Contents:
What is Dual cycle?
- Dual cycle is the air standard cycle which is combination of the Thermdynamic otto cycle & Diesel cycle.
- This thermodynamic power cycle consists of the two isenthalpic (constant volume) process & two adiabatic process & one isobaric process (Constant volume).
Dual cycle pv and ts diagram:
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
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`)
`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} )`
All parameters are summarized below:
Dual cycle application:
1] Heavy diesel engine.
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