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This section includes 16 Mcqs, each offering curated multiple-choice questions to sharpen your Distillation Design knowledge and support exam preparation. Choose a topic below to get started.
1. |
The given function f(x) = x3-3X2+ 4 root is? |
A. | Negative |
B. | Positive |
C. | Unity |
D. | Zero |
Answer» D. Zero | |
2. |
Total balance material balance is given as |
A. | αijSbj |
B. | αijHbj |
C. | αijDbj |
D. | αijVbj |
Answer» B. αijHbj | |
3. |
HJ is formulated as |
A. | H* /ϫHVJ |
B. | H*- ϫHVJ |
C. | H* ( ϫHVJ) |
D. | H* + ϫHVJ |
Answer» E. | |
4. |
HJ is defined as the |
A. | Ideal enthalpy |
B. | Total enthalpy |
C. | Final enthalpy |
D. | Pure enthalpy |
Answer» B. Total enthalpy | |
5. |
For Boston method, ln Kbj |
A. | Aj + B (1j/Tj – 1/T*) |
B. | Aj – B (1j/Tj – 1/T*) |
C. | Aj – B (1j/Tj + 1/T*) |
D. | (1j/Tj + 1/T*) |
Answer» C. Aj – B (1j/Tj + 1/T*) | |
6. |
The vapors side product ln Rvj is given as |
A. | ln (KbjLJ/VJ) |
B. | ln (WbjLJ/VJ) |
C. | ln (KbjLJ/WJ) |
D. | ln (WJ) |
Answer» C. ln (KbjLJ/WJ) | |
7. |
ln Sbj is represented as |
A. | ln (KbjLJ/VJ) |
B. | ln (Kbj/VJ) |
C. | ln( KbjLJ) |
D. | ln (LJ) |
Answer» B. ln (Kbj/VJ) | |
8. |
TOTAL_BALANCE_MATERIAL_BALANCE_IS_GIVEN_AS?$ |
A. | α<sub>ij</sub>S<sub>bj</sub> |
B. | α<sub>ij</sub>H<sub>bj</sub> |
C. | α<sub>ij</sub>D<sub>bj</sub> |
D. | α<sub>ij</sub>V<sub>bj</sub> |
Answer» B. ‚âà√≠¬¨¬±<sub>ij</sub>H<sub>bj</sub> | |
9. |
HJ is formulated a? |
A. | H* /ϫH<sub>VJ</sub> |
B. | H*- ϫH<sub>VJ</sub> |
C. | H* ( ϫH<sub>VJ</sub>) |
D. | H* + ϫH<sub>VJ</sub> |
Answer» E. | |
10. |
HJ is defined as the |
A. | Ideal enthalpy |
B. | Total enthalpy |
C. | Final enthalpy |
D. | Pure enthalpy |
Answer» B. Total enthalpy | |
11. |
For Boston method, ln Kbj |
A. | A<sub>j</sub> + B (1<sub>j</sub>/T<sub>j</sub> – 1/T*) |
B. | A<sub>j</sub> – B (1<sub>j</sub>/T<sub>j</sub> – 1/T*) |
C. | A<sub>j</sub> – B (1<sub>j</sub>/T<sub>j</sub> + 1/T*) |
D. | (1<sub>j</sub>/T<sub>j</sub> + 1/T*) |
Answer» C. A<sub>j</sub> ‚Äö√Ñ√∂‚àö√ë‚àö¬® B (1<sub>j</sub>/T<sub>j</sub> + 1/T*) | |
12. |
For Russell method, ln Kbj is defined as |
A. | Aj – (Bj/Tj) |
B. | – (Bj/Tj) |
C. | Aj (Bj/Tj) |
D. | Aj + (Bj/Tj) |
Answer» B. ‚Äö√Ñ√∂‚àö√ë‚àö¬® (Bj/Tj) | |
13. |
The liqiud side product ln RLj is given as |
A. | ln (K<sub>bj</sub>L<sub>J</sub>/V<sub>J</sub>) |
B. | ln (W<sub>bj</sub>/V<sub>J</sub>) |
C. | ln (K<sub>bj</sub>L<sub>J</sub>/W<sub>J</sub>) |
D. | ln (W<sub>J</sub>/L<sub>J</sub>) |
Answer» E. | |
14. |
The vapors side product ln Rvj is given as |
A. | ln (K<sub>bj</sub>L<sub>J</sub>/V<sub>J</sub>) |
B. | ln (W<sub>bj</sub>L<sub>J</sub>/V<sub>J</sub>) |
C. | ln (K<sub>bj</sub>L<sub>J</sub>/W<sub>J</sub>) |
D. | ln (W<sub>J</sub>) |
Answer» C. ln (K<sub>bj</sub>L<sub>J</sub>/W<sub>J</sub>) | |
15. |
ln Sbj is represented as |
A. | ln (K<sub>bj</sub>L<sub>J</sub>/V<sub>J</sub>) |
B. | ln (K<sub>bj</sub>/V<sub>J</sub>) |
C. | ln( K<sub>bj</sub>L<sub>J</sub>) |
D. | ln (L<sub>J</sub>) |
Answer» B. ln (K<sub>bj</sub>/V<sub>J</sub>) | |
16. |
Boston and Sullivan gave the concept of the |
A. | False method |
B. | Binary method |
C. | MESH Method |
D. | Inside Out Method |
Answer» E. | |