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Veseljchak [2.6K]
3 years ago
10

Considering the analogy between electrical circuit and thermal circuit, show your approach to derive an expression for the therm

al resistance for a wall

Engineering
1 answer:
Simora [160]3 years ago
5 0

Answer:

Thermal resistance for a wall depends on the material, the thickness of the wall and the cross-section area.

Explanation:

Current flow and heat flow are very similar when we are talking about 1-dimensional energy transfer. Attached you can see a picture we can use to describe the heat flow between the ends of the wall. First of all, a temperature difference is required to flow heat from one side to the other, just like voltage is required for current flow.  You can also see that R_{th} represents the thermal resistance. The next image explains more about the parameters which define the value of the thermal resistances which are the following:

  1. Wall Thickness.  More thickness, more thermal resistance.
  2. Material thermal conductivity (unique value for each material). More conductivity, less thermal resistance.
  3. Cross-section Area. More cross-section area, less thermal resistance.

A expression to define  the thermal resistance for the wall is as follows:  R_{th} =\frac{l}{Ak}, where  l is the distance between the tow sides of the wall, that is to say the wall thickness; A is the cross-section area and k is the material conducitivity.

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pychu [463]

Answer:

A) 88.42 ohms

c) 0.5145 and - 5.77 db

Explanation:

A)  Value of cut-off frequency = 12 kHz

capacitance of capacitor =0.15 microF

formula for cut-off frequency of a low-pass filter

Fc = \frac{1}{2\pi RC}   making R the subject of the formula in other to get the resistance of the low pass filter resistor

R = \frac{1}{2\pi (12*10^3)(0.15*10^{-6} )} = 88.42 OHMS

attached is the solution for B and C

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Assume that in orthogonal cutting the rake angle is 15° and the coefficient of friction is 0.15. a. Determine the percentage cha
alexira [117]

Answer:

Δr=20.45 %

Explanation:

Given that

Rake angle α =  15°

coefficient of friction ,μ = 0.15

The friction angle β

tanβ = μ

tanβ = 0.15

β=8.83°

2φ +  β - α  = 90°

φ=Shear angle

2φ + 8.833° - 15° = 90°

φ = 48.08°

Chip thickness r given as

r=\dfrac{tan\phi}{cos\alpha +sin\alpha\ tan\phi}

r=\dfrac{tan48.08^{\circ}}{cos15^{\circ} +sin15^{\circ}\ tan48.08^{\circ}}

r=0.88

New coefficient of friction ,μ'  = 0.3

tanβ' = μ'

tanβ' = 0.3

β'=16.69°

2φ' +  β' - α  = 90°

φ'=Shear angle

2φ' + 16.69° - 25° = 90°

φ' = 49.15°

Chip thickness r' given as

r'=\dfrac{tan\phi'}{cos\alpha +sin\alpha\ tan\phi'}

r'=\dfrac{tan44.15^{\circ}}{cos49.15^{\circ} +sin49.15^{\circ}\ tan44.15^{\circ}}

r'=0.70

Percentage change

\Delta r=\dfrac{r-r'}{r}\times 100

\Delta r=\dfrac{0.88-0.70}{0.88}\times 100

Δr=20.45 %

8 0
3 years ago
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