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:
1) The Kelvin temperature cannot be negative
2) The Kelvin degree is written as K, not ºK
Explanation:
The temperature of an object can be written using different temperature scales.
The two most important scales are:
- Celsius scale: the Celsius degree is indicated with ºC. It is based on the freezing point of water (placed at 0ºC) and the boiling point of water (100ºC).
- Kelvin scale: the Kelvin is indicated with K. it is based on the concept of "absolute zero" temperature, which is the temperature at which matter stops moving, and it is placed at zero Kelvin (0 K), so this scale cannot have negative temperatures, since 0 K is the lowest possible temperature.
The expression to convert from Celsius degrees to Kelvin is:
Therefore in this problem, since the student reported a temperature of -3.5 ºK, the errors done are:
1) The Kelvin temperature cannot be negative
2) The Kelvin degree is written as K, not ºK