coding for a finite iteration

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SITI AISHAH
SITI AISHAH 2019-10-5
评论： Walter Roberson 2019-10-5
% Specify the variable %
Temp_in = 23 ; % Sump Temperature (°C)
d = 0.150; % Shaft Journal Diameter (Meter, m)
r = (d/2); % Shaft Journal Radius (Meter, m)
N = (166.69/60); % Shaft Speed (revolution per second,rps)
W = 9000; % Radial Load (Newton, N)
L = 0.300 ; % Bearing Length (Meter, m)
c = 0.00065 ; % Clearance (Meter, m)
%-------------------------------------------------------------------------%
% Calculate the variable %
C_ratio = r/c; % Clearance ratio
P = W/(2*r*L); % Nominal Pressure (Pascal)
Slenderness = L/d ; % Bearing Aspect Ratio
%-------------------------------------------------------------------------%
% Calculation to find average temperature and viscosity %
Tf = Temp_in + 3; % Assumed Average Temperature (°C)
V = 0.0018*exp(-0.026*Tf); % Viscosity (Pa.s)
S = ((C_ratio)^2)*((V*N)/P); % Sommerfeld Numbert
if Slenderness == 1
delta_T = (0.349109 + (6.00940*S)+(0.047467*(S^2)))*(P*(10^-6))/0.120; % Temperature Rise(°C) at l/d=1
elseif Slenderness == 1/2
delta_T = (0.394552 + (6.392527*S)-(0.036013*(S^2)))*(P*(10^-6))/0.120; % Temperature Rise(°C) at l/d=1/2
elseif Slenderness == 1/4
delta_T = 0.933828 + (6.437512*S)- (0.011048*(S^2))*(P*(10^-6))/0.120; % Temperature Rise(°C) at l/d=1/4
else
delta_T = ((33405*(S)^6)-(51557*(S)^5)+(30669*(S)^4)-(8871.9*(S)^3)+(1297.4*(S)^2)-(85.088*(S))+2.6111)*(P*(10^-6))/0.120; % Temperature Rise(°C) at l/d=Infinity
end
T_average = Temp_in + (delta_T/2); % Average temperature (°C)
while abs (Tf-T_average)>0.001
delta = (Tf-T_average)/2;
Tf = Tf-delta;
V = 0.0018*exp(-0.026*Tf); % Viscosity (Pa.s)
S = (C_ratio)^2*((V*N)/P); % Sommerfeld Number
if Slenderness == 1
delta_T = (0.349109 + (6.00940*S)+(0.047467*(S^2)))*(P*(10^-6))/0.120; % Temperature Rise(°C) at l/d=1
elseif Slenderness == 1/2
delta_T = (0.394552 + (6.392527*S)-(0.036013*(S^2)))*(P*(10^-6))/0.120; % Temperature Rise(°C) at l/d=1/2
elseif Slenderness == 1/4
delta_T = (0.933828 + (6.437512*S)- (0.011048*(S^2)))*(P*(10^-6))/0.120; % Temperature Rise(°C) at l/d=1/4
elseif Slenderness > 1
delta_T = ((33405*(S)^6)-(51557*(S)^5)+(30669*(S)^4)-(8871.9*(S)^3)+(1297.4*(S)^2)-(85.088*(S))+2.6111)*(P*(10^-6))/0.120; % Temperature Rise(°C) at l/d=Infinity
end
T_average = Temp_in + (delta_T/2); % Average temperature (°C)end
disp ('Table for iteration');disp (' Tf V S deltaT T average ');
format shortG
disp ([Tf', V',S',delta_T', T_average',]);
end
I want to get the value of clearance when Tf-Taverage=0 for only when the iteration stop at the 36th iteration. Can somebody help me with this?Thank you.
  1 个评论
Walter Roberson
Walter Roberson 2019-10-5
What leads you to think that Tf-Taverage == 0 will ever hold? You are using floating point calculations and there will be round-off issues. It is entirely possible that evaluating at two adjacent representable numbers will lead to values that are negative for one of the two choices, and positive for the other choice, without it being possible to find an exact 0.
For example, if you were trying to solve x^2-2 == 0 then
>> (sqrt(2)*(1-eps))^2-2
ans =
-4.44089209850063e-16
>> (sqrt(2)*(1))^2-2
ans =
4.44089209850063e-16
>> (sqrt(2)*(1+eps))^2-2
ans =
8.88178419700125e-16
No exact solution

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