This set of Heat Transfer Operations Multiple Choice Questions & Answers (MCQs) focuses on “Evaporators – Multiple Effect Calculations”.

1. Which one of the following is the correct equation for enthalpy balance for the 1^{st} effect in a triple effect evaporator?

a) FH_{F}+S∇_{S}=EH_{E}+EH_{P}

b) FH_{F}+S∇_{S}=EH_{E}+PH_{P}

c) E_{1} H_{E}+P_{1} H_{P}=E_{2} H_{E}+P_{2} H_{P}

d) FH_{F}+S∇_{S}=E_{1} H_{E}+P_{1} H_{P}

View Answer

Explanation: The enthalpy balance for the 1

^{st}effect can be determined encircling the required data in inlet and outlet.

INLET: Feed F with enthalpy H

_{F}and Steam S with latent heat ∇

_{S}

OUTLET: Evaporate E

_{1}and Product P

_{1}.

2. Which one of the following is the correct equation for enthalpy balance for the 2^{nd} effect in a triple effect evaporator?

a) FH_{F}+S∇_{S}=EH_{E}+EH_{P}

b) FH_{F}+S∇_{S}=EH_{E}+PH_{P}

c) E_{1} ∇_{E1}+P_{1}H_{P1}=E_{2} H_{E2}+P_{2} H_{P2}

d) FH_{F}+S∇_{S}=E_{1} H_{E}+P_{1}H_{P}

View Answer

Explanation: The enthalpy balance for the 2

^{nd}effect can be determined encircling the required data in inlet and outlet.

INLET: Feed P

_{1}with enthalpy HP

_{1}and Steam E

_{1}with latent heat ∇

_{E1}

OUTLET: Evaporate E

_{2}and Product P

_{2}

E

_{1}∇

_{E1}+P

_{1}H

_{P1}=E

_{2}H

_{E2}+P

_{2}H

_{P2}.

3. In a triple effect evaporator, given the enthalpies as and the flow rates calculate the evaporation rate in the 2^{nd} effect.

Temperature | Latent heat(m) | Enthalpy |
---|---|---|

117 | 527 | 654 |

106 | 538 | 640 |

90 | 545 | 631 |

57 | 568 | 619 |

Feed rate = 50Kg/hr, Feed conc = 0.3, Product conc = 0.5

Find the evaporation rate in 2nd effect.

a) 25.7 kg/hr

b) 20 kg/hr

c) 15 kg/hr

d) 10 kg/hr

View Answer

Explanation: The enthalpy balance for the 2

^{nd}effect can be determined encircling the required data in inlet and outlet.

INLET: Feed P

_{1}with enthalpy H

_{P1}and Steam E

_{1}with latent heat ∇

_{E1}

OUTLET: Evaporate E

_{2}and Product P

_{2}

E

_{1}∇

_{E1}+P

_{1}H

_{P1}=E

_{2}H

_{E2}+P

_{2}H

_{P2}

Hence, P = X

_{F}F/X

_{p}= 50×0.3/0.5 = 30Kg/hr, E

_{1}= 50-30 =20kg/hr

E

_{1}∇

_{E1}+P

_{1}H

_{P1}=E

_{2}H

_{E2}+P

_{2}H

_{P2}

20×538+30×106 = (E)x631+(30-E)x90

Hence, the value of E = 25.7kg/hr.

4. In a triple effect evaporator, given the enthalpies as and the flow rates calculate the product formation rate in the 2^{nd} effect.

Feed rate = 25000kg/hr at 38℃, Feed conc = 0.10, Product conc = 0.50

Find the evaporate rate in 3^{rd} effect.

Temperature | Latent heat(m) | Enthalpy |
---|---|---|

117 | 527 | 654 |

106 | 538 | 640 |

90 | 545 | 631 |

57 | 568 | 619 |

a) 6630 kg/hr

b) 7190 kg/hr

c) 6180 kg/hr

d) 5000 kg/hr

View Answer

Explanation: The enthalpy balance for the 2

^{nd}effect can be determined encircling the required data in inlet and outlet.

INLET: Feed P

_{1}with enthalpy H

_{P1}and Steam E

_{1}with latent heat ∇

_{E1}

OUTLET: Evaporate E

_{2}and Product P

_{2}

FH

_{F}+S∇

_{S}=EH

_{E}+PH

_{P}

E

_{1}∇

_{E1}+P

_{1}H

_{P1}=E

_{2}H

_{E2}+P

_{2}H

_{P2}

Hence, P = X

_{F}F/Xp = 25000×0.1/0.5 = 5000Kg/hr, E = E

_{1}+ E

_{2}+ E

_{3}= 25000-5000 =20000kg/hr

E

_{1}∇

_{E1}+P

_{1}H

_{P1}=E

_{2}H

_{E2}+P

_{2}H

_{P2}

E

_{2}∇

_{E2}+P

_{2}H

_{P2}=EH

_{E}+PH

_{P}

We have the three equations with the unknowns E

_{1}, E

_{2}, E

_{3}, P

_{1}, P

_{2}, P = 5000 kg/hr.

Hence solving the equations using a scientific calculator we get, E

_{3}= 7190 kg/hr.

5. Which one of the following is the correct equation for enthalpy balance for the final effect in a triple effect evaporator?

a) E_{2} ∇_{E2}+P_{2} H_{P2}=EH_{E}+PH_{P}

b) FH_{F}+S∇_{S}=EH_{E}+PH_{P}

c) E_{1} ∇_{E1}+P_{1} H_{P1}=E_{2} H_{E2}+P_{2} H_{P2}

d) FH_{F}+S∇_{S}=E_{1} H_{E}+P_{1} H_{P}

View Answer

Explanation: The enthalpy balance for the 1

^{st}effect can be determined encircling the required data in inlet and outlet.

INLET: Feed P

_{2}with enthalpy H

_{P2}and Steam E

_{2}with latent heat ∇

_{E2}

OUTLET: Evaporate E and Product P

E

_{2}∇

_{E2}+P

_{2}H

_{P2}=EH

_{E}+PH

_{P}.

6. If we know the heat transfer coefficients of all the effects and the initial and final temperatures of the vapour, we can estimate the intermediate temperatures of the setup.

a) True

b) False

View Answer

Explanation: Yes the Statement is correct because the formula for heat transfer is = UAΔT

Hence if the area is same for all effects, U

_{1}A ΔT

_{1}= U

_{2}AΔT

_{2}= U

_{3}AΔT

_{3}, here we have 4 variables and three equations, hence anyone can be found out easily if one or more are available.

7. In a triple effect evaporator, given the enthalpies as and the flow rates calculate the product formation rate in the 2nd effect.

Feed rate = 25000kg/hr at 38℃, Feed conc = 0.10, Product conc = 0.50

Find the Area of condenser in 3rd effect.

U_{1}=2930 Kcal/hr^{2} K, U_{2}=1220 Kcal/hr^{2} K, U_{3}=610 Kcal/hr^{2} K

Temperature | Latent heat(m) | Enthalpy |
---|---|---|

117 | 527 | 654 |

106 | 538 | 640 |

90 | 545 | 631 |

57 | 568 | 619 |

a) 164m^{2}

b) 151m^{2}

c) 179.5m^{2}

d) 155m^{2}

View Answer

Explanation: The enthalpy balance for the 2

^{nd}effect can be determined encircling the required data in inlet and outlet.

INLET: Feed P

_{1}with enthalpy H

_{P1}and Steam E

_{1}with latent heat ∇

_{E1}

OUTLET: Evaporate E

_{2}and Product P

_{2}

FH

_{F}+S∇

_{S}=EH

_{E}+PH

_{P}

E

_{1}∇

_{E1}+P

_{1}H

_{P1}=E

_{2}H

_{E2}+P

_{2}H

_{P2}

Hence, P = X

_{F}F/X

_{p}= 25000×0.1/0.5 = 5000Kg/hr, E = E

_{1}+ E

_{2}+ E

_{3}= 25000-5000 =20000kg/hr

E

_{1}∇

_{E1}+P

_{1}H

_{P1}=E

_{2}H

_{E2}+P

_{2}H

_{P2}

E

_{2}∇

_{E2}+P

_{2}H

_{P2}=EH

_{E}+PH

_{P}

We have the three equations with the unknowns E

_{1}, E

_{2}, E

_{3}, P

_{1}, P

_{2}, P = 5000 kg/hr.

Hence solving the equations using a scientific calculator we get, E

_{2}= 6630 kg/hr

A

_{3}= E

_{2}∇

_{E2}/U∆T = 6630×545/610x(90 – 57) = 179.5

^{2}.

8. In a triple effect evaporator, given the enthalpies as and the flow rates calculate the product formation rate in the 2^{nd} effect.

Feed rate = 25000kg/hr at 38℃, Feed conc = 0.10, Product conc = 0.50

Find the evaporate rate in 2^{nd} effect.

Temperature | Latent heat(m) | Enthalpy |
---|---|---|

117 | 527 | 654 |

106 | 538 | 640 |

90 | 545 | 631 |

57 | 568 | 619 |

a) 6630 kg/hr

b) 7190 kg/hr

c) 6180 kg/hr

d) 5000 kg/hr

View Answer

Explanation: The enthalpy balance for the 2

^{nd}effect can be determined encircling the required data in inlet and outlet.

INLET: Feed P

_{1}with enthalpy H

_{P1}and Steam E

_{1}with latent heat ∇

_{E1}

OUTLET: Evaporate E

_{2}and Product P

_{2}

FH

_{F}+S∇

_{S}=EH

_{E}+PH

_{P}

E

_{1}∇

_{E1}+P

_{1}H

_{P1}=E

_{2}H

_{E2}+P

_{2}H

_{P2}

Hence, P = X

_{F}F/X

_{p}= 25000×0.1/0.5 = 5000Kg/hr, E = E

_{1}+ E

_{2}+ E

_{3}= 25000-5000 =20000kg/hr

E

_{1}∇

_{E1}+P

_{1}H

_{P1}=E

_{2}H

_{E2}+P

_{2}H

_{P2}

E

_{2}∇

_{E2}+P

_{2}H

_{P2}=EH

_{E}+PH

_{P}

We have the three equations with the unknowns E

_{1}, E

_{2}, E

_{3}, P

_{1}, P

_{2}, P = 5000 kg/hr.

Hence solving the equations using a scientific calculator we get, E

_{2}= 6630 kg/hr.

9. In a triple effect evaporator, given the enthalpies as and the flow rates

Feed rate = 25000kg/hr at 38℃, Feed conc = 0.10, Product conc = 0.50

Find the evaporate rate in 1^{st} effect.

Temperature | Latent heat(m) | Enthalpy |
---|---|---|

117 | 527 | 654 |

106 | 538 | 640 |

90 | 545 | 631 |

57 | 568 | 619 |

a) 6630 kg/hr

b) 7190 kg/hr

c) 6180 kg/hr

d) 5000 kg/hr

View Answer

Explanation: The enthalpy balance for the 2

^{nd}effect can be determined encircling the required data in inlet and outlet.

INLET: Feed P

_{1}with enthalpy H

_{P1}and Steam E

_{1}with latent heat ∇

_{E1}

OUTLET: Evaporate E

_{2}and Product P

_{2}

FH

_{F}+S∇

_{S}=EH

_{E}+PH

_{P}

E

_{1}∇

_{E1}+P

_{1}H

_{P1}=E

_{2}H

_{E2}+P

_{2}H

_{P2}

Hence, P = X

_{F}F/X

_{p}= 25000×0.1/0.5 = 5000Kg/hr, E = E

_{1}+ E

_{2}+ E

_{3}= 25000-5000 =20000kg/hr

E

_{1}∇

_{E1}+P

_{1}H

_{P1}=E

_{2}H

_{E2}+P

_{2}H

_{P2}

E

_{2}∇

_{E2}+P

_{2}H

_{P2}=EH

_{E}+PH

_{P}

We have the three equations with the unknowns E

_{1}, E

_{2}, E

_{3}, P

_{1}, P

_{2}, P= 5000 kg/hr.

Hence solving the equations using a scientific calculator we get, E

_{1}= 6180 kg/hr.

10. In a triple effect evaporator, given the enthalpies as and the flow rates calculate the product formation rate in the 2^{nd} effect.

Feed rate = 25000kg/hr at 38^{o}C, Feed conc = 0.10, Product conc = 0.50

Find the steam consumption rate.

Temperature | Latent heat(m) | Enthalpy |
---|---|---|

117 | 527 | 654 |

106 | 538 | 640 |

90 | 545 | 631 |

57 | 568 | 619 |

a) 6630 kg/hr

b) 7190 kg/hr

c) 6180 kg/hr

d) 9488 kg/hr

View Answer

Explanation: The enthalpy balance for the 2

^{nd}effect can be determined encircling the required data in inlet and outlet.

INLET: Feed P

_{1}with enthalpy H

_{P1}and Steam E

_{1}with latent heat ∇

_{E1}

OUTLET: Evaporate E

_{2}and Product P

_{2}

FH

_{F}+S∇

_{S}=EH

_{E}+PH

_{P}

E

_{1}∇

_{E1}+P

_{1}H

_{P1}=E

_{2}H

_{E2}+P

_{2}H

_{P2}

Hence, P = X

_{F}F/X

_{p}= 25000×0.1/0.5 = 5000Kg/hr, E = E

_{1}+ E

_{2}+ E

_{3}= 25000-5000 =20000kg/hr

E

_{1}∇

_{E1}+P

_{1}H

_{P1}=E

_{2}H

_{E2}+P

_{2}H

_{P2}

E

_{2}∇

_{E2}+P

_{2}H

_{P2}=EH

_{E}+PH

_{P}

We have the three equations with the unknowns E

_{1}, E

_{2}, E

_{3}, P

_{1}, P

_{2}, P = 5000 kg/hr.

Hence solving the equations using a scientific calculator we get, S = 9488 kg/hr.

11. In a triple effect evaporator, given the enthalpies as and the flow rates calculate the product formation rate in the 2^{nd} effect.

Feed rate = 25000kg/hr at 38℃, Feed conc = 0.10, Product conc = 0.50

Find the Area of condenser in 1_{st} effect.

U1=2930 Kcal/hr ^{2} K, U2=1220 Kcal/hr ^{2} K, U3=610 Kcal/hr ^{2} K

Temperature | Latent heat(m) | Enthalpy |
---|---|---|

117 | 527 | 654 |

106 | 538 | 640 |

90 | 545 | 631 |

57 | 568 | 619 |

a) 162^{2}

b) 151^{2}

c) 189^{2}

d) 155^{2}

View Answer

Explanation: The enthalpy balance for the 2

^{nd}effect can be determined encircling the required data in inlet and outlet.

INLET: Feed P

_{1}with enthalpy H

_{P1}and Steam E

_{1}with latent heat ∇

_{E1}

OUTLET:Evaporate E

_{2}and Product P

_{2}

FH

_{F}+S∇

_{S}=EH

_{E}+PH

_{P}

E

_{1}∇

_{E1}+P

_{1}H

_{P1}=E

_{2}H

_{E2}+P

_{2}H

_{P2}

Hence, P = X

_{F}F/X

_{p}= 25000×0.1/0.5 = 5000Kg/hr, E = E

_{1}+ E

_{2}+ E

_{3}= 25000-5000 =20000kg/hr

E

_{1}∇

_{E1}+P

_{1}H

_{P1}=E

_{2}H

_{E2}+P

_{2}H

_{P2}

E

_{2}∇

_{E2}+P

_{2}H

_{P2}=EH

_{E}+PH

_{P}

We have the three equations with the unknowns E

_{1}, E

_{2}, E

_{3}, P

_{1}, P

_{2}, P = 5000 kg/hr.

Hence solving the equations using a scientific calculator we get, E

_{1}= 6180 kg/hr

A

_{1}= S∇

_{S}/U∆T = 9488×527/2930x(117-106) = 155

^{2}.

12. In a triple effect evaporator, given the enthalpies as and the flow rates calculate the product formation rate in the 2^{nd} effect.

Feed rate = 25000kg/hr at 38oC, Feed conc = 0.10, Product conc = 0.50

Find the Area of condenser in 2nd effect.

U1=2930 Kcal/hr ^{2} K, U2=1220 Kcal/hr ^{2} K, U3=610 Kcal/hr ^{2} K

Temperature | Latent heat(m) | Enthalpy |
---|---|---|

117 | 527 | 654 |

106 | 538 | 640 |

90 | 545 | 631 |

57 | 568 | 619 |

a) 170^{2}

b) 151^{2}

c) 189^{2}

d) 155^{2}

View Answer

Explanation: The enthalpy balance for the 2

^{nd}effect can be determined encircling the required data in inlet and outlet.

INLET: Feed P

_{1}with enthalpy H

_{P1}and Steam E

_{1}with latent heat ∇

_{E1}

OUTLET: Evaporate E

_{2}and Product P

_{2}

FH

_{F}+S∇

_{S}=EH

_{E}+PH

_{P}

E

_{1}∇

_{E1}+P

_{1}H

_{P1}=E

_{2}H

_{E2}+P

_{2}H

_{P2}

Hence, P = X

_{F}F/X

_{p}= 25000×0.1/0.5 = 5000Kg/hr, E = E

_{1}+ E

_{2}+ E

_{3}= 25000-5000 =20000kg/hr

E

_{1}∇

_{E1}+P

_{1}H

_{P1}=E

_{2}H

_{E2}+P

_{2}H

_{P2}

E

_{2}∇

_{E2}+P

_{2}H

_{P2}=EH

_{E}+PH

_{P}

We have the three equations with the unknowns E

_{1}, E

_{2}, E

_{3}, P

_{1}, P

_{2}, P= 5000 kg/hr.

Hence solving the equations using a scientific calculator we get, E

_{1}= 6180 kg/hr

A

_{2}= E

_{1}∇

_{E1}/U∆T = 6180×538/1220x(106 -90) = 170

^{2}.

13. The three areas which we calculate from the steam consumption value, only the ________ one is used in calculation of number of tubes.

a) Maximum

b) Minimum

c) Average

d) Median

View Answer

Explanation: We consider only the maximum value of the area for calculations of number of tubes and other necessities such as the number of turns etc.

14. If the three areas obtained from steam calculation are 151^{2}, 162^{2}, and 189^{2}. If the diameter and the length of the tube are 5cm and 150cm respectively, find the number of tubes required in this multiple effect evaporator.

a) 697

b) 802

c) 892

d) 502

View Answer

Explanation: We consider only the maximum value of the area for calculations of number of tubes and other necessities. Hence the maximum value here is 189

^{2}which is our required answer.

The area of each tube is = πDo L hence we have πDo L N=189 or N=189/3.14×0.05×1.5=802.

15. If we have the areas calculated for the three effects as 151^{2}, 162^{2}, and 189^{2}. Then the value we consider to manufacture the three effects of this triple effect evaporator is ______________

a) 162^{2}

b) 151^{2}

c) 189^{2}

d) Their respective values

View Answer

Explanation: We consider only the maximum value of the area for calculations of number of tubes and other necessities. Hence the maximum value here is 189

^{2}.

**Sanfoundry Global Education & Learning Series – Heat Transfer Operations.**

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