# Dual cycle: Definition, Process, Formula, Pv & Ts Diagram, Application [with Pdf]

Contents

## What is Dual cycle?

1. Dual cycle is the air standard cycle which is combination of the Thermdynamic otto cycle & Diesel cycle.
1. This thermodynamic power cycle consists of the two isenthalpic (constant volume) process & two adiabatic process & one isobaric process (Constant volume).

## Processes:

• 1-2 :- Isentropic compression process
• 2-3 :- Constant volume process
• 3-4 :- Isobaric heat addition process
• 4-5 :- Isentropic expansion
• 5-1 :- Constant volume heat rejection 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)

therefore 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} )

In the form of temperature, it can be written as:-

W_{3-4}= R( T_{4}-T_{3} )

[ Because PV=RT ]

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} )

## Dual cycle application:

1] Heavy diesel engine.