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A single-effect evaporator is concentrating a feed of 9072 kg/h of a 10 wt % solution of NaOH in water at temperature of 288.8 K to a product of 50% solids. The pressure of the saturated steam used is 42 kPa (gage) and the pressure in the vapor space of the evaporator is 20 kPa (abs).



a)Neglecting boiling point raise and the heat of solution, calculate the steamused, the steam economy in kg vaporized/kg steam, required surface area ofthe evaporator.



b)Repeat the calculations taking into consideration boiling point raise and heatof solution

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temmydbrain

Answer:

Check the explanation

Explanation:

Going by the question above for parta, the feed flow rate is stated to be 9072g/hr.& result is offered for part a considering the same. if the unit is kg/hr, then kindly let me know & i shall make slight modification that might be required in solution which in that case, area & steam consumption answer shall differ.

In the b part of the question, reduced flow rate is higher than the initial flow rate therefore same has been considered in gram /hr.

Kindly check the attached images below to see the step by step solution to the above question.

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lhmarianateixeira

In this exercise we have to use the knowledge of mechanics to calculate the steamused of the solution, in this way we will have to:

a) [tex]M_S=\frac{H_S}{\lambda_S} = 8.43 KG/h[/tex] , [tex]A= 0.0526 m^2[/tex]

b) We can assume that all water shall evaporate and produd shell contain solid only.

So from the data informed in the exercise we have that:

  • [tex]F= 9072 g/h[/tex]
  • [tex]M_f=907,2 g/h[/tex]
  • [tex]M_L=1814.4 kg/h=L[/tex]

So apply overall mass balance, will be:

[tex]F=L+V\\9072=1814.4-V\\V=7.2576 kg/h[/tex]

Now, for enthalpy balance we shall make assumption that, all liquid phase enthalphy is that heet capacity of water is constant, so the data will be:

  • [tex]C=4.18 kJ/kg[/tex]
  • [tex]T_c=15.6[/tex]
  • [tex]P=20KPa[/tex]
  • [tex]T_P=60[/tex]

So, find the enthalpy of liquid, we have;

[tex]H_L= L*C_P*(T_P-T_F)\\=(1.8144)(4.18)(60.0586-15.6)\\H_v=V*(P(t_P-T_F)+\lambda)\\=(7.2576)(4.18(60.0586-15.6)+2357.5477))H_s=H_L+H_V\\H_S= (1.8144)(4.18)(60.0586-15.6)+(7.2576)(4.18(60.0586-15.6)+2357.5477))\\=18796.051 KJ/h[/tex]

Now, we can find the mass flow rate, that will be;

[tex]M_S=\frac{H_S}{\lambda_S} = 8.43 KG/h[/tex]

Now we can find the area of evaporator, that will be:

[tex]H_S=UA=(T_S-T_P)\\A=0.0526m^2[/tex]

See more about enthalpy at brainly.com/question/7827769

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