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HW3 5

This document presents a problem analyzing the equivalence ratios calculated from the fuel and air flow rates and the exhaust gas composition of an engine. The measured fuel flow rate was 0.4 g/s, air flow rate was 5.6 g/s, and the dry exhaust gas composition was 13.0% CO2, 2.8% CO, with zero O2. The problem involves calculating the equivalence ratios using the flow rates (φ1) and the exhaust gas composition (φ2), and showing that the values compare well. The calculations involve stoichiometric equations to determine the molar exhaust species fractions based on the measured exhaust composition. The results showed φ1 = 1.043 and φ2 = 1.012,

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207 views2 pages

HW3 5

This document presents a problem analyzing the equivalence ratios calculated from the fuel and air flow rates and the exhaust gas composition of an engine. The measured fuel flow rate was 0.4 g/s, air flow rate was 5.6 g/s, and the dry exhaust gas composition was 13.0% CO2, 2.8% CO, with zero O2. The problem involves calculating the equivalence ratios using the flow rates (φ1) and the exhaust gas composition (φ2), and showing that the values compare well. The calculations involve stoichiometric equations to determine the molar exhaust species fractions based on the measured exhaust composition. The results showed φ1 = 1.043 and φ2 = 1.012,

Uploaded by

owaiskhan
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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File:C:\Personal\Class\ME410\Heywood\HW3-5.

EES 9/30/02 10:43:28 AM Page 1


EES Ver. 6.577: #659: for use by Students and Faculty of the ME Dept, Rose-Hulman

Heywood Problem 3-5

The measured engine fuel flow rate is 0.4 g/s, air flow rate is 5.6 g/s, and the exhaust gas
composition (measured dry) is CO2=13.0%, CO=2.8%, with O2 essentially zero. Unburned hydrocarbon
emissions can be neglected. Compare the equivalence ratio calculated from the fuel and air
flow with the equivalence ratio calculated from the
exhaust gas composition. The fuel is gasoline with a H/C ratio of 1.87. Assume H2 concentration
equal to one third the CO concentration.

m fuel = 0.0004

m air = 0.0056

m air
AF =
m fuel

y = 1.87

4 + y
AFs =34.56
12.01 +1.008 y

AFs
1 =
AF

The stoichiometric equation looks like

CHy + 1/phi *(1+y/4) ( O2 + 3.773 N2) = n1 CO2 + n2 H2O + n3 CO + n4 H2 + n5 N2

Balance

1 =n1 + n3

y =2 n2 +2 n4

1 y
1 + 2 =2 n1 + n2 + n3
2 4

1 y
1 + 2 3.773 =2 n5
2 4

Nmdry =n1 + n3 + n4 + n5

n1
0.13 =
Nmdry

n3
0.028 =
Nmdry

Note that the two phi's compare well

Unit Settings: [kJ]/[K]/[kPa]/[kg]/[radians]


AF = 14 AFs = 14.6 mair = 0.0056 mfuel = 0.0004 n1 = 0.8228 n2 = 1.078
n3 = 0.1772 n4 = -0.1428 n5 = 5.472 Nmdry = 6.329 1 = 1.043 2 = 1.012
File:C:\Personal\Class\ME410\Heywood\HW3-5.EES 9/30/02 10:43:28 AM Page 2
EES Ver. 6.577: #659: for use by Students and Faculty of the ME Dept, Rose-Hulman

y = 1.87

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