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nadya68 [22]
3 years ago
15

Pressurized steam at 450 K flows through a long, thin-walled pipe of 0.5-m diameter. The pipe is enclosed in a concrete casing t

hat is of square cross section and 1.5 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 heat loss per unit length of pipe
Engineering
1 answer:
iris [78.8K]3 years ago
7 0

Answer:

Heat loss per unit length of pipe = q' = 1363.47 W/m

Explanation:

Heat loss per unit length would be in terms of W/m.

Assumptions:

• steady-state conditions

• negligible steam side convection resistance, pipe wall resistance and contact resistance (T1 = 450 K)

• constant properties

The thermal conductivity for concrete must be found.

Referring to the table of thermal conductivity of some common materials, the value for stone concrete will be used, which is 1.7.

The heat rate can be expressed as:

q = SkΔT₁₋₂ = Sk(T₁ − T₂)

The proper shape factor is needed for the solution and is found from the list of shape factors. The shape factor must be for a circular cylinder of length L centered in a square solid of equal length and is:

S = 2πL  ÷ [ln (1.08w/D) ]

Then,

q' =   q  ÷ L

The full equation is:

q' =   2πk(T₁ − T₂)  ÷ [ln (1.08w/D) ]

Substituting known values:

q' =   2π(1.7 W/mK)*(450 K − 300 K)  ÷ [ln [1.08(1.5 m)/0.5 m] ]

q' = 1363.47 W/m

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2. The following segment of carotid artery has an inlet velocity of 50 cm/s (diameter of 15 mm). The outlet has a diameter of 11
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This question is incomplete, the missing diagram is uploaded along this answer below.

Answer:

the forces required to keep the artery in place is 1.65 N

Explanation:

Given the data in the question;

Inlet velocity V₁ = 50 cm/s = 0.5 m/s

diameter d₁ = 15 mm = 0.015 m

radius r₁ = 0.0075 m

diameter d₂ = 11 mm = 0.011 m

radius r₂ = 0.0055 m

A₁ = πr² = 3.14( 0.0075 )² =  1.76625 × 10⁻⁴ m²

A₂ = πr² = 3.14( 0.0055 )² =  9.4985 × 10⁻⁵ m²

pressure at inlet P₁ = 110 mm of Hg = 14665.5 pascal

pressure at outlet P₂ = 95 mm of Hg = 12665.6 pascal

Inlet volumetric flowrate = A₁V₁ = 1.76625 × 10⁻⁴ × 0.5 = 8.83125 × 10⁻⁵ m³/s

given that; blood density is 1050 kg/m³

mass going in m' = 8.83125 × 10⁻⁵ m³/s × 1050 kg/m³ = 0.092728 kg/s

Now, using continuity equation

A₁V₁ = A₂V₂

V₂ = A₁V₁ / A₂ = (d₁/d₂)² × V₁

we substitute

V₂ =  (0.015 / 0.011 )² × 0.5

V₂ = 0.92975 m/s

from the diagram, force balance in x-direction;

0 - P₂A₂ × cos(60°) + Rₓ = m'( V₂cos(60°) - 0 )    

so we substitute in our values

0 - (12665.6 × 9.4985 × 10⁻⁵)  × cos(60°) + Rₓ = 0.092728( 0.92975 cos(60°) - 0 )    

0 - 0.6014925 + Rₓ =  0.043106929 - 0

Rₓ = 0.043106929 + 0.6014925

Rₓ = 0.6446 N

Also, we do the same force balance in y-direction;

P₁A₁ - P₂A₂ × sin(60°) + R_y = m'( V₂sin(60°) - 0.5 )  

we substitute

⇒ (14665.5 × 1.76625 × 10⁻⁴) - (12665.6 × 9.4985 × 10⁻⁵) × sin(60°) + R_y = 0.092728( 0.92975sin(60°) - 0.5 )

⇒ 1.5484 + R_y = 0.092728( 0.305187 )

⇒ 1.5484 + R_y = 0.028299    

R_y = 0.028299 - 1.5484

R_y = -1.52 N

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R = √( Rₓ² + R_y² )

we substitute

R = √( (0.6446)² + (-1.52)² )

R = √( 0.41550916 + 2.3104 )

R = √( 2.72590916 )

R = 1.65 N

Therefore, the forces required to keep the artery in place is 1.65 N

 

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