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Nata [24]
2 years ago
9

A dryer is shaped like a long semi-cylindrical duct of diameter 1.5 m. The base of the dryer is occupied with water-soaked mater

ials to be dried. The base is maintained at a temperature of 370K, while the dome of the dryer is maintained at 1000 K. If both surfaces behave as blackbody, determine the drying rate per unit length experienced by the wet materials.
Engineering
1 answer:
Anestetic [448]2 years ago
6 0

Answer:

0.0371 kg/s.m

Explanation:

From the given information, let's have an imaginative view of the semi-cylinder; (The image is shown below)

Assuming the base surface of both ends of the cylinder is denoted by:

A_1  \ and   \ A_2

Thus, using the summation rule, the view factor F_{11 and F_{12 is as follows:

F_{11}+F_{12}=1

Let assume the surface (1) is flat, the F_{11} = 0

Now:

0+F_{12}=1

F_{12}=1

However, using the reciprocity rule to determine the view factor from the dome-shaped cylinder A_2 to the flat base surface A_1; we have:

A_2F_{21} = A_{1}F_{12} \\ \\ F_{21} = \dfrac{A_1}{A_2}F_{12}

Suppose, we replace DL for A_1 and

A_2 =  \dfrac{\pi D}{2}

Then:

F_{21} = \dfrac{DL}{(\dfrac{\pi D}{2}) L} \times 1 \\ \\  =\dfrac{2}{\pi} \\ \\  =0.64

Now, we need to employ the use of energy balance formula to the dryer.

i.e.

Q_{21} = Q_{evaporation}

But, before that;  let's find the radian heat exchange occurring among the dome and the flat base surface:

Q_{21}= F_{21} A_2 \sigma (T_2^4-T_1^4) \\ \\ Q_{21} = F_{21} \times \dfrac{\pi D}{2} \sigma (T_2^4 -T_1^4)

where;

\sigma = Stefan \ Boltzmann's \ constant

T_1 = base \ temperature

T_2 = temperature  \ of  \ the  \ dome

∴

Q_{21} = 0.64 \times (\dfrac{\pi}{2}\times 1.5) \times 5.67 \times 10^4 \times (1000^4 -370^4)\\ \\ Q_{21} = 83899.15 \ W/m

Recall the energy balance formula;

Q_{21} = Q_{evaporation}

where;

Q_{evaporation} = mh_{fg}

here;

h_{fg} = enthalpy of vaporization

m = the water mass flow rate

∴

83899.15 = m \times 2257 \times 10^3  \\ \\  m = \dfrac{83899.15}{ 2257 \times 10^3 }\\ \\ \mathbf{m = 0.0371 \ kg/s.m}

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=> Expected Unit Weight = 3690.05 lb / 27 ft3 = 136.67 lb/ft3

Also, Concrete Yield = Weight of all components / Unit weight of concrete

=> Yield = 3690.05 / 136.67 = 27 ft3 or 1 yd3

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