Hello, I am currently working on the design of a device that measures thermal conductivity of sediments and I am not sure if i estimated heat losses and heat flux in the correct way. The device is a 0,0034 cubic meters (200 cubic inches aprox) box with a resistor (curved alloy wire) in contact with one face of the sample, connected to a power supply. The resistor is in the middle of one face of the sample and a heavy insulator, like glass wool (i am not sure if this is the proper term for the material).
This is not my work area, i'm a geologist, but given my current research i am in the need to resolve this issue
Please feel free to answer any of this questions, also any comment will be helpful.
My questions are:
- If it is ok to estimate heat losses considering an estimation of the temperature at a middle point in the box and thickness and the thermal conductivity at the five walls that surround the internal sample considering that one side of the device is in contact with the surrounding air
- If using a common insulator (say 0,02 to 0,04 W/m.C°) will yield an aproximate heat loss of less than 1W, considering 0,1 m wall thickness (4 inches aprox).
- If the sample will reach a quasi steady-state heat flux or will be far from it, considering the room temperature stays aproximately constant.
- How much heat will disipate the wire if the power output of the power supply is, say 6W. In other words what will be the heat flow at the resistor, through the first face of the sample near the resistor, given that supplied power.
Regards
Hello Matias! I want to help you but I have some questions to make you.
- You say that you want to measure the thermal conductivity of sediments. You speak on a box. Are the sedimets inside a box ? If so, do you know the thermal conductivity of the material of the box ? Cam you measure the thickness of the walls of the box ?
- The volume of 0.034 cubic meters is the volume of the sample or the volume including the box ? What is the thickness of you sample ?
- If you have a resistor on the lower face of your device you must study your resistor. You can use a mulimeter to measure ddp between the terminals of the resistor and the intensity of the current in the circuit. You can also measure the resistance of your resistor.
- You must use temperature sensors in your device. How much sensors are you planning to use. Where are yoy planning to colocate the sensors?
- You speak on a 10 cm wall thickness and 15 cm box length of a cubic box. What are the dimensions of your sample?
- The internal sample is surrounded by 5 walls. How can you have 5 walls?
I don't know if I am helping or not but must understand what you are planning to see if I can help you.
Maria Rosa Duque
Hello Maria, thank you in advance for your interest in this issue, i'll answer your questions bellow
- Sediments are inside the box. The thermal conductivity of the material of the box is 0,04 W/m.C°. The thickness of the walls is 0,1 meters.
- The volume of the sample is 0,034 m3. The thickness is 0,15 meters.
- The resistor will be a nicrom cable, i have some indications of the resistivity of the alloy, and some idea of the resistance of the cable from its longitude and area.
- I planned to meassure temperature of the in and out faces of the sample, i.e. the face that is in contact with the resistance and the face that is in contact with the air in the opposite side, with an infrarred termometer and with a thermocouple also, at the two sides. This will be done manually
- The total length of the box is 35 centimeters, the box has two blocks in the opposite sides of the sample that are removable
- The sample is a cube. Five faces of the box surrounds the sample, the sixth face of the box (removable) in the opposite end of the heat flux line is removed, so the face is in contact with the air
I hope i have clarified the device issues you asked, i'll also attach a diagram of the device,
Regards
Sorry Maria Rosa Alves Duque, I forgot to mention in my previous answer,
Regards
Hello Matias
Now I think that I understand what you are trying to do.
I think you can obtain a quasi-steady-state in your sample but you must wait some time to obtain it. As you have an insulator (your box) under the resistor you may have a problem with increasing temperature in that wall of the box and an increase of heat flowing in your sample.
You can obtain the heat lost through the walls of the box measuring temperature differences between inner and outer surfaces of the box P(lost) = (0,04) (Tin - Tout)/ 0,1. You can also have an idea about the heat lost through lateral sides using several termocouples in the upper surface of your sample located from the center to the lateral part of the samle.
The values obtained will depend on the sample you want to study. In principle if you are studying rock sediments you will obtain low thermal conductivity values (compared with common rock values ) but higher values than 0,04 and you will have heat flowing predominantely in the vertical direction.
Try to not use high temperature values in your experiment. Thermocouples are very good to use in this type of experiments because its volume is very small and they have a rapid answer.
As I said in my first message you are using Joule's effect in your experiment and the power dissipated is P= R I2 . The value o I depends on the value of R. You must know the ddp (voltage) at the ends of your resistor ((V=RI).
I think I unswer to your questions but if you have some question more you can contact me.
Best regards
Maria Rosa Duque
Maria Rosa Alves Duque I will follow your advice on not to use high temperatures and i hope i can install thermocouples between the sample and the walls to estimate heat losses (it puts me in the need of getting extra circuits and displays), also thank you for confirming me that the power dissipated is obtained from Joule's effect, but i wasn't aware of where to take the measurements, so to be sure, tell me if i understood properly: I have to measure the voltage at the ends of the resistor and then use ohm's law to calculate the intensity of the current given that the resistance of the resistor is known... I ask you this because i have the possibility to set the current intensity and voltage output values in the power supply
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