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3 years ago

What are the effects of applying an additional layer of insulation to a cylindrical pipe or a spherical shell?

Engineering

1 answer:

Brums [2.3K]3 years ago

4 0

Answer:

When an additional layer of insulation is applied to a cylindrical pipe or a spherical shell, the insulation layer works to increase its conduction resistance but at the same time, lowers the convection resistance of the surface.

Explanation:

Generally, when more insulation is added to a wall, the resulting effect is that heat transfer decreases. With increasing thickness of the insulation, the heat transfer rate becomes lesser. This is because the thermal resistance of the wall becomes more with the added insulation, wherein the heat transfer area and convection resistance aren't affected.

However, a different scenario occurs when an additional layer of insulation is applied to a cylindrical pipe or spherical shell. The conduction resistance increases, but the surface convection resistance decreases since the outer surface area for convection also increases and does not remain constant.

Overall, the heat transfer from the cylindrical pipe or spherical shell may increase or decrease, which depends on the effect that dominates.

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3 years ago

Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t

kodGreya [7K]

Answer:

$\frac{T-20}{130-20}= e^{-14.28*0.05}$

And if we solve for T we got:

$T= 20 + 110e^{-14.28*0.05} = 73.86 C$

The answer for this case would be $T = 73.86 C$ at 5cm from the base of the fin.

Explanation:

Data given

For this case we have the following data given:

$h = 20 \frac{W}{m^2 K}$ represent the heat transfer coefficient.

$p$ represent the perimeter for this case and would be given by:

$p = 2*0.05m +2*0.001m= 0.102m$

$k = 200 \frac{W}{m C}$ represent the thermal conductivity

$w = 5cm =0.05 m$ represent the width

$h = 1mm =0.001m$ represent the thickness

$A= wh= 0.05m *0.001m = 0.00005 m^2$

Solution to the problem

For this case we assume that we have steady conditions, the temperature of the fins varies just in one direction, the heat transfer coefficient not changes with the time and the thermal properties of the fin not change.

We can determine the temperature if the fin at x=5 cm=0.05 m from the base with the following formula:

$\frac{T-T_{\infty}}{T_b -T_{\infty}} = e^{-mx}$

Where m is a coefficient given by:

$m = \sqrt{\frac{hp}{kA}}=\sqrt{\frac{20 W/m^2 C 0.102 m}{200 W/ mC 0.00005 m^2}}= 14.28 m^{-1}$

The value of x for this case represent the distance $x =5 cm =0.05m$

$T_b =130 C$ represent the base temperature

$T_{\infty}= 20$ represent the temperature of the sorroundings or the ambient.

If we replace we have this:

$\frac{T-20}{130-20}= e^{-14.28*0.05}$

And if we solve for T we got:

$T= 20 + 110e^{-14.28*0.05} = 73.86 C$

The answer for this case would be $T = 73.86 C$ at 5cm from the base of the fin.

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Answer:A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

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