Two-Stage Air Compression and a Suspicious Discharge Temperature
A plant needs 1.2 kg/s of air at 700 kPa. Air is taken at 100 kPa and 300 K. Two options are compared: a single-stage reciprocating compressor with isentropic efficiency 0.85, and a two-stage machine with perfect intercooling to 300 K and the same stage isentropic efficiency. For air take cp = 1.005 kJ/(kg·K), γ = 1.4, and neglect pressure losses and kinetic-energy changes. The two-stage unit costs 35% more, and its intercooler rejects heat to cooling water at 30 °C. The single-stage discharge temperature is measured at 590 K, whereas the ideal-isentropic calculation gives about 523 K. The project engineer wants to know whether the measured temperature indicates an intercooler bypass fault or simply a lower compressor efficiency, and whether the two-stage option is worth the extra cost at an electricity price of 0.10 USD/kWh and 6000 h/year operation.
1. Calculate the single-stage discharge temperature and shaft power at η = 0.85. Compare with the measured 590 K and infer the implied efficiency.
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For a single stage the pressure ratio is 700/100 = 7. The isentropic temperature ratio is T2s/T1 = rp^((γ-1)/γ) = 7^0.2857 = 1.744, so T2s = 300 × 1.744 = 523.2 K. With η = 0.85, the actual temperature rise is (523.2 - 300)/0.85 = 262.6 K, giving T2 = 562.6 K. The specific work is cp(T2 - T1) = 1.005 × 262.6 = 263.9 kJ/kg, so the shaft power is 1.2 × 263.9 = 316.7 kW. The measured 590 K is higher than 562.6 K. If the machine were still single-stage with the same pressure ratio, the implied efficiency would be η = (523.2 - 300)/(590 - 300) = 223.2/290 = 0.770. Thus the measurement is consistent with a single-stage efficiency of about 77%, not with the specified 85%. It does not by itself prove an intercooler bypass, because a single-stage machine has no intercooler.
2. Calculate the two-stage power with perfect intercooling and the annual energy-cost saving versus the single-stage unit at η = 0.85. Is the 35% capital premium justified by energy alone?
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For minimum work with perfect intercooling, the intermediate pressure is the geometric mean: p_int = √(100 × 700) = 264.6 kPa, so each stage has rp = 2.646. The isentropic temperature rise per stage is T2s - 300 = 300(2.646^0.2857 - 1) = 300(1.320 - 1) = 96.0 K. With η = 0.85, the actual rise per stage is 96.0/0.85 = 112.9 K, so the stage discharge is 412.9 K and the intercooler returns the air to 300 K. The specific work is 2 × 1.005 × 112.9 = 226.9 kJ/kg, and the power is 1.2 × 226.9 = 272.3 kW. The saving versus 316.7 kW is 44.4 kW. Over 6000 h/year the energy saving is 44.4 × 6000 = 266,400 kWh/year, worth 26,640 USD/year at 0.10 USD/kWh. The 35% capital premium cannot be judged without the capital cost, but the energy saving gives a payback period of capital premium divided by 26,640 USD/year. The two-stage option is worth it if the premium is below roughly one to two years of savings; otherwise the single-stage unit is preferable on energy alone.
3. If the measured 590 K were caused by an intercooler bypass on the two-stage machine, what would you expect at the second-stage discharge? State one measurement that would distinguish a bypass fault from low efficiency.
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A bypass would send the first-stage discharge, not 300 K air, into the second stage. If the first stage still produced 590 K, the second-stage isentropic discharge would be 590 × 1.320 = 778.8 K, and with η = 0.85 the actual second-stage discharge would be 590 + (778.8 - 590)/0.85 = 812 K. That is far above 590 K. Therefore a measured 590 K at the single-stage discharge is not evidence of an intercooler bypass; it is more consistent with a lower single-stage efficiency. A distinguishing measurement is the intercooler outlet temperature: with a bypass it would be near the first-stage discharge temperature, whereas with normal intercooling it would be near 300 K. The intercooler cooling-water temperature rise and the pressure drop across the intercooler would also help confirm whether heat is actually being rejected.