Diesel cycle for engineering

1. Sal College Of Engineering
Prepared by:
Somesh Patani(131130119052)

2. Diesel Engine
• Also known as
Compression
Ignition Engine
(CI)
• Can this engine
“knock”?
• Difference from
Otto Cycle?

3. Early CI Engine Cycle and the Thermodynamic Diesel Cycle
A
I
R
Combustion
Products
Fuel injected
at TC
Intake
Stroke
Air
Air
BC
Compression
Stroke
Power
Stroke
Exhaust
Stroke
Qin Qout
Compression
Process
Const pressure
heat addition
Process
Expansion
Process
Const volume
heat rejection
Process
Actual
Cycle
Diesel
Cycle

4. Process a b
Isentropic compression
Process b 
c Constant pressure
heat addition
Process c  d
Isentropic expansion
Process d  a
Constant volume heat
rejection
- a=1,b=2,etc…for
book
Air-Standard Diesel cycle

rc 
vc
vb

v3
v2
(BOOK)
Cut-off ratio:

5.  
m
VVP
m
Q
uu in 232
23 )()(


AIR
23 Constant Pressure Heat Addition
now involves heat and work
)()( 222333 vPuvPu
m
Qin

)()( 2323 TTchh
m
Q
p
in

cr
v
v
T
T
v
RT
v
RT
P 
2
3
2
3
3
3
2
2
Qin
First Law Analysis of Diesel Cycle
Equations for processes 12, 41 are the same as those presented
for the Otto cycle

6. 23
1411
hh
uu
mQ
mQ
in
out
cycle
Diesel




Diesel
const cV
1
1
rk1
1
k

rc
k
1 
rc 1 








For cold air-standard the above reduces to:
Thermal Efficiency
1
1
1 
 kOtto
r
recall,
Note the term in the square bracket is always larger than one so for the
same compression ratio, r, the Diesel cycle has a lower thermal efficiency
than the Otto cycle
So why is a Diesel engine usually more efficient?

7. Typical CI Engines
15 < r < 20
When rc (= v3/v2)1 the Diesel cycle efficiency approaches the
efficiency of the Otto cycle
Thermal Efficiency
Higher efficiency is obtained by adding less heat per cycle, Qin,
 run engine at higher speed to get the same power.

8. k = 1.3
k = 1.3
The cut-off ratio is not a natural choice for the independent variable
a more suitable parameter is the heat input, the two are related by:
1
11
11
1 





 k
in
c
rVP
Q
k
k
r as Qin 0, rc1
MEP
Wnet
Vmax Vmin
- compares performance of
engines of the same size

9. Thank you…