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The purpose of this report is to evaluate the experiment findings from different sessions. This report discusses all four experiments performed with cumulative data. Data out of trend were discarded. The selected data were then used to discuss the experiment.
The air compressor is widely used as power tools, assembly line equipment in factories, and working valves in chemical plants due to its safety in wet surroundings. The aim of the compressor performance laboratory (T2) is to evaluate basic properties of the air compressor focusing on the relationship between compression ratio (rp), polytrophic index (n), and different compression speeds (RPM).
The polytropic index is found by using a fixed compression speed while increasing the delivery pressure from 2 bar to 10 bar in 2 bar increments.
This thermodynamic process is considered neither adiabatic (n=1.4) nor isothermal (n=1).
The experiment is conducted at P0 = 101.325 Kpa. Some of the values will be used in the following report, and these details are from the manufacturer's information.
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Manometer orifice discharge coefficient (Cd) is 0.62, D is 25.4 mm, d is 12.7mm, A is 0.000127 m2.
In the example analysis, we will use one set of data to show all the calculations. The following table shows the data that will be used in the example analysis section.
| RPM | P(abs) | Voltage | Current | T1 | Water flow rate | Torque | Manometer ΔP |
|---|---|---|---|---|---|---|---|
| 600 | 4.01 | 89 | 10 | 294.15 | 1 | 5.2 | 58 |
Calculate T3 from the polytrophic relationship and assuming the compressor operated in an adiabatic and reversible (the polytrophic index n=k), in this case n = k = 1.4:
P0/P1 = (V1/V0)k
P0 V0 = nRT1
P1 V1 = nRT3
T3 = T1 (P0/P1)((1-k)/k)
Substituting the data:
T3 = 294.15 ((101.325/(4.01*100)))((1-1.4)/1.4) = 435.7752 K
However, in reality, the compressor is not isentropic.
Thus, calculate the polytrophic index (n) by using the measured values T1, T3, and P1:
n = log10((4.01*100)/101.325)/(log10(294.15/438.15) + log10((4.01*100)/101.325))
n = 1.407787
Pressure ratio rp:
rp = P1/P0 = ((4.01*100))/101.325 = 3.957562
Find the dynamometer electrical power Welec:
Welec = VA = 89*10 = 890
Calculate Wmech:
Wmech = (τ*3*2π*ɷ)/60 = 980.1769 W
Using the first law of thermodynamics to find the heat lost from the compressor system:
Ẇ_in - Qout = mCp(T3-T1) = 602.712 J
The mass flow rate of air through the system can be determined from the pressure change of the manometer across:
mair mass flow = Cd*E*A √(2*ρ*ΔP) = 0.003
Where the density of air and manometer orifice discharge coefficient are determined by the following equations:
ρ = P0/(Rair*T1) = 1.20023
E = (1-(d/D)4)^(-0.5) = 1.0327956
The example analysis results can be summarized in the following table:
| T3 | n | rp | Welec | Wmech |
|---|---|---|---|---|
| 435.7752 | 1.407787 | 3.957562 | 890 | 980.1769 |
The aim of the laboratory is to evaluate the relationship between air compressor ratio (rp), polytrophic index (n), and different compression speed (RPM).
The compression ratio for every RPM decreased as the compression index increased. The graph showed that some experiments as the compression ratio decreased, the compression index went toward the ideal adiabatic index n=1.4. The purpose of increasing the RPM was to increase the compression index to the ideal 1.4 like assumed. The actual performance was not adiabatic due to the slow rotational speed ran during the sessions. Industrial compressors run at higher speeds more than 900 RPM, hence increasing the polytrophic index to the ideal adiabatic.
RPM is not the only factor preventing the ideal index. The heat released to the surroundings and friction affect the performance of the compressor. Friction may slow the RPM speed and lower the polytrophic index away from the ideal index.
The polytrophic index decreases while the compression ratio increases. This means the final temperature did not increase as fast as the final pressure. This may be because of the transfer of heat energy from the system to the surroundings. Thus, the environment is getting faster. Therefore, the final temperature does not rise as fast as the final pressure.
In addition, for the higher compression speed, the value of the polytrophic index (n) is closer to the ideal gas (n=1.4). This is because the period of each compression cycle is getting shorter as the compression speed RPM increases. As the time of each compression cycle gets shorter, a lesser amount of heat energy is able to transfer to the surrounding environment, and the final temperature becomes higher. So the polytrophic index (n) becomes closer to 1.4. This situation is also seen in Figure 5. Therefore, the process in 850 RPM laboratory is much closer to the ideal isentropic process.
Therefore, it could be concluded that:
For the laboratory T2, the polytrophic values decrease when the compression ratio increases. This is because of a faster rate of heat energy transfer to the surrounding environment if the compression ratio is higher (temperature cannot rise as fast as the pressure as heat energy leaks to the surrounding). Meanwhile, if RPM increased in a compressor, the time period of each revolution decreased. Therefore, there will be less heat energy transfer from the compressor to the surrounding environment for each cycle. Therefore, for a higher RPM, its process is much closer to the ideal isentropic process.
Air Compressor Performance Analysis. (2024, Jan 04). Retrieved from https://studymoose.com/document/air-compressor-performance-analysis
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