Air Compressors

An air compressor is a machine that raises the pressure of air by reducing its volume (positive displacement type) or by converting kinetic energy into pressure (dynamic type).

Classification

1. Principle of operation
   • Positive displacement: reciprocating (piston) and rotary (roots blower, sliding vane, screw, lobe).
   • Dynamic (non-positive displacement): centrifugal and axial flow compressors.
2. Number of stages: single-stage, two-stage and multi-stage.
3. Action: single-acting (one face of the piston) and double-acting.
4. Delivery pressure: low (up to 1 bar gauge), medium (1 to 10 bar), high (10 to 100 bar) and very high (above 100 bar).
5. Capacity: low (below 0.15 m³/s), medium (0.15 to 5 m³/s) and high (above 5 m³/s).
6. Cooling: air-cooled (fins) and water-cooled.
7. Cylinder arrangement: vertical, horizontal, V, W and radial.

Applications of compressed air

• Pneumatic tools: drills, hammers, rivetters, spray painting.
• Operating pneumatic brakes in trucks and trains.
• Starting diesel engines and supplying air to gas turbines.
• Cleaning, sand-blasting, air-lift pumps and mine ventilation.
• Process industries, food packaging and instrumentation air.

A reciprocating compressor draws air into a cylinder, compresses it by the movement of a piston, and delivers it at higher pressure.

            delivery valve   suction valve
               |                 |
         +-----v-----+-----------v----+
         |      cylinder head         |
         |  ______________________    |
  cooling|  |<-clearance->|       |   |cooling
  fins / |  |   piston |==|  rings|   |fins /
  jacket |  |__________|__|_______|   |jacket
         +---------------+------------+
                         |
                  connecting rod
                         |
                 crank --o-- crankshaft

Primary components

Working

1. Suction stroke: the piston moves away from the head; the pressure falls and the suction valve opens, air enters.
2. Compression stroke: both valves close; the air is compressed as the piston returns.
3. Delivery: when the pressure exceeds the receiver pressure, the delivery valve opens and air is delivered until the piston reaches the end. The clearance volume remains.
4. Re-expansion: the trapped clearance air expands as the piston moves back, until the suction valve opens again.

With negligible clearance, air is taken in at $p_1, V_1$, compressed to $p_2$ and delivered completely.

   p
   p2 |3 +--+ 2
      |  |   \
      |  |    \
      |  |     \_
   p1 |4 +--------- + 1
      +-------------------- V
        0  V2         V1

• 4-1: suction at constant $p_1$ (work done by atmosphere on piston $= p_1V_1$).
• 1-2: compression ($pV^n = C$ or $pV = C$).
• 2-3: delivery at constant $p_2$ (work done on the delivered air $= p_2V_2$).
• 3-4: pressure drop at zero volume (valve change-over).

Work done per cycle

The work per cycle is the area 4-1-2-3: suction work + compression work + delivery work, with the suction work $p_1V_1$ taken as a gain by the piston:

$$W = \int_1^2 p\,dV\Big|_{\text{work on gas}} + p_2V_2 - p_1V_1 = \int_{p_1}^{p_2} V\,dp$$

(a) Isothermal ($pV = p_1V_1 = C$)

$$\begin{aligned} W_{iso} &= \int_{p_1}^{p_2} \frac{p_1V_1}{p}\,dp\\ &= p_1V_1 \ln\frac{p_2}{p_1} = mRT_1 \ln\frac{p_2}{p_1} \end{aligned}$$

(b) Polytropic ($pV^n = C$)

Compression work (area under 1-2):

$$\int_1^2 p\,dV\ (\text{on gas}) = \frac{p_2V_2 - p_1V_1}{n-1}$$

Adding the delivery and subtracting the suction work:

$$\begin{aligned} W_{poly} &= \frac{p_2V_2 - p_1V_1}{n-1} + p_2V_2 - p_1V_1\\ &= \frac{n}{n-1}\left(p_2V_2 - p_1V_1\right) \end{aligned}$$

Since $p_2V_2 = p_1V_1 (p_2/p_1)^{(n-1)/n}$,

$$W_{poly} = \frac{n}{n-1}\,p_1V_1\left[\left(\frac{p_2}{p_1}\right)^{\frac{n-1}{n}} - 1\right] = \frac{n}{n-1}\,mR\,(T_2 - T_1)$$

Isothermal efficiency

$$\eta_{iso} = \frac{\text{isothermal work}}{\text{indicated (polytropic) work}} = \frac{W_{iso}}{W_{poly}}$$

Isothermal compression needs the least work, so $\eta_{iso}<1$. Better cooling lowers $n$ towards 1 and raises $\eta_{iso}$. Typical $n$ for air is 1.25 to 1.35.

Clearance volume $V_c$ is the volume left in the cylinder (including valve ports) when the piston is at the end of the delivery stroke. It is needed so that the piston does not strike the head and for valve movement. Clearance ratio $C = V_c/V_s$, where $V_s$ is the swept volume.

Effect of clearance

• The trapped air re-expands during the suction stroke, so suction starts only after the piston has moved some distance.
• The volume of fresh air drawn in is less than the swept volume, so the capacity falls.
• The work per kg of air delivered is not changed (compression and re-expansion work cancel for the same index), but the machine must be larger for a given output.
• Greater pressure ratio and larger clearance reduce the volume drawn in; at a high enough ratio no air is delivered.

Derivation of volumetric efficiency

   p
   p2 |   3+--+2
      |    |   \
      |     \   \
      |      \   \
   p1 |      4+---\--+1
      +-----+-+----+--+---- V
            Vc V4      V1
              <---Va--->

$V_s = V_1 - V_c$ (swept volume) and $V_a = V_1 - V_4$ (air drawn).

Effective suction volume $V_a = V_1 - V_4$, where $V_1 = V_c + V_s$ and $V_4$ is the re-expanded clearance volume.

The clearance air expands polytropically from $p_2$ to $p_1$: $p_2V_c^{\,n} = p_1V_4^{\,n}$, so

$$V_4 = V_c\left(\frac{p_2}{p_1}\right)^{1/n}$$ $$\begin{aligned} V_a &= V_s + V_c - V_c\left(\frac{p_2}{p_1}\right)^{1/n}\\ \eta_v &= \frac{V_a}{V_s} = 1 + \frac{V_c}{V_s} - \frac{V_c}{V_s}\left(\frac{p_2}{p_1}\right)^{1/n} \end{aligned}$$ $$\boxed{\eta_v = 1 + C - C\left(\frac{p_2}{p_1}\right)^{1/n}}$$

Hence $\eta_v$ decreases when the clearance ratio $C$ or the pressure ratio $p_2/p_1$ increases. This is one reason for multistage compression at high pressure ratios.

Given

$p_1 = 100$ kPa, $V_1 = 4$ m³/min $= 0.0667$ m³/s, $T_1 = 300$ K, $p_2 = 700$ kPa, $n = 1.3$.

$$p_1V_1 = 100 \times 0.06667 = 6.667\ \text{kW}$$

(a) Indicated (polytropic) power

Pressure-ratio factor: $(p_2/p_1)^{\frac{n-1}{n}} = 7^{0.2308} = 1.5668$

$$\begin{aligned} P &= \frac{n}{n-1}\,p_1V_1\left[\left(\frac{p_2}{p_1}\right)^{\frac{n-1}{n}} - 1\right]\\ &= \frac{1.3}{0.3} \times 6.667 \times (1.5668 - 1) = 16.38\ \text{kW} \end{aligned}$$

(b) Isothermal power and efficiency

$$P_{iso} = p_1V_1 \ln\frac{p_2}{p_1} = 6.667 \times \ln 7 = 12.97\ \text{kW}$$ $$\eta_{iso} = \frac{P_{iso}}{P} = \frac{12.97}{16.38} = 0.792$$

(c) Delivery temperature

$$T_2 = T_1\left(\frac{p_2}{p_1}\right)^{\frac{n-1}{n}} = 300 \times 1.5668 = 470\ \text{K} = 197\ \text{°C}$$

Answer: (a) 16.4 kW; (b) 12.97 kW, $\eta_{iso}$ = 79.2 %; (c) 470 K (about 197 °C).

Swept volume

$$V_s = \frac{\pi}{4}d^2L = \frac{\pi}{4}(0.15)^2(0.2) = 0.00353\ \text{m}^3$$

(a) Volumetric efficiency

Pressure ratio $= 8$, $(8)^{1/1.3} = 4.951$.

$$\eta_v = 1 + C - C\left(\frac{p_2}{p_1}\right)^{1/n} = 1 + 0.05 - 0.05 \times 4.951 = 0.8025$$

(b) Air drawn per minute

Effective suction volume per cycle: $V_a = \eta_v V_s = 0.8025 \times 0.00353 = 0.00284$ m³.

One suction per revolution (single-acting):

$$V = V_a N = 0.00284 \times 300 = 0.851\ \text{m}^3/\text{min}$$

(c) Indicated power

$(p_2/p_1)^{\frac{n-1}{n}} = 8^{0.2308} = 1.6159$ and $p_1V_1 = 100 \times 0.851/60$ kW.

$$P = \frac{1.3}{0.3} \times \frac{100 \times 0.851}{60} \times (1.6159 - 1) = 3.78\ \text{kW}$$

(The clearance air is compressed and re-expands over the same index, so only the effective volume $V_a$ is used in the work formula.)

Answer: (a) 80.3 %; (b) 0.851 m³/min; (c) about 3.78 kW.

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.