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frutty [35]
3 years ago
11

A 5 mm dia copper cable is insulated with a material of conductivity of 0.16 W/mK and is exposed to air at 30°C with a convecti

on coefficient of 20 W/m^2K. If the surface temperature of the wire can be 120°C, determine the insulation thickness for maximum heat flow and the heat dissipated per m length.
Engineering
1 answer:
Kipish [7]3 years ago
4 0

Answer:

t=5.5 mm

Heat dissipation per unit length = 90.477 W/m

Explanation:

Given that

Diameter d = 5 mm ⇒r = 2.5 mm

Conductivity of insulated material K = 0.16 W/mK

Heat transfer coefficient = 20 \frac{W}{m^2K}

When thickness reaches up to critical radius of insulation then heat dissipation will be maximum

We know that critical radius of insulation of wire is given as follow

r_{c}=\dfrac{K_{insulation}}{h_{surrounding}}

Now by putting the values

r_{c}=\dfrac{0.16}{20}

r_{c}=8 mm

So the thickness of insulation

t=8-2.5 mm

t=5.5 mm

As we know that heat transfer due to convection given as follows

Q = hAΔ T

Q=20 x 2 x π x 0.008 x (120-30)

Q = 90.477 W/m

So heat dissipation per unit length = 90.477 W/m

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Answer:

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Explanation:

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Who developed the process of blueprinting?
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Answer: C.) John Herschel
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3 years ago
An aluminum alloy tube with an outside diameter of 3.50 in. and a wall thickness of 0.30 in. is used as a 14 ft long column. Ass
slega [8]

Answer:

slenderness ratio = 147.8

buckling load = 13.62 kips

Explanation:

Given data:

outside diameter is 3.50 inc

wall thickness 0.30 inc

length of column is 14 ft

E = 10,000 ksi

moment of inertia = \frac{\pi}{64 (D_O^2 -D_i^2)}

I = \frac{\pi}{64}(3.5^2 -2.9^2) = 3.894 in^4

Area = \frac{\pi}{4} (3.5^2 -2.9^2) = 3.015 in^2

radius = \sqrt{\frac{I}{A}}

r = \sqrt{\frac{3.894}{3.015}

r = 1.136 in

slenderness ratio = \frac{L}{r}

                              = \frac{14 *12}{1.136} = 147.8

buckling load = P_cr = \frac{\pi^2 EI}}{l^2}

P_{cr} = \frac{\pi^2 *10,000*3.844}{( 14\times 12)^2}

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3 0
3 years ago
Which option identifies the next step in the following scenario?
Whitepunk [10]

Answer: The engineer will create a detailed sketch that labels all of the visual components.

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It should be noted that the reverse engineering is required for the replacement and the modification of an existing product.

With regards to the question, the correct answer is option A "The engineer will create a detailed sketch that labels all of the visual components".

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3 years ago
Consider a 50-mH inductor. a. Express the voltage across the inductor and then evaluate it at t = 0.25 s if iL (t) = 5e −2t + 3t
IrinaK [193]

Answer:

V=L(di/dt) where i is current, V=0.208

Explanation:

using expression iL(t)=5e-2t+3te-2t-2 and L=0.05H(50/1000)

V=0.05*d(5e-2t+3te-2t-2)/dt

since there is no power of e, I'll assume the power to be 1

V=0.05*(-2+3e-2)

at t=0.25

V=0.15e-0.2

V=0.208

3 0
3 years ago
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