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Heat Transfer Operations Multiple Choice Questions on “Evaporators – Multiple Effect Calculations”.
1. In a triple effect evaporator, given the enthalpies as and the flow rates calculate the evaporation rate in the 2nd 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
Answer: a
Clarification: The enthalpy balance for the 2nd effect can be determined encircling the required data in inlet and outlet.
INLET: Feed P1 with enthalpy HP1 and Steam E1 with latent heat ∇E1
OUTLET: Evaporate E2 and Product P2
E1 ∇E1+P1 HP1=E2 HE2+P2 HP2
Hence, P = XFF/Xp = 50×0.3/0.5 = 30Kg/hr, E1 = 50-30 =20kg/hr
E1 ∇E1+P1 HP1=E2 HE2+P2 HP2
20×538+30×106 = (E)x631+(30-E)x90
Hence, the value of E = 25.7kg/hr.
2. 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 evaporate rate in 3rd 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
Answer: b
Clarification: The enthalpy balance for the 2nd effect can be determined encircling the required data in inlet and outlet.
INLET: Feed P1 with enthalpy HP1 and Steam E1 with latent heat ∇E1
OUTLET: Evaporate E2 and Product P2
FHF+S∇S=EHE+PHP
E1 ∇E1+P1 HP1=E2 HE2+P2 HP2
Hence, P = XFF/Xp = 25000×0.1/0.5 = 5000Kg/hr, E = E1 + E2 + E3 = 25000-5000 =20000kg/hr
E1 ∇E1+P1 HP1=E2 HE2+P2 HP2
E2 ∇E2+P2 HP2=EHE+PHP
We have the three equations with the unknowns E1, E2, E3, P1, P2, P = 5000 kg/hr.
Hence solving the equations using a scientific calculator we get, E3 = 7190 kg/hr.
3. 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
Answer: a
Clarification: Yes the Statement is correct because the formula for heat transfer is = UAΔT
Hence if the area is same for all effects, U1A ΔT1 = U2AΔT2 = U3AΔT3, here we have 4 variables and three equations, hence anyone can be found out easily if one or more are available.
4. 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.
U1=2930 Kcal/hr2 K, U2=1220 Kcal/hr2 K, U3=610 Kcal/hr2 K
| Temperature | Latent heat(m) | Enthalpy |
|---|---|---|
| 117 | 527 | 654 |
| 106 | 538 | 640 |
| 90 | 545 | 631 |
| 57 | 568 | 619 |
a) 164m2
b) 151m2
c) 179.5m2
d) 155m2
Answer: c
Clarification: The enthalpy balance for the 2nd effect can be determined encircling the required data in inlet and outlet.
INLET: Feed P1 with enthalpy HP1 and Steam E1 with latent heat ∇E1
OUTLET: Evaporate E2 and Product P2
FHF+S∇S=EHE+PHP
E1 ∇E1+P1 HP1=E2 HE2+P2 HP2
Hence, P = XFF/Xp = 25000×0.1/0.5 = 5000Kg/hr, E = E1 + E2 + E3 = 25000-5000 =20000kg/hr
E1 ∇E1+P1 HP1=E2 HE2+P2 HP2
E2 ∇E2+P2 HP2=EHE+PHP
We have the three equations with the unknowns E1, E2, E3, P1, P2, P = 5000 kg/hr.
Hence solving the equations using a scientific calculator we get, E2 = 6630 kg/hr
A3 = E2 ∇E2/U∆T = 6630×545/610x(90 – 57) = 179.52.
5. 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 evaporate rate in 2nd 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
Answer: a
Clarification: The enthalpy balance for the 2nd effect can be determined encircling the required data in inlet and outlet.
INLET: Feed P1 with enthalpy HP1 and Steam E1 with latent heat ∇E1
OUTLET: Evaporate E2 and Product P2
FHF+S∇S=EHE+PHP
E1 ∇E1+P1 HP1=E2 HE2+P2 HP2
Hence, P = XFF/Xp = 25000×0.1/0.5 = 5000Kg/hr, E = E1 + E2 + E3 = 25000-5000 =20000kg/hr
E1 ∇E1+P1 HP1=E2 HE2+P2 HP2
E2 ∇E2+P2 HP2=EHE+PHP
We have the three equations with the unknowns E1, E2, E3, P1, P2, P = 5000 kg/hr.
Hence solving the equations using a scientific calculator we get, E2 = 6630 kg/hr.
6. 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 1st 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
Answer: c
Clarification: The enthalpy balance for the 2nd effect can be determined encircling the required data in inlet and outlet.
INLET: Feed P1 with enthalpy HP1 and Steam E1 with latent heat ∇E1
OUTLET: Evaporate E2 and Product P2
FHF+S∇S=EHE+PHP
E1 ∇E1+P1 HP1=E2 HE2+P2 HP2
Hence, P = XFF/Xp = 25000×0.1/0.5 = 5000Kg/hr, E = E1 + E2 + E3 = 25000-5000 =20000kg/hr
E1 ∇E1+P1 HP1=E2 HE2+P2 HP2
E2 ∇E2+P2 HP2=EHE+PHP
We have the three equations with the unknowns E1, E2, E3, P1, P2, P= 5000 kg/hr.
Hence solving the equations using a scientific calculator we get, E1 = 6180 kg/hr.
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 38oC, 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
Answer: d
Clarification: The enthalpy balance for the 2nd effect can be determined encircling the required data in inlet and outlet.
INLET: Feed P1 with enthalpy HP1 and Steam E1 with latent heat ∇E1
OUTLET: Evaporate E2 and Product P2
FHF+S∇S=EHE+PHP
E1 ∇E1+P1 HP1=E2 HE2+P2 HP2
Hence, P = XFF/Xp = 25000×0.1/0.5 = 5000Kg/hr, E = E1 + E2 + E3 = 25000-5000 =20000kg/hr
E1 ∇E1+P1 HP1=E2 HE2+P2 HP2
E2 ∇E2+P2 HP2=EHE+PHP
We have the three equations with the unknowns E1, E2, E3, P1, P2, P = 5000 kg/hr.
Hence solving the equations using a scientific calculator we get, S = 9488 kg/hr.
8. 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 1st 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) 1622
b) 1512
c) 1892
d) 1552
Answer: d
Clarification: The enthalpy balance for the 2nd effect can be determined encircling the required data in inlet and outlet.
INLET: Feed P1 with enthalpy HP1 and Steam E1 with latent heat ∇E1
OUTLET:Evaporate E2 and Product P2
FHF+S∇S=EHE+PHP
E1 ∇E1+P1 HP1=E2 HE2+P2 HP2
Hence, P = XFF/Xp = 25000×0.1/0.5 = 5000Kg/hr, E = E1 + E2 + E3 = 25000-5000 =20000kg/hr
E1 ∇E1+P1 HP1=E2 HE2+P2 HP2
E2 ∇E2+P2 HP2=EHE+PHP
We have the three equations with the unknowns E1, E2, E3, P1, P2, P = 5000 kg/hr.
Hence solving the equations using a scientific calculator we get, E1 = 6180 kg/hr
A1 = S∇S/U∆T = 9488×527/2930x(117-106) = 155 2.
9. 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 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) 1702
b) 1512
c) 1892
d) 1552
Answer: a
Clarification: The enthalpy balance for the 2nd effect can be determined encircling the required data in inlet and outlet.
INLET: Feed P1 with enthalpy HP1 and Steam E1 with latent heat ∇E1
OUTLET: Evaporate E2 and Product P2
FHF+S∇S=EHE+PHP
E1 ∇E1+P1 HP1=E2 HE2+P2 HP2
Hence, P = XFF/Xp = 25000×0.1/0.5 = 5000Kg/hr, E = E1 + E2 + E3 = 25000-5000 =20000kg/hr
E1 ∇E1+P1 HP1=E2 HE2+P2 HP2
E2 ∇E2+P2 HP2=EHE+PHP
We have the three equations with the unknowns E1, E2, E3, P1, P2, P= 5000 kg/hr.
Hence solving the equations using a scientific calculator we get, E1 = 6180 kg/hr
A2 = E1 ∇E1/U∆T = 6180×538/1220x(106 -90) = 170 2.
10. 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
Answer: a
Clarification: 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.
11. If the three areas obtained from steam calculation are 1512, 1622, and 1892. 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
Answer: b
Clarification: We consider only the maximum value of the area for calculations of number of tubes and other necessities. Hence the maximum value here is 1892 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.
12. If we have the areas calculated for the three effects as 1512, 1622, and 1892. Then the value we consider to manufacture the three effects of this triple effect evaporator is ______________
a) 1622
b) 1512
c) 1892
d) Their respective values
Answer: c
Clarification: We consider only the maximum value of the area for calculations of number of tubes and other necessities. Hence the maximum value here is 1892.