12-17-2024 02:16 PM
Hello,
I am currently working on a project where I want to measure changes in weight over time (in seconds). I have successfully measured the weight, but now I need help in tracking the change. Specifically, I am monitoring the change in weight as gas leaves the bottle. Based on that, I want to measure the mass flow rate of the gas leaving the bottle and calculate the amount of heat energy transferred to the environment during the combustion of that gas. If the weight has not changed, the calculated value should be zero. However, if the weight is less than the previous second, I want to display that difference as the rate of change of weight per second. Can anyone provide some guidance on how to achieve this?
12-17-2024 02:22 PM
Sir, this is a Wendy's.
😂 But seriously, All the physics descriptions are nice, but how are you actually doing this. What kind of sensors are you using? How are you reading their values?
12-17-2024 02:45 PM
"I don't see it as important to say which sensor I'm using because the problem is not with the sensor; the sensor is working. The issue is how to calculate the change in weight over time. The mass is always decreasing. If it's helpful, I'm using a load cell sensor with HX711.
12-17-2024 05:36 PM
So, you need to capture the timestamp along with reading from the load cell, so, each of your weight values will have an associated time, using this you can calculate the rate of change.
12-17-2024 07:50 PM
Since you posted here, I take a wild guess and assume you are using LabVIEW and have basic knowledge how to program with it. In order to repeatedly measure something over time, you need to use a loop that runs at a define rate and repeatedly takes measurements. Use a shift register to compare with the previous reading.
12-18-2024 03:36 AM
Tracking the change should be the easy part, measure once/sec in a loop and subtract the current from the previous value.
12-18-2024 05:58 PM
Thank you for taking the time to respond to my query. After rereading my initial post, I realize that I may not have fully explained the problem, and that could be my fault. In any case, I managed to resolve the issue— again thank you!