s7ussain
contestada

Consider a diesel-fired steam power plant that produces 400 MW of electric power. The turbine inlet conditions of 5 MPa and 500°C and a condenser pressure of 25 kPa. The state of the water at the inlet of the pump is saturated liquid. The steam has quality of (x = 0.88) at the inlet of the condenser. Water from a nearby river is used to cool the condenser. To prevent thermal pollution, the cooling water is not allowed to experience more than 8°C as it flows through the condenser. The diesel has a heating value (energy released when the fuel is burned) of 40,100 kJ/kg. Assuming that 80 percent of this energy is transferred to the steam in the boiler and that the electric generator has an efficiency of 95 percent, determine (a) the overall plant efficiency (the ratio of net electric power output to the energy input as fuel), (b) the required rate of diesel supply and (c) the minimum mass flow rate of the cooling water from the river.

Consider a dieselfired steam power plant that produces 400 MW of electric power The turbine inlet conditions of 5 MPa and 500C and a condenser pressure of 25 kP class=

Respuesta :

batolisis

Answer:

A) 26.45% efficiency

B) 37.732 kg/s

C)  98386.9259 kg/s

Explanation:

From the diagram attached

at state 1

condenser pressure = 25 kpa, steam quality (x) = 0

h1 = hf = 271.96 kj/kg ( from saturated pressure table)

s1 = sf = 0.8932 kj/kg-k ( from saturated pressure table )

at state 2

turbine inlet pressure = 5 Mpa , 0.8932 kj/kg-k

h2 = 276.24 kj/kg ( from compressed water table )

at state 3

5 Mpa , 5000 c

h3 = 3434.7 kj/kg ( from super heated steam table )

at state 4

25 kpa , x = 0.88

hf = 271.96 kj/kg (  from saturated pressure table )

hg = 2617.5 kj/kg ( from saturated pressure table )

therefore h4 = hf + x4 (hg - hf )

                     = 271.96 + 0.88 ( 2617.5 - 271.96) = 2336.03 kj/kg

work done by turbine ( Wturbine)

W electric gen / efficiency of gen (equation 1)

Welec = 400 MW

efficiency = 95%

equation 1 = 400 / 0.95  = 421.0526 Mw  = 421052.6 w

Note : Wturbine = m( h3 - h4)

           421052.6 = m ( 3434.7 - 2336.03) hence m = 383.2384 kg/s

Q2 - 3 = m ( h3 - h2 )

Q2 - 3 = 383.2384 ( 3434.7 - 276.24 )

Q2 - 3 = 383.2384 ( 3158.46)

Q2 - 3 = 1210443.1568 Kw

B ) calculate for the required rate of diesel supply ( mf )

n boiler = Q2 -3 / Qin

where Qin = mf * HV = mf * 40100( heating value of diesel)

 n boiler   = 1210443.1568 / ( mf * 40100 )

mf = 37.732 kg/s

A) the overall plant efficiency

n overall = Welec / Qin

               = 400000 / ( 37.732 * 40100 ) = 0.264 = 26.45% efficiency

C ) minimum mass flow rate

Qcond = m ( h4 - h1 )

           = 791030.8842 kw

Qcond = mass flow rate * Cp * change in T

mass flow rate = 791030.8842 / ( 1.005 * 8 ) = 98386.9259 kg/s

Otras preguntas