What happens in an heat flow apparatus when you reduce the length of t

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SUMMARY

Reducing the length of the second disk in a heat flow apparatus increases the temperature of the disk's lower surface and the temperature of the water exiting the system. The apparatus consists of a resistor, two metal disks with known and unknown thermal conductivity coefficients, and a cooling water flow. The heat flow is calculated using the formula Q = - λ A ΔΤ / d, where λ and A are constants. As the length of the second disk decreases, both the heat transfer rate and the temperature difference across the disk increase, resulting in higher outlet temperatures for both the disk and the water.

PREREQUISITES
  • Understanding of heat transfer principles
  • Familiarity with thermal conductivity coefficients
  • Knowledge of thermocouple temperature measurement
  • Basic grasp of the heat flow equation Q = - λ A ΔΤ / d
NEXT STEPS
  • Explore the effects of varying thermal conductivity on heat transfer rates
  • Investigate the role of insulation in heat flow apparatus
  • Learn about advanced thermocouple configurations for temperature measurement
  • Study the impact of fluid dynamics on heat exchange in cooling systems
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Engineers, physicists, and researchers involved in thermal management, heat transfer analysis, and the design of cooling systems will benefit from this discussion.

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What happens in an heat flow apparatus when you reduce the length of the disk?
The apparatus is this
1st level: A resistor to produce heat
2nd level: A metal disk with known thermal conductivity coefficient
3rd level: A metal disk with an unknown thermal conductivity coefficient
4th level: Flow of water for cooling
The device if insulated so all the heat (ideally) flow from the resistor through the disks to the water.
The disks carry 2 thermocouples each to measure temp difference at given distance for each
You use the 1st disk to calculate the heat flow (known length, area, coefficient, temp difference)
The heat flow in the 2nd disk is the same as the one in the 1st
You calculate 2nd disk's coefficient using the same formula (known heat flow, length, area, temp difference)

My question is: What happens if the 2nd disk's length (the one with the unknown coefficient) is reduced? Since Q= - λ A ΔΤ / d and λ, Α are definite constants, either ΔΤ or Q with change. My guess is both. The out temp of the will rise (will be closer to the in temp) AND the heat flow will be more (the heat will transfer faster)
 
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In order to be more clear

Say that the resistor gives off a temp 50C (constant)
In the original case: Let it be that the lower surface of the 2nd disk has a 30C temp and the running water's temp increases by 2C (from 20C to 22C) while touching that lower surface

1st level: Resistor at 50C
2nd level: 1st disk
3rd level: 2nd disk's upper surface
3rd level: 2nd disk's lower surface at 30C
4th level: Water comes in at steady 20C
4th level: Water takes heat from the lower surface of the 2nd disk
4th level: Water comes out at 22C

HEAT_FLOW_CASE_A

In the case that the 2nd disk has reduced length and everything else is the same (resistor 50C 1st disk length unchanged, water coming at 20C)

1st level: Resistor at 50C
2nd level: 1st disk
3rd level: 2nd disk's upper surface
3rd level: 2nd disk's lower surface at TEMP_A
4th level: Water comes in at steady 20C
4th level: Water takes heat from the lower surface of the 2nd disk
4th level: Water comes out at TEMP_B

HEAT_FLOW_CASE_B

Questions

HEAT_FLOW_CASE_A < HEAT_FLOW_CASE_B

TEMP_A > 30C (disk's lower surface has increased temp)

TEMP_B > 22C (water coming out has increased temp)

Basically the heat will travel faster but also the lower surface of the 2nd disk will be hotter and that the water will remove more heat (also hotter)
 

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