design of journal bearing

d=input('enter a diameter value:'); %journal diameter (m)
l=input('enter a length value:');   %bearing length (m)
r=d/2;                              %radius (m)
c=input('enter a radial clearance:'); % radial clearance (m)
T1=input('assign an inlet temperature:'); %inlet temperature (celcius)
delta_T=input('enter temperature rise value:'); %assumed temperature rise (celcius)
W=input('enter a radial load on journal:');     %radial load on journal (N)
n_s=input('enter journal speed value of shaft:');       %journal speed (rev/sec)
P=W/(l*d);                                        %bearing pressure (N/m^2)
Tavg_vector=[10:10:140];                        %average temperature vector 
abs_vis_vector=[500 190 90 45 28 18 12 9 6.5 5.2 4.2 3.2 2.8 2.6]; % absolute viscosity vector
Tavg=T1+(delta_T/2);                            %average temperature (celcius)
fabs_vis_vector=csapi(Tavg_vector,abs_vis_vector); %cubic spline interpolation 
absolute_viscosity_value=fnval(fabs_vis_vector,Tavg); %absolute viscosity value mPa.s
absolute_viscosity_value=absolute_viscosity_value*10^-3; %absolute viscosity value Pa.s
Sc=((absolute_viscosity_value*n_s)/P)*(r/c)^2;           %sommerfeld number
%performance parameters for corresponding l/d ratio
if l/d==1
    % for l/d=1 performance parameters
    S_1=[0.0188 0.0446 0.121 0.264 0.631 1.33 ];
    % the corresponding vector for the minimum film thickness variable,h0/c
    hoc_1=[0.1 0.2 0.4 0.6 0.8 0.9 ];
    % the corresponding vector for the coefficient of friction variable, Rf /c
    Rfc_1=[1.05 1.70 3.2 5.79 12.8 26.4];
    % the corresponding vector for the volumetric oil-flow rate, Q/ (RcnL)
    Qc_1=[4.74 4.62 4.33 3.99 3.59 3.37];
    % the corresponding vector for side leakage flow/total flow, Qs/Q
    Qsc_1=[0.919 0.842 0.680 0.497 0.280 0.150]; 
elseif l/d==0.5
    % for l/d=0.5 performance parameters
    S_2=[0.0313 0.0923 0.319 0.779 2.03 4.31];
    % the corresponding vector for the minimum film thickness variable,h0/c
    hoc_2=[0.1 0.2 0.4 0.6 0.8 0.9];
     % the corresponding vector for the coefficient of friction variable, Rf /c
    Rfc_2=[1.60 3.26 8.10 17.0 40.9 85.6];
    % the corresponding vector for the volumetric oil-flow rate, Q/ (RcnL)
    Qc_2=[5.69 5.41 4.85 4.29 3.72 3.43];
    % the corresponding vector for side leakage flow/total flow, Qs/Q
    Qsc_2=[0.939 0.874 0.730 0.552 0.318 0.173];
elseif l/d==0.25
    % for l/d=0.25 performance parameters
    S_3=[0.0736 0.261 1.07 2.83 7.57 16.2];
    % the corresponding vector for the minimum film thickness variable,h0/c
    hoc_3=[0.1 0.2 0.4 0.6 0.8 0.9];
    % the corresponding vector for the coefficient of friction variable, Rf /c
    Rfc_3=[ 3.50 8.8 26.7 61.1 153.0 322.0];
    % the corresponding vector for the volumetric oil-flow rate, Q/ (RcnL)
    Qc_3=[5.91 5.60 4.99 4.37 3.76 3.45 ];
    % the corresponding vector for side leakage flow/total flow, Qs/Q
    Qsc_3=[0.945 0.884 0.746 0.567 0.330 0.180 ];
elseif l/d==inf 
    % for l/d=inf performance parameters
    S_4=[0.0115 0.021 0.0389 0.0626 0.123 0.240];
    % the corresponding vector for the minimum film thickness variable,h0/c
    hoc_4=[0.1 0.2 0.4 0.6 0.8 0.9];
    % the corresponding vector for the coefficient of friction variable, Rf /c
    Rfc_4=[0.961 1.20 1.52 2.57 4.80];
    % the corresponding vector for the volumetric oil-flow rate, Q/ (RcnL)
    Qc_4=[0.760 1.56 2.26 2.83 3.03];
    % the corresponding vector for side leakage flow/total flow, Qs/Q
    Qsc_4=[0 0 0 0 0 0 0 0 0];
end
if l/d==1
% To interpolate the data, the MATLAB function csapi(X,Y) is used:
fhoc_1 = csapi( S_1, hoc_1 );
ffc_1 = csapi( S_1, Rfc_1);
fQR_1 = csapi( S_1, Qc_1);
fQs_1 = csapi( S_1, Qsc_1);
% The values of the interpolated data functions at Sc are calculated with:
hc = fnval(fhoc_1, Sc); % ho/c 
fc = fnval(ffc_1, Sc);  %(r/c)f
Qc = fnval(fQR_1, Sc);  % Q/(r c n L)
Qsc= fnval(fQs_1, Sc);  % Qs/Q 
elseif l/d==0.5
% To interpolate the data, the MATLAB function csapi(X,Y) is used:
fhoc_2 = csapi( S_2, hoc_2);
ffc_2 = csapi( S_2, Rfc_2);
fQR_2 = csapi( S_2, Qc_2);
fQs_2 = csapi( S_2, Qsc_2);
% The values of the interpolated data functions at Sc are calculated with:
hc = fnval(fhoc_2, Sc); % ho/c
fc = fnval(ffc_2, Sc);  %(r/c)f
Qc = fnval(fQR_2, Sc);  % Q/(r c n L)
Qsc= fnval(fQs_2, Sc);  % Qs/Q
elseif l/d==0.25
% To interpolate the data, the MATLAB function csapi(X,Y) is used:
fhoc_3= csapi( S_3, hoc_3 ); 
ffc_3= csapi( S_3, Rfc_3 );  
fQR_3= csapi( S_3, Qc_3 );   
fQs_3= csapi( S_3, Qsc_3 );  
% The values of the interpolated data functions at Sc are calculated with:
hc = fnval(fhoc_3, Sc);  % ho/c
fc = fnval(ffc_3, Sc);   %(r/c)f
Qc = fnval(fQR_3, Sc);   % Q/(r c n L)
Qsc= fnval(fQs_3, Sc);   % Qs/Q
else l/d==inf
% To interpolate the data, the MATLAB function csapi(X,Y) is used:
fhoc_4 = csapi( S_4, hoc_4 );
ffc_4 = csapi( S_4, Rfc_4 );
fQR_4 = csapi( S_4, Qc_4 );
fQs_4= csapi( S_4, Qsc_4 );
% The values of the interpolated data functions at Sc are calculated with:
hc = fnval(fhoc_4, Sc);  % ho/c
fc = fnval(ffc_4, Sc);   %(r/c)f
Qc = fnval(fQR_4, Sc);   % Q/(r c n L)
Qsc= fnval(fQs_4, Sc);   % Qs/Q
end 

4 types of l/d ratio is given on the chart as it seen. but when we write a l/d ratio which is not given on the chart we use interpolation to found it. for example if ı want to find l/d=0.60 how can ı write ıts matlab code and calculate it?

Answer 1

[Cris LaPierre]

I suggest having a look at this answer.

[matlab coder]

its not working on my question. ı have an interpolation equation

[Cris LaPierre]

The easy answer then is to use interp1.

That answer I sent you to was of a similar nature and that solution involved interpolation. You won't be able to just straight copy the solution provided there, but you should be able to apply it to your situation.

[matlab coder]

ı didnt copy and paste on my question :) ı said this is not working on my question because as it seen ı have an ınterpolatıon formula there for different kind of l/d ratio. so interp1 ıs not useful for my design. ı should use that equatıon given above. thank you ın advance

[matlab coder]

the point that ı am looking forward is not on the lines. ıts between the lines. thats why ı should use that equatıon. but ı dont know how to solve thıs problem. for example when S=0.20 and l/d=0.80 how can ı find the corresponding y axis value by matlab code?

[Alan Stevens]

You first need to define y1, y1/2, y1/4 and yinf as functions of S, using interp1 for each. e.g.

y1 = @(S) interp1(Svals, Flowvals,S);

where Svals and Flowvals are vectors of data either extracted from your graph, or already available somewhere.

Then you can define y as a function

y = @(S,L,D) (1/(L/D)^3)*( -1/8*(1-L/D)*(1-2*L/D)*(1-4*L/D)*yinf(S) + ...etc);

Then you can call this function with

y(S,L,D)

Note that the above assumes you are ony going to call y with single values of S, L and D at a time. If you are going to use multiple values in one go, then use

.* rather than * etc

for element by element multiplication etc.

[matlab coder]

ı dont have equation of this chart so ı cant create y1, y1/2, y1/4 and yinf as functions of S.

[Alan Stevens]

Then you will have to extract points from the chart itself. A little tedious perhaps, but not impossible.(If you had the equations you wouldn't need interp1).

[matlab coder]

% for l/d=1 performance parameters
    S_1=[0.0188 0.0446 0.121 0.264 0.631 1.33 ];
     % the corresponding vector for the volumetric oil-flow rate, Q/ (RcnL)
    Qc_1=[4.74 4.62 4.33 3.99 3.59 3.37];
    % for l/d=0.5 performance parameters
    S_2=[0.0313 0.0923 0.319 0.779 2.03 4.31];
    % the corresponding vector for the volumetric oil-flow rate, Q/ (RcnL)
    Qc_2=[5.69 5.41 4.85 4.29 3.72 3.43];
    % for l/d=0.25 performance parameters
    S_3=[0.0736 0.261 1.07 2.83 7.57 16.2];
    % the corresponding vector for the volumetric oil-flow rate, Q/ (RcnL)
    Qc_3=[5.91 5.60 4.99 4.37 3.76 3.45 ];
    % for l/d=inf performance parameters
    S_4=[0.0115 0.021 0.0389 0.0626 0.123 0.240];
    % the corresponding vector for the volumetric oil-flow rate, Q/ (RcnL)
    Qc_4=[0.760 1.56 2.26 2.83 3.03];
    

these are the exracted data from the chart but ı still cant solve the proble with your suggestion. can you show me on an example from these data sets. for example can you calculate for l=0.2 and d=0.5 and S=0.20. how can ı find the corresponding Qc value? thank you in advance ı am so glad that you help me

[Cris LaPierre]

The documentation is you friend. Here are 7 examples of how to use interp1.

[Cris LaPierre]

Ok, I put a little more time into this. With one of the lines having a ratio of inf, you can't use interp1 to get the final Qc value. Here's how I was able to do it.

First step is to find the Qc value for each of the 4 lines at your desired S value. You can do this visually or, if you have a decent vector of points representing each line, use interp1. These results represent the inputs , , and in the interpolation equation provided by the authors.

If you are needing to do this a lot, it is to your benefit to get a good vector of points for each line, as that will ultimately determine the accuracy of your interpolated value. Also note your Qc_4 vector only had 5 points. It appeared to be missing the first value. I used the graph to estimate a value of 0.4 @S=0.0115.

% Known values
% for l/d=1 performance parameters
S_1=[0.0188 0.0446 0.121 0.264 0.631 1.33 ];
Qc_1=[4.74 4.62 4.33 3.99 3.59 3.37];
% for l/d=0.5 performance parameters
S_2=[0.0313 0.0923 0.319 0.779 2.03 4.31];
Qc_2=[5.69 5.41 4.85 4.29 3.72 3.43];
% for l/d=0.25 performance parameters
S_3=[0.0736 0.261 1.07 2.83 7.57 16.2];
Qc_3=[5.91 5.60 4.99 4.37 3.76 3.45 ];
% for l/d=inf performance parameters
S_4=[0.0115 0.021 0.0389 0.0626 0.123 0.240];
Qc_4=[0.4 0.760 1.56 2.26 2.83 3.03];

% interpolate to get y value of each line at Sq
Sq = 0.20;
y1 = interp1(S_1,Qc_1,Sq);
y12 = interp1(S_2,Qc_2,Sq);
y14 = interp1(S_3,Qc_3,Sq);
yinf = interp1(S_4,Qc_4,Sq);

Now that you have the 4 y values for a specific S value, you can then use the interpolation equation to compute the Qc value for any ratio between infinity and 0.25. I defined the interpolation equation as an anonymous function. This allows me to easly calculate the Qc value by passing in the values for l,d,yinf,y1,y12,y14. This allows me to reuse the equation.

% desired ratio parameters
l=0.2;
d = 0.5;

% Create interpolatoin equation
ld=@(l,d,ying,y1,y12,y14) (-(1-l/d).*(1-2*l/d).*(1-4*l./d)*yinf/8 + (1-2*l./d).*(1-4*l./d)*y1/3 -(1-l./d).*(1-4*l./d).*y12/4 + (1-l./d).*(1-2*l./d)*y14/24)./(l/d).^3;
% Call interpolation equation with specific inputs
y=ld(l,d,yinf,y1,y12,y14)

y = 5.5067

[matlab coder]

thanks a lot man. ı really appreciate for your effort, ıt will help me on my project... thanks again

[matlab coder]

when ı changed Sq=0.20 to Sq=0.60 it gives NaN. ıs ıt normal? because it should be approximately 4.8.

[Cris LaPierre]

In this case, yes. The result of yinf is NaN because you have not defined values of Qc_4 beyond S_4=0.240. This causes the equation for y to return NaN.

When you go beyond your known data, that is called extrapolation. It is possible to exrapolate with interp1, but given the shape of your yinf line, I think it would be better to add more data points to your S_4 and Qc_4 vectors.

[matlab coder]

ok thank you ı will rearrange my data table.

Answer 2

[Alan Stevens]

Here's an example for l/d = 1/4. Note that because the S scale is logarithmic, it's best to interpolate logarithmically:

    S_3=[0.0736 0.261 1.07 2.83 7.57 16.2];
    % the corresponding vector for the volumetric oil-flow rate, Q/ (RcnL)
    Qc_3=[5.91 5.60 4.99 4.37 3.76 3.45 ];
   LOGS_3 = log10(S_3);
   yqtr = @(S) interp1(LOGS_3,Qc_3,log10(S));
   S = 0.08:0.01:10;
    semilogx(S_3,Qc_3,'o'),grid
    hold on
    semilogx(S,yqtr(S))

I'm not sure why you have a different set of values of S for each curve, as they all cover broadly the same range.

Also, I'd be inclined to pick off more than 6 pairs of points for each curve.

[matlab coder]

ı dont know how to express myself but ı guess ı couldnt describe my problem clearly. as it seen on the first code, when ı enter daiemeter value and length value and if they are 1, 1/2 or 1/4 ı calculate them with the data that ı have in tabular form using by "csapi" and "fnval" function. but when l/d=0.4 or 0.8 or 0.6, ı mean for the values that are not given on the chart how can ı find the corresponding Qc value? look, when ı run the code given below with a entered values the program calculate many parameters. you can check by hand when ı write l=0.20 and d=0.20 this means l/d=1. and S value is calculated on the program 0.4158. and as we can see on the chart for l/d=1 and S=0.4158 Qc value is about 3.75. and this program calculate it. and also many more parameters that i didnt share here. now my questıon is that how can ı solve if ı enter a l/d ratio which is not on the chart. ı mean differ from 1, 1/2 1/4 and inf. thank you

% purpose:
% To calculate performance parameters and temperature rise by user specified values
%        Date                   Programmer
%        ====                   ==========
%       31/12/20                M. AKGÖK
% some inputs specified by designer 

d=input('enter a diameter value:'); %journal diameter (m)
l=input('enter a length value:');   %bearing length (m)
r=d/2;                              %radius (m)
c=input('enter a radial clearance:'); % radial clearance (m)
T1=input('assign an inlet temperature:'); %inlet temperature (celcius)
delta_T=input('enter temperature rise value:'); %assumed temperature rise (celcius)
W=input('enter a radial load on journal:');     %radial load on journal (N)
n_s=input('enter journal speed value of shaft:');       %journal speed (rev/sec)
P=W/(l*d);                                        %bearing pressure (N/m^2)
Tavg_vector=[10:10:140];                        %average temperature vector 
abs_vis_vector=[500 190 90 45 28 18 12 9 6.5 5.2 4.2 3.2 2.8 2.6]; % absolute viscosity vector
Tavg=T1+(delta_T/2);                            %average temperature (celcius)
fabs_vis_vector=csapi(Tavg_vector,abs_vis_vector); %cubic spline interpolation 
absolute_viscosity_value=fnval(fabs_vis_vector,Tavg); %absolute viscosity value mPa.s
absolute_viscosity_value=absolute_viscosity_value*10^-3; %absolute viscosity value Pa.s
Sc=((absolute_viscosity_value*n_s)/P)*(r/c)^2;           %sommerfeld number
%performance parameters for corresponding l/d ratio
if l/d==1
    % for l/d=1 performance parameters
    S_1=[0.0188 0.0446 0.121 0.264 0.631 1.33 ];
    % the corresponding vector for the minimum film thickness variable,h0/c
    hoc_1=[0.1 0.2 0.4 0.6 0.8 0.9 ];
    % the corresponding vector for the coefficient of friction variable, Rf /c
    Rfc_1=[1.05 1.70 3.2 5.79 12.8 26.4];
    % the corresponding vector for the volumetric oil-flow rate, Q/ (RcnL)
    Qc_1=[4.74 4.62 4.33 3.99 3.59 3.37];
    % the corresponding vector for side leakage flow/total flow, Qs/Q
    Qsc_1=[0.919 0.842 0.680 0.497 0.280 0.150]; 
elseif l/d==0.5
    % for l/d=0.5 performance parameters
    S_2=[0.0313 0.0923 0.319 0.779 2.03 4.31];
    % the corresponding vector for the minimum film thickness variable,h0/c
    hoc_2=[0.1 0.2 0.4 0.6 0.8 0.9];
     % the corresponding vector for the coefficient of friction variable, Rf /c
    Rfc_2=[1.60 3.26 8.10 17.0 40.9 85.6];
    % the corresponding vector for the volumetric oil-flow rate, Q/ (RcnL)
    Qc_2=[5.69 5.41 4.85 4.29 3.72 3.43];
    % the corresponding vector for side leakage flow/total flow, Qs/Q
    Qsc_2=[0.939 0.874 0.730 0.552 0.318 0.173];
elseif l/d==0.25
    % for l/d=0.25 performance parameters
    S_3=[0.0736 0.261 1.07 2.83 7.57 16.2];
    % the corresponding vector for the minimum film thickness variable,h0/c
    hoc_3=[0.1 0.2 0.4 0.6 0.8 0.9];
    % the corresponding vector for the coefficient of friction variable, Rf /c
    Rfc_3=[ 3.50 8.8 26.7 61.1 153.0 322.0];
    % the corresponding vector for the volumetric oil-flow rate, Q/ (RcnL)
    Qc_3=[5.91 5.60 4.99 4.37 3.76 3.45 ];
    % the corresponding vector for side leakage flow/total flow, Qs/Q
    Qsc_3=[0.945 0.884 0.746 0.567 0.330 0.180 ];
elseif l/d==inf 
    % for l/d=inf performance parameters
    S_4=[0.0115 0.021 0.0389 0.0626 0.123 0.240];
    % the corresponding vector for the minimum film thickness variable,h0/c
    hoc_4=[0.1 0.2 0.4 0.6 0.8 0.9];
    % the corresponding vector for the coefficient of friction variable, Rf /c
    Rfc_4=[0.961 1.20 1.52 2.57 4.80];
    % the corresponding vector for the volumetric oil-flow rate, Q/ (RcnL)
    Qc_4=[0.760 1.56 2.26 2.83 3.03];
    % the corresponding vector for side leakage flow/total flow, Qs/Q
    Qsc_4=[0 0 0 0 0 0 0 0 0];
end
if l/d==1
% To interpolate the data, the MATLAB function csapi(X,Y) is used:
fhoc_1 = csapi( S_1, hoc_1 );
ffc_1 = csapi( S_1, Rfc_1);
fQR_1 = csapi( S_1, Qc_1);
fQs_1 = csapi( S_1, Qsc_1);
% The values of the interpolated data functions at Sc are calculated with:
hc = fnval(fhoc_1, Sc); % ho/c 
fc = fnval(ffc_1, Sc);  %(r/c)f
Qc = fnval(fQR_1, Sc);  % Q/(r c n L)
Qsc= fnval(fQs_1, Sc);  % Qs/Q 
elseif l/d==0.5
% To interpolate the data, the MATLAB function csapi(X,Y) is used:
fhoc_2 = csapi( S_2, hoc_2);
ffc_2 = csapi( S_2, Rfc_2);
fQR_2 = csapi( S_2, Qc_2);
fQs_2 = csapi( S_2, Qsc_2);
% The values of the interpolated data functions at Sc are calculated with:
hc = fnval(fhoc_2, Sc); % ho/c
fc = fnval(ffc_2, Sc);  %(r/c)f
Qc = fnval(fQR_2, Sc);  % Q/(r c n L)
Qsc= fnval(fQs_2, Sc);  % Qs/Q
elseif l/d==0.25
% To interpolate the data, the MATLAB function csapi(X,Y) is used:
fhoc_3= csapi( S_3, hoc_3 ); 
ffc_3= csapi( S_3, Rfc_3 );  
fQR_3= csapi( S_3, Qc_3 );   
fQs_3= csapi( S_3, Qsc_3 );  
% The values of the interpolated data functions at Sc are calculated with:
hc = fnval(fhoc_3, Sc);  % ho/c
fc = fnval(ffc_3, Sc);   %(r/c)f
Qc = fnval(fQR_3, Sc);   % Q/(r c n L)
Qsc= fnval(fQs_3, Sc);   % Qs/Q
else l/d==inf
% To interpolate the data, the MATLAB function csapi(X,Y) is used:
fhoc_4 = csapi( S_4, hoc_4 );
ffc_4 = csapi( S_4, Rfc_4 );
fQR_4 = csapi( S_4, Qc_4 );
fQs_4= csapi( S_4, Qsc_4 );
% The values of the interpolated data functions at Sc are calculated with:
hc = fnval(fhoc_4, Sc);  % ho/c
fc = fnval(ffc_4, Sc);   %(r/c)f
Qc = fnval(fQR_4, Sc);   % Q/(r c n L)
Qsc= fnval(fQs_4, Sc);   % Qs/Q
end 

Temperature_rise = (8.3*P*fc)/(10^6*Qc*(1-0.5*Qsc));

[Alan Stevens]

So , in the general expression for y, replace y1 by fnval(fQs_1, Sc);, y1/2 by fnval(fQs_2, Sc); etc.

[matlab coder]

thank you so much