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OverLord2011 [107]
3 years ago
11

Consider a pipe with an inner radius of 5cm and an outer radius of 7cm.The inner surface is kept at 100C, and the outer surface

at 80C. Determine the heat loss per meter length of the pipe if the pipe is made of pure copper [k=387W/m.oC], pore aluminum [k=200W/m.C], and pure iron [k=62W/m.C].
Engineering
1 answer:
Ivahew [28]3 years ago
7 0

Answer:

Explanation:

given data:

innner radius r_i = = 5cm

outer radius r_0 = 7 cm

temperature at outer surface = 80 degree celcius

temperature at inner surface = 100 degree celcius

thermal resistance per unit length is given as

R _{th} = \frac{1}{ 2\pi kl} ln \frac{ ro}{ri}

           = \frac{1}{ 2\pi k*1} ln \frac{ 7}{5}

           = \frac{0.053}{k}

heat loss rate per unit length q  = \frac{Ti-To}{R_{th}}

q = \frac{100-80}{\frac{0.053}{k}}

q = 373.474 K

1) for pure COPPER

k  = 387 W/m degree celcius

q = 373.474 * 387  = 144534.438 W

2) for pure ALUMINIUM

k  = 200 W/m degree celcius

q = 373.474 * 200  = 74694.84 W

3)

1) for pure IRON

k  = 62W/m degree celcius

q = 373.474 * 62  = 23155.416 W

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A mixing basin in a sewage filtration plant is stirred by a mechanical agitator with a power input/WF L T=. Other parameters de
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Answer: π= G[√(u.V/W)]

STEP 1

Given parameters:

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STEP 2

We start by expressing the velocity gradient G as a function of W, u, V

G= G(W,u,V)

To get the pii terms, we use the dimension number formula n=k - r

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n= 4-3=1

STEP 3:

We expressed the pii terms as

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The three fundamental F L T

We can write as

Fⁿ.Lⁿ.Tⁿ= 1/T. (FL/T)^a.(FT/L²)^b.(L³)

Using the exponential rule and by comparing coefficient on both sides;

Fⁿ.Lⁿ.Tⁿ= F^a+b. L^a-2b+3c. T^-a+b-1

Fⁿ= F^a+b = a+b= 0..............I

Lⁿ= L^a-2b+3c=0 = a-2b+3c=0...........ii

Tⁿ=L^-a+b-1=0. -a+b-1=0............iii

From the above equations we have,

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putting eq. iv into iii , we have

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substituting the above value of a into eq iv, we have

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substituting the value of b above into eq 2, we have,

-1/2-2(1/2)+3c=0

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Lastly, from the pii terms given above we can obtain dimensionless relationship,

π=G(W^-1/2.u^1/2.V^1/2)

We can write this as

π= G[ √1/W.√u. √1/2] = G[(√u.V/√W)] or G[√(u.V/W)].... final answer.

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