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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