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Digiron [165]
3 years ago
7

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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Answer:

h_{max} = 51.8 cm

Explanation:

given data:

height of tank = 60cm

diameter of tank =40cm

accelration = 4 m/s2

suppose x- axis - direction of motion

z -axis - vertical direction

\theta = water surface angle with horizontal surface

a_x =accelration in x direction

a_z =accelration in z direction

slope in xz plane is

tan\theta = \frac{a_x}{g +a_z}

tan\theta = \frac{4}{9.81+0}

tan\theta =0.4077

the maximum height of water surface at mid of inclination is

\Delta h = \frac{d}{2} tan\theta

            =\frac{0.4}{2}0.4077

\Delta h  0.082 cm

the maximu height of wwater to avoid spilling is

h_{max} = h_{tank} -\Delta h

            = 60 - 8.2

h_{max} = 51.8 cm

the height requird if no spill water is h_{max} = 51.8 cm

3 0
3 years ago
All of these are uses of microwaves except...
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Answer:D

Explanation:

5 0
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Find the current Lx in the figure
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Explanation:

\frac{1}{8}  +  \frac{1}{2}   \\ 1.6 + 1.4 = 3 \\  \frac{1}{3}  +  \frac{1}{9}   \\ 2.25 + 2 = 4.25 \: ohm

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\frac{9}{9 + 3}  \times 4 = 3\\   \frac{2}{2 + 8} \times 3 = 0.6a \\  = 0.6 \: milli \: amper

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Answer:

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