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A horizontal steel pipe having a diameter of 5 cm is maintained at a temperature of 50◦C in a large room where the air and wall temperature are at 20◦C. The surface emissivity of the steel may be taken as 0.8 and the heat-transfer coefficient for free convection with this geometry and air is he=16.5 W/m2 C. Calculate the total heat lost by the pipe per unit length.

Respuesta :

Answer: 1,580.75 w/m

Explanation:

The total heat loss will be the sum of the heat lost by radiation and convention.

Now lets find the heat lost by convention using the formula

q/L]conv =h(πd )( Tw - T∞ )= (16.5)( π)(50-20) = 1,555.71 w/m

But this pipe is a body surrounded by a large enclosure so the radiation heat

transfer can be calculated from this equation

q/L]rad =ε1(πd1)σ (T14– T24)

Here both T1 and T2 are raised to the power of 4, therefore

= (0.8 )( π )( 0.05) ( 5.669× 10raised to power of -8 ) (3234 – 2934) = 25.04 w/m

the total heat loss is therefore

q/L]tot = q/L]conv + q/L]rad =1,555.71 + 25.04 =1,580.75 w/m

oladayoademosu

The total heat lost by the pipe per unit length is 1027.96 W/m

The total heat loss in the horizontal steel pipe is due to radiation and convection.

The radiation heat loss, P = σεA(T₂⁴ - T₁⁴) where σ = stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²-K⁴, ε = surface emissivity of steel = 0.8, A = surface area of steel pipe = 2πrL where r = radius of steel pipe = 5 cm/2 = 2.5 = 0.025 m and L = length of steel pipe, T₁ = temperature of air and wall = 20 °C = 273 + 20 = 293 K and T₂ = temperature of steel pipe = 50 °C = 273 + 50 = 323 K

Also, the convection heat loss, P' = hA(T₂ - T₁) where h = heat- transfer coefficient for free convection = 16.5 W/m²-K, A = surface area of steel pipe, T₁ = temperature of air and wall = 20 °C and T₂ = temperature of steel pipe.

So, the total heat loss P" = P + P'

P" = σεA(T₂⁴ - T₁⁴) + hA(T₂ - T₁)

P" = σε2πrL (T₂⁴ - T₁⁴) + h2πrL(T₂ - T₁)

To find the total heat loss per unit length, we divide through by L. So,

P"/L = σε2πr(T₂⁴ - T₁⁴) + h2πr(T₂ - T₁)

P"/L = 2πr[σε(T₂⁴ - T₁⁴) + h(T₂ - T₁)]

Substituting the values of the variables into the equation, we have

P"/L = 2πr[σε(T₂⁴ - T₁⁴) + h(T₂ - T₁)]

P"/L = 2π(0.025 m)[5.67 × 10⁻⁸ W/m²-K⁴ × 0.8((323 K)⁴ - (293 K⁴) + 16.5 W/m²-K(323 K - 293 K)]

P"/L = 2π(0.025 m)[4.536 × 10⁻⁸ W/m²-K⁴(10884540241 K⁴ - 7370050801 K⁴) + 16.5 W/m²-K(323 K - 293 K)]

P"/L = π(0.05 m)[4.536 × 10⁻⁸ W/m²-K⁴ × 3514489440 K⁴ + 16.5 W/m²-K(30 K)]

P"/L = 0.5πm[15941724099.8 × 10⁻⁸ W/m² + 495 W/m²]

P"/L = 0.5πm[159.417240998W/m² + 495 W/m²]

P"/L = 0.5πm[654.417240998W/m²]

P"/L = π[327.208620499W/m]

P"/L = 1027.95619835 W/m

P"/L ≅ 1027.96 W/m

So, the total heat lost by the pipe per unit length is 1027.96 W/m

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