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Experimental identification of shaft misalignment in a turbo-generator system

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Abstract

Precise and authentic estimation of the dynamic features of rotating machines and prevention of failure requires accurate experimental characterisation of their critical components, for example, bearings and couplings. These are difficult to model theoretically, and they often suffer from uncertain parameters in their model. Especially, when there is a misalignment in the rotor system, dynamic characterisation of bearings and couplings changes drastically. In the present study, multiple fault parameters (MFPs) of critical components of turbo-generator, that is, bearing and coupling together with residual unbalances (RUs), are evaluated experimentally using model-based methodology. A test rig was developed and used for experimentation in which different levels of misalignment was introduced. After estimating the MFPs, then the accuracy was checked through an impact test on the rotor test rig. The effect of different levels of misalignments on estimated parameters was studied.

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Appendices

Appendix A: Timoshenko Beam Model [43]

1.1 A.1. Translational mass matrix

$$ [M_{t} ] = [M_{t} ]_{0} + \phi [M_{t} ]_{1} + \phi^{2} [M_{t} ]_{2} $$
(A.1)
$$ [M_{t} ]_{0} = \frac{\rho Al}{{420(1 + \phi )^{2} }}\left[ {\begin{array}{*{20}c} {156} & 0 & 0 & {22l} & {54} & 0 & 0 & { - 13l} \\ {} & {156} & { - 22l} & 0 & 0 & {54} & {13l} & 0 \\ {} & {} & {4l^{2} } & 0 & 0 & { - 13l} & { - 3l^{2} } & 0 \\ {} & {} & {} & {4l^{2} } & {13l} & 0 & 0 & { - 3l^{2} } \\ {} & {} & {} & {} & {156} & 0 & 0 & { - 22l} \\ {} & {} & {} & {} & {} & {156} & {22l} & 0 \\ {} & {\text{Sym}} & {} & {} & {} & {} & {4l^{2} } & 0 \\ {} & {} & {} & {} & {} & {} & {} & {4l^{2} } \\ \end{array} } \right] $$
(A.2)
$$ [M_{t} ]_{1} = \frac{\rho Al}{{420(1 + \phi )^{2} }}\left[ {\begin{array}{*{20}c} {294} & 0 & 0 & {38.5l} & {126} & 0 & 0 & { - 31.5l} \\ {} & {294} & { - 38.5l} & 0 & 0 & {126} & {31.5l} & 0 \\ {} & {} & {7l^{2} } & 0 & 0 & { - 31.5l} & { - 7l^{2} } & 0 \\ {} & {} & {} & {7l^{2} } & {31.5l} & 0 & 0 & { - 7l^{2} } \\ {} & {} & {} & {} & {294} & 0 & 0 & { - 38.5l} \\ {} & {} & {} & {} & {} & {294} & {38.5l} & 0 \\ {} & {\text{Sym}} & {} & {} & {} & {} & {7l^{2} } & 0 \\ {} & {} & {} & {} & {} & {} & {} & {7l^{2} } \\ \end{array} } \right] $$
(A.3)
$$ [M_{t} ]_{2} = \frac{\rho Al}{{420(1 + \phi )^{2} }}\left[ {\begin{array}{*{20}c} {140} & 0 & 0 & {17.5l} & {70} & 0 & 0 & { - 17.5l} \\ {} & {140} & { - 17.5l} & 0 & 0 & {70} & {17.5l} & 0 \\ {} & {} & {3.5l^{2} } & 0 & 0 & { - 17.5l} & { - 3.5l^{2} } & 0 \\ {} & {} & {} & {3.5l^{2} } & {17.5l} & 0 & 0 & { - 3.5l^{2} } \\ {} & {} & {} & {} & {140} & 0 & 0 & { - 17.5l} \\ {} & {} & {} & {} & {} & {140} & {17.5l} & 0 \\ {} & {\text{Sym}} & {} & {} & {} & {} & {3.5l^{2} } & 0 \\ {} & {} & {} & {} & {} & {} & {} & {3.5l^{2} } \\ \end{array} } \right] $$
(A.4)

with

$$ \phi = \frac{12EI}{{k_{sc} GAl^{2} }},\quad k_{sc} = \frac{6(1 + \upsilon )}{(7 + 6\upsilon )} $$

Displacement vector \( \left\{ \eta \right\}^{s} = \left\{ {x_{i} ,y_{i} ,\varphi_{{x_{i} }} ,\varphi_{{y_{i} }} ,x_{i + 1} ,y_{i + 1} ,\varphi_{{x_{i + 1} }} ,\varphi_{{y_{i + 1} }} } \right\}^{T} \), where sub-script i is element number

1.2 A.2. Rotational mass matrix

$$ [M_{r} ] = [M_{r} ]_{0} + \phi [M_{r} ]_{1} + \phi^{2} [M_{r} ]_{2} $$
(A.5)
$$ [M_{r} ]_{0} = \frac{\rho Al}{{(1 + \phi )^{2} }}\left[ {\begin{array}{*{20}c} {6/5l} & 0 & 0 & {1/10} & { - 6/5l} & 0 & 0 & {1/10} \\ {} & {6/5l} & { - 1/10} & 0 & 0 & { - 6/5l} & { - 1/10} & 0 \\ {} & {} & {2l/15} & 0 & 0 & {1/10} & { - l/30} & 0 \\ {} & {} & {} & {2l/15} & { - 1/10} & 0 & 0 & { - l/30} \\ {} & {} & {} & {} & {6/5l} & 0 & 0 & { - 1/10} \\ {} & {} & {} & {} & {} & {6/5l} & {1/10} & 0 \\ {} & {\text{Sym}} & {} & {} & {} & {} & {2l/15} & 0 \\ {} & {} & {} & {} & {} & {} & {} & {2l/15} \\ \end{array} } \right] $$
(A.6)
$$ [M_{r} ]_{1} = \frac{\rho Al}{{(1 + \phi )^{2} }}\left[ {\begin{array}{*{20}c} 0 & 0 & 0 & { - 1/2} & 0 & 0 & 0 & { - 1/2} \\ {} & 0 & {1/2} & 0 & 0 & 0 & {1/2} & 0 \\ {} & {} & {l/6} & 0 & 0 & { - 1/2} & { - l/6} & 0 \\ {} & {} & {} & {l/6} & {1/2} & 0 & 0 & { - l/6} \\ {} & {} & {} & {} & 0 & 0 & 0 & {1/2} \\ {} & {} & {} & {} & {} & 0 & { - 1/2} & 0 \\ {} & {\text{Sym}} & {} & {} & {} & {} & {l/6} & 0 \\ {} & {} & {} & {} & {} & {} & {} & {l/6} \\ \end{array} } \right] $$
(A.7)
$$ [M_{r} ]_{2} = \frac{\rho Al}{{(1 + \phi )^{2} }}\left[ {\begin{array}{*{20}c} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ {} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ {} & {} & {l/3} & 0 & 0 & 0 & {l/6} & 0 \\ {} & {} & {} & {l/3} & 0 & 0 & 0 & {l/6} \\ {} & {} & {} & {} & 0 & 0 & 0 & 0 \\ {} & {} & {} & {} & {} & 0 & 0 & 0 \\ {} & {\text{Sym}} & {} & {} & {} & {} & {l/3} & 0 \\ {} & {} & {} & {} & {} & {} & {} & {l/3} \\ \end{array} } \right] $$
(A.8)

1.3 A.3. Stiffness matrix

$$ \left[ K \right] = \left[ K \right]_{0} + \phi \left[ K \right]_{1} $$
(A.9)
$$ \left[ K \right]_{0} = \frac{EI}{{(1 + \phi )l^{3} }}\left[ {\begin{array}{*{20}c} {12} & 0 & 0 & {6l} & { - 12} & 0 & 0 & {6l} \\ {} & {12} & { - 6l} & 0 & 0 & { - 12} & {6l} & 0 \\ {} & {} & {4l^{2} } & 0 & 0 & {6l} & {2l^{2} } & 0 \\ {} & {} & {} & {4l^{2} } & { - 6l} & 0 & 0 & {2l^{2} } \\ {} & {} & {} & {} & {12} & 0 & 0 & { - 6l} \\ {} & {} & {} & {} & {} & {12} & {6l} & 0 \\ {} & {\text{Sym}} & {} & {} & {} & {} & {4l^{2} } & 0 \\ {} & {} & {} & {} & {} & {} & {} & {4l^{2} } \\ \end{array} } \right] $$
(A.10)
$$ \left[ K \right]_{1} = \frac{EI}{{(1 + \phi )l^{3} }}\left[ {\begin{array}{*{20}c} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ {} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ {} & {} & {l^{2} } & 0 & 0 & 0 & { - l^{2} } & 0 \\ {} & {} & {} & {l^{2} } & 0 & 0 & 0 & { - l^{2} } \\ {} & {} & {} & {} & 0 & 0 & 0 & 0 \\ {} & {} & {} & {} & {} & 0 & 0 & 0 \\ {} & {\text{Sym}} & {} & {} & {} & {} & {l^{2} } & 0 \\ {} & {} & {} & {} & {} & {} & {} & {l^{2} } \\ \end{array} } \right] $$
(A.11)

1.4 A.4. Gyroscopic matrix

$$ \left[ G \right] = \left[ G \right]_{0} + \phi \left[ G \right]_{1} + \phi^{2} \left[ G \right]_{2} $$
(A.12)
$$ \left[ G \right]_{0} = \frac{{\rho Ar^{2} }}{{60(1 + \phi )^{2} l}}\left[ {\begin{array}{*{20}c} 0 & {36} & { - 3l} & 0 & 0 & { - 36} & { - 3l} & 0 \\ {} & 0 & 0 & { - 3l} & {36} & 0 & 0 & { - 3l} \\ {} & {} & 0 & {4l^{2} } & { - 3l} & 0 & 0 & { - l^{2} } \\ {} & {} & {} & 0 & 0 & { - 3l} & {l^{2} } & 0 \\ {} & {} & {} & {} & 0 & {36} & {3l} & 0 \\ {} & {} & {} & {} & {} & 0 & 0 & {3l} \\ {} & {\text{Skew}} & {\text{Sym}} & {} & {} & {} & 0 & {4l^{2} } \\ {} & {} & {} & {} & {} & {} & {} & 0 \\ \end{array} } \right] $$
(A.13)
$$ \left[ G \right]_{1} = \frac{{\rho Ar^{2} }}{{60(1 + \phi )^{2} l}}\left[ {\begin{array}{*{20}c} 0 & 0 & {15l} & 0 & 0 & 0 & {15l} & 0 \\ {} & 0 & 0 & {15l} & 0 & 0 & 0 & {15l} \\ {} & {} & 0 & {5l^{2} } & {15l} & 0 & 0 & { - 5l^{2} } \\ {} & {} & {} & 0 & 0 & {15l} & {5l^{2} } & 0 \\ {} & {} & {} & {} & 0 & 0 & 0 & { - 15l} \\ {} & {} & {} & {} & {} & 0 & 0 & {5l^{2} } \\ {} & {\text{Skew}} & {\text{Sym}} & {} & {} & {} & 0 & {5l^{2} } \\ {} & {} & {} & {} & {} & {} & {} & 0 \\ \end{array} } \right] $$
(A.14)
$$ \left[ G \right]_{2} = \frac{{\rho Ar^{2} }}{{60(1 + \phi )^{2} l}}\left[ {\begin{array}{*{20}c} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ {} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ {} & {} & {10l^{2} } & 0 & 0 & 0 & 0 & {5l^{2} } \\ {} & {} & {} & 0 & 0 & 0 & {5l^{2} } & 0 \\ {} & {} & {} & {} & 0 & 0 & 0 & 0 \\ {} & {} & {} & {} & {} & 0 & 0 & 0 \\ {} & {\text{Skew}} & {\text{Sym}} & {} & {} & {} & 0 & {10l^{2} } \\ {} & {} & {} & {} & {} & {} & {} & 0 \\ \end{array} } \right] $$
(A.15)

1.5 A.5. Rigid disc model

Mass matrix

$$ [M]^{d} = \left[ {\begin{array}{*{20}c} {m_{d} } & 0 & 0 & 0 \\ 0 & {m_{d} } & 0 & 0 \\ 0 & 0 & {I_{d} } & 0 \\ 0 & 0 & 0 & {I_{d} } \\ \end{array} } \right] $$
(A.16)

Gyroscopic matrix

$$ [G]^{d} = \left[ {\begin{array}{*{20}c} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & {I_{p} } \\ 0 & 0 & { - I_{p} } & 0 \\ \end{array} } \right] $$
(A.17)

Displacement vector

$$ \left\{ \eta \right\}^{d} = \left\{ {x,y,\varphi_{x} ,\varphi_{y} } \right\}^{T} $$
(A.18)

Appendix B: Natural Frequencies for Simply Supported Rotor Systems

Closed form solution is provided to obtain natural frequency of a simply supported rotor system with massless shaft and a disc at mid-span is given as

$$ \omega_{nf} = \frac{1}{2\pi }\sqrt {\frac{k}{m}} \,{\text{Hz}} $$
(B.1)

where

$$ k = \frac{48EI}{{L^{3} }} $$
(B.2)

Material properties and rotor dimensions are taken from table 1 and natural frequencies are summarised in table B1.

Table B1 Natural frequencies with simply supported boundary conditions.

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Lal, M., Tiwari, R. Experimental identification of shaft misalignment in a turbo-generator system. Sādhanā 43, 80 (2018). https://doi.org/10.1007/s12046-018-0859-1

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