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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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Marysya12 [62]

Answer:

8 to 10 times

Explanation:

For dry road

u= 15 mph        ( 1 mph = 0.44 m/s)

u= 6.7 m/s

Let take coefficient of friction( μ) of dry road is 0.7

So the de acceleration a = μ g

a= 0.7 x 10  m/s ²                         ( g=10 m/s ²)

a= 7 m/s ²

We know that

v= u - a t

Final speed ,v=0

0 = 6.7 - 7 x t

t= 0.95 s

For snow road

μ = 0.4

de acceleration a = μ g

a = 0.4 x 10 = 4 m/s ²

u= 30 mph= 13.41 m/s

v= u - a t

Final speed ,v=0

0 = 30 - 4 x t'

t'=7.5 s

t'=7.8 t

We can say that it will take 8 to 10 times more time as compare to dry road for stopping the vehicle.

8 to 10 times

7 0
3 years ago
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EMB agar is a medium used in the identification and isolation of pathogenic bacteria. It contains digested meat proteins as a so
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Answer:

A selective medium, a differential medium, and a complex medium.

Explanation:

A selective media is a microbiological media which only support the growth of a particular specie or types of species of microorganisms,this media acts in such a way to inhibit or hinder the growth of other microorganisms.

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At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho
victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0
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What is the shape of the output signal on a rigexpert analyzer?
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