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mamaluj [8]
3 years ago
9

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
Engineering
1 answer:
iris [78.8K]3 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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Answer:

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Explanation:

8 0
3 years ago
A fluid at 300 K flows through a long, thin-walled pipe of 0.2-m diameter. The pipe is enclosed in a concrete casing that is of
andrew-mc [135]

Answer:

The correct answer is "1341.288 W/m".

Explanation:

Given that:

T₁ = 300 K

T₂ = 500 K

Diameter,

d = 0.2 m

Length,

l = 1 m

As we know,

The shape factor will be:

⇒ SF=\frac{2 \pi l}{ln[\frac{1.08 b }{d} ]}

By putting the value, we get

⇒       =\frac{2 \pi l}{ln[\frac{1.08\times 1}{0.2} ]}

⇒       =3.7258 \ l

hence,

The heat loss will be:

⇒ Q=SF\times K(T_2-T_1)

       =3.7258\times 1\times 1.8\times (500-300)

       =3.7258\times 1.8\times (200)

       =1341.288 \ W/m

3 0
3 years ago
Homes may be heated by pumping hot water through radiators. What mass of water (in g) will provide the same amount of heat when
Nitella [24]

Answer:

a mass of water required is mw= 1273.26 gr = 1.27376 Kg

Explanation:

Assuming that the steam also gives out latent heat, the heat provided should be same for cooling the hot water than cooling the steam and condense it completely:

Q = mw * cw * ΔTw = ms * cs * ΔTw + ms * L

where m = mass , c= specific heat , ΔT=temperature change, L = latent heat of condensation

therefore

mw = ( ms * cs * ΔTw + ms * L )/ (cw * ΔTw )

replacing values

mw = [182g * 2.078 J/g°C*(118°C-100°C) + 118 g * 2260 J/g ] /[4.187 J/g°C * (90.7°C-39.4°C)] = 1273.26 gr = 1.27376 Kg

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Radioactive wastes are temporarily stored in a spherical container, the center of which is buried a distance of 10 m below the e
a_sh-v [17]

Answer:

Outside temperature =88.03°C

Explanation:

Conductivity of air-soil from standard table

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To find temperature we need to balance energy

Heat generation=Heat dissipation

Now find the value

We know that for sphere

q=\dfrac{2\pi DK}{1-\dfrac{D}{4H}}(T_1-T_2)

Given that q=500 W

so

500=\dfrac{2\pi 2\times .6}{1-\dfrac{2}{4\times 10}}(T_1-25)

By solving that equation we get

T_2=88.03°C

So outside temperature =88.03°C

6 0
3 years ago
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