College of Engineering and Computer Science
Mechanical Engineering Department
Spring 2010 Number: 17724 Instructor: Larry Caretto
1. The ultimate analysis of a coal with a heating value of 12,000 Btu/lb m
is 69.76%C, .4.65%H,
Nitrogen 1.24%, Sulfur 3.50%, Oxygen 8.14%, and Ash 12.71%, determine the air fuel ratio for 25% excess air (λ = 1.25) combustion. Assuming that all the sulfur in the coal reacts to
SO
2
, determine the emission rate of SO
2
in pounds per million Btu of heat input. What are the CO
2
emissions per Btu of heat input for this coal?
We can use the formulas from the combustion notes and lecture to compute the subscripts in the computational fuel formula, C x
H y
S z
O w
N v
and the resulting product species information. x = wt%C/12.0107 = 69.76/12.0107 = 5.808, y = wt%H/1.00794 = 4.65/1.00794 = 4.613 z = wt%S/32.065 = 3.50/32.066 = 0.109, w = wt%O/16.0004 = 8.14/16.0004 = 0.509 v = wt%N/14.0067 = 1.25/14.0067 = 0.089
The stoichiometric oxygen requirement for C x
H y
S z
O w
N v
is A = x + y/4 + z – w/2 moles O
2
= 5.808
+ 4.613/4 + 0.109 – 0.509/2 = 6.816. We can now compute the air fuel ratio on a mass basis.
We have assumed 100 mass units of fuel in computing the fuel molecule and the corresponding mass of air is given as 138.28

A on slide 13 of the February 1 lecture presentation. With the value of A just found and the given value of 1.25 for

, we find
Air
Fuel

138 .
28

A
100

138 .
28

1 .
25

6 .
816
100

11 .
78
Each pound of coal contains 0.035 pounds of sulfur. If all the sulfur reacts to SO
2
, the mass of
SO
2
formed will be 0035 M
SO2
/M
S
= 0.035(64)/(32) = 0.07 lb SO
2
/lb fuel. Since each pound of fuel has 12000 Btu = 0.012 MMBtu, the SO
2
production per MMBtu is found as follows:
E
SO
2

0 .
035 lbS lb fuel
2 lbSO
2 lbS lb fuel
0 .
012 MMBtu

5 .
83 lb SO
2
MMBtu
Similarly, each pound of coal contains 0.6976 pounds of carbon. If all the carbon reacts to CO
2
, the mass of CO
2
formed will be 0.6976 M
CO2
/M
C
= 0.6976(44)/(12) = 2.558 lb SO
2
/lb fuel. Since each pound of fuel has 12000 Btu = 0.012 MMBtu, the CO
2
production per MMBtu is found as follows:
E
CO
2

0 .
6976 lbC lb fuel
44 lbCO
2
12 lbC lb fuel
0 .
012 MMBtu

213 lb CO
2
MMBtu
2. When the coal of problem one is burned in a power plant, the measured NOx concentration is 10 ppm in the dry exhaust. Assuming that the oxygen content of the exhaust is the same as that computed for complete combustion (at the 25% excess air specified in problem one), determine the emission rate of NOx in pounds per million Btu.
In order to compute the emission rate per unit heat input we have to find the partial mass density of the species in question, NOx, the F-factor and the mole percent of O
2
in the dry exhaust. With these three items, we can use in the equation below, which is from slide 16 of the February 1 lecture (with a slight modification to spell out the subscripts).
Jacaranda Hall (Engineering) 3333
Email: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062

February 8 homework solutions ME 483, L. S. Caretto, Spring 2010 Page 2
E species , dry
  species , dry
F d
20 .
9
20 .
9

% O
2 , d
The first term is the partial density of NO x
in the dry exhaust gases. To find this we start with the molar density given by the ideal gas equation, using a standard pressure of 14.696 psia and a standard temperature of 68 o F = 527.67 R. n
V

P
RT

10 .
73

14 .
696 psia
 ft
3 lbmole

R psia


527 .
67 R


0 .
0025956 lbmole ft
3
Of these total moles, 10 ppm will be NOx; this is equivalent to a mole fraction of 10x10 -6 of NOx which has a molecular weight of 46, so the partial mass density of NOx in the dry exhaust gas will be

NOx
 n y
NOx
M
NOx
V

46 lb lbmol
NOx 10 x 10

6 lbmol
NOx lbmol
NOx 0 .
0025956 lbmole ft 3

1 .
194 x 10

6 ft 3 lb
The second term is the F-factor. We can use the F-factor for bituminous coal from the notes page of slide 16 on the February 1 lecture: 9780 dscf/MMBtu or alternatively we could compute the F-factor from the ultimate analysis using the K factors from the notes page of slide 17 from the February 1 lecture. The latter approach gives the following result.
F d

K

K
C
% C

K
H
% H

K
O
% O
Q c

K
S
% S

K
N
% N


10
6
Btu
MMBtu


1 .
53 lb m scf

%
69 .
76 %

3 .
64 lb m scf

%
4 .
65 %


0 .
46 lb m scf

%
12000 Btu
8 .
14 %

0 .
57 lb m scf

%
3 .
50 %

0 .
14 scf lb m

%
1 .
24 %

 lb m
This gives a value of F d
= 9635 dscf/MMBtu; use of the standard factor would result in only a
1.5% error.
The emission rate formula requires the dry exhaust oxygen mole percentage. This can be found from the information on slide 14 of the February 1 lecture. The total dry moles in the exhaust is found from the equation, D = x + z + (
 – 1)A + 3.77

A + v/2 = 5.808 + 0.109 + (1.25 – 1)( 6.816)
+ 3.77(6.816) + 0.089/2 = 33.32. The moles of O
2
in the exhaust are computed as (

-1)A = (1.25
– 1)(6.816) = 1.704 so the dry mole fraction of oxygen is 1.704/33.32 = 5.114%.
We now have all the information we need to compute the NOx emission rate per unit heat input.
E
NO x
, dry
 
NOx , dry
F d
20 .
9
20 .
9

% O
2 , dry

1 .
194 x 10

6 dscf lb m
9635 dscf
MMBtu 20 .
9
20 .
9

5 .
114
E
NOx,dry
= 0.015 lb m
/MMBtu

February 8 homework solutions ME 483, L. S. Caretto, Spring 2010 Page 3
3. Problem 6 of chapter 5 in the text. If the heat input were provided by a natural gas with a (lower) heating value of
20,000 Btu.lb
m
, what air/fuel ratio is required?
GIVEN: Gas turbine cycle with data as shown.
2
Combustor
3 T
3
= 1200 o C
= 1473.15 K
Compressor
P
2
/P
1
= 20
 c
= 85%
Turbine
 t
= 90%
FIND: (a) compressor work, (b) turbine work, (c) net work, (d) turbine exit temp-temperature and (e) the thermal efficiency of the cycle.
1
T
1
= 20 o C =
293.15 K
4
APPROACH: this solution uses the equations in Hodge, which are slightly different in form from those in a typical engineering thermodynamics class.
The equations in Hodge use the equations for an ideal gas with constant heat capacity. Hodge accounts for the temperature of the heat capacity by using different values of heat capacity different components in the system. The Hodge equations assume that each component in the gas turbine cycle is steady flow with negligible kinetic and potential energy changes. The turbine and the compressor each have one inlet and one outlet so that the general first law for these components is q = w u
+ h out
– h in
. We assume all devices are adiabatic. In the combustor, all the chemical energy of the fuel is converted to thermal energy in the exhaust stream, with no external heat loss.
We can use equation (5-14) in Hodge to find the outlet temperature of the compressor.
T
2

T
1

1

1
 c






P
2
P
1

 k

1 k

1







303 .
15 K

1

1
0 .
85

 20

1 .
4
1 .
4

1


1 


760 .
0 K
Similarly, equation (5-20) in Hodge gives the outlet temperature of the turbine. Here we use
Hodge’s value of k = 1.334 for the turbine. Since we are not given data on the combustor pressure loss we will assume an ideal cycle in which P
3
/ P
4
= P
2
/P
1
.
T
4

T
3



1
  t






P
4
P
3

 k

1 k

1








1473 .
15 K


1

0 .
88




1
20
1 .
334

1
1 .
334 
1


 



T
4
= 773.6 K
The specific work for the compressor and turbine are found, respectively, from equations (5-12) and (5-18) in Hodge, again using the values for mean heat capacity. w c
 c p , compressor

T
1

T
2


1 .
004 kJ kg

K

293 .
15 K

760 .
0 K



468 .
7 kJ kg

February 8 homework solutions ME 483, L. S. Caretto, Spring 2010 Page 4 w t
 c p , turbine

T
3

T
4


1 .
148 kJ kg

K

1473 .
15 K

853 .
2 K


803 .
1 kJ kg
If we neglect the difference in mass flow rates between the compressor and turbine, the new work is simply the algebraic sum of the two work terms. w net
 w t
 w c

803 .
1 kJ kg


468 .
7 kJ kg

334 .
4 kJ kg
The combustor heat addition can be inferred from equations (5-23) or (5-24) in Hodge, who uses a value of 1.076 kJ/kg

K for the combustor. q
 c p , combustor

T
3

T
2


1 .
076 kJ kg

K

1473 .
15 K

529 .
6 K


767 .
4 kJ kg
The thermal efficiency is simply the ratio of the net work to the combustor heat input.
334 .
4 kJ
  w net q combustor
 kg
767 .
4 kJ kg

= 43.6%
The fuel air ratio can be found from equation (5-22) in Hodge for an adiabatic combustor (the value found above for the combustor heat addition). fuel
 air
 q combustor

H v
767 .
4 kJ kg

K
20000 Btu 1 .
055056 kJ lb m
Btu
0 .
45359 lb m kg fuel
 air

0 .
01650
Although it was not required we can find the net thermal efficiency accounting for the fuel flow in the turbine.
The thermal efficiency is the net power output divided by the heat input from the fuel. The calculation below uses the intermediate result from the previous equation that H v
= 46.52 MJ/kg.

 
 net fuel
H v kg
46 .
52 MJ


 air
MJ
1000 kJ
 fuel
 w t
 fuel
H v




1
0 .
01699
 air

1 w c 
1
H v
482 .
0 kJ kg









 air fuel
1

1
0 .
01699 w t
 


 air
277 .
4 kg fuel w c kJ










= 45.3%