Ask question

Ask question

All categories

3 years ago

6

A thermocouple junction is used to measure the temperature of a solid material. The junction is inserted into a small circular h

ole and is held in place by epoxy. Identify the heat transfer processes associated with the junction. Will the junction sense a temperature less than, equal to, or greater than the solid temperature? How will the thermal conductivity of the epoxy affect the junction temperature?

1 answer:

Ratling [72]3 years ago

4 0

Answer:

Conduction,

Less temperature

Epoxy for effective heat transfer by conduction

Explanation:

(1) CONDUCTION is the heat transfer process associated with the junction.

(2) The junction will experience a temperature that is LESS than the temperature of the solid

(3) Epoxy has a high conductivity degree, meaning it transfers heat effectively without much resistance to the heat flow. Hence, to answer the question, the epoxy is used to ensure the effective heat transfer from the solid to the junction.

You might be interested in

Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a

Elena L [17]

Answer:

$\mathbf{\tau_c =5.675 \ MPa}$

Explanation:

Given that:

The direction of the applied tensile stress =[001]

direction of the slip plane = [ $\bar 1$ 01]

normal to the slip plane = [111]

Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:

$cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big]$

where;

$[d_1\ e_1 \ f_1]$ = directional indices for tensile stress

$[d_2 \ e_2 \ f_2]$ = slip direction

replacing their values;

i.e $d_1$ = 0 , $e_1$ = 0 $f_1$ = 1 & $d_2$ = -1 , $e_2$ = 0 , $f_2$ = 1

$cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big]$

$cos \ \lambda = \dfrac{1}{\sqrt{2}}$

Also, to find the angle $\phi$ between the stress [001] & normal slip plane [111]

Then;

$cos \ \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big]$

replacing their values;

i.e $d_1$ = 0 , $e_1$ = 0 $f_1$ = 1 & $d_3$ = 1 , $e_3$ = 1 , $f_3$ = 1

$cos \ \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big]$

$cos \phi= \dfrac{1} {\sqrt{3} }$

However, the critical resolved SS(shear stress) $\mathbf{\tau_c}$ can be computed using the formula:

$\tau_c = (\sigma )(cos \phi )(cos \lambda)$

where;

applied tensile stress $\sigma =$ 13.9 MPa

∴

$\tau_c =13.9\times ( \dfrac{1}{\sqrt{2}} )( \dfrac{1}{\sqrt{3}})$

$\mathbf{\tau_c =5.675 \ MPa}$

3 0

3 years ago

You can safely place a jack on a floor pan to keep a vehicle steady.

Elis [28]

Answer: Yes

Explanation:

7 0

3 years ago

Read 2 more answers

Engineers need to be open-ended when dealing with their designs. Why?

melomori [17]

I think the answer would be A if its wrong I’m sorry

7 0

3 years ago

Grace [21]

I need help my self lol XD

5 0

3 years ago

Read 2 more answers

YO CUTIE! HELP MEEEEEE! Hazardous waste is a category that includes all of the following EXCEPT

madam [21]

your answer should be <u>B. Soluble</u>

3 0

3 years ago

Read 2 more answers