Instructor Name:
Prof. Dr. Ramadan Youssef Sakr
Benha University
Faculty of Engineering at Shoubra
Mechanical Engineering Dept.
Heat Engines & Combustion (B)
June 2011
Time Allowed: 3 hours
3
rd
year Mech. Power
It is allowed to use combustion table
s
Please: Attempt All Questions:
1a)
Define
:
Octane number
–
equivalence ratio

flame surface

turbulence macoscale.
1
b
)
Acetylene gas (C
2
H
2
) is burned completely with
20 percent excess air during a steady

flow
combustion
process. The fuel and the ai
r enter the combustion chamber
separately at 25
°
C and 1 atm,
and heat is being lost from the
combustion chamber to the surroundings at 25
°
C at a rate of
300,000
kJ/kmol C
2
H
2
. The combustion products leave the
combustion chamber at 1 atm pressure. Determine
(a) the
temperature of the products, (b) the total entropy change per
kmol of C
2
H
2
, and (c) the exergy
destruction during this
process.
(15 Marks)
2a)
A jet of ethylene C
2
H
4
exits a 12 mm diameter nozzle into still air at 300 K and 1 atm. Comp
are the
spreading angles and the axial locations where the jet centerline mass fraction drops to the
stoich
iometric value for initial jet velocities of 10 cm/s
and 1 cm/s.
T
he viscosity of ethylene at 300 K
is 102.3x10

7
N.s/m
2
.
2
b
)
Make
a heat balance, c
alculate the thermal efficiency and comment on the result obtained for a
continuous slab reheating furnaces which receives 100 m
3
of coke oven gas (net calorific value =
16720 kJ/m
3
) per ton slab heated combustion air is preheated to 410
o
C and is consumed
by the fuel
gas at air/gas ratio of 5 and the combustion products/fuel gas volume is 5.6.
The
waste gases leave the
furnace
chamber at 1100
o
C, having entered at ambient 10
o
C. Water cooling losses amount 8% of the
total heat input and external losses
4
%
.
Du
ring the heating process 1.6 % of the metal is oxidized and
the mill scale produced evolves 6372 kJ/kg of metal. Slab discharge temperature is 1180
o
C.
The
mean specific heats (kJ/m
3
.
o
C) of air and combustion products are 1.25 and 1.48 respectively and tha
t
for steel is 0.65 kJ/kg.
o
C. Assume complete combustion of fuel.
(15 Marks)
3a)
Why is the criterion for chemical equilibrium
expressed in terms of the Gibbs function instead of
entropy?
3b)
Write three different KP relations for reacting idea
l
gas
mixtures, and state when each relation should
be used.
3
c
)
One k
mol of carbon monoxide, CO, reacts with kmol of oxygen, O
2
, to form an equilibrium mixture
of CO
2
, CO, and
O
2
at 2500 K and 1 atm.
determine
the equilibrium composition in terms of mole
fractions.
(15 Marks)
4a) What the physical significance of Lewis number? What role does the Le=1 assumption play in the
analysis of laminar flame propagation?
4b) A premixed propane

air mixture emerges from
a
round nozzle with a uniform veloci
ty of 75 cm/s.
T
he laminar flame speed of propane

air mixture is 35 cm/s.
A
flame is lit at the nozzle exit. What is
the cone angle of this flame? What principle determines the cone angle?
4c)
D
rive the theoretical flame shape for a premixed flame sp
eed
stabilized over a circular tube, assuming
the unburned mixture velocity profile is parabolic
:
v(r) = v
o
(1

r
2
/R
2
), where v
o
is the centerline
velocity and R is the tube burner radius. Ignore the region close to the tube wall where S
L
is greater
than v
(r). Discuss your result.
(15 Marks)
Good luck
Question 1:
Octane number:
it
is defined as the isooctane fraction of the comparison fuel
.
Equivalence ratio:
(A/F
theo
)/(A/F
act
)
Flame surface:
locus of points where the equivalence ration, equ
als unity
Turbulence macoscale:
l
o
the mean size of the large eddies in a turbulent flow.
S O L U T I O N
Known: A system initially consisting of 1 kmol of CO and kmol of O
2
reacts to form an equilibrium
mixture of CO
2
, CO,
and O
2
. The temperature of the m
ixture is 2500 K and the pressure is 1 atm.
Find: Determine the equilibrium composition in terms of mole fractions.
Assumption: The equilibrium mixture is modeled as an ideal gas mixture.
Analysis:
The above equation
relates temperature, pressure, and co
mposition for an ideal gas mixture at
equilibrium. If any two
are known, the third can be determined using this equation. In the present case, T
and p are known, and the composition is
unknown.
Applying conservation of mass, the overall balanced chemical
reaction equation is
where z is the amount of CO, in kmol, present in the equilibrium mixture. Note that
The total number of moles n in the equilibrium mixture is
Accordingly, the molar analysis of the equilibrium mixture is
At equilibrium, the ten
dency of CO and O2 to form CO2 is just balanced by the tendency of CO2 to form
CO and O2, so
CO
2
==== CO + 1/2 O
2
we have Accordingly, Eq. 14.35 takes the form
At 2500 K, Table A

27 gives log10K
=

1.44. Thus, K
=
0.0363. Inserting this value into t
he last
expression
By trial and errors,
z = 0.129
and then t
he equilibrium composition in terms of mole
fractions
.
Question 2:
a)
Replace
only
the value of R= 0.006 instead of 0.005 in
Reynolds number calculations
:
b)
Solution:
Basis per to
n of slab (1000kg)
Heat input through
Heat (kJ)
i)
Fuel (coke oven gas) = 100
x
16720
=
1672000
ii)
Air =
100
x
5x
1.25
x
(410

10) =
250000
iii)
Scale
= 1000
x
0.01
6
x
6372 =
101952
Total ============================================= 2023952
Heat output by
i)
cooling water =
0.08 x
2023952
= 161916
ii)
external losses =
0.04 x 2023952
= 80958
iii)
waste gases =
100x5.6x1.48x1100 =
911680
iv)
steel =
1000x0.984x0.65x1180 = 754728
v)
unaccounted for (by difference)
= 114670
Thermal efficiency of the furnace = (Heat in steel /Total
heat input) x100 =
37.3%
Comment
The unaccounted is about 5.6% which is critically accepted.
Question
3
:
2
a) Criterion for
chemical equilibrium
expressed in terms of the Gibbs function instead of
entropy
because it suitable for adiabatic and non

adi
abatic systems but entropy is valid only for adiabatic
systems.
3
b)
3c)
Question
4
:
a) It is the ratio between the thermal diffusivity and the mass diffusivity and Le=1 assumption simplifies
greatly the energy equation.
b)
u
L
v
S
1
sin
o
30
75
35
sin
1
The principle that determine the cone angle is "The flame speed must equal to the speed of the normal
component of unburned gas at each location"
c) Put
:
dz/dr = tan
β where β=90

and integrate the equation to get
z=z(r), assuming SL as a fraction of
v
o
or v
u
.
Enter the password to open this PDF file:
File name:

File size:

Title:

Author:

Subject:

Keywords:

Creation Date:

Modification Date:

Creator:

PDF Producer:

PDF Version:

Page Count:

Preparing document for printing…
0%
Comments 0
Log in to post a comment