Two-stage compressor with perfect intercooling

Need for multistage compression with intercooling

• A single stage at a high pressure ratio gives a very high delivery temperature (lubricating oil breaks down, risk of fire) and a low volumetric efficiency.
• The work is reduced because cooling between stages brings the process nearer to isothermal.
• Smaller, lighter cylinders and better mechanical balance.

Data

$p_1 = 100$ kPa, $V_1 = 3$ m³/min $= 0.05$ m³/s, $T_1 = 293.15$ K, $p_3 = 1500$ kPa, $n = 1.3$, $R = 0.287$ kJ/kg K.

$$p_1V_1 = 100 \times 0.05 = 5.00\ \text{kW}$$

(a) Intermediate pressure

For minimum work with perfect intercooling: $p_2 = \sqrt{p_1p_3}$

$$p_2 = \sqrt{1 \times 15} = 3.873\ \text{bar}$$

(b) Total power

Each stage has the same pressure ratio $p_2/p_1 = 3.873$, so $(p_2/p_1)^{0.2308} = 1.3668$.

$$\begin{aligned} P &= 2\,\frac{n}{n-1}\,p_1V_1\left[\left(\frac{p_2}{p_1}\right)^{\frac{n-1}{n}} - 1\right]\\ &= 2 \times 4.3333 \times 5.00 \times (1.3668 - 1) = 15.89\ \text{kW} \end{aligned}$$

(c) Saving over single stage

$(15)^{0.2308} = 1.8681$

$$P_{single} = 4.3333 \times 5.00 \times (1.8681 - 1) = 18.81\ \text{kW}$$ $$\text{Saving} = \frac{18.81 - 15.89}{18.81} \times 100 = 15.5\ \%$$

Single-stage delivery temperature would be $293.15 \times 1.8681 = 548$ K (about 274 °C).

(d) Heat rejected in the intercooler

Air leaving low-pressure cylinder: $T_2 = 293.15 \times 1.3668 = 400.7$ K. Perfect intercooling cools it back to $T_1$.

$$\dot m = \frac{p_1V_1}{RT_1} = \frac{5.00}{0.287 \times 293.15} = 0.0594\ \text{kg/s}$$ $$Q = \dot m\,c_p\,(T_2 - T_1) = 0.0594 \times 1.005 \times (400.7 - 293.15) = 6.42\ \text{kW}$$

Answer: (a) 3.87 bar; (b) 15.9 kW; (c) 15.5 % saving; (d) about 6.4 kW.