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2 years ago

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Pressurized steam at 400 K flows through a long, thin-walled pipe of 0.6-m diameter. The pipe is enclosed in a concrete (stone m

ix) casing that is of square cross section and 1.75 m on a side. The axis of the pipe is centered in the casing, and the outer surfaces of the casing are maintained at 300 K. What is the rate of heat loss per unit length of pipe, in W/m

1 answer:

iris [78.8K]2 years ago

4 0

Answer:

Rate of heat loss per unit length of pipe, q' = 767.01 W/m

Explanation:

Let q' be the Rate of heat loss per unit length

Let q be the Rate of heat loss

q' = q/L

Where L is the length of the pipe

Diameter, D= 0.6m

The rate of heat loss q is given by the formula: q = Sk(T₂ - T₁)

Where k is the thermal conductivity of the concrete at 300 K

k = 1.4 Wb/m-K (at 300K)

And S is the shape factor given by the formula:

S = 2πL/ ln(1.08w/D)

S = (2π*L) / ln(1.08*1.75/0.6))

S = (2π*L) / 1.147

S = 5.48 L

q = 5.48L*1.4(400-300)

q = 767.01 L

q' = q/L

q' = 767.01L/L

q' = 767.01 W/m

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Andreyy89

Answer:

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Explanation:

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2 years ago

A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h

hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

$D_{1}$ = 150 mm

$A_{1}= \frac{\pi }{4}\times 150^{2}$

= 17671.45 $mm^{2}$

$D_{2}$ = 250 mm

$A_{2}= \frac{\pi }{4}\times 250^{2}$

= 49087.78 $mm^{2}$

The centrifugal force acting on the flywheel is fiven by

F = M ( $R_{2}$ - $R_{1}$ ) x $w^{2}$ ------------(1)

Here F = ( -UTS x $A_{1}$ + UCS x $A_{2}$ )

Since density, $\rho = \frac{M}{V}$

$\rho = \frac{M}{A\times t}$

$M = \rho \times A\times t$ $M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t$

$M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37$

$M = 8252963901$

∴ $R_{2}$ - $R_{1}$ = 50 mm

∴ F = $763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}$

F = 33618968.38 N --------(2)

Now comparing (1) and (2)

$33618968.38 = 8252963901\times 50\times \omega ^{2}$

∴ ω = 4036.61

We know

$\omega = \frac{2\pi N}{60}$

$4036.61 = \frac{2\pi N}{60}$

∴ N = 38546.82 rpm

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Answer:

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2 years ago

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makkiz [27]

Answer:

The correct answer is

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3 years ago

Help please if you don't know don't give wrong answer please

mrs_skeptik [129]

4.75cm * 5.22cm is 24.795cm²

but let's do this with m² instead:

0.0475m * 0.0522m = 0.0024795

now we can compare it with the 49780 much easier.

devide. that's the <u>roughly</u> 20,000,000fold of the plot, but it's still squared, so let's take the root

so the representative fraction is 1:4,480.

the exact value is in the screens

have a nice day:)

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3 years ago