Thermal engineering

1. :
MGM College of Engineering and Technology
Department of Mechanical Engineering
By-Dr. Subhash Kamal
Thermal Engineering
MEC502
Module -6
Engine Testing and Performance

2. Module -6 topics:
6.1 Engine Testing and Performance:
Measurement of various performance parameters, Performance
characteristic of SI and CI Engine, Effect of load and speed on
performance parameters, Heat balance sheet.
6.2 Engine Emission and Control:
Sources of Engine Emissions, Constituents of S.I. and C.I. Engine
exhaust and their effects on environment and health. Study of
emission (Euro & Bharat stage) norms, Control methods for S.I
and C I engine emissions.

3. Engine Efficiencies and Performance
Indicated Power (IP)
Indicated power is defined as the power developed inside the
engine cylinder. In measuring the indicated power,

4. Brake Power (BP)
It is useful power available at the crank shaft or clutch shaft. The brake
power is less than indicated power because of the following losses as
power flows from the cylinder to the crank shaft.
i) Friction between the cylinder surface and piston rings, in bearings,
gears, valve mechanism etc.
ii) Resistance of air to fly wheel rotation
iii) Power required to drive auxiliaries – fuel pump, lubrication pump,
radiator circulation pump etc.

5. Indicate power
developed
Break
Power(BP)

6. Friction Power
F.P. = I.P. - B.P.
Friction Power = Indicated Power – Brake Power

9. Problem 1:
A 2-S cycle internal combustion engine has a mean effective pressure of 6
bar. The speed of the engine is 1000 r.p.m. If the diameter of piston and
stroke are 110 mm and 140 mm respectively, find the indicated power
developed.
Given
Actual mean effective pressure, pam = 6 bar= 600KPa;
Engine speed, N = 1000 r.p.m;
Diameter of the piston, D = 110 mm = 0.11 m;
Stroke length, L = 140 mm = 0.14 m;
No. of cylinders = 1
I.P = 13.305 kW

10. Problem 2
A 4-cylinder four-stroke petrol engine develops 14.7 kW at 1000 r.p.m. The
mean effective pressure is 5.5 bar, Calculate the bore and stroke of the
engine, if the length of stroke is 1.5 times the bore.
Given:
Number of cylinder, x = 4;
Power developed, I.P. = 14.7 kW;
Engine speed, N = 1000 r. p. m.;
Actual mean effective pressure,
Pam = 5.5 bar = 550KPa
Length of stroke, L = 1.5 D (bore)
D= 0.0879 or 87.9 mm
L= 1.5 D = 1.5x 87.9
L = 131.8 mm

11. Problem 3
A rope brake was used to measure the brake power of a single cylinder,
four stroke cycle petrol engine. It was found that the torque due to brake
load is 175 N-m and the engine makes 500 r. p.m. Determine the brake
power developed by the engine.
Given:
Torque due to brake load, T = 175 N-m .
Engine speed, N = 500 r.p.m.
Determine the brake power, B.P.

12. Problem 4
A single cylinder, four stroke cycle oil engine is fitted with a rope brake.
The diameter of the brake wheel is 600 mm and the rope diameter is 26
mm. The dead load on the brake is 200 N and the spring balance reads 30
N. If the engine runs at 450 r.p.m. what will be the brake power of the
engine?
Given:
Diameter of the brake wheel, Db = 600 mm = 0.6 m
Rope diameter, d = 26 mm = 0.026 m
Dead load on the brake, W= 200 N
Spring balance reading, S = 30 N
Engine speed, N = 450 r.p.m.

13. Problem 5:
Following data refers to an oil engine working on Otto four stroke cycle
Brake power = 14.7 kW
Suction pressure = 0.9 bar
Mechanical efficiency = 80%
Ratio of compression = 5
Index of compression curve = 1.35
Index of expansion curve = l.3
Maximum explosion pressure = 24 bar
Engine speed = 1000 r.p.m.
Ratio of stroke: bore (L/D) = 1.5
Determine the diameter and stroke of the piston?
Given:
Brake power, B.P. = 14.7 kW
Suction pressure, p1 = 0.9 bar =90KPa
Mechanical efficiency,
ηmech = 80%
Index of compression curve, nc = 1.35;
Index of expansion curve, ne = l.3;
Maxi. explosion pressure, p3 = 24 bar =2400KPa
Engine speed, N = 1000 r.p.m.;
Ratio of stroke: bore, L/D = 1.5
we have L = 1.5D;
Compression ratio = rc

14. Finding unknown, pm and I.P.
Finding unknown, pm
pm = Mean effective pressure

17. length of the stroke, L = 1.5D
L = 1.5 x 0.142
L = 0.213 m

18. Problem 6:
The following readings were taken during the test of a single cylinder four stroke
oil engine:
Cylinder diameter = 250 mm, Stroke length = 400 mm,
Gross m.e.p. = 7 bar
Pumping m.e.p. = 0.5 bar, Engine speed = 250 r. p. m.,
Net load on the brake = 1080 N,
Effective diameter of the brake = 1.5 m; Fuel used per hour = 10 kg
Calorific value of fuel = 44300 kJ/kg
Calculate:
(i)Indicated power
(ii)Brake power
(iii) Mechanical efficiency
(iv) Indicated thermal efficiency.
Given
Net Pam = (pam)gross- (pam)pumping = 7 – 0.5 = 6.5 bar
Net load on the brake, (W-S) = 1080 N

21. Home work problems
Problem 1:
The following observations were recorded during a test on a 4-stroke
engine. Bore = 25cm, stroke=40cm, crank speed=250 rpm, net load on the
brake drum=700N, diameter of brake drum=2m, indicated mean effective
pressure=6bar, fuel consumption=0.0013kg/s, specific gravity of
fuel=0.78, calorific value of fuel=43900kJ/kg.
Determine (i) BP, (ii) IP, (iii) FP, and (iv) mechanical efficiency (v) indicated
and brake thermal efficiency.
Problem 2:
The following are the details of a 4-stroke petrol engine. (i) diameter of
brake drum=60.03cm, (ii) full brake load on drum=250N, (iii) brake drum
speed = 450 rpm, (iv) calorific value of petrol = 40MJ/kg, (v) brake thermal
efficiency=32%, (vi) mechanical efficiency=80%, specific gravity of
petrol=0.82.
Determine (i) brake power, (ii) indicated power, (iii) fuel consumption in
liter per second, and (iv) indicated thermal efficiency

22. Problem 3:
A person conducted a test on a single cylinder two-stroke petrol engine
and found that the mechanical efficiency and brake thermal efficiency of
the engine are 0.7 and 0.2 respectively. The engine with a mean effective
pressure of 6bar ran at 300 rev/min consuming fuel at a rate of 2.2kg/hr.
Given that the calorific value of fuel is 42500 kJ/kg and that the stroke to
bore ratio of the engine is 1.2,
Find the bore and stroke of the engine

23. Heat balance sheet
The real picture of overall performance of an engine can be made clear by
its heat balance sheet. Heat balance sheet of an engine is prepared by
running it under steady state conditions i.e. at constant load and r.p.m.
The work produced is measured with the help of P-V diagram drawn by
indicator mechanism. Other parameters noted are: quantity of fuel used in
a given time, Calorific value of fuel, cooling water flow rate and its inlet
and outlet temperatures, weight of exhaust gases and their temperature.
Calculations can be done based on this data to find account of all heat
input. A sample of heat balance sheet is given:
(a) Total heat input
It can be calculated as the product of rate of mass of fuel used and its
calorific value. Hf = Mf x C V
(b) Total output (BP)
i)Heat absorbed in producing Indicated Power or
the heat energy converted to work.
It can be calculated by indicator mechanism

24. ii) Heat taken away by Cooling Water:
It can be calculated as Hw = Mw x Cpw x (ti – to)
Where,
Mw=Cooling Water flow rate
Cw = Specific heat of cooling water
ti = Inlet temperature of Cooling Water
t0 = Outlet temperature of Cooling Water
iii) Heat taken away by exhaust gases:-
It can also be calculated likewise as Hg= Mg x Cpg x (te – tr)
Where,
Me = Mass of exhaust gases produced in a given time
Cpg = Specific heat of exhaust gases
te = temperature of exhaust gases
tr = room temperature
iv) Heat unaccounted for:- Ha = Hf- (BP+Hw+Hg)
It is the heat energy which is not accounted and is difference of
total heat input minus sum of above three components of output.
It is due to loss of heat by radiation and other unaccounted
losses.

25. Problem 1:
An I.C. engine uses 6 kg fuel having calorific value 44000
kJ/kg. in one hour. The brake power developed is 18kW. The
temperature of 11.5 kg of cooling water found to rise through
25 0C per minute. The temperature of 4.2 kg of exhaust gas
with specific heat 1 kJ/kg K was found to rise though 220oC.
Draw heat balance sheet for the engine.
Mass of Fuel = 6 Kg/hr = 6/60 = 0.1 Kg/min.
BP = 18 KW CV = 44000 kJ/kg
Mass of cooling water Mw = 11.5 Kg/min
Cpw = 4.187 kJ/kg K
Temp rise of cooling water Δtw= 25oC
Mass of exhaust gas Meg = 4.2 Kg/hr =4.2/60 Kg/min = 0.7 Kg/min
Temp rise of gas Δtg= 220 0C
Cpg = 1 kJ/kg K
i)Heat equivalent in Fuel (Hf )
Hf = Mf x C V = 0.1 x 44000
Hf = 4400 Kg/min

26. ii) Heat converted in B P (Hb )
Hb = B P x 60 = 18 x 60
Hb= 1080 Kg/min
iii) Heat carried by cooling water (Hw)
Hw = Mw x Cpw x Δtw
Hw = 11.5 x 4.187 x 25
Hw = 1203.76 Kg/min
iv) Heat in Exhaust Gas (Hg )
Hg= Mg x Cpg x Δtg
Hg = 0.7 x 1 x 220 = 154 Kg/min
v) Heat lost as Unaccounted (Ha)
Ha= Hf- (Hb+Hw+Hg)
Ha = 4400 – (1080+1203.76+154)
Ha= 1962.24 Kg/min

28. Problem-2
The following observation where recorded in a test of hour duration a
single cylinder oil engine on four stroke cycle.
Bore = 300 mm
Stroke = 450 mm
Fuel used = 8.8 kg
Calorific value = 41800 kJ/kg
Avg. speed = 200 rpm
Brake load = 1860 N
MEP = 5.8 bar
Quantity of cooling water = 650 kg
Temperature rise in water = 22oC
Diameter of brake wheel = 1.22 m
Calculate
1. Mech. efficiency,
2. Break thermal η also
3. Draw heat balance sheet

29. D = 0.3m
ΔTw = 22oC
L = 0.45m
Dw = 1.22m
mf = 8.8 kg/hr
R =1.22/2 =0.61 m
CV = 41800 kJ/kg
N = 200 rpm
W = 1860 N
pm = 5.8 bar
mw = 650 kg/hr
Given data, SC, 4S, SI engine
Brake Torque (T) = W.R
= 1860x0.61 N-m
= 1134.6 J,
I.P = 30.74 x60 = 1844.4 kJ/min

30. For the Heat Balance Sheet
Total Heat supply by the fuel(Qs)
(Qs) = mf x CV
mf = 8.8 Kg/hr
CV = 41800 kJ/kg
Qs = 8.8 x41800
Qs = 8678400 kJ/hr = 8678400/60 = 6130.68 KJ/min

31. I.P = 30.75KW = 30.75x 60 = 1844.4 KJ/min.
Heat rejected to cooling water
QCW = mcw Cpw (Two - Tci)
QCW = mcw Cpw (ΔT)cw
QCW = 650/60 x4.2 (2)
QCW = 997.66 KJ/min
Uncountable Heat
Quc = Qs - (IP + Qcw)
Quc = 6130.68 - (1844.4 + 997.66)
Quc = 3288.58

32. SI Engine

35. 6.2 Engine Emission and Control:
topics
1.Sources of Engine Emissions,
1.Constituents of S.I. and C.I. Engine exhaust
and their effects on environment and health.
1.Study of emission (Euro & Bharat stage)
norms,
1.Control methods for S.I and C I engine
emissions.

37. Sources of Engine Emissions,
There are four main sources of these emissions:
1. Engine exhaust,
2. Crankcase,
3. Fuel tank and carburetor
1. Engine exhaust,
The exhaust pipe discharges burned and unburned hydrocarbons, carbon
monoxide (CO), oxides of nitrogen and sulfur(Nox & Sox), and traces of various
acids, alcohols, and phenols. (60%)
2. Crankcase ventilation (20%)
The crankcase is a secondary source of unburned
hydrocarbons and, to a lesser extent, carbon monoxide
3. Fuel tank and Carburetor. (20%)
In the fuel tank, hydrocarbons that are continuously
evaporating from petrol constitute, but its very minor

39. 2. Constituents of S.I. and C.I. Engine exhaust and their
effects on environment and health.

40. SOX and NOX effects on human health
On the skin it produces burns. Other health effects include headache,
general discomfort and anxiety. People with impaired heart or lung
function and asthmatics are at increased risk. Repeated or prolonged
exposure to moderate concentrations may cause inflammation of the
respiratory tract, wheezing and lung damage
Carbon monoxide (CO)
It is a toxic air pollutant produced largely from vehicle emissions.
Breathing CO at high concentrations leads to reduced oxygen transport
by hemoglobin, which has health effects that include
1. impaired reaction timing,
2. headaches,
3. Lightheadedness,
4. nausea,
5. vomiting,
6. weakness, c
7. louding of consciousness,
8. coma, and,
at high enough concentrations and long enough exposure to death.

41. S.N Name of pollutant Symbol Effects on human health
1 Carbon monoxide CO Have great effect on the oxygen delivery to the body’s
organs and tissues, and may cause death
2 Nitrogen dioxides NOX Linked to a wide range of respiratory problems; cough
and sore throat.
3 Ozone O3 Can cause chest pain, coughing and shortness of breath
4 Heavy metals e.g. Pb Exposure will lead to irreversible damage to brain and
in some cases, also lead to premature deaths.
5 Particulate matter PM Acute exposure will cause death, increased respiratory
symptoms, decreased lung function, and pulmonary
inflammation. Chronic exposure will increased mortality
rates, chronic cardiopulmonary disease, and reduce lung
function
6 Hydrocarbons HC HC, NOX and sunlight can generate the photochemical
smog. Some hydrocarbon compounds caused irritation
to eye and damage lungs such as benzene (C6H6). At
high concentration it causes asthma and death.
7 Polycyclic
aromatic
hydrocarbons
PAHs Mutagenic and carcinogenic…causes caners

44. It was in 1991 that first time emission norms were introduced in India for
petrol cars; diesel cars followed in 1992. From then onwards,
new cars manufactured in India had to adhere to these standards; their
exhaust fumes could not contain more than specified quantity of
pollutants.
These standards were compounded with the implementation of
mandatory catalytic converters in 1995 for the 4 Metro cities, thus
reducing pollution further
From 2000, India introduced stricter Emission standards modeled on the
European ones. This meant the birth of Bharat Norms, with the first set of
norms known as Bharat Stage II, followed by BS III, and BS IV (BS I was
the earlier, Indian standard).
The tables given below give details of Emission norms at different stages
and area of implementation. Here we are focusing on petrol engines, but
diesel engines have similar norms also.
3. Study of emission norms, (Bharat stage)

46. Indian emission standards (4-wheel vehicles)

49. Excessive HC emissions may be
caused by
1.Ignition system misfiring
2.Improper ignition timing
3.Excessively lean or rich air/Fuel ratio
4.Low cylinder compression
5.Defective valves, guides, or filters
6.Defective rings, pistons or cylinders
7.Vacuum leaks
Excessive CO emissions are caused
by
1.Rich air/fuel mixtures
2.Dirty air filter
3.Faulty injectors
4.Higher than normal fuel pressures
5.Defective system input sensor
Excessive HC and CO emissions
caused by
1.Plugged Positive Crank case
Ventilation system (PCV system)
2.Excessively rich air/Fuel ratio
3.Stuck open heat riser valve
4.AIR pump inoperative or disconnected
5.Engine oil diluted with gasoline
Higher than normal NOX emissions may
be caused by
1.An overheated engine
2.Lean air/fuel mixtures
3.Vacuum leaks
4.Over advanced ignition timing
5.Defective EGR system
Causes of emission

50. Euro Emission Standards
The first European exhaust emissions standard for passenger cars was
introduced in 1970

52. 4. Emission Control methods for S.I engine
emissions.
1. Crankcase Emission Control (PCV System)
2. Evaporative Emission Control
3. Exhaust Gas Recirculation
1. Crankcase Emission Control (Positive
Crankcase Ventilation PCV System)
A small amount of charge in the cylinder
leaks past piston rings into crankcase of
the reciprocating engines. Near top dead
centre (TDC) when the rings change their
position in the grooves at the end of
compression stroke, combustion has
already begun and the cylinder pressures
are high. A significant part of charge
stored in the piston- ring-cylinder crevice
leaks into the crankcase. These gases are
known as ‘crankcase blow by and their
flow rate increases as the engine is worn
out
and the piston - cylinder clearances and ring gaps increase. In the
homogeneous charge engines, the crankcase blow by gas is high in HC
concentration. Only a small fraction of the gas stored in the ring crevices

53. and hence blow by gasses gases may consist of partially burnt mixture.
This source contributes about 20 percent of total hydrocarbons emitted
by an uncontrolled car. The crankcase blowby gases in the uncontrolled
engines were ventilated to atmosphere under the effect of pressure
difference occurring naturally between the crankcase and atmosphere.
For control of crankcase emissions, the blowby gases are recycled back
to the engine assisted by a positive pressure drop between the crankcase
and intake manifold. When engine is running and intake charge is
throttled the intake manifold is at a lower pressure than the crankcase.
The blow-by gases mix with the intake charge to be burned inside the
engine cylinder to CO2 and H2O. A typical PCV system is shown in Fig. A
tube connects crankcase or cylinder head cover to the intake manifold
below throttle valve, which leads the blowby gases back to the engine.
Due to suction effect of intake manifold as the pressure in the crankcase
falls, ventilation air from the air cleaner is drawn into the crankcase that
continuously purges it. A one-way valve (PCV valve) is used to control the
flow of blowby gases PCV valve restricts flow of blowby gases during
idling and very light loads which otherwise would cause excessive
leaning of the charge by ventilation air.
https://www.youtube.com/watch?v=JotiMO3R8bQ

54. 2. Evaporative Emission Control
In the uncontrolled vehicles, fuel vapours from the fuel tank and
carburettor were vented into the atmosphere that constituted about 20%
of all hydrocarbon emissions from a gasoline passenger car. From 1970,
evaporative emission control was required to be employed on production
gasoline vehicles in the USA.
The evaporative emission control system consists of a device to store
fuel vapours produced in the fuel system due to evaporation.

55. A canister containing activated charcoal is used to store the fuel vapours.
The vapours produced in the fuel tank normally collect in the fuel tank
itself and are vented to the charcoal canister when fuel vapour pressure
becomes excessive. The fuel vapours from the tank and carburetor led to
and adsorbed into the charcoal. In the Port Fuel Injection (PFI) engines
only the fuel tank is connected to the canister.
When engine is running, the vacuum created in the intake manifold is
used to draw fuel vapours from the canister into the engine. Purging air is
sucked through the canister which leads the fuel vapours from canister to
the engine. An electronically controlled purge valve is used.
During engine acceleration additional mixture enrichment can be
tolerated and under these operating conditions the stored fuel vapours
are usually purged into the intake manifold.
This system is a fully closed system. A sealed fuel tank filler cap is used
and a stable fuel tank pressure is maintained by the purging process of
the canister.
https://www.youtube.com/watch?v=10F3f8Ytpv8

56. 3. Exhaust Gas Recirculation
An EGR control valve is used to
regulate flow of EGR depending upon
engine operating conditions. The intake
manifold pressure or exhaust back
pressure may be used to control EGR
rate as these parameters vary with
engine load. In the modern engines,
EGR rate is controlled by the engine
electronic control unit. A pressure
sensor in the exhaust or intake provides
signal to the electronic control module
of the engine, which in its turn regulates
the operation of the EGR valve.
Electronically controlled EGR valves actuated by high-response stepper
motor are being used on modern engines. Their fast response during
transient operation makes it possible to reduce NOx more than what is
obtained by use of a mechanically controlled EGR valves. Effectively a
lower rate of EGR can be employed to obtain the same reduction in NOx
that results in a lower penalty on HC emissions
https://www.youtube.com/watch?v=neU7RudLg-c

58. 4. Emission Control methods for C I engine
emissions.
3. Exhaust Gas Recirculation
Same as back

59. The End
Module -6