Let assume the **ideal Brayton cycle** that describes the workings of a **constant pressure** **heat engine**. **Modern gas turbine** engines and **airbreathing 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 drawn into the compressor, pressurized (1 → 2). The work required for the compressor is given by**W**_{C}= H_{2}– H_{1}.**Isobaric heat addition**– the compressed air then runs through a combustion chamber, burning fuel, and air or another medium is heated (2 → 3). It is a constant-pressure process since the chamber is open to flow in and out. The net heat added is given by**Q**_{add}= H_{3 }– H_{2}**Isentropic expansion**– the heated, pressurized air then expands on a turbine, gives up its energy. The work done by the turbine is given by**W**_{T}= H_{4}– H_{3}**Isobaric heat rejection**– the residual heat must be rejected to close the cycle. The net heat rejected is given by**Q**_{re}= H_{4 }– H_{1}

Assume an **isobaric heat addition** (2 → 3) in a heat exchanger. In typical gas turbines, the high-pressure stage receives gas (point 3 at the figure; p_{3 }= **6.7 MPa**; T_{3} = 1190 K (917°C)) from a heat exchanger. Moreover, we know that the compressor receives gas (point 1 at the figure; p_{1 }= **2.78 MPa**; T_{1} = 299 K (26°C)), and we know that the isentropic efficiency of the compressor is **η**_{K}** = 0.87 (87%)**.

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

**Solution:**

From the **first law of thermodynamics**, the net heat added is given by **Q**_{add}** = H**_{3 }**– H**** _{2}** or

**Q**but we do not know the temperature (T

_{add}= C_{p}.(T_{3}-T_{2s}),_{2s}) at the outlet of the compressor. We will solve this problem in intensive variables. We have to rewrite the previous equation (to include

**η**

**) using the term (+**

_{K}**h**

_{1}**– h**

**) to:**

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

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

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

In this equation, the factor for helium is equal to **=c _{p}/c_{v}=1.66**. The previous equation follows that the compressor outlet temperature, T

_{2s}, is:

From 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 into units of **J/kg K via:**

*c*_{p}* = C*_{p}* . 1/M (molar weight of helium) = 20.8 x 4.10*^{-3}* = 5200 J/kg K*

Using this temperature and the isentropic compressor efficiency we can calculate the heat added by the heat exchanger:

**Q _{add }= **c

_{p}(T

_{3}-T

_{1}) – (c

_{p}(T

_{2s}-T

_{1})

**/**

**η**

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

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