 In this class, we are going to discuss about some numericals on noise, vibration and data acquisition. As by now, we would have understood the fundamentals of vibration, then little introduction into noise and of course, we need to acquire data so that we can analyze them on the computer. So, in this class, I am just going to give you six example problems, which normally we encounter while doing machinery condition monitoring and this through these examples, I will tell you certain simple tricks of the trade which has to be used while doing a machinery condition monitoring. So, the first problem is actually a very simple problem like what is the sound pressure level in decibel or dB of a plane wave at the end of 1 meter, if it is excited by a sound source of one Pascal at the other end. So, let me explain you this problem in the sense we have a duct, this is a long duct it could be of any length L. So, usually to give a source at one end, which is a plane wave source because the diameter of this duct D such that the frequency below which the plane waves propagate are given by this equation. So, 0.86 times speed of sound by the diameter of the duct. So, for example, our typical duct of diameter about 35 mm, this close comes to about 4000 hertz, this is typical. So, that means if I have a plane wave tube of duct diameter 35 mm. So, for any source whose frequency is less than 4000, I will be having plane waves travelling in this waveform, by plane waves I mean they are longitudinal, they are one directional. So, pressure at any cross section is the same, any point in the cross section is the same. So, this is a little tricky problem in the sense the sound wave, plane wave if it is propagating in a duct as long as the duct diameter satisfies this equation and the frequencies are less than 4000 hertz, I am going to have plane waves at the other end. So, this is the other end and thus the pressure will be the same if the inlet pressure is equal to one Pascal and the out load pressure p outlet will be also equal to one Pascal. So, in order now to express this Pascal's in decibel way you will use the formula as P L is nothing but 20 logarithm to the base 10 of P by P reference. So, P reference is 2 into 10 to the power minus 5 Pascal, if I plug P as one Pascal in this equation I will come up with an value of 94 decibel. This has a practical significance in the sense that in any problem or any special duct applications for example, I have an HVAC system where I have a blower giving in a conditioned air to a point say inside a room I have diffuser. So, if this diameter of this duct obeys this relationship that plane waves will propagate. So, any pressure here in Pascal will be the same sound pressure here in terms of this P is the sound pressure. So, we do not require any complication complicated formula to estimate the sound pressure level at the end of the duct as long as we know that it is a plane wave and plane wave conditions are being satisfied. The reason I told you about this problem is many a times in the industry we will come across such scenarios where noise is generated at one end and then we have to guess or estimate what is the sound pressure level at the other end and if you just by measuring the diameter of will ensure that the these equations obeyed where see the speed of sound in air at 20 degree Celsius I will know that this if these 35 mm and you can somehow calculate f is 4000 hertz. So, this can be calculated ok. So, now we will move over to the second problem which is again a very simple problem. This is an handheld accelerometer calibrator provides 10 meters per second square of peak vibration in terms of acceleration at a circular frequency of 1000 radians per second calculate the linear frequency and the RMS amplitudes of velocity and displacement provided by the calibrator. Usually to calibrate an accelerometer in the field it is actually mounted on a calibrator this is the handheld calibrator and this will give certain voltage or charge signal corresponding to this mechanical vibration is subjected and this vibration is typically about 10 meters per second square at a circular frequency of 1000 radians per second. The question is what is the velocity and what is the acceleration? As you know this is a pure tone of 1000 radians per second its frequency omega is sorry frequency linear frequency f is equal to nothing, but omega by 2 pi and this comes down to about 159.1 hertz ok. But I know the velocity magnitude for a harmonic wave is nothing, but the acceleration magnitude divided by omega and the displacement magnitude is nothing, but the velocity magnitude divided by omega R which is equal to the acceleration magnitude divided by omega square ok. So in this case we have our acceleration magnitude is 10 meters per second square and it says it is peak. So that means and if I was to plot the time history of such a signal. So this value a is nothing, but 10 meters per second square this is the acceleration waveform ok. So in this case the velocity is will be nothing, but 10 divided by omega is 1000. So this will come to meters per second. So this will correspond to 10 millimeter per second. So I am putting them in the box to explain to you later and similarly the displacement magnitude will be nothing, but 10 divided by 1000 into 1000 meters that is equal to 10 micro meter. So if my acceleration magnitude is 10 meters per second square for the same calibrator at 1000 radians per second I have velocity as 10 millimeters per second displacement as 10 kilometers. So to remember a good rule of thumb in the field is this is 10 this is 10 this is 10 only thing that we have to be careful about the units 1 is in meters per second square other is in millimeters per second another is in micron. So this problem has a second problem as to calculate the RMS amplitude and we know for a harmonic wave any general x is equal to A sin omega t the RMS amplitude x RMS is nothing, but A by root 2 that is 0.707 A. So to answer the second part of the problem find out the RMS amplitudes of velocity and displacement provided by the calibrator the V RMS would be therefore 0.707 times 10 millimeters per second that is 7.07 millimeters per second and the displacement RMS A RMS is equal to nothing, but 0.707 times 10 micron that is 7.07 micron. This is a certain handy numbers to remember in the field for example whenever we go to the field with this handheld calibrator which I had shown in one of my previous classes as to how this calibrator is actually used to calibrate accelerometers in the field all we know is we give a mechanical signal of 10 meters per second square at 1000 radians per second. We will just see the voltage output from this accelerometer and if we are going to calibrate in the displacement mode or in the velocity mode or in the accelerometer mode we will apply these actual mechanical values to this signal. So as to calibrate what is the voltage level corresponding to or what is the mechanical value corresponding to a particular voltage we would have obtained. Now we will move over to the next problem question 3 which says for a machinery having 90 decibel of sound power level compute the sound pressure level in dB and Pascal at a distance of 1 meter in the free field condition ok. Usually sound pressure level is denoted by SPL and sound power level by SWL and the general equation for the sound pressure level as with relation with sound power level is given by this equation that SPL is equal to SWL plus 10 logarithm based on of q theta by 4 pi r square plus 4 by r r c. This is a very important formula and I will explain you what these terms mean here. This is the sound pressure level we all know that this is the sound power level theta naught or q theta is the directivity r is the distance from the source r c is the room constant which depends on the amount of sound absorption in the room. So r c is given by this formula as alpha average by 1 minus alpha average in meter square where s is the total surface area of the room alpha average is the average sound absorption coefficient of the room. So in a room suppose I take a wall and I divide them into small patches and each one has a different alpha i and area of this is s i. So the average alpha average is nothing but s i alpha i is equal to 1 to n and then or the total area ok. Now this could be of different materials and they all could be having different alpha i for example this could be gypsum, this could be fiberglass, this could be jute of course nowadays you know we have found out that jute and its derivatives are good sound absorbers also. So anyway just to come back to the problem which we are discussing so we can have the room constant like this. So if the room was totally absorbing or there was no sound which is generated so for example this is a room and I have a sound source and these are walls which are treated with sound absorbers and then we have a good amount of absorption in this walls. So and this was in the center of the room a typical case in this case your absorption is the maximum so your RC is actually close to infinite ok. So the second term 4 by RC because the room constant actually disappears in a case where it is totally absorbing that means there are no echoes or an anechoic condition have been obtained and this is essentially what is known as the free field condition. So only in the case of a free field condition this problem will become so SPL will become SWL plus 10 log 10 q theta by 4 pi r square plus 4 by RC. So this will become 0 in the case of for free field condition and if it is in the center of the room if the source is in the center of a room if it is here in the center of the room q theta is equal to 1 in the center of the room is equal to 2 in on the floor of the center of the room is equal to 4 if it is on the center of the this is 2 this is 4 this is 1 floor this is on the center of edge between wall and floor and this is equal to 8 it is the corner between walls and floor. So by this I mean the vice versa also you would have realized in because from the reciprocity relationship suppose you are standing in sorry suppose you are standing in a room in the center of a room I will just draw one plane suppose you are standing here and I move your loud speaker to this location here the directivity will increase and you will hear it to be louder then I put it in the middle of the room because of the directivity issues you know this is just explain. So in other words if I come back to this problem in this case SPL is equal to SWL plus 10 log 10 q theta by 4 pi r square plus 4 by r c it has been told this is free field condition. So free field condition means this this will vanish and the center of the room if I take so q theta will be equal to 1 excuse me in the center of the room excuse me and at the distance r r is equal to 1 1 meter my expression will come up to SPL is equal to SWL minus 10 log 10 of 4 pi. Now some of the important things we have to observe from this case because here I have taken q theta is equal to 1 so log of 1 is actually 0 so it will disappear and this value will be approximately equal to 10 decibel. Again a good rule of thumb, thumb rule which we should follow is in any free field condition now what are typical examples of free field conditions one is an echoic chamber other is here. An open field in fact many times we do lot of noise destructing in open field wherein we do not have any reflection there has to be no reflections. So you can pretty much understand the SPL at 1 meter away from the source will be actually SWL minus 10 decibel is equal to SPL. So SPL is something constant to a particular machine it is the inherent noise quality of the machine which is fixed. So at 1 meter away from the sound pressure level from the source the sound pressure level will be whatever the value you have in SWL in decibels you have to subtract it by 10 dB ok. Of course, we have to keep in mind the SWL dB and SPL dB they have different references and there is something you have to find out. So SWL is actually 10 log 10 of W by W reference where W reference is actually 10 power minus 12 watts you have to keep this in mind. So in this problem the machinery was having 90 dB as SWL the SPL would be close to about 80 decibel in free field conditions at distance of 1 meter. So through this example I just introduce it to few concepts of room constants which is very important in machinery environment as to we will be called to monitor the noise of machines which are having certain defects. So we have to be careful about the what conditions in which we measure the sound pressure level because if the environment changes the sound pressure level will obviously change like we saw it depends on the directivity it depends on the room constant. So whenever we want to monitor the health of a machine by noise monitoring it is always recommended that we actually always do the measurements in the same noise environment otherwise there will be no apple to apple or orange to orange comparison unless we follow the simple rule. And that is why through this example I just showed you how the room constant the distance from the source and the directivity does affect the sound pressure level which have been measured. Now we will move over to the next problem this is again a simple problem on vibration that is question number 4. A machine can be modeled as a single degree of freedom spring mass dash pod system if the machine is operating at 1200 rpm and has a mass of 100 kg and is supported on 4 linear springs each of stiffness 25 Newton per meter and an overall damping ratio of 0.03 determine the damped natural frequency of the machine. See this is a machine a mass and supported on 4 springs I am just showing a 2D plane view so I am not showing the other 2 springs. So the effective springs because there are 4 springs k effective nothing but 4 times k and the value of k is 100. So this becomes 400 Newton per meter and the natural frequency of this system omega n is equal to root over k by n and that happens to be 400 and mass is also I just did a mistake here this will be actually k is 25 so this is 100 so 4 times 25 so this will be 100 divided by 100 so this will be 1 radians per second. So now the damped natural frequency of the system is given by this expression so we have found out the natural frequency as 1 radians per second so damped natural frequency is 1 times 1 minus 0.03 square and this will be somewhere close to about 0.95 radians per second. It is a very simple example where we are just used the formula to find out the natural frequency of a system and the damped natural frequency of the system. Now many a times what happens that of course there is a fact that in this machine was running at 1200 rpm. So if this machine is running at 1200 rpm the forcing frequency omega forcing is nothing but the frequency of forcing frequency is nothing but 1200 by 60 Hertz 20 Hertz and omega is 2 pi into 20 this is 40 pi radians per second. Many a times just by estimating the mass of a machine and estimating the springs the stiffness of the springs we can get some idea as to what is the natural frequency of the machine. Particularly in many machinery condition monitoring situations the defects occur because unknowingly or unknowingly while I would say unknowingly somebody is operating the machine at its resonant frequency or the designers did not take care of certain issues in mounting which would have been introduced. So that the stiffness change the damping change and then the natural frequency of the system happens to be close to the forcing frequency and then a condition of resonance occurs and because of resonance will have large motions and then there will be failure in the machine. So a good way to estimate the natural frequency of a system is through experimental model analysis which we would have discussed in one of these lectures. But just by knowing the mass stiffness and usually some guess on the damping we can estimate the natural frequency because vibration is such a phenomenon which is dependent on these three parameters and we cannot neglect it and the phenomena of vibration is actually very much frequency dependent. This is the dynamic magnificence in factor x by x naught and r is nothing but omega by omega n where omega could be the forcing frequency and this is the natural frequency. So the dynamic magnification factor is always given by 1 by 1 minus r square whole square plus twice zeta r square where zeta is the damping ratio. In this case zeta was given as 0.03. As you see the phenomena of damping is very much dependent on r and r is nothing but the forcing frequency ratio of the forcing frequency and the natural frequency. So these simple formula helps you know the dynamic response of a machine knows helps you to estimate the damped natural frequency and you can also see how the machine is going to behave as a function of frequency by plotting the dynamic magnification factor. So these are simple things which one needs to keep in mind while doing machinery condition monitoring or doing machinery fault detection. So just to summarize you know we just saw two examples each from the classes of vibration and noise as to how simple equations gives us insight into handling few numbers be it in noise be it in vibration and how these numbers can help us do a better estimation or better assessment of the machine's health condition. And these are a very very preliminary and rudimentary problem which one needs to be absolutely sure about because later on in machinery condition monitoring we will be building on some of these very very simple equations and if you remember this you know this will help you go a long way in doing actually machinery troubleshooting. Now we will come to two different other classes of problem which is this on computer aided data acquisition in the acquisition. So coming to the fifth problem this says a pure tone signal of 2200 hertz is to be digitally acquired by a 16 bit computer aided data acquisition system. So what is the minimum amount of onboard RAM that is random access memory required to store the digital data in the digital data acquisition system so that the frequency resolution of one hertz can be obtained by performing analysis on the acquired time domain signal. As you know this equation there are two three concepts which are being in this problem we require two three concepts to solve this problem. For example, I have a pure tone by a pure tone I mean a signal of a single frequency and this pure tone signal is of 200 hertz. So by Shannon's sampling theorem without worrying about over sampling and all that my sampling frequency f s should be equal to at least twice of f max. So in this case it is 400 hertz is my sampling frequency because that is something I have to keep in mind and in the next part of the system it is a 16 bit computer that means one digital data point will be stored in 16 bit spaces or that corresponds to 2 bytes because 8 bit is equal to 1 byte ok. Now the question is this signal pure tone signal which has been sampled will be processed digitally into the frequency domain so that the frequency domain resolution delta f is equal to one hertz ok. But we know there is a relationship between the delta f and the block of time data which you have taken because this data which has been digitized so I take this digital data I take n data points this total time is nothing but n times delta t where delta t is nothing but one of the one by the sampling frequency this is known as the sampling interval this value is delta t. So in the frequency resolution or in the frequency spectrum because this is a pure tone these values we get certain digital value and then we join them. So typical spacing between any two consecutive point is delta f certain amplitude ok. So there is a inverse relationship between the time and frequency and that is the most fundamental relationship this is a very very important relationship. So in this example we had the sampling frequency so delta f is nothing but f s by n since delta f is equal to one hertz f s because of the sampling frequency requirement of sanums this comes to be four hertz or n is nothing but f s by delta f that is 400 by 1 that is equal to 400 data points. Here of course we have done an assumption which is normally not true that we have just f f t on one block of data because no information is given whether we are going to take more blocks of data and so on. So just assuming that one block of data is being used we have to take 400 data points. So each data point requires two bytes of space so total space total space is equal to twice two times 400 that is equal to 800 bytes but as you know memory is available in 2 to the power some power. So the closest highest power will be actually 2 to the power 12 or 2 to the power 10 that will be 1000 to 4 bytes or which we known as 1 kilobyte. So to analyze such a signal I will require 1 kilobyte of RAM space many times in a students give answer as 800 bytes and I would normally prefer that the answer is actually 1 kilobyte and close not one not 800 bytes but 1 kilobyte because you will not get 800 bytes of memory to my in the market. So this is how you can decide on the number of the RAM requirement and you would have known in the FFT analysis if we buy the signal analyzer they are sometimes specified by the lines of FFT. So you say 400 lines 800 lines ok this actually correspond to 400 by 2.56 this corresponds to about times 2.56 about 1024 data points. So more lines in more data points means more memories and you would have seen when you buy of course nowadays you know we have system with 6400 lines available the present day systems. So many times some of these FFT analyzers are actually on chips or on IC chips or processors thus this kind of an example helps us understand what is the numerical requirement in terms of the memory storage for data equation. As you know in data equation system two most important parameters are the one is the sampling frequency the resolution of the card in terms of you know whether it is 16 bit 12 bit etcetera. So this are very important I mean if you have a mistake in the sampling frequency you have erroneous sampling frequency you have the problem of signal aliasing and if you have do not have a adequate resolution you have what is known as the quantization error. So looking back to the this previous problem so this means if you want to have one hertz resolution we have to have more spaces. Next question is suppose I wanted to have 0.1 hertz you will see how your memory size will increase n is equal to FS by delta F. So in this case it will 400 by 0.1 so this will increase by about 4000 ok. If you want better resolution we have to have more memory space ok so this will this is what is going to change isn't it. So we have to keep that in mind ok. So better resolution means more memory space and these are there deleted. Now I will come to the next and the last problem on this topics that is what is the gain in a signal conditioning analog amplifier required so that a 12 bit data equation system with a maximum input analog voltage range of 10 volt can acquire a signal of 1 micro volt amplitude. So this comes the problem because see I have a machine I put a sensor and it gives me some analog voltage ok voltage. Now I have a A to D system and then which gives me digital data which goes into a computer. Now obviously this A to D system has a minimum voltage which it can sense that is known as the resolution and this resolution is given by what is known as the maximum voltage it can sense the range by 2 to the power bit size where this is the bit size. So that is the lowest voltage it can sense. So in this case if this range is 10 volts I have the bit size as 12 so 2 to the power 12 and this corresponds to about 2.44 and 10 to the power minus 3 volt that is 2.44 milli volt. So in other words a 12 bit DAQ system with 10 volt range can only sense or detect minimum of 2.44 milli volt. So I have a big problem in hand my I have been asked to acquire 1 micro volt that is my actual analog signal a very low temperature thermocouple signal will give you this kind of a value and suppose you want to detect such a system and you want to acquire it remotely or through a computer aided data processing the problem is is your 12 bit system good enough to sense it. Obviously you see because to 1 micro volt is much much less than 2.44 milli volt it cannot sense. There are couple of ways we can do it I can increase the bit size instead of 2 to the power 12 I can make it 2 to the power 24 in a 24 bit system that calls of course for a new hardware new DAQ system sometimes that may not be possible but nowadays 24 bit DAQ systems are available and so to have good enough amplitude resolutions I have to always increase the bit size if that is not possible in between what I can do is I can have a very nice analog amplifier with high signal to noise ratio so that I give my 1 micro volt and I make it at least 2.44 milli volt or more. So, I have to amplify I have to multiply with some gain. So, this analog amplifier the question is what is this gain so that a 12 volt system is going to register 1 micro volt obviously. So, the gain of the amplifier would be nothing gain of the amplifier is nothing but 2.44 into 10 to the power minus 3 volt divide by 1 micro volt 10 to the power minus 6. So, this is equal to 2.44 into 10 to the power 3 is 2444. So, this is the gain. That means, if I have a 1 micro volt signal I multiply with this value of gain I will get 2.44 into 10 to the power sorry this is plus 3 minus 3 volt which can be detected and sensed by the DAQ. So, we have to keep this in mind while we have to select on a certain data acquisition system. So, in summary on this class we just looked into 6 example problems typical example problems couple of them from noise couple of them from vibration and couple of them from DAQ system as to making us familiar on some of the important concepts on data acquisition signal aliasing quantization error conversion of SPL to in P Pascals to decibel. How do you relate sound pressure level to an estimate sound pressure level from sound power level? What is the meaning of dynamic magnification factor? How do you estimate natural frequency damp natural frequency? So, with this kind of an understanding we are ready to go to the field to do some actually troubleshooting and in the subsequent classes actually will be discussing about how machine monitoring is done through vibration how faults are detected by vibration monitoring. Thank you.