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 21 is connected to a dynamometer which give a brake output torque reading of 205 Nm at 3600 RPM. At this speed, air enters the cylinder at 85 kpa and 600C 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.
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 Summer '19
 Combustion, Internal combustion engine