What is the process
What is the isobaric process - definition
Example of an isobaric process - isobaric addition of heat
Let's take thatideal Brayton cycle , which shows how aHeat engine withconstant pressure describes .Modern gas turbine engines andair-breathing jet engines also follow the Brayton cycle.
The ideal Brayton cycle consists of four thermodynamic processes. Two isentropic processes and two isobaric processes.
- isentropic compression - Ambient air is sucked into the compressor and pressurized there (1 → 2). The work required for the compressor is given byW.C. = H2 - H1 .
- isobaric heat supply - The compressed air then flows through a combustion chamber in which fuel is burned and air or another medium is heated (2 → 3). It is a constant pressure process as the chamber is open for inflow and outflow. The added net heat is given byQadd = H3 - H2
- isentropic expansion - The heated compressed air then expands on the turbine and releases its energy. The work done by the turbine is given byW.T = H4 - H3
- isobaric heat release - The residual heat must be dissipated in order to close the circuit. The net heat emitted is given byQre = H4 - H1
Take oneisobaric heat supply (2 → 3) in a heat exchanger. In typical gas turbines, the high pressure stage receives gas (point 3 in the figure; p3 = 6.7 MPa ; T3 = 1190 K (917 ° C)) from a heat exchanger. In addition, we know that the compressor is receiving gas (point 1 in the figure; p1 = 2.78 MPa ; T1 = 299 K (26 ° C)) and we know that the isentropic efficiency of the compressorηK = 0,87 (87) amounts to%) .
Calculate the heat supplied by the heat exchanger (between 2 → 3).
After thisfirst law of thermodynamics is the added net heat given byQadd = H3 - H2 orQadd = Cp (T3 -T2s ), but in this case we know the temperature (T2s not) ) at the outlet of the compressor. We will solve this problem in intensive variables. We need to use the previous equation (umηK to include ) using the term (+H1 - H1 ) rewrite to:
Qadd = H3 - H2 = h3 - H1 - (H2s - H1 ) / ηK.
Qadd up= cp (T3 -T1 ) - (cp (T2s -T1 ) / ηK )
Then we calculate the temperature T2s using thep, V, T relationship for the adiabatic process between (1 → 2).
In this equation, the factor for helium is the same= cp / cv = 1,66 . From the previous equation it follows that the compressor outlet temperature T2s is:
From the ideal gas law we know that the molar specific heat of a monatomic ideal gas is:
C.v = 3 / 2R = 12.5 J / mol K and C.p = Cv + R = 5 / 2R = 20.8 J / mol K.
We transfer the specific heat capacities in units ofJ / kg K over:
cp = Cp . 1 / M (molar weight of helium) = 20.8 x 4.10 &min;³ = 5200 J / kg K.
With this temperature and the efficiency of the isentropic compressor, we can calculate the heat supplied by the heat exchanger:
Qadd = cp (T3 -T1 ) - (cp (T2s -T1 ) / ηK ) = 5200th (1190-299) - 5200th (424-299) / 0.87 = 4.633 MJ / kg - 0.747 MJ / kg =3.886 MJ / kg
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