performance of the four strokes diesel engine

Saif al-din ali Madhi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
25/6/2020 1 | P a g e
[heat lab 4]
University of Baghdad
Name: - Saif Al-din Ali -B-

25/6/2020 2 | P a g e
TABLE OF CONTENTS
1. Objective
2. Engine Performance
3. Starting the Engine
4. Test Procedure
5. Stopping the Engine
6. Calculations
7. Results
8. Discussion

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Experiment name
Performance of The Four Strokes Diesel Engine
1. Objective: The objective of this test is to study the effect of the speed
engine on its performance and another parameter
Figure 1 Schematic Diagram of The Test Rig
2. Engine Performance:
Engine performance is more precisely defined by:
1. The maximum power (or the maximum torque) available at each speed
within the useful engine operating range.
2. The range of speed and power over which engine operation is
satisfactory.
The following performance definitions are commonly used:
Maximum rated power. The highest power an engine is allowed to develop
for short periods of operation.

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Normal rated power. The highest power an engine is allowed to develop
in continuous operation. Once ignited and burnt, the combustion products
(hot gases) have more available thermal energy than the original
compressed fuel-air mixture (which had higher chemical energy). The
available energy is manifested as high temperature and pressure that can
be translated into work by the engine. In a reciprocating engine, the high-
pressure gases inside the cylinders drive the engine's pistons.
Once the available energy has been removed, the remaining hot gases
are vented (often by opening a valve or exposing the exhaust outlet) and
this allows the piston to return to its previous position (top dead center, or
TDC). The piston can then proceed to the next phase of its cycle, which
varies between engines. Any heat that isn't translated into work is normally
considered a waste product and is removed from the engine either by an air
or liquid cooling system.
Engine efficiency can be discussed in a number of ways but it usually
involves a comparison of the total chemical energy in the fuels, and the
useful energy extracted from the fuels in the form of kinetic energy. The
most fundamental and abstract discussion of engine efficiency is the
thermodynamic limit for extracting energy from the fuel defined by a
thermodynamic cycle. The most comprehensive is the empirical fuel
efficiency of the total engine system for accomplishing a desired task; for
example, the miles per gallon accumulated.
Engines using the Diesel cycle are usually more efficient, although the
Diesel cycle itself is less efficient at equal compression ratios. Since diesel
Engines use much higher compression ratios (the heat of compression is
used to ignite the slow-burning diesel fuel), that higher ratio more than
compensates for the lower intrinsic cycle efficiency, and allows the diesel
engine to be more efficient. The most efficient type, direct injection
Diesels, are able to reach an efficiency of about 40% in the engine speed
range of idle to about 1,800 rpm. Beyond this speed, efficiency begins to
decline due to air pumping losses within the engine.

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3. Starting the Engine:
The method of starting diesel engines as follow:
1. Turn On the tap of the fuel tank.
2. If the fuel system has been changed, air will have entered the fuel
pipes and will prevent the injector pump from working. The air must
be removed from the system. To do this loosens the bleed hex screw
on the top of the fuel pump until clean, bubble-free fuel leaks out.
Retighten the screw.
3. Move the Low/High lever to the High position. This sets the injector
to pump an excess of fuel into the engine for starting and is analogous
to the choke of a petrol engine.
4. If the engine is cold, it is essential to prime the engine for hand
starting.
a. Remove the priming plunger.
b. Fill the priming chamber with engine oil - not fuel. This is to
reduce leakage past the piston rings and to raise the
compression ratio.
c. Replace the priming plunger and press down.
5. Pull the starting rope to start the engine.
6. Should the engine fire and then stop, prime again (if cold) before
attempting to start again.
7. Allow the engine to warm up for about 5 minutes. The excess fuel
device resets automatically.
4. Test Procedure:
1. Advance the throttle or rack control to its maximum position.
2. Note the maximum speed of the engine. (The dynamometer water
flow should still be the trickle used for starting).
3. Keep the throttle or rack open and slowly adjust the needle valve to
increase the flow of water through the dynamometer until the needle
valve is fully open. Note the engine speed.
4. Assume, for the time being, that the engine will be tested over the
speed range just established. Choose at least five speeds between the
two extremes at which to take readings of engine performance.
5. Keep the throttle open and reduce the water flow to a trickle, so that
the engine returns to its maximum speed.

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6. When the engine has settled down to a steady output, record the
readings of speed, torque, exhaust temperature and air consumption.
Operate the fuel tap beneath the pipette so that the engine takes its
fuel from the pipette. Time the consumption of 8 ml of fuel. Turn the
tap so that the pipette again fills. Enter the results in Table 1.
7. Check that the temperature of the water flowing out of the
dynamometer is less than 80°C. If the temperature is higher than this,
increase the water flow to cool the dynamometer bearing seals.
8. Increase the flow of water into the dynamo meter until the engine
speed drops to the next highest selected value. Because the time
response of the dynamometer is fairly slow, the needle valve has to be
operated slowly.
9. Allow time for the engine speed to stabilize before taking another set
of results. If the dynamometer is too sensitive to obtain the desired
speed, it will help if the drain tap is partially closed. Do not close
fully.
10.Repeat Step (8) until the dynamometer needle valve is fully open.
11.Study the torque results. Engines normally produce a maximum
torque at a certain speed. If your results suggest that the maximum is
at a lower speed than you have reached, restrict the water flow from
the dynamometer.
5. Stopping the Engine
1. Reduce the water flow through the dynamometer to a trickle.
2. Close the throttle or rack control until the engine is running at a fast
idling speed.
3. Allow the engine to run for a few minutes.
4. Completely close the throttle or rack control. If the engine is a diesel,
move the Stop/Run lever to the Stop position.
5. Close the main fuel tap at the fuel tank
6. Obtain the values of ambient temperature and pressure and record in

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6. Calculations:
Where:
𝑚̇ 𝑎= Air mass flow rate (kg/hr).
𝑇 = Torque (Nm).
𝑁 = Speed Engine (Rev/min).
𝑆 = Specific Gravity.
QCV = Calorific Value of Fuel (kJ/kg).
Vs = Swept Volume (m3).
𝜌 𝑎= Air Density (kg/m3).
𝑇 𝑎= Air Temperature (°C).
𝑇 𝑒= Exhaust Temperature (°C).
𝑝 𝑜, 𝑇 𝑜= Ambient Pressure (kPa) and Temperature (°C).
𝑝 𝑠, 𝑇 𝑠= Standard Atmospheric Pressure (1 bar) and Temperature (25 °C).

25/6/2020 8 | P a g e
N(rpm) Te (C) P (mmH2o) t (s)
1250 200 6.5 59.7
1500 215 13 53.1
1750 233 15.5 49.2
2000 248 14 42.4
2250 256 17.5 32
Figure 2 Viscous Flow Meter Calibration
Air mass flow rate ( 𝒎̇ 𝒂 ) can be calculated from fig. 2 by intersect the
vertical line from manometer reading (mm H2O) with the calibration
line.
p (mmH2o) ma kg/hr
6.5 7.5
13 14.9
15.5 16.5
14 15.3
17.5 19

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7. Results
1.
∆P = 6.5 mmH2O = 0.000637 bar → 𝑃𝑂 = 1.000637bar
𝑃𝐵 =
2𝜋𝑁𝑇
60000
=
2𝜋×1250×6
60000
→ 𝑃𝐵 = 0.785 kW
𝑚. 𝑓 =
𝑆 ×8×3.6
𝑡
=
0.83 ×8×3.6
59.7
→ 𝑚. 𝑓 = 0.400402 kg/hr
𝑠𝑓𝑐 =
𝑚. 𝑓×1000
𝑃 𝐵
=
0.400402 ×1000
0.785
→ 𝑠𝑓𝑐 =510.0662548 g/kWh
AFR =
𝑚. 𝑎
𝑚. 𝑓
=
7.5
0.400402
→ AFR = 18.73117
𝜂 𝑣 =
𝑚. 𝑎
60×𝑁×𝜌 𝑎 ×𝑉 𝑆
=
7.5
60×1250×1.2 ×230×10−6
→ 𝜂 𝑣 = 36.21388 %
𝜂 𝐵 =
𝑃 𝐵×3600
𝑚. 𝑓×𝐶𝑉
=
0.785 ×3600
0.400402 ×40850
→ 𝜂 𝐵 =36.23188 %
Loss% =
100×(1+18.73117)(𝑇𝑒 −𝑇𝑎 )
𝐶𝑉
=
100×(1+0.005203)(200 −25)
40850
→
Loss = 8.452768 %
𝑃 𝐵(𝐶𝑜𝑟𝑟𝑒𝑐𝑡𝑒𝑑) = 𝑃 𝐵(𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑) ×
𝑃𝑆
𝑃𝑂
×
(𝑇𝑂 + 273)
(𝑇𝑆 + 273)
= 0.785 ×
1
1.00637
×
(20+273)
(25+273)
→
𝑃 𝐵(𝐶𝑜𝑟𝑟𝑒𝑐𝑡𝑒𝑑)= 0.771337517 kW
𝑚. 𝑎(𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑒𝑑) = 𝑚. 𝑎(𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑)×
𝑃𝑆
𝑃𝑂
×
(𝑇𝑂 + 273)
(𝑇𝑆 + 273)
= 7.5 ×
1
1.00637
×
(20+273)
(25+273)
→
𝑚. 𝑎(𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑒𝑑) = 7.369466724 kg/hr

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2.
∆P = 13 mmH2O = 0.00127 bar → 𝑃𝑂 = 1.00127 bar
𝑃𝐵 =
2𝜋𝑁𝑇
60000
=
2𝜋×1500×6
60000
→ 𝑃𝐵 = 0.942 kW
𝑚. 𝑓 =
𝑆 ×8×3.6
𝑡
=
0.83 ×8×3.6
53.1
→ 𝑚. 𝑓 = 0.4501695 kg/hr
𝑠𝑓𝑐 =
𝑚. 𝑓×1000
𝑃 𝐵
=
0.4501695×1000
0.942
→ 𝑠𝑓𝑐 =477.8869337 g/kWh
AFR =
𝑚. 𝑎
𝑚. 𝑓
=
14.9
0.4501695
→ AFR = 33.09864
𝜂 𝑣 =
𝑚. 𝑎
60×𝑁×𝜌 𝑎 ×𝑉 𝑆
=
14.9
60×1500×1.2 ×230×10−6
→ 𝜂 𝑣 = 59.9839 %
𝜂 𝐵 =
𝑃 𝐵×3600
=
0.942×3600
0.4501695 ×40850
→ 𝜂 𝐵 =18.44103464 %
Loss% =
100×(1+𝐴𝐹𝑅)(𝑇𝑒 −𝑇𝑎 )
𝐶𝑉
=
100×(1+33.09864)(215 −25)
40850
→
Loss = 15.85983 %
𝑃𝑆
𝑃𝑂
×
(𝑇𝑂 + 273)
(𝑇𝑆 + 273)
= 0.942 ×
1
1.00127
×
(20+273)
(25+273)
→
𝑃𝑆
𝑃𝑂
×
(𝑇𝑂 + 273)
(𝑇𝑆 + 273)
= 14.9 ×
1
1.00127
×
(20+273)
(25+273)
→

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3.
∆P = 15.5 mmH2O = 0.0015 bar → 𝑃𝑂 = 1.0015 bar
𝑃𝐵 =
2𝜋𝑁𝑇
60000
=
2𝜋×1750×6
60000
→ 𝑃𝐵 = 1.099 kW
𝑚. 𝑓 =
𝑆 ×8×3.6
𝑡
=
0.83 ×8×3.60
49.2
→ 𝑚. 𝑓 = 0.4858537 kg/hr
𝑠𝑓𝑐 =
𝑚. 𝑓×1000
𝑃 𝐵
=
0.4858537×1000
1.099
→ 𝑠𝑓𝑐 =442.0870414g/kWh
AFR =
𝑚. 𝑎
𝑚. 𝑓
=
16.5
0.4858537
→ AFR = 33.96084
𝜂 𝑣 =
𝑚. 𝑎
60×𝑁×𝜌 𝑎 ×𝑉 𝑆
=
16.5
60×1750×1.2 ×230×10−6
→ 𝜂 𝑣 = 56.93582 %
𝜂 𝐵 =
𝑃 𝐵×3600
=
1.099 ×3600
0.4858537×40850
→ 𝜂 𝐵 =19.93437643 %
Loss% =
100×(1+𝐴𝐹𝑅)(𝑇𝑒 −𝑇𝑎 )
𝐶𝑉
=
100×(1+33.96084)(233 −25)
40850
→
Loss = 17.80136 %
𝑃𝑆
𝑃𝑂
×
(𝑇𝑂 + 273)
(𝑇𝑆 + 273)
= 1.099 ×
1
1.00637
×
(20+273)
(25+273)
→
𝑃𝑆
𝑃𝑂
×
(𝑇𝑂 + 273)
(𝑇𝑆 + 273)
= 16.5 ×
1
1.0015
×
(20+273)
(25+273)
→

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4.
bar𝑃𝑂 = 1.00137→P = 14 mmH2O = 0.00137 bar∆
𝑃𝐵 =
2𝜋𝑁𝑇
60000
=
2𝜋×2000×6
60000
→ 𝑃𝐵 = 1.256 kW
𝑚. 𝑓 =
𝑆 ×8×3.6
𝑡
=
0.83 ×8×3.6
42.4
→ 𝑚. 𝑓 = 0.5637736 kg/hr
𝑠𝑓𝑐 =
𝑚. 𝑓×1000
𝑃 𝐵
=
0.5637736 ×1000
1.256
→ 𝑠𝑓𝑐 =448.8643192 g/kWh
AFR =
𝑚. 𝑎
𝑚. 𝑓
=
15.3
0.5637736
→ AFR = 27.13855
𝜂 𝑣 =
𝑚. 𝑎
60×𝑁×𝜌 𝑎 ×𝑉 𝑆
=
15.3
60×2000×1.2 ×230×10−6
→ 𝜂 𝑣 = 46.19565 %
𝜂 𝐵 =
𝑃 𝐵×3600
=
1.256 ×3600
0.5637736 ×40850
→ 𝜂 𝐵 =19.63339281 %
Loss% =
100×(1+𝐴𝐹𝑅)(𝑇𝑒 −𝑇𝑎 )
𝐶𝑉
=
100×(1+27.13855 )(248−25)
40850
→
Loss = 15.36083 %
𝑃𝑆
𝑃𝑂
×
(𝑇𝑂 + 273)
(𝑇𝑆 + 273)
= 1.256 ×
1
1.00137
×
(20+273)
(25+273)
→
𝑃𝑆
𝑃𝑂
×
(𝑇𝑂 + 273)
(𝑇𝑆 + 273)
= 15.3 ×
1
1.00137
×
(20+273)
(25+273)
→

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5.
∆P = 17.5 mmH2O = 0.0017 bar → 𝑃𝑂 = 1.0017 bar
𝑃𝐵 =
2𝜋𝑁𝑇
60000
=
2𝜋×2250×6
60000
→ 𝑃𝐵 = 1.413 kW
𝑚. 𝑓 =
𝑆 ×8×3.6
𝑡
=
0.83 ×8×3.6
32
→ 𝑚. 𝑓 = 0.747 kg/hr
𝑠𝑓𝑐 =
𝑚. 𝑓×1000
𝑃 𝐵
=
0.747 ×1000
1.413
→ 𝑠𝑓𝑐 =528.6624204 g/kWh
AFR =
𝑚. 𝑎
𝑚. 𝑓
=
19
0.747
→ AFR = 25.43507
𝜂 𝑣 =
𝑚. 𝑎
60×𝑁×𝜌 𝑎 ×𝑉 𝑆
=
19
60×2250×1.2 ×230×10−6
→ 𝜂 𝑣 = 50.9930 %
𝜂 𝐵 =
𝑃 𝐵×3600
=
1.413 ×3600
0.747 ×40850
→ 𝜂 𝐵 =16.66986182 %
Loss% =
100×(1+𝐴𝐹𝑅)(𝑇𝑒 −𝑇𝑎 )
𝐶𝑉
=
100×(1+ 25.43507)(256−25)
40850
→
Loss = 14.9486 %
𝑃𝑆
𝑃𝑂
×
(𝑇𝑂 + 273)
(𝑇𝑆 + 273)
= 1.413 ×
1
1.0017
×
(20+273)
(25+273)
→
𝑃𝑆
𝑃𝑂
×
(𝑇𝑂 + 273)
(𝑇𝑆 + 273)
= 19 ×
1
1.0017
×
(20+273)
(25+273)
→
After correcting both the ma and PB values, the calculation process
was repeated, based on the complete program written in Excel

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N
(rpm)
Te
(C)
Dp
(mmH2o)
t (s) ma
Po
(bar)
PB
(Kw)
m.f
(kg/hr)
Sfc
g/kWh
AFR eff v eff B loss
PB
CORR
(KW)
m.a
corr
(kg/hr)
1250 200 6.5 59.7 7.5 1.000637 0.785 0.400402 510.06 18.73117 36.23188 17.2776 8.452768 0.7713 7.36946
1500 215 13 53.1 14.9 1.00127 0.942 0.4501695 477.886 33.09864 59.9839 18.4414 15.85983 0.925 14.63
1750 233 15.5 49.2 16.5 1.0015 1.099 0.4858537 442.0870 33.96084 56.93582 19.93643 17.80136 1.0789 16.1988
2000 248 14 42.4 15.3 1.00137 1.256 0.5637736 448.864 27.13855 46.19565 19.633 15.36083 1.2332 15.0228
2250 256 17.5 32 19 1.0017 1.413 0.747 528.662 25.43507 50.99302 16.682 14.9486 1.3869 18.649
N
(rpm)
Te
(C)
dp
(mmH2o)
t (s)
m.a
Corr
(kg/hr)
Po
(bar)
PB
CORR
(KW)
m.f
(kg/hr)
Sfc
g/kWh
PB
CORR
(KW)
m.a corr
(kg/hr)
1250 200 6.5 59.7 7.3694667 1.000637 0.77133 0.400402 519.100 18.40517 35.60129 16.976 8.313108 0.7579 7.241205
1500 215 13 53.1 14.631418 1.00127 0.925056 0.4501695 486.659 32.50202 58.90265 18.108 15.58234 0.9083 14.3676
1750 233 15.5 49.2 16.198856 1.0015 1.07899 0.4858537 450.305 33.34102 55.89667 19.5702 17.48576 1.0593 15.9038
2000 248 14 42.4 15.022707 1.00137 1.2332 0.5637736 457.14 26.6467 45.35842 19.271 15.09233 1.2109 14.75053
2250 256 17.5 32 18.649504 1.0017 1.38698 0.747 538.598 24.96587 50.05235 16.3616 14.68327 1.3613 18.3045
N
(rpm)
Te
(C)
Dp
(mmH2o)
t (s)
m.a
Corr
(kg/hr)
Po
(bar)
PB
CORR
(KW)
m.f
(kg/hr)
sfc
g/kWh
PB
CORR
(KW)
m.a corr
(kg/hr)
1250 200 6.5 59.7 7.2412053 1.000637 0.7579 0.400402 528.295 18.08484 34.98167 16.68143 8.175879 0.7447 7.11505
1500 215 13 53.1 14.367678 1.00127 0.9089 0.4501695 495.5926 31.91615 57.84089 17.7829 15.30984 0.892 14.101
1750 233 15.5 49.2 15.903208 1.0015 1.0592 0.4858537 458.677 32.73251 54.8765 19.2162 17.17592 1.0399 15.6129
2000 248 14 42.4 14.750441 1.00137 1.21088 0.5637736 465.5877 26.16377 44.53635 18.98255 14.82869 1.1889 14.4836
2250 256 17.5 32 18.305473 1.0017 1.36134 0.747 548.720 24.50532 49.12902 16.05121 14.42284 1.3362 17.9639

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8.Discussion
Note that the relationship is direct
Assuming the science is linear, we notice that the relationship is direct
between efficiency and speed
0
100
200
300
400
500
600
0 500 1000 1500 2000 2500
sfcg/kWh
N(rpm)
sfc g/kWh sfc corr sfc corr 1
Linear (sfc g/kWh) Linear (sfc corr) Linear (sfc corr 1)
0
10
20
30
40
50
60
70
0 500 1000 1500 2000 2500
effv%
N(rpm)
eff v eff v corr eff v corr
Linear (eff v) Linear (eff v corr) Linear (eff v corr )

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We note that the relationship between temperature and velocity is
direct
We note that the relationship between temperature and velocity is
direct
0
50
100
150
200
250
300
0 500 1000 1500 2000 2500
Te(C)
N(rpm)
Te (C) Linear (Te (C))
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
0 500 1000 1500 2000 2500
PB(Kw)
N(rpm)
PB (Kw) PB CORR (KW) PB CORR (KW)

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Inverse relationship, assuming it is linear
A direct relationship with the assumption that it is linear
0
0.00001
0.00002
0.00003
0.00004
0.00005
0.00006
0 500 1000 1500 2000 2500
effB%
N(rpm)
eff B eff B corr eff B corr 1
Linear (eff B) Linear (eff B corr) Linear (eff B corr 1)
0
0.002
0.004
0.006
0.008
0.01
0.012
0 500 1000 1500 2000 2500
AFR
N(rpm)
AFR ARF corr ARF CORR 1
Linear (AFR) Linear (ARF corr) Linear (ARF CORR 1)

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