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Answer:
a)31%
b)34MW
Explanation:
A rankine cycle is a generation cycle using water as a working fluid, when heat enters the boiler the water undergoes a series of changes in state and energy until generating power through the turbine.
This cycle is composed of four main components, the boiler, the pump, the turbine and the condenser as shown in the attached image
To solve any problem regarding the rankine cycle, enthalpies in all states must be calculated using the thermodynamic tables and taking into account the following.
• The pressure of state 1 and 4 are equal
• The pressure of state 2 and 3 are equal
• State 1 is superheated steam
• State 2 is in saturation state
• State 3 is saturated liquid at the lowest pressure
• State 4 is equal to state 3 because the work of the pump is negligible.
Once all enthalpies are found, the following equations are used using the first law of thermodynamics
Wout = m (h1-h2)
Qin = m (h1-h4)
Win = m (h4-h3)
Qout = m (h2-h1)
The efficiency is calculated as the power obtained on the heat that enters
Efficiency = Wout / Qin
Efficiency = (h1-h2) / (h1-h4)
For this problem, we will first find the enthalpies in all states
h1=3231kJ/Kg
h2=2310kJ/Kg
h3=h4=272kJ/Kg
A) using the eficiency ecuation
Efficiency = (h1-h2) / (h1-h4)
Efficiency =(3231-2310)/(3231-272)=0.31=<u>31%</u>
b)using ecuation for Wout
Wout = m (h1-h2)
Wout=37(3231-2310)=34077KW=<u>34.077MW</u>
Answer:
2285kw
Explanation:
since it is an isentropic process, we can conclude that it is a reversible adiabatic process. Hence the energy must be conserve i.e the total inflow of energy must be equal to the total outflow of energy.
Mathematically,
Note: from the question we have only one source of inflow and two source of outflow (the exhaust at a pressure of 50kpa and the feedwater at a pressure of 5ookpa). Also the power produce is another source of outgoing energy .
Where are the mass flow rate and the enthalpies at the inlet at a pressure of 3Mpa
,
are the mass flow rate and the enthalpies at the outlet 2 where we have a pressure of 500kpa respectively.
,
and are the mass flow rate and the enthalpies at the outlet 3 where we have a pressure of 50kpa respectively.
,
We can now express write out the required equation by substituting the new expression for the energies
from the above equation, the unknown are the enthalpy values and the mass flow rate.
first let us determine the enthalpy values at the inlet and the out let using the Superheated water table.
It is more convenient to start from outlet 3 were we have a temperature and pressure value of (50kpa or 0.05Mpa ). using double interpolation method on the superheated water table to determine the enthalpy value with careful calculation we have
= 2682.4 KJ/KG , at this point also from the table the entropy value ,
value is 7.6953 KJ/Kg.K.
Next we determine the enthalphy value at outlet 2. But in this case, we don't have a temperature value, hence we use the entrophy value since the entropy is constant at all inlet and outlet.
So, from the superheated water table again, at a pressure of 500kpa (0.5Mpa) and entropy value of 7.6953 KJ/Kg.K with careful interpolation we arrive at a enthalpy value of 3206.5KJ/Kg.
Finally for inlet one at a pressure of 3Mpa, interpolting with an entropy value of 7.6953KJ/Kg.K we arrive at enthalpy value of 3851.2KJ/Kg.
Now we determine the mass flow rate at each inlet and outlet. since mass must also be balance, i.e
From the question the, the mass flow rate at the inlet is 2Kg/s
Since 5% flow is delivered into the feedwater heating,
Also for the outlet 3 the remaining 95% will flow out. Hence
Now, from
we substitute values
.
Hence the power produced is 2285kW