GAS POWER CYCLES 629
dharm
\M-therm\Th13-2.pm5
Putting the value of ρ in ηDiesel, we get
ηDiesel = 1 –^11
1
1
1
1
1
()r
r
r
r
r
γ
γ
− γ
×
F
HG
I
KJ
−
−
L
N
M
M
M
M
M
O
Q
P
P
P
P
P
From above equation, we observe
r
r 1 > 1
Let r 1 = r – δ, where δ is a small quantity.
Then
r
r 1 =
r
r
r
r
r
−
=
F −
HG
I
KJ
δ 1 δ =
1
1
F −
HG
I
KJ
δ−
r
= 1 +
δδ δ
r rr
++
2
2
3
3 + ......
and r
r 1
F
HG
I
KJ
γ
=
r
r
r
r
γ
γ
γ
γ
δ
δ
1
1
F −
HG
I
KJ
=−F
HG
I
KJ
−
= 1 +
γδ γ γ δ
r r
+
()+
!
.
1
2
2
2 + ......
∴ηDiesel = 1 –
11
1
2
1
2
2
2
2
()
.()
!
. ......
r ......
r r
r r
γ γ
γδ γγ δ
− δδ
×
+ + +
++
L
N
M
M
M
M
O
Q
P
P
P
P
= 1 –
1
1
2
1
2
2
2
2
()
r ......
r r
r r
γ
δγ δ
− δδ
+ + +
++
L
N
M
M
M
M
O
Q
P
P
P
P
The ratio inside the bracket is greater than 1 since the co-efficients of terms δ^2 /r^2 is greater
than 1 in the numerator. Its means that something more is subtracted in case of diesel cycle than
in Otto cycle.
Hence, for same compression ratioηηotto> diesel.
13.5. Constant Pressure or Diesel Cycle
This cycle was introduced by Dr. R. Diesel in 1897. It differs from Otto cycle in that heat is
supplied at constant pressure instead of at constant volume. Fig. 13.15 (a and b) shows the p-v
and T-s diagrams of this cycle respectively.
This cycle comprises of the following operations :
(i) 1-2......Adiabatic compression.
(ii) 2-3......Addition of heat at constant pressure.
(iii) 3-4......Adiabatic expansion.
(iv) 4-1......Rejection of heat at constant volume.
Point 1 represents that the cylinder is full of air. Let p 1 , V 1 and T 1 be the corresponding
pressure, volume and absolute temperature. The piston then compresses the air adiabatically (i.e.,
pVγ = constant) till the values become p 2 , V 2 and T 2 respectively (at the end of the stroke) at point
- Heat is then added from a hot body at a constant pressure. During this addition of heat let