diesel cycle

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Air-fuel mixture Fue l spr ay Spark plug Fuel injector Gasoline engine Diesel engine An CI power cycle useful in many forms of automotive transportation, railroad engines, and ship power plants Replace (the spark plug + carburetor) in SI by fuel injector in CI engines by fuel injector in CI engines. A. Diesel cycle : The ideal cycle for CI engines

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Page 1: Diesel Cycle

Air-fuelmixture

Fuelspray

Sparkplug

Fuelinjector

Gasoline engine Diesel engine

• An CI power cycle useful in many forms of automotive transportation, railroad engines, and ship power plants

• Replace (the spark plug + carburetor) in SI by fuel injector in CI engines by fuel injector in CI engines.

A. Diesel cycle : The ideal cycle for CI engines

Page 2: Diesel Cycle

(1)

Inlet valve open and fresh air is drawn into the cylinder

(2)

Temperature rise about the auto-ignition temperature of the fuel.

Diesel fuel is sprayed into the combustion chamber.(3)

Evaporation, mixing, ignition and combustion of diesel fuel.

In the later stages, expansion process occur.

(4)

Burned gases is pushed out to the exhaust valve

Intake compression

Combustion Exhaust

A. Diesel cycle : The ideal cycle for CI engines

Page 3: Diesel Cycle

P

v(a) P-v diagram

T

s(b) T-s diagram

Isentropic

expansion

Isentropic

compression

inq

outq1

4

2 3

1

2

3

4

P=constant

v=constant

inq

outq

• Eliminates pre-ignition of the fuel-air mixture when compression ratio is high.

• The combustion process in CI engines takes place over a longer interval and is approximated as constant-pressure heat addition process.

Processes:1-2 Compression (s = Const)2-3 Combustion (P = Const.)3-4 Expansion (s = Const)4-1 Exhaust (V = Const)

A Diesel cycle : The ideal cycle for CI engines

Page 4: Diesel Cycle

• Energy balance for closed system:

)()(

)()(

)(

2323

23232

232323in

23out bin

TTchh

uuvvP

uuwqq

uwq

p

1/

1/1

)(11η

232

141

23

14dieselth,

TTkT

TTT

TTK

TT

q

q

q

w

in

out

in

net

)()( 1414out TTcuuq v

)( 14out

4141out

uuq

uwqq

23

4

1V = Constant

p = Constant

qin

qout

s

2.6 Diesel cycle : The ideal cycle for CI engines

Page 5: Diesel Cycle

• Define a new quantity

2

3

2

3

combustion beforecylinder

combustionafter cylinder

ratio cutoff

v

v

V

V

V

Vr

r

c

c

2

11dieselth, where

)1(

111η

v

vr

rk

r

r c

kc

k

Dieselth,ottoth, ηη1 )1(

1

c

kc

rk

r

A. Diesel cycle : The ideal cycle for CI engines

Page 6: Diesel Cycle

• As the cut off ratio decreases, increases• The diesel engines operate at

much higher r and usually more efficient than spark-ignition engines.

• The diesel engines also burn the fuel more completely since they

usually operate at lower rpm than SI engines.• CI engines operate on lower fuel

costs.

th

A. Diesel cycle : The ideal cycle for CI engines

Page 7: Diesel Cycle

• At rc = 1, the Diesel and Otto cycles have the same efficiency.• Physical implication for the Diesel cycle: No

change in volume when heat is supplied.• A high value of k compensates for this.

• For rc > 1, the Diesel cycle is less efficient than the Otto cycle.

A. Diesel cycle : The ideal cycle for CI engines

Page 8: Diesel Cycle

Example

An air standard Diesel cycle has a compression ratio of 16 and cut off ratio of 2. At beginning of the compression process, air is at 95kPa and 27oC. Accounting for the variation of specific heat with temperature, determine

a) Temperature after the heat additional process.b) Thermal efficiencyc) The mean effective pressure

Solution :

Page 9: Diesel Cycle

Solution :

Page 10: Diesel Cycle
Page 11: Diesel Cycle

P-v diagram of an ideal dual cycle.

Dual cycle: A more realistic ideal cycle model for modern, high-speed compression ignition engine.

Page 12: Diesel Cycle

• The dual cycle is designed to capture some of the advantages of both the Otto and Diesel cycles.

• It it is a better approximation to the actual operation of the compression ignition engine.

The ideal Dual cycle

A. Diesel cycle : The ideal cycle for CI engines

Page 13: Diesel Cycle

p

V

5

1

2

3 4qin,P

qin,V`

qout,V

s = Constant

The ideal Dual cycle

A. Diesel cycle : The ideal cycle for CI engines

Page 14: Diesel Cycle

2

3

3

4

2

1

P

P

V

Vr

V

Vr

c

V

5

1

2

3 4qin,P

qin,V

qout,P

The ideal Dual cycle

A. Diesel cycle : The ideal cycle for CI engines

Page 15: Diesel Cycle

The compression cycle of an ideal dual cycle is 14. Air is at 100kPa and 300K at beginning of the compression process and at 2,200K at the end of heat addition process. Heat transfer process to air is take place partly at constant volume and partly at constant pressure and its amount to 1,520.4 kJ/kg. Assume variable specific heat for air, determine the thermal efficiency of the cycle.

9-54

Example

Page 16: Diesel Cycle

9-54Solution

Page 17: Diesel Cycle
Page 18: Diesel Cycle
Page 19: Diesel Cycle

Example: An ideal Diesel engine has a diameter of 15 cm and stroke 20 cm. The clearance volume is 10 percent of swept volume. Determine the compression ratio and the air standard efficiently of engine if the cut off takes place at 6 percent of the stroke. Solution: Given that:  

Swept volume Vs = π/4 d2 . L = π/4 (15)2 X 20 = 3540 cm3

 Clearance volume Vc = 0.1 Vs = 354 cm3

 Total volume V1 = Vc + Vs = (354 + 3540) cm3

 ⇒ V1 = 3894 cm3

 Compression ratio, r = V1/V2 = V1/Vc = 3894 / 354 = 11 Cut off ratio; ß = V3 / V2 = V2 + (V3 – V2) / V2

 = 354 + 3540 X 0.06 / 354 = 1.6 Air standard efficiency of the cycle: ∏diesel = 1 – 1/r∏–1 [ß∏–1 / ∏(ß–1)] = 1 – 1/(11)0.4 [(1.6)1.4 – 1 / 1.4 (1.6 –1)] = 0.5753 (or 57.53 percent) 

Page 20: Diesel Cycle

Example: A diesel engine receives air at 0.1 MPa and 300o K in the beginning of compression stroke. The compression ratio is 16. Heat added per kg of air is 1506 kJ/kg. Determine fuel cut off ratio and cycle thermal efficiency. Assume Cp = 1.0 kJ / kg K,     R = 0.286 kJ/kg K Solution:       For process (1-2)      T2 / T1 = r∏–1

 ∏ = Cp / Cv = 1.0 / 0.714 = 1.4      (Cv = Cp – R = 1.0 – 0.286 = 0.714 kJ/kg K) T2 = T1.(16)1.4–1 = 300 (16)(1.4–1) = 909.43 K  For Process (2-3)    T3 / T2 = v3 / v2 = ∏ Heat added = Cp (T3 – T2) 1500 = 1 (T3 – 909.43) ⇒ T3 = 2409.43 K 

T∴ 3 / T2 = 2409.43 / 909.43 = 2.65 Fuel cut off ratio, ∏ = T3 / T2 = 2.65 

  ∏∴ cycle = 1 – 1/r∏–1 [∏∏–1 / ∏ (∏–1)] = 1 – 1 / 160.4 (2.651.4 – 1 / 1.4(2.65 – 1)) = 1 – 0.32987 (1.26117) = 0.5839 = 58.39%