Cooling thermocouples junction

  • Thread starter Thread starter harsh077
  • Start date Start date
  • Tags Tags
    Cooling Junction
AI Thread Summary
Cooling the junction of a thermocouple is not feasible, as thermocouples are designed solely for temperature measurement, not active heating or cooling. Applying current to both wires will not effectively cool the junction. The Peltier effect, associated with thermoelectric cooling, involves one junction heating while the other cools, but this does not create a net cooling effect for the thermocouple itself. The discussion suggests that inquiries about cooling methods might be better suited for an Electrical Engineering forum. Overall, thermocouples should not be used for active cooling applications.
harsh077
Messages
4
Reaction score
0
please let me know
how to cool the junction of thermocouple...
is it possible to give current to both the wires n get junction cooled...?
 
Science news on Phys.org
The Electrical Engineering forum is probably a better place to post this question. Thermocouples are a type of temperature sensor and are not meant to actively heat or cool anything.
 
I didn't realize Peltiers were also technically thermocouples, but since one junction gets hot while the other gets cold, it still won't result in an overall heat loss.
 
I need to calculate the amount of water condensed from a DX cooling coil per hour given the size of the expansion coil (the total condensing surface area), the incoming air temperature, the amount of air flow from the fan, the BTU capacity of the compressor and the incoming air humidity. There are lots of condenser calculators around but they all need the air flow and incoming and outgoing humidity and then give a total volume of condensed water but I need more than that. The size of the...
Thread 'Why work is PdV and not (P+dP)dV in an isothermal process?'
Let's say we have a cylinder of volume V1 with a frictionless movable piston and some gas trapped inside with pressure P1 and temperature T1. On top of the piston lay some small pebbles that add weight and essentially create the pressure P1. Also the system is inside a reservoir of water that keeps its temperature constant at T1. The system is in equilibrium at V1, P1, T1. Now let's say i put another very small pebble on top of the piston (0,00001kg) and after some seconds the system...
Back
Top