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
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
2 answers:
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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Answer:
The velocity of the particle is 2 m/s,
Explanation:
Kinetic energy is defined as energy of the body due to its motion. It is given by :
Where :
m = mass of the object
v = velocity of the object
We have , particle with mass m and its kinetic energy is twice its mass.
And unit of velocity are m/s , so the velocity of the particle is 2 m/s.