A two-phase flow approach to cavitation modelling in journal bearings

Abstract

Cavitation has been extensively treated in numerical models for lubrication using boundary conditions in the pressure equation, and several criteria are available. However, an inappropriate choice can lead to imprecise results, thus having serious implications for performance prediction. This work proposes the numerical solution for lubrication analysing the changes suffered by the lubricant along a journal bearing, considering the release of gas from the liquid and the existence of a two-phase flow. Results obtained are compared with those using the Reynolds, or Swift-Steiber, boundary condition. The influence of fluid properties on the main parameters of bearing operation is also discussed.

Appendix A: Calculation of physical properties

Empiricism is the most common procedure to determine physical properties for oil–refrigerant mixtures, very often adjusting curves from experimental data that has been made available. In this work, properties for the mixture of refrigerant R134a and oil ICI EMKARATE RL10H were calculated using mainly data provided in graphical form by the oil manufacturer. For pure refrigerant, data were obtained from the software REFPROP [34]. This appendix presents the numerical correlations adopted to calculate the physical properties required. For further discussion on the properties of the fluids and the behaviour of this specific mixture, reference is made to Silva [36].

A.1. Solubility

The solubility of R134a in the polyol ester oil ICI EMKARATE RL10H was provided by the oil manufacturer in a diagram and adjusted by curve fitting for the interval 0 < p < 1000 kPa and 0<T<60 °C as,

$$ w_{\rm sat} = {{a_1 + b_1 p + c_1 T + d_1 p^2 + e_1 T^2 + f_1 Tp}\over {a_2 + b_2 p + c_2 T + d_2 p^2 + e_2 T^2 + f_2 Tp}} $$

(A.1)

where a ₁=0.6825, b ₁=0.0701, c ₁=0.0699, d ₁=−1.2087× 10⁻⁴, e ₁=−1.7157× 10⁻³, f ₁=2.4124× 10⁻³, a ₂=1.0, b ₂=−3.1315× 10⁻³, c ₂=0.0503, d ₂=1.0541× 10⁻⁶, e ₂=1.3645× 10⁻³, f ₂=−6.4074× 10⁻⁵.

A.2. Density

The density for the mixture R134a-EMKARATE RL10H is calculated using the additive law of mixtures. Considering an ideal mixture, the result is presented in equation (A2),

$$ \rho _{\rm l} = {{\rho _{\rm oil} }\over {1 + w_{\rm r} \left( {{{\rho _{\rm oil} }\over {\rho _{\rm lr} }} - 1} \right)}} $$

(A.2)

where ρ_l is the density of the liquid mixture, ρ_oil the density of the pure oil, ρ_lr the density of the liquid refrigerant, and w _r is the refrigerant mass fraction.

The oil density, provided by the manufacturer and adjusted in the range 20<T<120 °C is given by

$$ \rho _{\rm oil} = a_3 + b_3 T + c_3 T^2 $$

(A.3)

where a ₃=966.4364, b ₃=−0.5739, c ₃=−2.447× 10⁻⁴, and ρ_oil the density in kg/m³.

The density of the liquid refrigerant is obtained from the software REFPROP [34] and validated for the interval −5<T<50 °C as follows

$$ \rho _{\rm lr} = a_4 + b_4 T + c_4 T^2 $$

(A.4)

where a ₄=1294.6790, b ₄=−3.2213, c ₄=−0.0123, and ρ_lr the density in kg/m³.

A.3. Dynamic viscosity

The viscosity of the liquid mixture R134a and the polyol ester oil was provided by the oil manufacturer and the following fit is proposed for the interval 0<T<60 °C and 0<w _r<1,

$$ \mu _{\rm l} = {{a_5 + b_5 T + c_5 w_{\rm r} + d_5 T^2 + e_5 w^2 _{\rm r} + f_5 Tw_{\rm r} }\over {a_6 + b_6 T + c_6 w_{\rm r} + d_6 T^2 + e_6 w^2 _{\rm r} + f_6 Tw_{\rm r} }} $$

(A.5)

where a ₅=0.0371, b ₅=9.1603× 10⁻⁵, c ₅=−0.0800, d ₅=−2.7390× 10⁻⁷, e ₅=−0.0435, f ₅=−6.0485× 10⁻⁵, a ₆=1.0, b ₆=0.0531, c ₆=2.2309, d ₆=1.1656× 10⁻³, e ₆=−0.3053, f ₆=0.0334; and μ_l the viscosity (Pa s).

A.4. Properties for the refrigerant in gas phase

The properties of the gas were obtained using the software REFPROP [34], and for the interval 0<p<1600 kPa and 0<T<60 °C the following fits are proposed for density ρ_g (kg/m³) and viscosity μ_g (Pa s), respectively,

$$ \rho_{\rm g} = {{a_7 + b_7 p + c_7 T + d_7 p^2 + e_7 T^2 + f_7 Tp}\over {a_8 + b_8 p + c_8 T + d_8 p^2 + e_8 T^2 + f_8 Tp}} $$

(A.6)

where a ₇=4.2473× 10⁻³, b ₇=−1.9077× 10⁻⁴, c ₇=0.0448, d ₇=3.4605× 10⁻⁵, e ₇=−2.4624× 10⁻⁵, f ₇=5.3830× 10⁻⁴, a ₈=1.0, b ₈=0.0155, c ₈=−8.2500× 10⁻⁴, d ₈=4.5680× 10⁻⁵, e ₈=6.9326× 10⁻⁸, f ₈=−2.1388× 10⁻⁶.

$$ \mu _g = {{a_9 + b_9 p + c_9 T + d_9 T^2 + e_9 p^3 }\over {a_{10} + b_{10} p + c_{10} p^2 + d_{10} T}} \times 10^{ - 6} $$

(A.7)

where a ₉=10.8186, b ₉=−2.6052× 10⁻³, c ₉=0.1451, d ₉=3.7658× 10⁻⁴, e ₉=−2.0170× 10⁻⁷, a ₉=1.0, b ₁₀=−2.1278× 10⁻⁴, c ₁₀=−7.752× 10⁻⁹, d ₁₀=9.6695× 10⁻³.