NAZARIN B. NORDIN nazarin@icam.edu.my

forestercuckooMechanics

Oct 27, 2013 (3 years and 9 months ago)

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NAZARIN B. NORDIN

nazarin@icam.edu.my

What you will learn:


First law of thermodynamics


Isothermal process, adiabatic process,
combustion process for petrol/diesel engines


Volumetric Efficiency; spark ignition/
compression; ignition process and tests


First Law of Thermodynamics

Conservation of Energy for Thermal
Systems

Joule Equivalent of Heat



James Joule showed that mechanical energy
could be converted to heat and arrived at the
conclusion that heat was another form of
energy.


He showed that 1 calorie of heat was
equivalent to 4.184 J of work.








1 cal = 4.184 J

Energy


Mechanical Energy: KE, PE, E


Work is done by energy transfer.


Heat is another form of energy.


Need to expand the conservation of energy
principle to accommodate thermal systems.

1
st

Law of Thermodynamics


Consider an
example system of
a piston and
cylinder with an
enclosed dilute
gas characterized
by P,V,T & n.

1
st

Law of Thermodynamics


What happens to
the gas if the
piston is moved
inwards?

1
st

Law of Thermodynamics


If the container is
insulated the
temperature will
rise, the atoms
move faster and the
pressure rises.


Is there more
internal energy in
the gas?

1
st

Law of Thermodynamics


External agent
did work in
pushing the
piston inward.


W = Fd



=(PA)
D
x


W =P
D
V


D
x

1
st

Law of Thermodynamics


Work done on
the gas equals
the change in the
gases internal
energy,

W =
D
U


D
x

1
st

Law of TD



Let’s change the
situation:


Keep the piston fixed
at its original location.


Place the cylinder on a
hot plate.


What happens to gas?

Heat flows into the gas.

Atoms move faster,
internal energy
increases.

Q = heat in Joules

D
U = change in internal
energy in Joules.


Q =
D
U

1
st

Law of TD


What if we added
heat and pushed
the piston in at
the same time?

F

1
st

Law of TD


Work is done on the
gas, heat is added to
the gas and the
internal energy of the
gas increases!



Q = W +
D
U

F

1
st

Law of TD

Some conventions:

For the gases perspective:


heat added is positive, heat removed is
negative.


Work done on the gas is positive, work done
by the gas is negative.


Temperature increase means internal energy
change is positive.

1
st

Law of TD


Example: 25 L of gas is enclosed in a
cylinder/piston apparatus at 2 atm of pressure
and 300 K. If 100 kg of mass is placed on the
piston causing the gas to compress to 20 L at
constant pressure. This is done by allowing
heat to flow out of the gas. What is the work
done on the gas? What is the change in
internal energy of the gas? How much heat
flowed out of the gas?


P
o

= 202,600 Pa, V
o

= 0.025 m
3
, T
o

= 300 K, P
f

=
202,600 Pa, V
f
=0.020 m
3
, T
f
=




n = PV/RT.




W =
-
P
D
V




D
U = 3/2 nR
D
T




Q = W +
D
U

W =
-
P
D
V =
-
202,600 Pa (0.020


0.025)m
3



=1013 J energy added to the gas.

D
U =3/2 nR
D
T=1.5(2.03)(8.31)(
-
60)=
-
1518 J

Q = W +
D
U = 1013


1518 =
-
505 J heat out


Performance Factors

Volumetric Efficiency

1a. Indicated Power.

Indicated Power (IP) : Power obtained at the cylinder. Obtained
from the indicator diagram. Given by:

IP = P
i
LANn/60x in Watts



where P
i

is the indicated mean effective pressure, in
N/m
2
,



L is the stroke length, in m




A is the area of cross section of the piston, m
2
,




N is the engine speed in rev/min,




n is the number of cylinders and




x =1 for 2 stroke and 2 for 4 stroke engine.

1b. Brake Power

Brake Power (BP) : Power obtained at the shaft.
Obtained from the engine dynamometer.

Given by:

BP = 2

NT/60 in Watts

where T is the brake torque, in Nm, given by

T = W.L

where W is the load applied on the shaft by the
dynamometer, in N and


L is the length of the arm where the load is
applied, in m


N is the engine speed, in rev/min

1c. Friction Power

Friction Power (FP) : Power dissipated as
friction. Obtained by various methods like
Morse test for multi
-
cylinder engine, Willan’s
line method for a diesel engine, and
Retardation test and Motoring test for all
types of engines. Given in terms of IP and BP
by:

FP = IP


BP in Watts

2. Mean Effective Pressure.

Indicated Mean Effective Pressure (IMEP). This is also denoted by
P
i

and is given by

P
i

= (Net work of cycle)/Swept Volume in N/m
2

The net work of cycle is the area under the P
-
V diagram.

Brake Mean Effective Pressure (BMEP). This is also denoted by P
b

and is given by

P
b

= 60.BPx/(LANn) N/m
2


This is also the brake power per unit swept volume of the
engine.

Friction Mean Effective Pressure (FMEP). This is also denoted by
P
f

and is given by

P
f

= P
i

-

P
b

N/m
2


3. Efficiencies.

Indicated Thermal Efficiency (

i
) given by


i

= IP/(m
f

. Q
cv
)


m
f

is the mass of fuel taken into the engine in kg/s


Q
cv

is the calorific value of the fuel in J/kg

Brake Thermal Efficiency (

b
) given by


b

= BP/(m
f

. Q
cv
)

Indicated Relative Efficiency (

i,r
) given by


i,r

=

i
/ASE

ASE is the efficiency of the corresponding air standard cycle

Brake Relative Efficiency (

b,r
) given by


b,r

=

b
/ASE

Mechanical Efficiency (

m
) given by


m

= BP/IP = P
b
/P
i

=

b
/

i

=

b,r
/

I,r


Specific Fuel Consumption (sfc or SFC)

This is the fuel consumed per unit power.

Brake Specific Fuel Consumption (bsfc). This is given by

bsfc = m
f
/BP kg/J

if BP is in W and m
f

is in kg/s

bsfc is usually quoted in kg/kWh. This is possible if BP is in kW
and m
f

is in kg/h.

Indicated Specific Fuel Consumption (isfc). This is given by

isfc = m
f
/IP kg/J

if IP is in W and m
f

is in kg/s

isfc is also usually quoted in kg/kWh. This is possible if IP is in
kW and m
f

is in kg/h.

Mechanical Efficiency in terms of the sfc values is given by


m

= isfc/bsfc

Specific Energy Consumption (sec or
SEC).

This is the energy consumed per unit power.

Brake Specific Energy Consumption (bsec). This
is given by

bsec = bsfc.Q
cv


We can similarly define indicated specific energy
consumption (isec) and based on the two
quantities also we can define mechanical
efficiency.

Air Capacity of Four
-
stroke cycle
Engines


The power, P, developed by an engine is given
by




Power will depend on air capacity if the
quantity in the bracket is maximized.


Plot of power versus air flow rate is normally a
straight line.

Volumetric Efficiency

Indicates air capacity of a 4 stroke engine. Given by








Mi is the mass flow rate of fresh mixture.


N is the engine speed in rev/unit time.


V
s

is the piston displacement (swept volume).


ρ
i

is the inlet density.

Volumetric Efficiency

Can be measured:


At the inlet port


Intake of the engine


Any suitable location in the intake manifold

If measured at the intake of the engine, it is also
called the overall volumetric efficiency.

Volumetric Efficiency Based on Dry
Air

Since there is a linear relationship between
indicated output (power) and air capacity
(airflow rate), it is more appropriate to express
volumetric efficiency in terms of airflow rate
(which is the mass of dry air per unit time).

Since fuel, air and water vapor occupy the same
volume

V
a

= V
f

= V
w

= V
i

Thus we have:

Here ρ
a

is the density of dry air or the mass of dry air per unit
volume of fresh mixture.

Thus, since

Also V
d

= A
p
L


s = 2LN

L is the piston stroke and s is the piston
speed.

Measurement of Volumetric
Efficiency in Engines

The volumetric efficiency of an engine can be
evaluated at any given set of operating
conditions provided and ρ
a

can be
accurately measured.

Measurement of Air Flow

Airflow into the engine can be measured with the
help of a suitable airflow meter. The
fluctuations in the airflow can be reduced with
the help of surge tanks placed between the
engine and the airflow meter.

Measurement of Inlet Air Density

By Dalton’s Law of partial pressures:

p
i

= p
a

+ p
f

+ p
w

In this case p
i

is the total pressure of the fresh mixture,


p
a

is the partial pressure of air in the mixture,


p
f

is the partial pressure of fuel in the mixture,


p
w

is the partial pressure of water vapor in the air.

Since each constituent is assumed to behave as a perfect gas, we
can write


M indicates mass of the substance,


29 is the molecular weight of air,

m
f

is the molecular weight of the fuel, and

18 is the molecular weight of water vapor.

F
i

is the ratio of mass of fuel vapor to that of dry air and h is the
ratio of mass of water vapor to that of dry air at the point where
p
i

and T
i

are measured.


This

indicates

that

the

density

of

air

in

the

mixture

is

equal

to

the

density

of

air

at

p
i

and

T
i

multiplied

by

a

correction

factor,

that

is,

the

quantity

in

the

parentheses
.

The value of h depends on the humidity ratio of the air and is
obtained from psychrometric charts.


For conventional hydrocarbon fuels, the correction factor is
usually around 0.98, which is within experimental error. For
diesel engines and GDI engines, F
i

is zero.

In practice, with spark ignition engines using gasoline and with
diesel engines the volumetric efficiency, neglecting the terms in
the parentheses, is given by

If we do not neglect the terms in the parentheses we get the
following relation for volumetric efficiency:

If

the

humidity

is

high

or

a

low

molecular

weight

fuel

is

used


in

a

carbureted

engine,

the

correction

factor

cannot

be

ignored
.


For

example,

with

methanol

at

stoichiometric

conditions


and

h

=

0
.
02
,

the

correction

factor

is

0
.
85
.

Volumetric Efficiency, Power and
Mean Effective Pressure

Since


and

For

an

engine,

the

mean

effective

pressure,

mep,

is

given

by


Ways to increase power and mep


The mean effective pressure may be indicated or
brake, depending on whether η is indicated or brake
thermal efficiency. Thus, the mean effective pressure
is proportional to the product of the inlet density
and volumetric efficiency when the product of the
thermal efficiency, the fuel
-
air ratio, and the heat of
combustion of the fuel is constant.


From the preceding two expressions we can figure
out ways to increase the power and mep of an
engine.



OTTO
CYCLE
-
THE IDEAL CYCLE FOR

SPARK
-
IGNITION
ENGINES



The Otto cycle is the ideal cycle for spark
-
ignition reciprocating engines. It is
namedafter

Nikolaus

A. Otto, who built a successful four
-
stroke engine in 1876 in
Germany using the cycle proposed by Frenchman Beau de
Rochas

in 1862. In most
spark
-
ignition engines, the piston executes four complete strokes (two mechanical
cycles) within the cylinder, and the crankshaft completes 2 revolutions for each
thermodynamic cycle. These engines are called FOUR
-
STROKE internal combustion
engines
.