 : 7 MPa , 50 C

specific volume ( v1) = 0.00100905458 m^3 / kg
specific enthalpy (h1) =     215.35500890247 KJ / kg
specific entropy (s1) = 0.70056678956696 KJ / kg K


For state 2 : 7 MPa , 600 C

specific volume ( v2) = 0.0556640505 m^3 / kg
specific enthalpy (h2) =     3650.6193111309 KJ / kg
specific entropy (s2) = 7.0909298102198 KJ / kg K

For state 3 : 10 KPa , 90% quality

T_sat = 45.807548207024 C
v_sat_vap = 14.6705584919 m^3 / kg
v_sat_liq = 0.0010102605727 m^3 / kg

h_sat_vap = 2583.8869371734 KJ/kg
h_sat_liq = 191.81229519356 KJ/kg

s_sat_vap = 8.1488932823443 KJ/kg K
s_sat_liq = 0.64921808302364 KJ/kg K

qquality = 0.9

so
v3 = 0.0010102605727 + (0.9*(14.6705584919 - 0.0010102605727 )) = 13.20360366876727 m^3 / kg
h3 = 191.81229519356 + (0.9*(2583.8869371734 - 191.81229519356)) = 2344.679473 KJ/kg
s3 = 0.64921808302364 + ( 0.9* ( 8.1488932823443 - 0.64921808302364) ) = 7.398925762 KJ/kg K


A)
heat work done by steam in the turbine = h2 - h3 = 3650.6193111309- 2344.679473 = 1305.9398381309 KJ/kg
total work output = 31.6 MW = 31600 KJ/sec
so.. mass flow rate of steam = 31600/1305.9398381309 = 24.197133 kg/sec

heat transfered in steam generator = (h2 - h1)*mass flow rate = (3650.6193111309 - 215.35500890247)*24.197133 = 3435.26430222843*24.197133 KJ/kg = 83123.547334 KW = 83.123547 MW

Let the mass flow rate of the combustible gases be x ..

specific heat capacity of ideal gas at 600 K = 1.0117 KJ/kg K

so.. x * 1.0117 * ( 730 - 170 ) = 83123.547334
so.. x = 146.7183 kg/sec

B) The rate of energy generation bentween the gases and the vapour in steam generator = 83.123547 MW

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.