# Heat Transfer Operations Questions and Answers – Enthalpy for a Single Evaporator

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This set of Heat Transfer Operations written test Questions & Answers focuses on “Enthalpy for a Single Evaporator”.

1. 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 –

1. Flow of non-condensable gas is negligible
2. The superheat and sub-cooling conditions of the condensable steam is negligible
3. No solid should precipitate out from the concentrating solution.

2. 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 –

1. Flow of non-condensable gas is negligible
2. The superheat and sub-cooling conditions of the condensable steam is negligible
3. No solid should precipitate out from the concentrating solution

Hence as the flow of non-condensable is assumed negligible, the given statement is wrong.

3. 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.

4. 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.

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

Explanation: The heat transfer rate is equal to the heat loss in feed and heat supplied by the steam, which is Qs = Qf = UAW ΔT.

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 of 25℃ to a final temperature of 150℃?
Latent heat of vaporization of water = 2,260 kJ/kg
Specific heat capacity of solution = 4 KJ/Kg K
a) 52600 KJ
b) 56200 KJ
c) 2272.6 MJ
d) 47600 KJ

Explanation: The total energy required is = mTSΔT + mL = 50×4×125 + 10×2260 = 47600 KJ.

7. What is the overall heat transfer coefficient of an evaporator setup of internal diameter is 30cm and 1m length cylinder in order to evaporate 10kg of water from a feed of 50kg at a temperature of 25℃ to a final temperature of 150℃? Saturated temperature of steam = 160℃
Latent heat of vaporization of water = 2,260 kJ/kg
Specific heat capacity of solution = 4 KJ/Kg K
a) 503 kJ/m2 K
b) 553 kJ/m2 K
c) 5553 kJ/m2 K
d) 5053 kJ/m2 K

Explanation: The heat transferred can be calculated as Q = mTSΔT + mL = 50×4×125 + 10×2260 = 47600 KJ = Q = UAW ΔT = U(πDL)(160-150)=9.42U kJ or U=$$\frac{47600}{9.42}$$=5053 kJ/m2K.

8. Given the overall heat transfer coefficient of the wall is 4053 kJ/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 of 25℃ to a final temperature of 150℃. What is the heat transfer area of the setup?
Saturated temperature of steam = 160℃
Latent heat of vaporization of water = 2,260 kJ/kg
Specific heat capacity of solution = 4 KJ/Kg K
a) 1.10m2
b) 1.27m2
c) 2.17m2
d) 1.17m2

Explanation: The heat transferred can be calculated as Q = mTSΔT + mL = 50×4×125 + 10×2260 = 47600 KJ = Q = UAW ΔT = U(A)(160-150)=40530A kJ or A=$$\frac{47600}{40530}$$=1.17m2.

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

Explanation: Yes, the given statement “Because the area is on a diameter between outer and inner diameter” is correct because the equivalent heat transfer area for the steam part is larger than the feed part, hence the calculation gives us an intermediate value.

10. What is the correct representation of the solution temperature as the evaporation proceeds in an evaporator?
a)
b)
c)
d)

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.

11. If the mole fraction of solutes in the feed is XF and in product is XP, then which one of the following relation is correct?
E = evaporate release rate
P = product formation rate
F = Feed supply rate
a) E = $$\frac{P (X_P – X_F)}{X_F}$$
b) E = $$\frac{F (X_P – X_F)}{X_F}$$
c) E = $$\frac{P (X_F – X_P)}{X_P}$$
d) E = $$\frac{F (X_F – X_P)}{X_P}$$

Explanation: The enthalpy mole fraction balance yields us XFE = P(XF – XP) or E = $$\frac{P (X_P – X_F)}{X_F}$$.

12. If the mole fraction of solutes in the feed is 0.2 and in product is 0.75, 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
Explanation: The enthalpy mole fraction balance yields us E = $$\frac{P(X_F – X_P)}{X_F}$$ = 15 (0.75 – 0.2)/0.2 = 41.25 kg.