1. Part B
Given:
• ShaftSpeed(N) = 2600 rpm
• Roller/race diameterratio(r) = 0.25
• Numberof rollingelement(n) =10
• Contact angle (ϴ) = 0°
• ClassII (ISO2372)
• Frequency:x(t) =A sin(2πƒt)
• Sensitivity(S) =1 V/g
For Shaft
Runningfrequency (f) =N/60,
f = 2600/60 = 43.33 Hz.
AngularFrequency, ω= 2πƒ or ω = 2π/T
Displacement(X); x(t) =A sin(ωt)……… ……………...(1)
Velocity(V); ẋ(t) =- Aω cos (ωt)………………………….(2)
Acceleration(α); ẍ(t) =- Aω² sin(ωt) ………………..(3)
For good case;AssumingVelocity(V) =0.71 mm/sfrom(ISO2372)
Substitutioninequation(2); 0.71 = A (2π ) (43.33)
A = 0.71/ 2π (43.33), therefore;A =0.0026 mm
For peakacceleration; α = Aω² So;α = 0.0026 × (2π X 43.33)²
Therefore;α = 192.71 mm/s²or α = 0.19271 m/s²
For Amax = α/ g So,Amax = 0.19271 / 9.81 therefore,Amax =0.0196 g
For OutputVoltage (v); v= (S) * (Amax) So,v = (1) (0.0196)
Therefore,v= 0.0196 v
2. For acceptable case; AssumingVelocity(V) =1.12 mm/sfrom (ISO2372)
Substitutioninequation(2); 1.12 = A (2π ) (43.33)
A = 1.12/ 2π (43.33), therefore;A =0.00411 mm
For peakacceleration; α = Aω² So;α = 0.00411 (2π X 43.33)²
Therefore;α = 304.63 mm/s²or α = 0.30463 m/s²
Amax = α/g So, Amax = 0.30463 / 9.81 therefore,Amax =0.0311 g
For OutputVoltage (v);v= (S) * (Amax)
v = (1) (0.0311)
Therefore,v= 0.0311 v
For Inner Race
InnerRace (BPFI) = (n/2) (N/60) (1 + r. cos ϴ)………………………………………(4)
So,BPFI = (10/2) (2600/60) (1 + 0.25. cos (0°))
Therefore,BPFI=270.833 Hz
For good case;AssumingVelocity(V) =0.71 mm/sfrom(ISO2372)
Substitutioninequation(2); 0.71 = A (2π ) (270.833)
A = 0.71/ 2π (270.833) = 0.00042 mm
For peakacceleration; α = Aω² So;α = 0.00042 (2π X 270.833)²
Therefore;α = 1187.3 mm/s²or α = 1.1873 m/s²
Amax = α/g So, Amax = 1.1873 / 9.81 = 0.121 g
For OutputVoltage (v);v= (S) * (Amax) So,v= (1) (0.121)
Therefore,v= 0.121 v
3. For acceptable case; AssumingVelocity(V) =1.12 mm/sfrom (ISO2372)
Substitutioninequation (2); 1.12 = A (2π ) (270.833)
A = 1.12/ 2π (270.833), therefore;A = 0.00066 mm
For peakacceleration; α = Aω² So;α = 0. 00066 (2π X 270.833)²
Therefore;α = 1911.2 mm/s²or α = 1.9112 m/s²
Amax = α/g = 1.9112 / 9.81 = 0.195 g
For OutputVoltage (v);v= (S) * (Amax) So,v= (1) (0.195)
Therefore,v= 0.195 v
4. Part C
From part B results;
• f (shaft) = 43.33 Hz
• f (inner) =270.833 Hz
• v (shaft,good) = 0.0196 v
• v (shaft,acceptable) =0.0311 v
• v (inner,good) =0.121 v
• v (inner,acceptable) =0.195 v
For suitable resolution(R); R= range/2n or R = Vmax – Vmin/2n
So,R (shaft) =(0.0311) – (0.0196) / 210
Therefore,R(shaft) =11.23 x 10-6
Similarly,R(inner)=(0.195) – (0.121) / 210
Therefore,R(inner) =72.27 x 10-6
For samplingfrequency(fs);fs= 2×f
So,fs (shaft) = 2 (43.33),
Therefore,fs(shaft) =86.66 Hz
Similarly,fs(inner) =2 (270.833)
Therefore,fs(inner) =541.666 Hz
5. For numberof samples(Ns);
Ns = 2n
Ns = 210
Therefore,Ns=1024
Betterresolutioncouldbe alsoachievedbyusingop-ampstoamplifythe signal forclearer
precise sampling.Addedtothat,low andhighpass filterscouldalsobe usedtoeliminate noise
and unwantedsignals.
