In the given equation, what do you understand by the enthalpy balance, which element is it applied for?
a) Feed
b) Steam
c) Vapour
d) Evaporator wall
Q.2.
Which one of the following is not an assumption taken before applying enthalpy balance on the evaporator?
a) No solid should precipitate out from the concentrating solution
b) The superheat and sub-cooling conditions of the condensable steam is negligible
c) No fouling should occur on the evaporator surface
d) Flow of non-condensable gas is negligible
Q.3.
The flow of vapour in an evaporator is assumed to be non-condensable else the calculations would differ for the evaporator if the gas condenses.
a) True
b) False
Q.4.
The amount of heat transferred through the evaporator wall is actually equal to which one of the following?
a) The steam condensing on the outer wall
b) The heat carried by the vapour leaving the evaporator
c) The total heat of increasing the feed temperature to boiling point
d) The total heat of increasing the feed temperature to boiling point and the vapour
Q.5.
If Qs is heat supplied by the steam and Qf is the heat evolved in the feed, then the overall heat transfer coefficient can be represented as? Here AF = Area of film and AW = Area of wall.
a) U = Qf/AW ΔT
b) U = Qs/AF ΔT
c) U = (Qs – Qf)/AW ΔT
d) U = (Qs – Qf)/AF ΔT
Q.6.
What is the amount of heat that needs to be supplied in order to evaporate 10kg of water from a feed of 50kg at a temperature ofto a final temperature of 150℃?
a) 52600 KJ
b) 56200 KJ
c) 2272.6 MJ
d) 47600 KJ
Q.7.
What is the overall heat transfer coefficient of an evaporator setup of internal diameter is 30cm andlength cylinder in order to evaporate 10kg of water from a feed of 50kg at a temperature ofto a final temperature of 150℃? Saturated temperature of steam = 160℃
a) 503 kJ/m2 K
b) 553 kJ/m2 K
c) 5553 kJ/m2 K
d) 5053 kJ/m2 K
Q.8.
Given the overall heat transfer coefficient of the wall iskJ/m2 K, the internal diameter is 30cm and the length of evaporator is 1m. The evaporator is used to evaporate 10kg of water from a feed of 50kg at a temperature ofto a final temperature of 150℃. What is the heat transfer area of the setup?
a) 1.10m2
b) 1.27m2
c) 2.17m2
d) 1.17m2
Q.9.
Why is the calculated heat transfer area more than the inner surface heat transfer area of the evaporator?
a) Because the area is outer surface area
b) Because the area is equivalent to a large diameter
c) Because the area is on a diameter between outer and inner diameter
d) Area is more than outer surface area
Q.10.
What is the correct representation of the solution temperature as the evaporation proceeds in an evaporator?
a)
b)
c)
d)
Q.11.
If the mole fraction of solutes in the feed is 0.2 and in product is 0.then what is the evaporate formation rate if product formation rate is = 15kg/hr?
a) 45.25 kg
b) 44.25 kg
c) 41.25 kg
d) 43.25 kg
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