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Question 11.13: Calculate the plate efficiency for the plate design consider...

Calculate the plate efficiency for the plate design considered in Examples 11.11 and 11.12, using Van Winkle’s correlation.

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From Examples 11.12 and 11.11:

 

ρL=925kg/m3,\rho_{L}=925 kg / m ^{3},

 

ρv=1.35kg/m3\rho_{v}=1.35 kg / m ^{3}

 

μL=0.34×10−3Ns/m2\mu_{L}=0.34 \times 10^{-3} Ns / m ^{2}

 

μv=10.0×10−6Ns/m2\mu_{v}=10.0 \times 10^{-6} Ns / m ^{2}

 

DLK=DL=4.64×10−9m2/sD_{L K}=D_{L}=4.64 \times 10^{-9} m ^{2} / s

 

hw=50mmh_{w}=50 mm

 

FA( fractional area )=Ah/Ac=0.0380.50=0.076F A(\text { fractional area })=A_{h} / A_{c}=\frac{0.038}{0.50}=0.076

 

uv= superficial vapour velocity =0.810.50=1.62m/su_{v}=\text { superficial vapour velocity }=\frac{0.81}{0.50}=1.62 m / s

 

σL=60×10−3N/m\sigma_{L}=60 \times 10^{-3} N / m

 

Dg⁡=(60×10−30.34×10−3×1.62)=109\operatorname{Dg}=\left(\frac{60 \times 10^{-3}}{0.34 \times 10^{-3} \times 1.62}\right)=109

 

Sc=(0.34×10−3925×4.64×10−9)=79Sc =\left(\frac{0.34 \times 10^{-3}}{925 \times 4.64 \times 10^{-9}}\right)=79

 

Re⁡=(50×10−3×1.62×1.350.34×10−3×0.076)=4232\operatorname{Re}=\left(\frac{50 \times 10^{-3} \times 1.62 \times 1.35}{0.34 \times 10^{-3} \times 0.076}\right)=4232

 

EmV=0.07(109)0.14(79)0.25(4232)0.08E_{m V}=0.07(109)^{0.14}(79)^{0.25}(4232)^{0.08} = 0.79 (79 per cent)

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