## What is Dual cycle?

Dual cycle is the air standard cycle, which is a combination of the thermodynamic Otto cycle and Diesel cycle.

This thermodynamic power cycle consists of the two constant volume processes, two adiabatic processes, and one isobaric process (Constant pressure).

Contents:

## Dual cycle PV and TS diagram:

The cycle consists of following processes:

**Process 1-2 (Isentropic compression):** In this process, the air is compressed in isentropic (reversible adiabatic) manner from state 1 to 2.

**Process 2-3 (Constant volume heat addition):** During this process, the volume of the air is kept constant and heat is added externally to the air.

**Process 3-4 (Constant pressure heat addition):** In this process, the heat is added to the air by keeping the pressure constant.

**Process 4-5 (Isentropic expansion):** The high pressure air is expanded isentropically to obtain work.

**Process 5-1 (Constant volume heat rejection):** In this process, the heat is rejected by the air in adiabatic manner.

## Thermodynamic analysis of cycle:

Let’s analyze each of the process in detail.

**Process 1-2:** Reversible adiabatic compression

For reversible adiabatic process,

Q_{1-2} = 0

By compression of air from 1 to 2 temperature of air increases from T_{1} to T_{2} hence change in internal energy is given by :-

du_{1-2} = C_{v}( T_{2} – T_{1})

Now by First law of thermodynamics,

Q_{1-2} = du_{1-2} +W_{1-2}

0 = C_{v}( T_{2} – T_{1}) + W_{1-2}

W_{1-2}= C_{v}( T_{1} – T_{2} )

W_{1-2} = C_{v}( T_{1} – T_{2} )

**Process 2-3: **Constant volume heat addition:

For the isochoric process, Change in volume (`delta v=0`)

∴ W_{2-3} = 0

Here, temperature rises from T_{2} to T_{3}. Hence, change in internal energy is given by,

du_{2-3} = C_{v}( T_{3}-T_{2} )

Now to find heat transfer (Q), By first law of thermodynamics,

Q_{2-3} = du_{2-3} + W_{2-3}

Q_{2-3} = C_{v}( T_{3} – T_{2} ) + 0

Q_{2-3} = C_{v}( T_{3} – T_{2} )

**Process 3-4: **Constant pressure heat addition:

For constant pressure process,

W_{3-4} = P(V_{4} – V_{3})

As PV=RT, the above equation can write in the form of temperature as follows,

W_{3-4} = R(T_{4} – T_{3})

The temperature rises from T_{3} to T_{4}. Hence internal energy change is given by,

**du _{3-4} = C_{v}(T_{4} – T_{3})**

Now to find rate of heat transfer by first law of thermodynamics,

Q_{3-4} =du_{3-4} + W_{3-4}

Q_{3-4} = C_{v}( T_{4} – T_{3}) + R(T_{4} – T_{3})

Q_{3-4} = (C_{v} + R)(T_{4} – T_{3})

Q_{3-4} = (C_{v} + C_{p} – C_{v})(T_{4} – T_{3})

Q_{3-4} = C_{p}( T_{4} – T_{3})

**Process 4-5:** Reversible expansion for the reversible adiabatic process:

Q_{4-5} = 0

By expansion of air from 4 to 5, the temperature of the air decreases from T_{4} to T_{5}. Hence, the change in internal energy is given by,

du_{4-5} = C_{v}(T_{5} -T_{4})

Now, by the first law of thermodynamics,

Q_{4-5} =du_{4-5} + W_{4-5}

0 = C_{v}(T_{5} – T_{4}) + W_{4-5}

W_{4-5} = C_{v}(T_{4} – T_{5})

**Process 5-1: **Constant volume heat rejection:

For constant volume process,

W_{5-1} = 0

In this process temperature of air decreases from T_{5} to T_{1}. Hence, change in internal energy is given by,

du_{5-1} = C_{v}(T_{1} – T_{5})

Now by first law of thermodynamics,

Q_{5-1} =du_{5-1} + W_{5-1}

Q_{5-1} = C_{v}(T_{1} – T_{5}) + 0

Q_{5-1} = C_{v}( T_{1} – T_{5})

**All parameters are summarized below:**

## Dual cycle application:

1] Heavy diesel engine.

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