1. In the given formula for overall heat transfer coefficient in Rising film evaporators, What is the term hOD?
\(\frac{1}{U}=\frac{1}{ho}+\frac{1}{h_{OD}}+(\frac{Do}{Di})(\frac{1}{h_{ID}})+ (\frac{Do}{Di})(\frac{1}{hi})\)
a) Outside dirt coefficient
b) Inside dirt coefficient
c) Outside film heat transfer coefficient
d) Inside film heat transfer coefficient
Explanation: The given equation represents the overall heat transfer coefficient of a rising film evaporator under the condition of fouling and hence the terms hID is inside dirt factor and hOD as the outside dirt factor.
2. Find the value of Residence time if we have a feed of 30kg/hr at 25℃ and with evaporator consumption as 10kg/hr?
Evaporator Volume = 5L
Latent heat of vaporization of water = 2,260 kJ/kg
Specific heat capacity = 6 KJ/Kg K
Average density of solution = 1.2Kg/L
a) 0.2hr
b) 0.3hr
c) 0.4hr
d) 0.5hr
Explanation: The mass flow rate = 30kg/hr, hence the volume flow rate = 30/1.2 = 25L/hr. Thus, as the volume of the evaporator is 5L, residence time = 5/25 = 0.2hr.
3. When the flow velocity of the fluid increases the temperature plot maxima shifts towards __________
a) Steam exit
b) Middle of the tube
c) Feed inlet
d) Feed exit
Explanation:As the flow velocity of the fluid increases, the liquid temperature reaches its maxima when the liquid is about to leave hence the maxima is at the feed exit. At higher velocity the temperature raise is less and the liquid boils near the top of the tube.
4. The liquid temperature in the tubes increases up to certain height and then the temperature decreases due to ___________
a) Loss of superheat
b) Loss of pressure
c) Concentration increase
d) Concentration decrease
Explanation:As the liquid reaches its final stage in the evaporator, its concentration is so high that is hardly releases any vapour and hence its temperature decreases. The steam too has by then condensed completely.
5. What is the real temperature driving force of the evaporator in the given diagram?
a) TBP – Tw
b) TW – TBP
c) TW – Tsteam
d) Tsteam – TBP
Explanation: The real temperature driving force is Tsteam – TBP because Tw is just the pure water boiling point which is not involved in the process, rather the temperature of steam and fluid is all that is of design concern.
6.When the flow rate of the fluid decreases the temperature plot shifts towards __________
a) Steam exit
b) Middle of the tube
c) Feed inlet
d) Feed exit
Explanation: As the flow velocity of the fluid decreases, the liquid temperature reaches its maxima when the liquid is at about the middle of the tube hence the maxima is at the middle little bit near the steam inlet. At higher velocity the temperature raise is less and the liquid boils near the top of the tube
7. 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
Explanation: The assumptions taken before applying enthalpy balance are summarised as –
- Flow of non-condensable gas is negligible
- The superheat and sub-cooling conditions of the condensable steam is negligible
- No solid should precipitate out from the concentrating solution.
8. 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
Explanation:The assumptions taken before applying enthalpy balance are summarised as –
- Flow of non-condensable gas is negligible
- The superheat and sub-cooling conditions of the condensable steam is negligible
- No solid should precipitate out from the concentrating solution
Hence as the flow of non-condensable is assumed negligible, the given statement is wrong.
9. 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
Explanation:The heat provided to the evaporator goes to heating the feed to boiling point and then vaporising the liquid. The outside steam is super-heated, hence the heat to cool it and then condensation is what it supplies.
10. In the given equation, what do you understand by the enthalpy balance, which element is it applied for?
\(\dot{q}_l=(\dot{m}_{fl} – \dot{m}_{cl})h_v-\dot{m}_{fl}h_{fl} + \dot{m}_{cl}h_{cl}\)
a) Feed
b) Steam
c) Vapour
d) Evaporator wall
Explanation: The given equation \(\dot{q}_l=(\dot{m}_{fl} – \dot{m}_{cl})h_v-\dot{m}_{fl}h_{fl} + \dot{m}_{cl}h_{cl}\) is the enthalpy balance for the feed side, the terms used are
QL = rate of heat transfer from heating surface to the liquid
hv = specific enthalpy of vapour
hcl = specific enthalpy of concentrated liquid
hfl = specific enthalpy of liquid feed
MFL = flow rate of liquid feed
MCL = flow rate of concentrated liquid.