Answer :
Here is the full question.
Saturated ethylene glycol at 1 atm is heated by a chromium-plated surface which is circular in shape and has a diameter of 200-mm and is maintained at 480 K.
At 470 K the properties of the saturated liquid are mu = 0.38 * 10-3 N. s/m^2, Cp = 3280 J/kg. K and Pr = 8.7. The saturated vapour density is p= 1.66 kg/m^3. Take the liquid to surface constants to be Cnb = 0.010 and m=4.1.
Estimate the heating power requirement and the rate of evaporation
What fraction is the power requirement of the maximum power associated with the critical heat flux
Answer:
The heating power requirement = 559.2 W
The rate of evaporation = [tex]6.89*10^{-4}kg/s[/tex]
The fraction of the power requirement of operating heat flux to the maximum power associated with the critical heat flux is = 0.026
Explanation:
From the thermodynamics tables; we deduced the value for enthalpy at the pressure 1 atm and [tex]T_{sat}[/tex] = 470 K for the saturated ethylene glycol.
Value for enthalpy of formation [tex]h_{fg}[/tex] = 812 kJ/kg
Density of saturated ethylene glycol [tex]\rho___l[/tex] = 1111 kg/m³
Surface tension [tex]\sigma = 32.7*10^{-3}N/m[/tex]
The heat flux can be calculated by using the formula:
[tex]q"s = \mu___l}}}h_{fg}{[\frac{g(\rho{__l}- \rho{__v} }{\sigma} ]^{1/2} [\frac{C_p*\delta T_c}{C_{sf}*h_{fg}P_r} ]^3[/tex]
[tex]= [0.38*10^{-3}\frac{NS}{m^2} *812*10^3\frac{J}{kg} (\frac{9.81m/s^2*(1111-1.66)kg/m^3}{32.7*10^{-3}N/m} )^{1/2}*(\frac{3280J/kg.K(480-470)K}{0.01*812*10^3\frac{J}{kg}*(8.7)^1 } )][/tex]
= 308.56 × 576.6 × 0.1
= 1.78 × 10⁴ W/m²
Now; to find the heating power requirement; we have:
[tex]q_{boil} = q__s }*A S[/tex]
= [tex]1.78*10^4 \frac{W}{m^2}*(\frac{\pi}{4}*(0.2m))^2[/tex]
Thus, the heating power requirement = 559.2 W
The rate of evaporation is given as:
[tex]m= \frac{q_{boil}}{h_[fg}}[/tex]
= [tex]\frac{559.2}{812*10^3}[/tex]
= [tex]6.89*10^{-4}kg/s[/tex]
Thus, the rate of evaporation = [tex]6.89*10^{-4}kg/s[/tex]
To determine to what fraction in the power requirement of the maximum power is associated with the critical total flux ; we needed to first calculate the critical heat flux.
So, the calculation for the critical heat is given as:[tex]q"max = 0.149*h_{fg}}* \rho{___l}}}}[ \frac{\sigma_g (\rho_l - \rho_v}{\rho_v^2} ]^{1/4}[/tex]
= [tex]q"max = 0.149*812810^3* 1.66[ \frac{32.7*10^{-3}*9.8 (1111- 1.66}{1.66^2} ]^{1/4}[/tex]
= 200840.08 × 3.37
= 6.77 × 10⁵ W/m²
Finally, the fraction of the power requirement of operating heat flux to the maximum power associated with the critical heat flux is as follows:
= [tex]\frac{q''s}{q''max}[/tex]
= [tex]\frac{1.78*10^4}{6.77*10^5}[/tex]
= 0.026
Thus, the fraction of the power requirement of operating heat flux to the maximum power associated with the critical heat flux is = 0.026