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lord [1]
3 years ago
10

A 2 in. diameter pipe supplying steam at 300°F is enclosed in a 1 ft square duct at 70°F. The outside of the duct is perfectly i

nsulated. Assume the emissivity of pipe surface and duct surfaces as 0.79 and 0.276, respectively. Estimate the radiation heat transfer per foot length.
Engineering
1 answer:
Shkiper50 [21]3 years ago
4 0

Answer:

The value of heat transferred watt per foot length Q = 54.78 Watt per foot length.

Explanation:

Diameter of pipe = 2 in = 0.0508 m

Steam temperature T_{1} = 300 F  = 422.04 K

Duct temperature T_{2} = 70 F = 294.26 K

Emmisivity of surface 1 = 0.79

Emmisivity of surface 2 = 0.276

Net emmisivity of both surfaces ∈ = 0.25

Stefan volazman constant \sigma = 5.67 × 10^{-8} \frac{W}{m^{2} K^{4}  }

Heat transfer  per foot length is given by

Q = ∈ \sigma A ( T_{1}^{4} - T_{2} ^{4} ) ------ (1)

Put all the values in equation (1) , we get

Q = 0.25 × 5.67 × 10^{-8} × 3.14 × 0.0508 × 1 × ( 422.04^{4} - 294.26^{4} )

Q = 54.78 Watt per foot.

This is the value of heat transferred watt per foot length.

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Explanation:

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                           F_n = m*a_n

Where, a_n is the centripetal acceleration or normal component of acceleration as follows:

                           a_n = v^2_2 / r

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                          F_n = m*v^2_2 / r

Where, F_n normal force acting on block by the surface is:

                          F_n = W*cos(θ)

- Substitute:

                          W*cos(θ) = m*v^2_2 / r

                          v_2 = sqrt ( r*g*cos(θ) )

- Plug in the values:

                          v^2_2 = 2*32.2*cos(60)

                          v^2_2 = 32.2 (ft/s)^2

- Apply the conservation of energy between points A and B where θ= 60° :

                      T_A + V_A = T_B + V_B

Where,

                      T_A : Kinetic energy of the block at inital position = 0

                      V_A: potential energy of the block inital position

                      V_A = 0.5*k*x_A^2

                      x_A = 2*pi - L            ..... ( L is the original length )

                      V_A = 0.5*2*(2*pi - L)^2 =(2*pi - L)^2

                      T_B = 0.5*W/g*v_2^2 = 0.5*2 / 32.2 *32.2 = 1

                      V_B = 0.5*k*x_B^2 + W*2*cos(60)

                      x_B = 2*0.75*pi - L            ..... ( L is the original length )

                      V_B = 0.5*2*(1.5*pi - L)^2 + 2*1 = 2 + ( 1.5*pi - L )^2

- Input the respective energies back in to the conservation expression:

                      0 + (2*pi - L)^2 = 1 + 2 + ( 1.5*pi - L )^2

                      4pi^2 - 4*pi*L + L^2 = 3 + 2.25*pi^2 - 3*pi*L + L^2

                      pi*L = 1.75*pi^2 - 3

                         L = 4.574 ft

                         

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