Force act through a distance equal to the length of

m
v

21
Ind Thermal Eff = ƞ
ti
=
wf= fuel supplied per hour in m
3
for gaseous fuel and kg for liquid fuel
HHV = high heating volume of fuel KJ/kg or KJ/m
3
The brake thermal efficiency flows what fraction of heat supplied was actually delivered
by the engine shaft
Brake Thermal Eff = ƞ
tb =
Conversion: 575
Example 2 -2
The engine in problem 2-1 is connected to a dynamometer which give a brake output
torque reading of 205 N-m at 3600 RPM. At this speed, air enters the cylinder at 85 kpa and
60
0
C and the Mech Eff of the engine is 85%. Calculate :
a) Brake power (KW) e) fmep
b) Indicated Power (kW) f) Power lost to friction
c) Bmep g) brake work per unit mass
d) Imep
Solution :

22
a.)
N
m
=
; = =
= 90.9 kw
b.) B
mep
= =
B
mep
= 859 000 = 859 Kpa
c.) I
mep
=
= =
1010 Kpa
d.) f
mep
= I
mep
- B
mep
= 1010 - 859 = 151 kpa
e.) Power lost =
f = i - b
= 90.9 – 77.3 = 13.6 kw
f.)
b = where m
a
is mass of air
Brake work for one cylinder
B
mep
=
Wb
=
B
mep
x Vd = 859 x 0.0005m
3
= 0.43 kj
Since gas is air :
m
a
= where V is volume at BDC

3
c
t
=

=
=
=

25
Standard values of surrounding air pressure and temperature can be used to find
density
Po (std) = 101.3 kpa or 14.7 psia
To (std) = 25
o
c 0r 298 k
P
o
= pressure of surrounding air
T
o
= temperature of surrounding air
R = gas constant of air – 0.287
At standard conductions:
ρ
air
= 1.181 ; ρ
a
=
Typical values of Nv for an engine at WOT are in a range of 75% to 90% going
down to much lower value as the throttle is closed.