I am trying to calculate venting time of a pressurized tank filled with saturated water at 800 PSIG.
The vent from the tank is a 2 inch line 12 ft long which goes into 6 inch line 10 ft long and then it goes into 8 inch line 150 ft long. These lines vent into a flash tank (Operating at 50 PSIG), which then is released into atmosphere.
My questions would be:
Velocity of water into tank?
Water pressure coming into the flash tank?
I have attached sample calculation regarding how I am approaching this problem. Any guidance would be appreciated.
I am using Bernoulli equations, Darcy Weisbach equation to balance pressure
Calculations that I am performing:
Related
I've got a total of 9 sensors in the ground, which measure the water content of the soil. 1-3 are in a depth of 1m, 4-6 are in a depth of 2m and sensors 7-9 are in a depth of 3m.
My dataset also contains the precipiation of the location. It is hourly data:
Time
Sensor-ID
Precipitation
Soil Water Content
2022-01-01 11:00
1
74
120
2022-01-01 11:00
2
74
100
2022-01-01 11:00
3
74
110
...
...
...
...
2022-01-01 11:00
9
74
30
The goal now is to find out if the different ground / soil depths behave differently regarding the water content after raining (over time).
I thought about a clustering method to find out if the sensors can be clustered based on the data and confirm this. Since I'm not very experienced in data science, would that be the right approach and is it even possible to analyse it with clustering?
For clustering, you can add a new column with three new classes to your data - for 1-3 sensors : Class 1, for 4-6 sensors : Class 2, for 7-9 sensors : Class 3 and perform your analysis using the new classes. Either can be done using Python, Power BI or Excel.
You should start by analyzing different variables w.r.t to the sensors at different ground depths: Use univariate, Bi-Variate and Multi-Variate plots to derive your goal.
I have a table as follows
Day Savings
1 : 251
2 : 722
3 : 1132
4 : 1929
5 : 3006
6 : 4522
7 : 8467
...
14 : x
These savings are growing day by day, I want to find a formula to expect the final value of day 14 which is x!
I didn't look at the data in any detail, but it seems like an exponential growth situation. If that's the case, then you can estimate the growth rate by fitting an exponential curve to the data using least squares approximation to get an estimated interest rate, r. If you find the data not conducive to that, you could try fitting it to some other curve. You can then use the estimated interest rate to compute the expected funds using the standard a = p*exp(r*i) where p is the initial principal and i is the elapsed time.
This all assumes compounding interest which is an exponential growth situation. If that assumption is incorrect, this approach is probably not going to work for you.
I have a Q4X Stainless Steel Analog Laser Sensor to calculate the distance with analog output voltage.
It does display distance on Laser Sensor display and I am trying to do the same thing in my code using scale factor but its not matches with Laser Sensor display value.
here is my scaling factor.
#define A2D_SCL_LASER ( 11.81f / ( 10.0f - 0.0f ) ) // inches per volt
Devices specs:
Supply Voltage (Vcc)
12 to 30 V dc
Sensing Range—Threaded Barrel Models
500 mm models: 25 mm to 500 mm (0.98 in to 19.68 in)
300 mm models: 25 mm to 300 mm (0.98 in to 11.81 in)
100 mm models: 25 mm to 100 mm (0.98 in to 3.94 in)
Comparison of distance calculation in my code and sensor display
My values(mm) sensor display value(mm)
1V 29.9974 52
2V 59.944 80
3V 89.916 107
4V 119.888 134
5V 150.114 162
6V 179.832 190
7V 209.804 217
8V 240.03 245
9V 270.002 272
10V 300 300
Ref:
http://info.bannerengineering.com/cs/groups/public/documents/literature/185623.pdf
I simply fitted the sensor displayed value with the voltage you give in your question. The R²=1 means that the fit is perfect (or near perfect) and this is a good sign.
The equation you are searching for is
Distance(mm) = 27.533 x Volt + 24.467
I am using data collected from two different instruments which have different resolution because of the sampling rate of each instrument. For a specific time, one of the sets have >10k entries while the other has ~2.5k. They however capture data over the same time interval, and I want to plot them on top of each other even though they have different resolution in data. The minimum and maximum x of both sets are the same however one of them have more entries.
Simplified it could look like this:
1st set from instrument with higher sampling rate:
time(s) value
0.0 10
0.2 11
0.4 12
0.6 13
0.8 14
... ..
100 50
2nd set from instrument with lower sampling rate:
time(s) value
0 100
1 120
2 125
3 128
4 130
. ...
100 430
They are measuring different things, but I would like to display them in the same plot. How can I accomplish this?
I found the mistake.. I was trying to plot both datasets using the time data from the first instrument. Of course they need to be plotted with their respective time data and I put the first time data in the second plot by mistake..
i am new here, I am using MERRA monthly solar radiation data. I want to convert w/M^2 to MJ/m^2
I am bit confused, how to convert solar radiation monthly average data W/m^2 to MJ/m^2
so far i understood by reading different sources,
Firstly i have to convert w/m^2 to kw/m^2
after kw/m^2 to mj/m^2 .......
Am i doing correctly
Just i am taking one instance:
For may month i have value 294 w/m^2
So 294 * 0.001 = 0.294 kw/m^2
0.294 * 24 (kw to kwh (m^/day)) = 7.056 kwh/m^2/day
7.056 * 3.6 (kwh to mj) = 25.40 mj/day
i am confused i am doing right or wrong .
Not sure why you would take the kWh step in between.
Your panels do 294 Watt per m², i.e. 294 Joule per sec per m². So that's 24*60*60 * 294 = 25401600 Joule per m² per day, or 25.4016 MJ per m² per day.
So if:
1 W/m2 = 1 J/m2 s
Then:
294 W/m2 = 294 J/m2 s
if you want it in days then:
1 day = 60s * 60min *24h = 86400s
294 J/m2 s x 86000s/1day = 25284000 J/m2 day
25284000 J/m2 day x 1MJ/1000000J = 25.284 MJ/m2 day
all together:
294 W/m2 = 294/(1000000/86400) = 25.4016 MJ/m2 day
A watt is the unit of power and Joules are the units of energy, they are related by time. 1 watt is 1 Joule per second 1W = 1 J/s. So the extension of that equation is that 1J = 1w x 1second. 1J = 1Ws. A loose analogy is if you say Litre is a unit of volume and L/S is a unit of flow. So your calculation needs to consider how long you are gathering the solar energy. So the number of Joules, if the sunlight shines at 90degrees to the solar panel for 1 hour is 294W/m2 x 3600s and would give ~1 x 10^7 joules per square metre. Of course as the inclination [the angle of light] varies away from 90 degrees, this will cause the effective power and hence the energy absorbed to drop, as a function of the sine of the angle to the sun. 90 degrees gives a sine of 1 and is full power.