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 materials 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.

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:

[tex]A_1 \ and \ A_2[/tex]

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

[tex]F_{11}+F_{12}=1[/tex]

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

Now:

[tex]0+F_{12}=1[/tex]

[tex]F_{12}=1[/tex]

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

[tex]A_2F_{21} = A_{1}F_{12} \\ \\ F_{21} = \dfrac{A_1}{A_2}F_{12}[/tex]

Suppose, we replace DL for [tex]A_1[/tex] and

[tex]A_2[/tex] =  [tex]\dfrac{\pi D}{2}[/tex]

Then:

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

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

i.e.

[tex]Q_{21} = Q_{evaporation}[/tex]

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

[tex]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)[/tex]

where;

[tex]\sigma = Stefan \ Boltzmann's \ constant[/tex]

[tex]T_1 = base \ temperature[/tex]

[tex]T_2 = temperature \ of \ the \ dome[/tex]

∴

[tex]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[/tex]

Recall the energy balance formula;

[tex]Q_{21} = Q_{evaporation}[/tex]

where;

[tex]Q_{evaporation} = mh_{fg}[/tex]

here;

[tex]h_{fg}[/tex] = enthalpy of vaporization

m = the water mass flow rate

∴

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

The drying rate per unit length is 0.037 kg/S.m

Given data;

From the attached diagram, the surface 1 is flat, it is a view factor, f[tex]_1_1[/tex] = 0.

Applying summation rule and solving the view factor from the base surface A[tex]_1[/tex] to the cylindrical dome A[tex]_2[/tex].

[tex]f_1_1+f_1_2=1[/tex]

Put F[tex]_1_2=0[/tex]

[tex]0+f_1_2=1[/tex]

This makes [tex]f_1_2=1[/tex]

Applying reciprocal rule and solving the view factor from the cylindrical dome A[tex]_2[/tex] to the base surface A[tex]_1[/tex].

[tex]A_2F_2_1=A_1F_1_2\\F_2_1=(\frac{A_1}{A_2})F_1_2[/tex]

Where A is the area of the surface.

Substitute DL for A[tex]_1[/tex] and [tex]\frac{\pi D}{2}[/tex] for A[tex]_2[/tex]

[tex]F_2_1 = \frac{DL}{(\frac{\pi D}{2}})L *1 = \frac{2 }{\pi } =2/3.14 = 0.64[/tex]

Using the energy balance equation to the dryer,

[tex]Q_2_1=Q_e_v_a_p[/tex]

Let's calculate the radiation heat exchange between the dome and the base surface per unit length by using the equation below

[tex]Q_2_1=F_2_1A_2[/tex]δ[tex](T_2^4-T_1^4)[/tex]

[tex]Q_2_1= F_2_1 * \frac{\pi D}{2}[/tex]δ[tex](T_2^4-T_1^4)[/tex]

substitute the respective values into the equation

[tex]Q_2_1=0.64*(\frac{\pi }{2}*1.5)*5.67*10^-^8*(1000^4-370^4)\\Q_2_1=8.3899.15W/m[/tex]

Let's calculate the mass flow rate of water using the amount of heat required for drying up.

[tex]Q_2_1=Q_e_v_a_p\\Q_e_v_a_p=mh_f_g\\[/tex]

where [tex]h_f_g= 2257*10^3J/kg[/tex]

and this is the enthalpy of vaporization and mass flow rate of water.

[tex]83899.15=m*2257*10^3\\m=0.037kg/S.m[/tex]

The drying rate per unit length is 0.037kg/S.m

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