13.3 Air Compressors, Fans & Blowers

13.3 Air Compressors, Fans & Blowers

Air compressors, fans, and blowers increase gas pressure to deliver pneumatic power or move high volume flows in industrial plant systems. Understanding compressor thermodynamic cycles, intercooling benefits, clearance volumetric efficiency, and fan pressure heads is essential for MELE licensure candidates.

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1. Air Compressors

Compressors increase the pressure of gas by reducing its specific volume.

Classification

• Positive Displacement Compressors: Reciprocating piston (single and double acting), rotary screw, sliding vane, and scroll compressors.
• Dynamic Compressors: Centrifugal (radial) and axial flow compressors.

Thermodynamic Compression Work

For gas compressed from suction state ($P_1, V_1, T_1$) to discharge pressure ($P_2$):

1. Isothermal Compression ($n = 1$): Constant temperature process achieved with ideal heat removal. Represents minimum theoretical work: $W_{iso} = P_1 V_1 \ln\left(\frac{P_2}{P_1}\right) = m R T_1 \ln\left(\frac{P_2}{P_1}\right)$
2. Polytropic Compression ($1 < n < k$): Real compression with partial heat transfer: $W_{poly} = \frac{n}{n-1} P_1 V_1 \left[ \left(\frac{P_2}{P_1}\right)^{\frac{n-1}{n}} - 1 \right] = \frac{n}{n-1} m R T_1 \left[ \left(\frac{P_2}{P_1}\right)^{\frac{n-1}{n}} - 1 \right]$
3. Isentropic (Adiabatic) Compression ($n = k = 1.4$ for air): Zero heat transfer during compression: $W_{isen} = \frac{k}{k-1} P_1 V_1 \left[ \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}} - 1 \right]$

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2. Multi-Stage Compression & Intercooling

When compressing air to high overall pressure ratios ($P_2/P_1 > 4$), single-stage compression yields excessively high discharge temperatures (leading to lubricant degradation) and reduced volumetric efficiency.

Intercooling Principles

Air is compressed in a Low Pressure (LP) cylinder, passed through an intercooler heat exchanger to cool back to intake temperature ($T_1$), and then compressed in a High Pressure (HP) cylinder.

Optimum Intermediate Pressure ($P_x$)

For minimum total compression work in a 2-stage compressor with perfect intercooling: $P_x = \sqrt{P_1 \cdot P_2} \quad \implies \quad \frac{P_x}{P_1} = \frac{P_2}{P_x} = \sqrt{\frac{P_2}{P_1}}$

• Equal pressure ratio per stage: $r_{p,stage} = \sqrt{P_2 / P_1}$.
• Equal work distribution between LP and HP stages ($W_{LP} = W_{HP}$).
• Total minimum 2-stage polytropic work: $W_{total, 2-stage} = 2 \times \frac{n}{n-1} P_1 V_1 \left[ \left(\frac{P_x}{P_1}\right)^{\frac{n-1}{n}} - 1 \right]$

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3. Clearance Volumetric Efficiency ($\eta_v$)

In reciprocating compressors, clearance volume ($V_c$) is necessary to prevent the piston from striking the cylinder head. The clearance ratio is defined as: $c = \frac{V_c}{V_D}$ Where $V_D$ is piston displacement (swept volume).

Clearance Volumetric Efficiency Formula

High-pressure air trapped in $V_c$ re-expands during the suction stroke, delaying fresh air suction intake: $\eta_v = 1 + c - c \left(\frac{P_2}{P_1}\right)^{1/n}$

• Free Air Delivery ($FAD$): Actual volume rate of air delivered, reduced by volumetric efficiency: $V_{actual} = V_D \cdot \eta_v$

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4. Fans & Blowers

Fans move large gas volumes at low static pressures, whereas blowers operate at intermediate pressure ranges.

Pressure Range Classification

• Fans: Differential pressure $\Delta P \le 3.45\text{ kPa}$ ($34.5\text{ mbar}$ or $14\text{ in. w.g.}$). Density changes are neglected (incompressible flow assumption).
• Blowers: $3.45\text{ kPa} < \Delta P \le 103.4\text{ kPa}$ ($1\text{ to } 30\text{ psig}$).
• Compressors: $\Delta P > 103.4\text{ kPa}$ ($> 30\text{ psig}$). Compressibility effects must be considered.

Centrifugal vs. Axial Fans

1. Centrifugal Fans:
   • Backward-Curved Vanes (BC): Highest efficiency (up to 85%), self-limiting power characteristic (prevents motor overload at maximum flow rate).
   • Forward-Curved Vanes (FC): Delivers large air volumes at low rotational speed; power rises steeply with flow rate (risk of motor overloading).
   • Radial / Straight Vanes: Rugged, self-cleaning blades for industrial dust and particle handling.
2. Axial Fans: Propeller, tubeaxial, and vaneaxial fans designed for straight-through axial air delivery.

Fan Performance Equations

• Velocity Pressure ($VP_d$): $VP_d = \frac{1}{2} \rho_{air} V_d^2 \quad \implies \quad h_v = \frac{V_d^2}{2g}$
• Fan Total Head ($FTH$): $FTH = FSH + VP_d$ Where $FSH$ is Fan Static Head measured across the fan casing.
• Fan Air Power ($FAP$): $FAP = Q \cdot FTH = Q \cdot (FSH + VP_d)$ (In SI units, $FAP\text{ (kW)} = \frac{Q\text{ (m}^3\text{/s)} \times FTH\text{ (Pa)}}{1000}$. In US units, $FAP\text{ (HP)} = \frac{Q\text{ (cfm)} \times FTH\text{ (in. w.g.)}}{6356}$).
• Fan Brake Power ($FBP$): $FBP = \frac{FAP}{\eta_{fan}}$

Fan Affinity Laws

Same relationships as centrifugal pumps: Flow rate $Q \propto N$, Total Head $FTH \propto N^2$, and Power $P \propto N^3$.

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Worked Air Machinery Calculation

Problem: A two-stage reciprocating air compressor compresses $V_1 = 0.15\text{ m}^3/\text{s}$ of free air from suction pressure $P_1 = 100\text{ kPa}$ and $T_1 = 300\text{ K}$ to a final discharge pressure $P_2 = 900\text{ kPa}$. The polytropic expansion/compression index is $n = 1.3$. Perfect intercooling returns the air to $300\text{ K}$ prior to the second stage. The clearance ratio is $c = 0.05$ (5%). Determine (a) optimum intermediate pressure $P_x$, (b) total compressor polytropic power, (c) power saved compared to single-stage compression, and (d) clearance volumetric efficiency of the first stage.

Step-by-Step Solution:

1. Calculate Optimum Intermediate Pressure ($P_x$): $P_x = \sqrt{P_1 \cdot P_2} = \sqrt{100\text{ kPa} \times 900\text{ kPa}} = \sqrt{90,000} = 300\text{ kPa}$ Stage Pressure Ratio: $r_{p,stage} = \frac{300}{100} = 3.0$

2. Calculate Work Per Stage ($W_{stage}$): $W_{stage} = \frac{n}{n-1} P_1 V_1 \left[ \left(\frac{P_x}{P_1}\right)^{\frac{n-1}{n}} - 1 \right] = \frac{1.3}{0.3} \times 100\text{ kPa} \times 0.15\text{ m}^3/\text{s} \left[ (3.0)^{\frac{0.3}{1.3}} - 1 \right]$ $(3.0)^{0.23077} = 1.2886$ $W_{stage} = 4.3333 \times 15 \times (1.2886 - 1) = 65.0 \times 0.2886 = 18.759\text{ kW}$

3. Calculate Total Two-Stage Power Output ($W_{total}$): $W_{total} = 2 \times W_{stage} = 2 \times 18.759\text{ kW} = 37.518\text{ kW}$

4. Calculate Single-Stage Power Requirement ($W_{single}$): $W_{single} = \frac{1.3}{0.3} \times 100 \times 0.15 \left[ \left(\frac{900}{100}\right)^{\frac{0.3}{1.3}} - 1 \right] = 65.0 \times \left[ (9.0)^{0.23077} - 1 \right]$ $(9.0)^{0.23077} = 1.6603$ $W_{single} = 65.0 \times (1.6603 - 1) = 65.0 \times 0.6603 = 42.920\text{ kW}$ Power Saved by 2-Stage Intercooling: $\Delta P = 42.920\text{ kW} - 37.518\text{ kW} = 5.402\text{ kW}$ (or a 12.59% energy saving).

5. Calculate Clearance Volumetric Efficiency ($\eta_v$): $\eta_v = 1 + c - c \left(\frac{P_x}{P_1}\right)^{1/n} = 1 + 0.05 - 0.05 \times (3.0)^{\frac{1}{1.3}}$ $(3.0)^{0.76923} = 2.3276$ $\eta_v = 1.05 - 0.05 \times 2.3276 = 1.05 - 0.1164 = 0.9336 = 93.36\%$