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Compressor

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Submitted By shiyasbasheer
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Dublin Institute of Technology | Combustion | Thermodynamics Laboratory | |
Name: Shiyas Basheer
Student Number: D10119909
Programme/Course: DT022/3
Group: A
Date of lab: 09/04/2013

Abstract

Table of Contents Abstract 1 Introduction and Procedure 3 Calculations 3 Theory 3 Mass Flow rate analysis 4 Exhaust Gas 6 Results 7 Conclusion 8 Reference 9

Equivalent ratio of a gas turbine
Introduction and Procedure
For this experiment, a two shaft gas turbine (ET792) was used. At the core of ET792 are a so-called gas generator and a free-running power turbine. The gas generator consists of a radial compressor, a combustion chamber and a radial turbine. The compressor and turbine are mounted on a shaft. The gas turbine works as an open cyclic process, with the ambient air being drawn out and fed back in. Intake and exhaust silencers reduce the noise in operation of the power turbine. The use of propane as the combustion gas ensures clean, odorless operation. A start-up fan is used to start the gas turbine. The air is compressed and fed to the combustion chamber. Before it enters the combustion chamber the air is slowed by means of a diffuser. The air is then split into primary and secondary.
The primary air mixes and burns with the fuel and the secondary air cool the exhaust gases. Relevant measuring values are recorded by sensors and indicated on the display and control panel. The energy of the exhaust gas stream is either converted into mechanical energy in the free-running power turbine or accelerated and transformed into thrust via a nozzle.
Here, first, the air to fuel ratio (stoichiometric) was calculated using theoretical values. Then, volume flow rate of the air and mass flow rate of the fuel was read from the display. This was used to find the air to fuel ratio. Hence, the equivalent ratio was calculated by comparing this to the stoichiometric value. Later, to analyze the exhaust gas, the composition of air was read form the gas turbine and the air to fuel ratio was calculated. This was then used to calculate the second equivalent ratio by comparing with the stoichiometric value.
Calculations
Theory
The first step in the lab was to determine the air/fuel ratio using stoichiometric conditions.
The air is assumed to contain 21% O2 and 79%N2 in molar terms (Stoichiometric conditions, no excess air).
Balancing the chemical equation:
C3H8+aO2+3.76N2=bCO+cH20+dN2
Determine values for a, b, c and d by equating coefficients from the reactants to the products.
Ccarbon⇒3=b
Hhydrogen⇒8=2c⇒c=4
Ooxygen⇒2a=c+2b⇒a=5
Nnitrogen⇒3.76=d⇒d=18.8
Substituting the values for a, b, c and d, the equation becomes:
C3H8+5O2+3.76N2=3CO+4H20+18.8N2
From this equation the air/fuel ratio could be determined, after calculating the molar masses below:
Molar mass of air = 28.96 kg/kmol
Molar mass of carbon = 12.011 kg/kmol
Molar mass of hydrogen = 2.016 kg/kmol
AF=mairmfuel= 5 1+3.7628.96312.011+42.016=15.630
Mass Flow rate analysis
The mass flow rate of propane and the volume flow rate of air from the experiment were measured to be:
MFuel=4.25kghour=0.001181kg/s
VAir=60.1dm3/s=0.0601m3/s
To determine the Air/Fuel ratio, the mass flow rate of air must be determined. For that first density of air at standard condition must be determined. This is done using the ideal gas equation:
Pυ=RT
The pressure was measured to be 1.009mbar =100,900 Pa
The temperature was assumed to be 20˚C
R=0.287J/kgK
ρ=1υ=PRT=100,9000.287*293=1.2kg/m3
The mass of rate of the air was then calculated:
MAir=VAir*ρ=0.0601*1.2=0.073kg/s
The Air/Fuel ratio was then calculated as follows:
AF= Mass flow rate of AirMass flow rate of Fuel=0.0730.001181=61.81
To determine the equivalent ratio, this was then compared to previously determine stoichiometric value for the A/F ratio:
Equivalent ratio=A/FexpA/Fstoic=61.8115.630=3.955
Now,
%Theoretical air=Equivalent ratio*100
%Theoretical air=3.955*100=395.5%
Therefore,
%Excess air=%Theoretical air-100
%Excess air=395.5-100=295.5%
Exhaust Gas
Composition of exhaust gases and air was read from the apparatus as below: Exhaust gases | Percentage | Composition | index | CO2 | 1.0% | 0.01 | b | O2 | 19.4% | 0.194 | e | CO | 9ppm=0.0009% | 0.000009 | d | N2 | 79.59% | 0.7959 | f |

Air | Percentage | CO2 | 0.0% | O2 | 20.9% | CO | 0.0% |

Dry air was removed from it during the experiment and hence no H2O in the above table.
Balancing the chemical equation:
Here, x is being used as the value of fuel is unknown. xC2H8+aO2+3.76N2→bCO2+cH2O+dCO+eO2+fN2 Where,
‘a’ is the stoichiometric coefficient
Note: On a molar basis air contains 21% O2 and 79%N2. Therefore, for every 1Kmol of O2, there will be 3.76 Kmol N2
Now, by equating the reactants to the products the values of a, c and x (molar masses) can be found. The values of b, e, d and f are been used from the exhaust table above.
C ⇒3x=b+d ⇒x=0.01+0.0000093=0.00336
H ⇒8x=2c⇒c=8*0.003362=0.01345
O⇒2a=2b+c+d+2e⇒a=0.01+0.013452+0.0000092+0.194=0.2174
Substituting the values for a, c and x, the equation becomes:
0.00336C2H8+0.2174O2+3.76N2→0.01CO2+0.01345H2O+0.000009CO+0.194O2+0.7959N2
The Air/fuel (A/F) ratio can now be calculated as follows,
AF=mass of airmass of fuel=0.21741+3.76*290.003363*12.011+8*1=300.1479=202.77
To determine the equivalent ratio, this was compared to the previously calculated stoichiometric value for A/F ratio:
Equivalent ratio=A/FexpA/Fstoic=202.7715.630=12.97
Now,
%Theoretical air=Equivalent ratio*100
%Theoretical air=12.97*100=1297%
Therefore,
%Excess air=%Theoretical air-100
%Excess air=1297-100=1197%
Calculation of Q + W
Exhaust temperature=833 K Stoichiometric | @ 25 degrees | | | Reactants | N | hf | Δh | N(hf + Δh) | C3H8 | 1 | -103847 | 0 | -103847 | O2 | 5 | 0 | 0 | 0 | N2 | 18.8 | 0 | 0 | 0 | | | | HR | -103847 |

Products | N | hf | Δh | N(hf + Δh) | CO | 3 | -110530 | 0 | -331590 | H2O (L) | 4 | -285820 | 0 | -1143280 | N2 | 18.8 | 0 | 0 | 0 | | | | HP | -1474870 |

Q+W=Hp-HR
Q+W= -147870--103847=-1371023kJkmolof fuel

Actual | @ 25 degrees | | | | Reactants | N | hf | Δh | N(hf + Δh) | C3H8 | 0.00336 | -103847 | 0 | -348.92592 | O2 | 0.21 | 0 | 0 | 0 | N2 | 0.7896 | 0 | 0 | 0 | | | | HR | -348.92592 |

| @833 K | | | | Products | N | hf | Δh | N(hf + Δh) | CO2 | 0.01 | -393520 | 24562.35 | -3689.5765 | H2O (G) | 0.0133 | -241830 | 19308.86 | -2959.531162 | CO | 0.000009 | -110530 | 16249.32 | -0.84852612 | O2 | 0.194 | 0 | 0 | 0 | N2 | 0.7959 | 0 | 0 | 0 | | | | HP | -6649.956188 |

Q+W=Hp-HR
Q+W = -6649956--348.9592=-6301.0302kJkmolof fuel

Results
Stoichiometric A/F ratio:
AF=mairmfuel=15.630
Mass flow analysis A/F ratio:
AF= mairmfuel=61.81
Equivalent ratio:
Equivalent ratio=A/FexpA/Fstoic=3.955
%Excess air=395.5-100=295.5%
Exhaust gas analysis A/F ratio:
AF=mass of airmass of fuel=202.77
Equivalent ratio:
Equivalent ratio=A/FexpA/Fstoic=12.97
%Excess air=1297-100=1197%
Q + W:
Stoichiometric:
Q+W= -147870--103847=-1371023kJkmolof fuel
Actual:
Q+W = -6649956--348.9592=-6301.0302kJkmolof fuel
Conclusion
The equivalent ratio between the theory of combustion and experimental values were expected to be close to each other. Unfortunately the two equivalent ratios for the exhaust gas analysis and mass flow rate analysis were 12.97 and 3.96 respectively, which was nowhere near to close to each other. This simply means that there were some significant experimental errors. Some of them might be due to as follows: * The assumptions made to calculate the density could be incorrect. * The ideal gas equation used to calculate the density of air is subject to ideal gas assumptions. * The sensors for measuring the exhaust gases could be faulty. * The equipment could be faulty. * Assumptions made about the composition of air to be 79% nitrogen and 21%oxygen could be incorrect. There is also a possibility of argon in the air.
Reference

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