Screw Compressors- Mathematical Modelling And Performance Calculation

Screw Compressors: Mathematical Modelling and Performance Calculation

Modern industrial systems rely heavily on screw compressors for efficient gas compression in applications ranging from refrigeration to natural gas processing. The transition from intuitive design to high-performance machinery was driven by sophisticated mathematical modelling and performance calculation. 1. Mathematical Foundations of Rotor Geometry

The performance of a screw compressor is fundamentally dictated by its rotor profile. Mathematical modelling begins by defining the coordinate systems for the male (lobe) and female (groove) rotors.

Coordinate Systems: A right-handed system is typically attached to each rotor ( -axis along the rotor axis, -axis perpendicular).

Profile Generation: Modern asymmetric rotor profiles are designed using enveloping theory to minimize the "blow-hole" area—the primary source of internal leakage. Volume Calculation: The instantaneous working volume is a function of the rotation angle

. This volume decreases as the rotors mesh, leading to compression. 2. Thermodynamic Modelling of the Compression Process Example output from the feature: At pressure ratio = 4

The core of performance calculation involves solving a set of coupled differential equations derived from the conservation of mass and energy. Screw Compressors - Springer Nature

Mathematical modelling and performance calculation are the cornerstones of modern screw compressor design, transitioning the industry from empirical "trial-and-error" methods to precise computer-aided engineering

. This analytical approach is essential for optimizing complex rotor profiles and predicting performance across varying operating conditions. Springer Nature Link 1. Geometric Modelling

The foundation of any screw compressor model is the geometric definition of the rotors and their intermeshing cycle. Screw Compressors - Springer Nature 14 Oct 2010 —

Here’s a solid feature you can include in a project, thesis, or technical paper on “Screw Compressors – Mathematical Modelling and Performance Calculation”: Would you like a sample MATLAB/Python code structure

Example output from the feature:

At pressure ratio = 4.5, speed = 3000 rpm:
- Volumetric efficiency = 82.3%
- Adiabatic efficiency = 76.1%
- Leakage fraction: blowhole = 8.2%, radial = 5.4%

Would you like a sample MATLAB/Python code structure for implementing this feature, or a mathematical derivation of the leakage model?

4. Performance Calculation – Key Metrics

From the thermodynamic model, the following performance parameters are extracted.

2.1 Key Geometric Parameters

| Parameter | Symbol | Description | |-----------|--------|-------------| | Rotor length | L | Axial length of rotors | | Male rotor lobe number | $z_1$ | Typically 4–6 | | Female rotor lobe number | $z_2$ | Typically 5–7 | | Rotor outer diameter | $D$ | Tip diameter | | Center distance | $C$ | Between rotor axes | | Wrap angle | $\theta_w$ | Helix angle twist | | Lead | $P$ | Axial advance per turn |

The volume index ($V_i$) defines the built-in volume ratio: $$ V_i = \fracV_suctionV_discharge = \fracV_maxV_min $$

2.4 Multi-stage or segmented models

For screw compressors, internal compression pockets evolve continuously; treat compression as series of incremental polytropic steps along axial position x (0→L_axial):
- Subdivide axial travel into N slices, compute local pocket volume V(x), leakages, heat transfer per slice, and integrate p–V work numerically: W = ∑ ∫ p dV over each slice.
- This approach captures effects of non-uniform heat transfer, re-expansion from leakage, and pressure equalization events.

4.3 Shaft Power and Mechanical Efficiency

Friction losses (bearings, oil shear, rotor meshing) are modelled as torque losses ( T_loss ): (This is a simplified estimate

[ P_shaft = P_ind + T_loss \cdot \omega ]

Mechanical efficiency:

[ \eta_m = \fracP_indP_shaft ]

2.3 Adiabatic (isentropic) compression

pV^γ = constant (γ = cp/cv)
Work: W = (p2V2 − p1V1)/(1 − γ)
Temperature relation: T2/T1 = (p2/p1)^((γ−1)/γ)

11. Worked Example (Simplified Lumped Model)

Given:

Inlet p1 = 1.013 bar (101300 Pa), T1 = 300 K
Discharge p2 = 8 bar (800000 Pa)
Displacement per rev V_d = 5e-4 m^3/rev
Speed n = 3000 rpm = 50 rps
Gas: air, R = 287 J/kg·K, γ = 1.4
Polytropic exponent assumed n = 1.25
Volumetric efficiency η_v = 0.85
Mechanical efficiency η_mech = 0.95

Steps:

Q_th = V_d × n = 5e-4 m^3/rev × 3000 rev/min × (1/60) min/s = 0.025 m^3/s
ρ1 = p1/(R T1) = 101300/(287×300) ≈ 1.176 kg/m^3
m_dot_th = Q_th × ρ1 = 0.025 × 1.176 ≈ 0.0294 kg/s
m_dot = η_v × m_dot_th = 0.85 × 0.0294 = 0.0250 kg/s
Polytropic work per kg: w = (R T1 /(1 − n)) [(p2/p1)^(1−n)/n − 1] Compute exponent (1−n)/n = (1 − 1.25)/1.25 = −0.2 (p2/p1)^−0.2 = (800000/101300)^−0.2 ≈ (7.897)^−0.2 ≈ 0.645 So w = (287×300 /(1 − 1.25)) [0.645 − 1] = (86100 / (−0.25)) [−0.355] ≈ (−344400) × (−0.355) ≈ 122,232 J/kg ≈ 122.2 kJ/kg
Compressor power P_comp = m_dot × w = 0.0250 × 122,232 ≈ 3056 W ≈ 3.06 kW
Shaft power (accounting mechanical losses): P_shaft = P_comp / η_mech ≈ 3.06 / 0.95 ≈ 3.22 kW
Discharge temperature: T2 = T1 × (p2/p1)^(n−1)/n with (n−1)/n = 0.25/1.25 = 0.2 T2 = 300 × (7.897)^0.2 ≈ 300 × 1.55 ≈ 465 K

(This is a simplified estimate; real machine will differ due to leakage, transient effects, and oil cooling.)