Short Laboratory Report
Experiment T2
Two-stage Air
Compressor
Tuesday 10
th
March 2015
Experiments and
Statistics
William Haynes
1324410
Group 14
Introduction
Two-stage air compressors have an advantage over the same single stage process because they save on the compression work W =
∫
p dV
. This is shown in Figure 1 by the labelled area on the idealized p-V diagram: the total work is the area enclosed by the diagram. The reason for this advantage is that the air is cooled in an inter-cooler between the two stages of compression, which reduces the mass of the air delivered to the second compressor while keeping the pressure constant. This is why the high-pressure part of the p-V diagram: above p1, p2; has a smaller area, which represents less work done, however the same mass of air is delivered in total.In the experiment, the air compressor was tested under varying receiver
pressures (p3) by adjusting an outlet valve. Readings of various values at each pressure were taken and used to test the hypothesis and get values for the overall efficiency (%) of the system, volumetric efficiency (%), Free Air Delivery (m3 /s), polytropic index of compression n.
Figure 1: Idealised p-V diagram for a two-stage compression cycle with intercooling.
Objectives
To gain experience of operating a two-stage air compressor with an inter-cooler and other equipment first-hand.
To explore which parameters affect the performance characteristics of the compressor, focussing on the index of compression and work done in compression.
Method and Results
Figure 2: Schematic Diagram of the Two-Stage Intercooled Compressor System Firstly, the speed of each stage was set to 750rpm, and this was kept constant through the whole experiment. Then the air flow valve was adjusted to set the receiver pressure at P3 (see figure 2) to a constant 10 bar (gauge). We then waited ten minutes for the compressor outlet temperatures (T3 and T6) and the pressures to stabilize. Once the conditions were stable we recorded the values of all the parameters in Tables 1&2. Then the previous two steps were repeated with the receiver pressure set at approximately 9,8,7,6 and 5 bar gauge. All the recorded values are shown below in Tables 1&2.
Table 1: Receiv er Pressu re P3 Air at Inlet T1 Air to Interco oler T3 Air from Interco oler T5 Air at Delive ry T6 1st Stage Delive ry Pressu re P0 2nd Stage Deliver y Pressur e P2 1st Stage Compr. Speed 2nd Stage Compr. Speed bar gauge °C °C °C °C bar gauge bar gauge rpm rpm 10.0 24.0 99.0 38.0 187.0 1.1 10.0 750.0 750.0
9.5 24.0 94.5 35.0 180.0 1.1 9.5 750.0 750.0 8.0 25.0 95.5 35.0 170.0 1.1 8.0 750.0 750.0 7.0 25.0 95.0 34.0 160.0 1.0 7.0 750.0 750.0 6.0 24.0 94.0 33.0 148.0 1.0 6.0 750.0 750.0 5.0 25.0 93.5 32.0 138.0 1.0 5.0 750.0 750.0 Table 2: Receiv er Pressu re P3 Orifice Compressor Pressure Drop Temperat ure T1 Current A1 Current A2 Voltage V1 Voltage V2 bar
gauge mmH2O °C Amp Amp Volt Volt
10.0 127.0 24.0 6.0 10.0 120.00 120.00 9.5 130.0 24.0 6.5 10.0 120.00 122.00 8.0 132.0 25.0 6.0 8.5 120.00 120.00 7.0 133.0 25.0 6.0 8.0 120.00 120.00 6.0 132.0 24.0 6.0 7.2 120.00 121.00 5.0 132.0 25.0 5.8 7.0 120.00 120.00
Analysis
This section analyses the results from the experiment by using them to calculate certain performance characteristics of the compressor plant.
The first performance characteristic to calculate was the Air Flow Rate at STP conditions (Standard Temperature and Pressure), referred to as Free Air Delivery (m3 /s). To do this we first had to determine values for the mass flow
rate (kg/s). The compressed air passes through an orifice of known proportions, therefore flowrate can be calculated with the formula
m
´
a=
C
dEA
√
2 ∆ pρ
(1), where:m
´
a = mass flow rate (kg/s);C
d = 0.62, orifice discharge coefficient;ρ
=p
0/
R
airT
1 , air density (kg/m3);∆ p
= pressure drop (mmH2O); A =
πd2/4, orifice area (m2); E = (1-β4)-0.5, where β = d/D; T
1 = absolute temperature
of air downstream (K). The known values for the orifice are Cd = 0.62, d =
12.7mm, D = 25.4mm, therefore E = 1.0328 and A = 0.000127m2. Having
plugged all the values into eq. (1), we could then convert the values for mass flowrate
m
´
a (kg/s) into volumetric flowrate at STP (or FAD) (m3/s) using theformula V = ´m´ aRairT / p (2), where: p = 101300 N/m2 (abs.); T = 288K (15oC); Rair = 287 J/kg K. The values calculated using equations (1) & (2) are shown in
Table 3. Table 3: Receiv er Pressur e P3 Pressur e Drop ∆ p Temperat ure T1 Mass Flow Rate
´
m
a Free Air Delivery ´ V bar gauge mmH2O °C kg/s m3/s 10.0 127.0 24.0 0.00437 0.00356 9.5 130.0 24.0 0.00442 0.00361 8.0 132.0 25.0 0.00445 0.00363 7.0 133.0 25.0 0.00447 0.00365 6.0 132.0 24.0 0.00445 0.00363 5.0 132.0 25.0 0.00445 0.00363Figure 3: A graph of Free Air Delivery plotted against a base of delivery pressure.
Free Air Delivery
Figure 3 shows that the FDA does not vary significantly as the receiver pressure changes. There is a slight negative gradient to the trend line, however we assumed this to be negligible.
The second performance characteristic to calculate was Electrical Power input (W) for stage 1 and stage 2 compressors and overall. This was done using the formula P = VI (3), where: P is electrical power (W); V is voltage (V); I is current (A). Eq. (3) was used to calculate the electrical input power (W) at each stage, then the overall electrical input power (W) is simply the sum of the two. The answers from these calculations are shown in Table 4 below.
Table 4: Receiv er Pressur e P3 Curre nt A1 Curre nt A2 Volta ge V1 Volta ge V2 Electrical Input Power Stage 1 Electrical Input Power Stage 2 Electrical Input Overall
bar g Amp Amp Volt Volt W W W
10.0 6.0 10.0 120.0 120.0 720.0 1200.0 1920.0 9.5 6.5 10.0 120.0 122.0 780.0 1220.0 2000.0 8.0 6.0 8.5 120.0 120.0 720.0 1020.0 1740.0 7.0 6.0 8.0 120.0 120.0 720.0 960.0 1680.0 6.0 6.0 7.2 120.0 121.0 720.0 871.2 1591.2 5.0 5.8 7.0 120.0 120.0 696.0 840.0 1536.0
The next performance characteristic to be calculated was Air Indicated Power (W) for stages 1 and 2, and overall. The formula for this is given as
P
R n−1 n−1
¿
´
W
i= ´
m
aR
airT
i(
n
n−1
)
¿
(4), where: PR = pressure ratio in absolute values; Ti =
initial air temperature in Kelvin. The values were calculated using eq. (4) for each stage and then again the overall power is simply the sum of the two stages. The calculated values are shown in Table 5 below.
Table 5: Recei ver Press ure P3 Air at Inlet T1 Mass Flow Rate
´
m
a Press ure ratio (abs.) Air Indicated Power for Stage 1 Air Indicated Power for Stage 2 Overall Air Indicate d Power bar g °C kg/s W W W 10.0 24.0 0.00437 5.24 798.9 944.3 1743.22 9.5 24.0 0.00442 5.00 779.7 912.0 1691.66 8.0 25.0 0.00445 4.29 695.2 817.3 1512.566.0 24.0 0.00445 3.50 577.4 683.4 1260.74
5.0 25.0 0.00445 3.00 495.7 587.2 1082.93
Figure 4: A graph Electrical Power (W) and Air Indicated Power (W) for stages 1 & 2 and overall against a base of Receiver Pressure (bar gauge).
Electrical Power and Air Indicated Power
Electrical Input Power Stage 2 Linear (Electrical Input Power Stage 2 )
Electrical Input Power Stage 1 Linear (Electrical Input Power Stage 1 )
Air Indicated Power for Stage 1 Linear (Air Indicated Power for Stage 1)
Air Indicated Power for Stage 2 Linear (Air Indicated Power for Stage 2)
Electrical Input Overall Linear (Electrical Input Overall )
Overall Air Indicated Power Linear (Overall Air Indicated Power)
Figure 4 shows a positive correlation between electrical power and air indicated power at stages 1 & 2 and overall, and receiver pressure.
There two important values for the efficiency of the system that were calculated: The overall efficiency of the compressor system can be defined as:
η
overall=
Air Indicated Power
Electrical Input Power
.The volumetric efficiency can be defined as:
η
V=
(
actual air delivered /second)
(
maximum possibledelivery / second)
, where the actual air delivered is theFree Air Delivery (m3/s), which has already been calculated and the maximum
possible air delivery rate is assumed to be the swept volume of the first stage piston, which we know to be 192.4cm3. Therefore volumetric efficiency for the
system can be found using the formula
η
V=
(
FAD /second)
The calculated values for both of these efficiencies at each pressure are shown below in Table 6. Table 6: Receive r Pressur e P3 Overall Electrical Input Power Overall Air Indicated Power Overall Efficienc y Free Air Delivery Volume tric Efficien cy bar g W W % m3/s % 10.0 1920.00 1743.22 90.79 0.00356 74.09 9.5 2000.00 1691.66 84.58 0.00361 74.96 8.0 1740.00 1512.56 86.93 0.00363 75.53 7.0 1680.00 1431.23 85.19 0.00365 75.82 6.0 1591.20 1260.74 79.23 0.00363 75.53 5.0 1536.00 1082.93 70.50 0.00363 75.53
Figure 5: A graph showing overall efficiency (%) against a base of delivery pressure.
Efficiency
Overall Efficiency Linear ( Overall Efficiency ) Volumetric Efficiency
Linear (Volumetric Efficiency)
It can be seen from Figure 5 that the system has a greater overall efficiency at higher delivery pressures. However it is evident that the volumetric efficiency does not vary as pressure does, because it is proportional to Free Air Delivery, which we have already seen is independent of delivery pressure.
The final performance characteristic that was calculated was the polytropic index of compression for each stage and overall. For a polytropic process of an ideal gas, the relationship between pressure and volume is given as pVn =
constant, so between two states 1&2, p1V1n = p2V2n (5), where n is the
polytropic index. Eq. 5, can be rearranged to get
V
V
12
=
(
p
2p
1)
1 n (6). The idealgas law states pV = nRT. Substituting this into eq. 6 gives the expression
T
1T
2=
(
p
1p
2)
n−1 n(7). We know T1, T2, p1 and p2 for stage 1 and 2 of the compressor
logarithms, eq 7, can be rearranged to get
n=
ln
(
p
2p
1)
ln
(
T
1T
2)
+
ln
(
p
2p
1)
(8). Thepolytropic index for receiver pressure 10 bar (gauge), at stage 1, stage 2 and overall were calculated and the values are shown in Table 7.
Table 7: Receiver
Pressure 1st Stageindex 2nd Stageindex Overallindex
bar g - -
-10.0 1.430 1.308 1.222
The final piece of analysis was to sketch a p-V diagram, similar to fig. 1, using the measured values of temperatures and pressures for receiver pressure 10 bar (gauge).
Discussion
From these results and subsequent calculations we are able to see which performance characteristics are affected by changing the pressure at the receiver. We can see that, in this system, flowrate and Free Air Delivery do not change significantly when the pressure changes, because there is negligible gradient to the line in figure 3. Figure 4 clearly shows that a higher electrical power input at each stage is required to reach higher pressure at the receiver. This is because more work in compression must be done to reach higher
pressures. The system has a higher overall efficiency when the receiver pressure is higher (fig.5), but volumetric efficiency is constant, because it is proportional to Free Air Delivery which we have already seen does not vary.
Table 7 shows the effect of intercooling on the polytropic index of the system. The overall index is significantly lower than it would be without intercooling, because it is closer to being isothermal (n=1), i.e. the difference between the inlet and outlet temperatures is less. Adding more stages with more intercooling would reduce the index further, increasing the compression work saved. Figure 6 confirms that the two-stage system in the experiment with an intercooler did save on total work of compression. The work saved by inter-cooling is labelled in fig 6. This can maximised by finding the ideal inter-stage pressure. Of the
different pressures we tested, 10 bar g (shown in fig. 6) was the optimal.
Conclusion
The experiment was successfully carried out without any major issues that could’ve caused the results not to be valid. There is, however, a degree of small uncertainty in all the values due to a number of factors: precision errors in the initial readings; human error in the setting and stabilizing of the pressures or in the reading of the dials; assumptions that are not perfectly accurate eg. Air density, Vapour content of air; etc.
Overall, the experiment was successful in giving us experience of operating a two-stage air compressor with an inter-cooler, in allowing us to explore how some performance characteristics of the compression are affected by changing pressure, and understand why multiple stages and intercooling save on
