It is allowed to use combustion tables Please: Attempt All Questions:

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22 Φεβ 2014 (πριν από 3 χρόνια και 5 μήνες)

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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
.