Physics, 23.03.2020 20:01 antmike31ox1v8e

Saturated ethylene glycol at 1 atm is heated by a horizontal chromiumplated surface which has a diameter of 200 mm and is maintained at 480 K. 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

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Answer from: cheyennegenevie9833

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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

The heating power requirement = 559.2 W

The rate of evaporation = $6.89*10^{-4}kg/s$

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 $T_{sat}$ = 470 K for the saturated ethylene glycol.

Value for enthalpy of formation $h_{fg}$ = 812 kJ/kg

Density of saturated ethylene glycol $\rho___l$ = 1111 kg/m³

Surface tension $\sigma = 32.7*10^{-3}N/m$

The heat flux can be calculated by using the formula:

$q"s = \mu___l}}}h_{fg}{[\frac{g(\rho{__l}- \rho{__v} }{\sigma} ]^{1/2} [\frac{C_p*\delta T_c}{C_{sf}*h_{fg}P_r} ]^3$

$= [0.38*10^{-3}\frac{NS}{m^2} *812*10^3\frac{J}{kg} (\frac{9.81m/s^2*(1111-1.66)kg/m^3}{32.7*10^{-3}N/m} )^{1/2}*(\frac{3280J/kg.K(480-470)K}{0.01*812*10^3\frac{J}{kg}*(8.7)^1 } )]$

= 308.56 × 576.6 × 0.1

= 1.78 × 10⁴ W/m²

Now; to find the heating power requirement; we have:

$q_{boil} = q__s }*A S$

= $1.78*10^4 \frac{W}{m^2}*(\frac{\pi}{4}*(0.2m))^2$

Thus, the heating power requirement = 559.2 W

The rate of evaporation is given as:

$m= \frac{q_{boil}}{h_[fg}}$

= $\frac{559.2}{812*10^3}$

= $6.89*10^{-4}kg/s$

Thus, the rate of evaporation = $6.89*10^{-4}kg/s$

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: $q"max = 0.149*h_{fg}}* \rho{___l}}}}[ \frac{\sigma_g (\rho_l - \rho_v}{\rho_v^2} ]^{1/4}$

= $q"max = 0.149*812810^3* 1.66[ \frac{32.7*10^{-3}*9.8 (1111- 1.66}{1.66^2} ]^{1/4}$

= 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:

= $\frac{q''s}{q''max}$

= $\frac{1.78*10^4}{6.77*10^5}$

= 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

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