634 ENGINEERING THERMODYNAMICS
dharm
\M-therm\Th13-3.pm5
= 1 – 0.248 × 1.563 = 0.612 or 61.2%
When the fuel is cut-off at 8%, we have
ρ−
−1
r 1
=^8
100
or ρ−
−(^1) =
14 1
8
100
= 0.08
∴ρ = 1 + 1.04 = 2.04
ηdiesel = 1 –
11
1
1 1
414
04 1
(^1141) 04 1
14
γ
ρ
γ ρ
γ
() ( )
()
.
.
r −−
−
−
L
N
M
M
O
Q
P
P
=−
×
−
−
L
N
M
M
O
Q
P
- P
2.
2.= 1 – 0.248 × 1.647 = 0.591 or 59.1%.
Hence percentage loss in efficiency due to delay in fuel cut off
= 61.2 – 59.1 = 2.1%. (Ans.)
Example 13.20. The mean effective pressure of a Diesel cycle is 7.5 bar and compression
ratio is 12.5. Find the percentage cut-off of the cycle if its initial pressure is 1 bar.
Solution. Mean effective pressure, pm = 7.5 bar
Compression ratio, r = 12.5
Initial pressure, p 1 = 1 bar
Refer Fig. 13.15.
The mean effective pressure is given by
pm =
pr r
r1
111
11γγγρ ργ
γ[( ) ( )]
()()−− −
−−−7.5 =
1 12.5 4 1 12.5 1
4 1 12.5 1×−^14 − −1 14^14
−−()[( )() (− )]
()( )..1..
1.ρρ7.5 = 34.33 4 1 4 0^0
6[.1. .364^14. .364]
4.ρρ−− +7.5 = 7.46 (1.4 ρ – 1.036 – 0.364 ρ1.4)
1.005 = 1.4 ρ – 1.036 – 0.364 ρ1.4
or 2.04 = 1.4 ρ – 0.364 ρ1.4 or 0.346 ρ1.4 – 1.4 ρ + 2.04 = 0
Solving by trial and error method, we get
ρ = 2.24
∴ % cut-off =
ρ−
−1
r 1
× 100 =2.24 1
12.5 1−
− × 100 = 10.78%. (Ans.)
+Example 13.21. An engine with 200 mm cylinder diameter and 300 mm stroke works
on theoretical Diesel cycle. The initial pressure and temperature of air used are 1 bar and 27°C.
The cut-off is 8% of the stroke. Determine :
(i)Pressures and temperatures at all salient points.
(ii)Theoretical air standard efficiency.
(iii)Mean effective pressure.
(iv)Power of the engine if the working cycles per minute are 380.
Assume that compression ratio is 15 and working fluid is air.
Consider all conditions to be ideal.