# What is the process

## What is the isobaric process - definition

### Example of an isobaric process - isobaric addition of heat

Let's take that**ideal Brayton cycle** , which shows how a**Heat engine with****constant pressure** describes .**Modern gas turbine engines** and**air-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 by**W.**_{C.}= H_{2}- H_{1}.**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 by**Q**_{add}= H_{3 }- H_{2}**isentropic expansion**- The heated compressed air then expands on the turbine and releases its energy. The work done by the turbine is given by**W.**_{T}= H_{4}- H_{3}**isobaric heat release**- The residual heat must be dissipated in order to close the circuit. The net heat emitted is given by**Q**_{re}= H_{4 }- H_{1}

Take one**isobaric heat supply** (2 → 3) in a heat exchanger. In typical gas turbines, the high pressure stage receives gas (point 3 in the figure; p_{3} = **6.7 MPa** ; T_{3} = 1190 K (917 ° C)) from a heat exchanger. In addition, we know that the compressor is receiving gas (point 1 in the figure; p_{1} = **2.78 MPa** ; T_{1} = 299 K (26 ° C)) and we know that the isentropic efficiency of the compressor**η**_{K}** = 0,87 (87)** amounts to**%)** .

**Calculate the heat supplied by the heat exchanger (between 2 → 3).**

**Solution:**

After this**first law of thermodynamics** is the added net heat given by**Q**_{add}** = H**_{3}** - H**** _{2}** or

**Q**but in this case we know the temperature (T

_{add}= C_{p}(T_{3}-T_{2s}),_{2s not)}) at the outlet of the compressor. We will solve this problem in intensive variables. We need to use the previous equation (um

**η**

_{K}**to include**) using the term (+

**H**

_{1}**- H**

**) rewrite to:**

_{1}Q_{add} = **H**_{3}** - H**_{2}** = h**_{3}** - H**_{1}** - (H**_{2s}** - H**_{1}** ) / ****η**_{K.}

Q_{add up}**=** c_{p} (T_{3} -T_{1} ) - (c_{p} (T_{2s} -T_{1} ) **/ ****η**** _{K}** )

Then we calculate the temperature T_{2s} using the**p, V, T relationship** for the adiabatic process between (1 → 2).

In this equation, the factor for helium is the same**= c _{p} / c_{v} = 1,66** . From the previous equation it follows that the compressor outlet temperature T

_{2s}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}

*= C*

_{v}

*+ R = 5 / 2R = 20.8 J / mol K.*We transfer the specific heat capacities in units of**J / kg K over:**

*c*_{p}* = C*_{p}* . 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:

**Q _{add} =** c

_{p}(T

_{3}-T

_{1}) - (c

_{p}(T

_{2s}-T

_{1})

**/**

**η**

**) = 5200th (1190-299) - 5200th (424-299) / 0.87 = 4.633 MJ / kg - 0.747 MJ / kg =**

_{K}**3.886 MJ / kg**

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