# Dual cycle: Definition, Process, Formula, PV & TS graph, Application

## 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.

Q1-2 = 0

By compression of air from 1 to 2 temperature of air increases from T1 to T2 hence change in internal energy is given by :-

du1-2 = Cv( T2 – T1)

Now by First law of thermodynamics,

Q1-2 = du1-2 +W1-2

0 = Cv( T2 – T1) + W1-2

W1-2= Cv( T1 – T2 )

W1-2 = Cv( T1 – T2 )

Process 2-3: Constant volume heat addition:

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

∴ W2-3 = 0

Here, temperature rises from T2 to T3. Hence, change in internal energy is given by,

du2-3 = Cv( T3-T2 )

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

Q2-3 = du2-3 + W2-3

Q2-3 = Cv( T3 – T2 ) + 0

Q2-3 = Cv( T3 – T2 )

Process 3-4: Constant pressure heat addition:

For constant pressure process,

W3-4 = P(V4 – V3)

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

W3-4 = R(T4 – T3)

The temperature rises from T3 to T4. Hence internal energy change is given by,

du3-4 = Cv(T4 – T3)

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

Q3-4 =du3-4 + W3-4

Q3-4 = Cv( T4 – T3) + R(T4 – T3)

Q3-4 = (Cv + R)(T4 – T3)

Q3-4 = (Cv + Cp – Cv)(T4 – T3)

Q3-4 = Cp( T4 – T3)

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

Q4-5 = 0

By expansion of air from 4 to 5, the temperature of the air decreases from T4 to T5. Hence, the change in internal energy is given by,

du4-5 = Cv(T5 -T4)

Now, by the first law of thermodynamics,

Q4-5 =du4-5 + W4-5

0 = Cv(T5 – T4) + W4-5

W4-5 = Cv(T4 – T5)

Process 5-1: Constant volume heat rejection:

For constant volume process,

W5-1 = 0

In this process temperature of air decreases from T5 to T1. Hence, change in internal energy is given by,

du5-1 = Cv(T1 – T5)

Now by first law of thermodynamics,

Q5-1 =du5-1 + W5-1

Q5-1 = Cv(T1 – T5) + 0

Q5-1 = Cv( T1 – T5)

All parameters are summarized below:

## Dual cycle application:

1] Heavy diesel engine.