MCQOPTIONS
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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. | |