Below is a collection of problems from my Maintainability Engineering course. These are mainly for me to reference in the future – but feel free to use these as a resource. If you find anything wrong with any of the answers please contact me using the contact form. Thanks!
- The signal X(t) from an accelerometer, mounted on a pump bearing, contains frequency components up to 250 Hz. The pump shaft rotational speed is 1,185 RPM.
- Express the pump shaft rotational frequency in Hz.
- At what rate must X(t) be sampled in order to reconstruct it without error? Choose one answer from below and explain your choice.
- A.Less than 500 Hz. B. Greater than 500 Hz. C. Greater than 400 Hz.
Let’s Dive into the solution:
What is an Accelerometer?
An Accelerometer, in short, is a device that measures the change in acceleration on its own instantaneous rest frame. For example, it would only measure gravity at rest, straight upwards. In free fall, it would measure an acceleration of 9.81 m/s^2 toward the center of mass of the Earth.
Expressing the pump shaft rotational frequency in terms of Hertz (Hz). Firstly, a Hertz is defined as one cycle per second. To convert between RPM to Hertz, we need the cycles per second. An RPM is a revolution (or “cycle”) per minute. Therefore, we must divide RPM by the number of seconds in a minute.
# of Hertz = RPM / 60 = 1185/60 = 19.75 Hz.
For the second question, according to the Nyquist Sampling Theorem, a sampling frequency must at least be two times bigger than the largest frequency of the equipment being measured. Given the maximum is 250 Hz for the equipment being measured, we know that the sampling frequency needs to be at least 500Hz.
(a) An acoustic signal picked up from a microphone is sampled at 10 kHz. Calculate the maximum information frequency of the digitized signal.
(b)It was later determined that this signal contained an unexpected frequency component at 5.5 kHz. Calculate the frequency at which aliasing effect is seen in the digitized signal.
(c) State two approaches you may use to avoid the aliasing problem in this case.
A. From Nyquist’s theorem above, we can apply the following equation:
F_s >= 2 * F_max
Which simply states, that the sampling frequency needs to be two times the maximum frequency. Since we are given the sample in this case, we can find the F_max of the information frequency. This value is 10 kHz/2 = 5kHz.
B. Now we introduce a problem into our system. Whenever the informational frequency becomes greater than half the sample frequency, a new effect happens in the sample which makes it harder to get accurate values. This effect is called aliasing.
Aliasing is what happens in a sample measurement whenever signals aren’t able to be distinguished from other samples. I’ve included an example of this effect to the left (Or above if browsing on mobile.) Notice that most of the photo looks fairly decent until you begin to look at my shirt. (I’m the guy on the left.) This photo was taken on an older phone and does a decent job “sampling” in most areas of the photo. However, whenever it comes to the small checkered pattern of my shirt, the frequency needed to capture this “signal” is not achievable on the phone used to take the picture. Therefore, the image is distorted and the “information sample” cannot be accurately represented by the signal frequency used to capture the image.
The problem asks us which frequencies we would then expect the aliasing effect to occur at. This can be done through the use of two equations.
First, we need to determine the frequency of the sampling equipment. In this case it is given to us as 10kHz. F_s = 10kHz.
Then we need the frequency of the information signal. It is given as 5.5kHz.
Aliasing will therefore occur at all frequencies where the sampling frequency adds or subtracts to the information frequency. Equationally:
10kHz ( + / -) X = 5.5kHz where X is all frequencies of where aliasing will occur. Therefore we can easily see aliasing will occur at 4.5 kHz in this situation.
Finally, are asked two solutions to this problem in part C. Right off hand, one could safely say that if we increased sampling frequency above twice the maximum information frequency, we would remove the aliasing problem. Additionally, we could apply an aliasing filter to the system which would remove the aliasing signals from the measurements.
- State four important and distinct causes of vibration in rotating machinery.
What causes machines to begin to vibrate? A lot of vibrations from my experience in industry have been a result of two major components: Machine failure over time and improper installation. Simple mistakes can leave a machine slowly vibrating out of control until total failure happens. Here is a brief list of cases leading to vibrating:
- Mechanical looseness or machine imbalance.
- Faults in gearboxes due to wear/fatigue
- inadequate lubrication materials. (Missing oil, grease, etc.)
- Instability in Bearings
The rotational speed of an electric motor is 3555 RPM. Calculate the frequency of rotation in Hz.
From previously, we know that that RPM stands for revolutions per minute and Hertz is a unit of “cycles” per second. They are both units of revolutions or cycles with the difference being in the time in which they are being counted. To find Hertz from RPM, we divide RPM count by 60 seconds to make the unit “cycles” / second.
3555 RPM / 60 Seconds = 59.25 Hertz.
Give an example for each of the following: (a) a harmonic frequency of the rotating speed, (b) a sub-synchronous frequency of the rotating speed, (c) a non-synchronous frequency of the rotating speed.
In this scenario, we are just asked for definitions of each of the above three harmonics.
In short, a harmonic frequency is just one of a positive whole number integer (ex. 2,3,4 etc).
A sub synchronous frequency occurs whenever the frequency is at a value below 1. (ex. .45)
A non-synchronous frequency occurs whenever the value is greater than one but not an integer but a floating number. (Example 1.354)
- The rotational speed of a steam turbine shaft is 3600 RPM. Calculate the shaft rotational speed in Hz.
At what basic frequency (Hz) in the spectral domain do you expect to see energy peaks
- Due to imbalance
- Due to bent shaft
- Due to parallel misalignment?
- State the directional placement of the accelerometer to measure each of these anomalies.
Hz = RPM /60 = 3600/60 = 60 Hz