I'm trying to design a kind of water valve with inexpensive materials as a first prototype.
The water flow from the PVC pipe (1) reach the body of the valve and pass through an aluminum grid (3) to the water tank. When the water level goes up pushes the float closing the water intake at point (2).
enter image description here
How can I calculate the buoyancy force needed to stop the water flow? And, what will be the mass of the float?
Let’s back to the basics; here I present the problem and some math that I been doing, I would like your opinion:
Connected to the PVC pipe (2) I have a garden hose whit a water flow pressure of, let's say...49 kPa (I need to check this with a manometer), and I attached a 25 diameter and 0.5 meters long PVC pipe. Let’s pretend that the float seals the other side of the PVC pipe, so I need to calculate the force generate by the water flow pressure against the float.
Please take in consideration that I'm not a fluid mechanic expert.
When I open the garden hose, the PVC pipe starts to fill, so based on this situation:
enter image description here
P1+ρgh_1+(v^1 ρ)/2=P2+ρgh_2+(v^2 ρ)/2
If I took the height of P1 as the reference, h=0, and the diameter of the PVC pipe and the garden hose pipe are the same (25 mm), the water flow velocity at those points are equals:
P1=P2+ρgh_2
So, if the garden hose pressure is 49 Kpa:
49000 kg/(m s^2 )=P2+9.8 m/(s^2 ) x 1000 kg/m^3 x 0.5 m
P2=53900 kg/(m s^2 )
P2=53.9 Kpa
Ok, assuming this math is correct…now I have to calculate the force against the bottom of the PVC pipe at point 2:
P=F/A
In order to simplify this example, I took the diameter of the PVC pipe as the contact area.
A=πr^2=π(〖0.025〗^2 )=0.002 m^2
F=107.8 N
If the pressure of the water flow generates a force of 107.8 N, I need an opposite force with a higher value to counteract it. Is that correct?
My goal is to find a material (mass; area) that generate enough buoyancy force to stop the water flow through the valve and seal the water intake, and when the water level goes down, the float valve will let pass the water flow to continue to fill the water tank.
I agree with the previous comment : the pressure at the contact is the total pressure at the line feed (corrected for gravity at height 1). When water is flowing, some of that pressure is converted to dynamic pressure, meaning you will measure a lower pressure at point 1; the total sum at p+1/2 v^2 should remain more or less the same irrespective of v - if we neglecting head losses running up to point 1 which depend on the flow rate.
Anyway, when the valve is closed, the flow is stopping anyway, so you it is even more obvious to work with the static pressure measured in the absence of flow.
You need to choose the density and form of your floater such that the buoyancy force, given by $(density of water - density of floater material)*(submerged volume at chosen reservoir height)$ is equal to $p2 * A$.
Related
I want to create some simple heart rate monitor in LabVIEW.
I have sensor which gives me heart workflow (upper graph): Waveform
On second graph (lower graph) is amount of hills (0 - valley, 1 - hill) and that hills are heart beats (that is voltage waveform). From this I want to get amount of those hills, then multiply this number by 6 and I'll get heart rate per minute.
Measuring card I use: NI USB-6009.
Any idea how to do that?
I can sent a VI file if anyone will be able to help me.
You could use Threshold Peak Detector VI
This VI does not identify the locations or the amplitudes of peaks
with great accuracy, but the VI does give an idea of where and how
often a signal crosses above a certain threshold value.
You could also use Waveform Peak Detection VI
The Waveform Peak Detection VI operates like the array-based Peak
Detector VI. The difference is that this VI's input is a waveform data
type, and the VI has error cluster input and output terminals.
Locations displays the output array of the peaks or valleys, which is
still in terms of the indices of the input waveform. For example, if
one element of Locations is 100, that means that there is a peak or
valley located at index 100 in the data array of the input waveform.
Figure 6 shows you a method for determining the times at which peaks
or valleys occur.
NI have a great tutorial that should answer all your questions, it can be found here:
I had some fun recreating some of your exercise here. I simulated a squarewave. In my sample of the square wave, I know how many samples I have and the sampling frequency. As a result, I calculate how much time my data sample represents. I then count the number of positive edges in the sample. I do some division to calculate beats/second and multiplication for beats/minute. The sampling frequency, Fs, and number of samples, N or #s are required to calculate your beats per minute metric. Their uses are shown below.
The contrived VI
Does that lead you to a solution for your application?
At first blush this presumably means -
(1) looking only at lower IR frequencies,
(2) select a IR frequency cut-off for low frequency buckets of the u/v FFT grid
(3) Once we have that, derive the power distribution - squares of amplitudes - for that IR range of frequency buckets the camera supports.
(4) Fit that distribution against the Rayleigh-Jones classical Black Box radiation formula:
(https://en.wikipedia.org/wiki/Rayleigh%E2%80%93Jeans_law#Other_forms_of_Rayleigh%E2%80%93Jeans_law)
(5) Assign a Temperature of 'best fit'.
The units for B(ν,T) are Power per unit frequency per unit surface area at equilibrium Temperature
Of course, this leaves many details out, such as (6) cancelling background, etc, but one could perhaps use the opposite facing camera to assist in that. Where buckets do not straddle the temperature of interest, (7) use a one-sided distribution to derive an inferred Gaussian curve to fit the Rayleigh-Jeans curve at that derived central frequency ν, for measured temperature T.
Finally (8) check if this procedure can consistently detect a high vs low surface temperature (9) check if it can consistently identify a 'fever' temperature (say, 101 Fahrenheit / 38 Celcius) pointing at a forehead.
If all that can be done, (10) Voila! a body fever detector
So those who are capable can fill us in on whether this is possible to do so for eventual posting at an app store as a free Covid19 safe body temperature app? I have a strong sense there's quite a few out there who can verify this in a week or two!
It appears that the analog signal assumed in (1) and (2) are not available in the Android digital Camera2 interface.
Android RAW image stream, that is uncompressed YUV, is already encoded Y green monochrome, and U,V are blue and red shifts from zero for converting green monochrome to color.
The original analog frequency / energy signal is not immediately accessible. So adaptation is not possible (yet).
When the cross section of the flow tube decreases, the flow speed increases, and therefore the pressure decreases.
can someone explain to me why this is true, i would think that as the cross section decreases the pressure would also increase .
This is related to "Continuity Equation" of fluid mechanisam.(Assuming fluid as incompressible)
if we have two cross-sections of areas A1 and A2 having velocities V1 and V2 respectively .Then according to continuity equation
A1*V1=A2*V2 or we can write
V2=(A1*V1)÷A2
V2 Is inversly proportional to the A2.
so velocity increases as the area decreases.
further we have a theorem in fluid mechanics called "Bernouli's theorem".
which states that the sum of all energies at any cross-section is constant.
So if the velocity(i.e kinetic energy) increases at any section there will be decrease in pressure(i.e pressure energy)
Think of it this way, what is pressure in the first place.!
Well pressure is the force acting perpendicular to a unit area write ?
So the fluid whatsoever particles are exerting force on that unit area, that's fine..
image five 10 people standing in an elevator standing next to each other, these guys are too much to fit inside the width of the elevator, thus they would push themselves towards the wall of the elevator making huge amount of force on the adjacent walls write ? what was that again ? aha!! huge force per area which are in this examples the walls of the elevator. hence they are too much pressure, okay.. now imagine that these people instead of standing 10 next to each others they formed themselves as groups of twos, so 5 rows of twos instead one of ten, i bet that they will feel more comfortable right? they won't push themselves that much to the wall and hence the wall will have small force on it and then small pressure, that was an example for proving that physics isn't just some numbers that define what is going to happen, Bernoulli's equation predicted that the pressure will decrease based on logic. science works :D
To my understanding, the noise floor for each USRP may be different. I want to know how I can calculate the noise floor without physically going into the fft and spotting it out manually. I want to know if there is a block in GNU Radio that will calculate this, or if there is a stream of blocks I can use to calculate it. Please provide a block diagram in your answer ( block 1 ---> block 2 ---> ...etc.).
For my application, let's say I have a QT GUI frequency sink that is showing all noise at the moment. I want to calculate the noise floor so that I have a value that represents "no signal present" ie. noise. Once I have this value, I plan to set a threshold 5dB higher than the noise floor to indicate that a signal has been detected. I've been able to kind of eye ball the average noise value from the QT GUI Frequency Sink but that's not good enough for me. I want to be able to calculate it and not have to look into the plot every time to update the noise value every time I change USRPs.
For instance:
You can see the average noise value for this is around -55dB. I want to calculate this without having to eye ball it. This way, when a signal gets transmitted at (in this example) 0Hz, then the power of the signal will increase and I can see if a signal was detected.
I am implementing the ValveLinear model from the Modelica standard fluid library into a model of mine using Dymola. I have some questions regarding its parameters which I can hopefully clear up:
The key parameters for this valve are as follows:
parameter Medium.MassFlowRate m_flow_nominal
"Nominal mass flowrate at full opening";
final parameter Types.HydraulicConductance k = m_flow_nominal/dp_nominal
"Hydraulic conductance at full opening";
Modelica.Blocks.Interfaces.RealInput opening(min=0,max=1)
"=1: completely open, =0: completely closed"
The mass flow over the valve is then caclulated as
m_flow = opening*k*dp;
Am I right in assuming that m_flow_nominal is the maximum mass flow rate with a linear drop off in mass_flow down to zero as opening goes from 1 to 0?
Furthermore is dp_nominal the corresponding minimum pressure drop across the valve? (i.e. at fully open). Therefore would we see a linear increase in dp from dp_nominal to some maximum value as opening goes from 1 to 0?
The answer may seem trivial but I have run some examples with valves in Dymola so far and in some cases it seems that dp remains constant across the valve as the opening in varied which doesn't make sense to me.
The nominal mass flow rate and pressure drop are just design values used to calculate the valve coefficient k (fixed relation between pressure drop and mass flow). Since no "nominal opening degree" can be specified in ValveLinear the valve opening in the design point is assumed to be one (fully open valve).
The mass flow rate through the valve is not limited to m_flow_nominal. If you double the pressure drop the mass flow through the valve will double, regardless of the nominal mass flow rate.
An example model is shown below:
m_flow_nominal is 5 kg/s and dp_nominal is 10 bar.
At time = 0 s the (fixed) pressure drop over the valve is 10 bar and the valve is fully open. Therefore, the mass flow through the valve is 5 kg/s.
At time = 1 s the pressure drop over the valve is increased by 50
pct (from 10 to 15 bar). The mass flow increases with 50 pct as well
(to 7.5 kg/s).
At time = 3 s the valve opening is reduced by 50 % (from fully to
half open). The pressure drop remains at 15 bar (of course, since
it's a boundary value) while the mass flow rate is reduced to 50 pct (= 3.75 kg/s).
Regarding your second question. The pressure drop is not limited. If the mass flow through the valve is given as a boundary condition (e.g. if source in the model is replaced with a MassFlowSource_T) and the mass flow rate is reduced to half of the nominal value (from 5 to 2.5 kg/s) the pressure drop will also be reduced to half of the nominal value (10 to 5 bar). If the mass flow rate is zero, so will the pressure drop be.
If, on the other hand, you fix the mass flow rate to a value > 0 kg/s and ramp the valve opening towards zero, the pressure drop will approach infinity.
Best regards,
Rene Just Nielsen