1. LECTURE UNIT NO. 3
I. Diesel Cycle (SPSV)
- Basis of comparison for compression-ignition engines (Diesel engines)
Flow Diagram:
QA QA QR QR
P=C V=C
ma 2
TDC
ma
X
3
ma
ma ma
L
Y
S=C S=C
1 4 5
BDC
PV diagram:
TDC BDC
P VC L
X Y
2 3
4
1
Pm
V
V2 VD=V1 – V2
V1 – V4
VC = Clearance Volume = V2 = cVD VCo = cut-off volume = V3 – V2 = co VD
Where: c = % clearance
Clearance Ratio = c = V2 / VD = VC / VD
TS diagram:
3
4
2
1
Seek Input
Whatever your idea of success, conduct a "sanity check" throughout the process of reaching your goal. This should be done
with someone you trust and who is themselves successful. Ask them to provide honest feedback about your success and as you
move through different milestones, bounce concerns or new ideas off them to help keep you on the right track.

2. EQUATIONS:
1. PVT Relationships:
Process: 1 – 2 Isentropic Compression Process (S = C)
T2 / T1 = (P2 / P1) k-1 / k = (V1 / V2) k-1
(T2 / T1) 1/ k-1 = (P2 / P1) 1 / k = V1 / V2 = rk (S=C)
Isentropic compression ratio
rk (S=C) = V1 / V2 = (V2 + VD) / V2 = (cVD + VD) / cVD
rk (S=C) = (c + 1) / c
Where: c = % clearance
Process: 2 - 3 Constant Pressure Heat Addition Process (P = C)
P3V3 / T3 = P2V2 / T2
T3 / T2 = V3 / V2 = rC (P=C)
Cut-off ratio
rC (P=C) = V3 / V2 = (V2 + VCo) / V2 = (cVD + coVD) / cVD
rC (P=C) = (c + co) / c
Where: co = % cut-off
Process: 3 - 4 Isentropic Expansion Process (S = C)
T4 / T3 = (P4 / P3) k-1 / k = (V3 / V4) k-1
(T4 / T3) 1/ k-1 = (P4 / P3) 1 / k = V3 / V4
(T3 / T4) 1/ k-1 = (P3 / P4) 1 / k = V4 / V3 = re (S=C)
Isentropic expansion ratio
re (S=C) = V4 / V3 = (V2 + VD) / (V2 + VCo) = (cVD + VD) / co VD
re (S=C) = (c + 1) / (c + co)
Process: 4 – 1 Constant Volume Heat Rejection Process (V = C)
P1V1 / T1 = P4V4 / T4
T4 / T1 = P4 / P1 = rP (V=C)
2. Heat Added, QA = Σ + Q
Process: 2 – 3 (P=C)
Q2-3 = mCP ΔT
Q2-3 = mCP (T3 – T2)
Where: CP air = 1.0062 kJ/kg-K
3. Heat Rejected, QR = Σ - Q
Process: 4 – 1 (V=C)
Q4-1 = mCV ΔT
= mCV (T1 – T4)
Q4-1 = - mCV (T4 – T1)
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Good listening takes time to learn but in the end, it will be your greatest tool.

3. 4. Network, Wnet
Wnet = | QA | - | QR |
= | mCP (T3 - T2) | - | - mCV (T4 – T1) |
Wnet = mCP(T3 – T2) – mCV (T4 - T1)
Or by cyclic integration of W = PdV,
Wnet = (P2V2 – P1V1) /1-k + (P3V3 – P2V2) + (P4V4 – P3V3) /1-k
Wnet = mR(T2 – T1)/1-k + mR(T3 – T2) + mR(T4 – T3)/1-k
5. Diesel Cycle Thermal Efficiency, eDC
eDC = Wnet / QA x 100 %
= mCP (T3 – T2) – mCV (T4 – T1) x 100%
mCP (T3 – T2)
= 1 – CV (T4 – T1) x 100%
CP (T3 – T2)
Note: k = CP / CV
eDC = 1 – _(T4 – T1)_ x 100%
k (T3 – T2)
Or by knowing that:
T4 = T3 (rC/rk)k-1, T3 = T2 (rC), T2 = T1 (rk)k-1 and T3 = T1 (rk)k-1 rC
eDC = 1 - __1___ _(rC)k – 1_ x 100%
rkk-1 k (rC – 1)
6. Diesel Cycle Mean Effective Pressure, Pm
Pm = Wnet / VD
Pm = eDC x QA / (V1 – V2)
Or from Power Plant Engineering by Morse
SEATWORK:
1. An ideal Diesel Engine operates on 0.5 kg/s of air with suction state of 100 kPa and 45 °C. The pressure at the
end of compression is 3.25 MPa, and the cut-off is at 6% of the stroke from head-end dead center position.
Determine
a. The compression ratio
b. The percent clearance
c. The network
d. The thermal efficiency
e. The mean effective pressure
2. An engine operates on the air standard diesel cycle with compression ratio of 8. The pressure and temperature
at the beginning of compression ratio are 120 kPa and 43 °C. The maximum temperature is 1992 K and the
heat added is 1274 kJ/kg. Determine
a. The maximum pressure
b. The temperature at the beginning of heat addition
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