MCQOPTIONS
Saved Bookmarks
MCQOPTIONS
This section includes 14 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. |
Given ε>0. Thena) f(a)f(b)>0b) f(a)f(b)=0c) f(a)f(b)=1d) f( |
| A. | f(a)f(b)>0b) f(a)f(b)=0c) f(a)f(b)=1d) f(a)f( |
| B. | >0b) f(a)f(b)=0 |
| C. | f(a)f(b)=1 |
| D. | f(a)f(b)<0 |
| Answer» E. | |
| 2. |
Newton’s method is a fixed point iteration method, As Pna) g(Pn-1)b) (Pn-1)c) gd) Pn 9) Newton’s method cannot continue, if for some Pn-1 |
| A. | g(Pn-1)b) (Pn-1)c) gd) Pn 9) Newton’s method cannot continue, if for some Pn-1a) f’(Pn+1) |
| B. | (Pn-1)c) gd) Pn 9) Newton’s method cannot continue, if for some Pn-1a) f’(Pn+1)b) f’(P) |
| C. | gd) Pn 9) Newton’s method cannot continue, if for some Pn-1a) f’(Pn+1)b) f’(P)c) f’(Pn-1) |
| D. | Pn 9) Newton’s method cannot continue, if for some Pn-1a) f’(Pn+1)b) f’(P)c) f’(Pn-1)d) PView Answer |
| Answer» B. (Pn-1)c) gd) Pn 9) Newton’s method cannot continue, if for some Pn-1a) f’(Pn+1)b) f’(P) | |
| 3. |
The θj factor is a |
| A. | Subtracted |
| B. | Additive |
| C. | Multiplier |
| D. | Divider |
| Answer» D. Divider | |
| 4. |
XIJ and Yij are the component balances of |
| A. | Equilibrium relation |
| B. | Constant relation |
| C. | Final relation |
| D. | Direct relation |
| Answer» B. Constant relation | |
| 5. |
NEWTON‚ÄÖ√Ñ√∂‚ÀÖ√Ë‚ÀÖ¬•S_METHOD_CANNOT_CONTINUE,_IF_FOR_SOME_PN-1?$# |
| A. | f’(P<sub>n+1</sub>) |
| B. | f’(P) |
| C. | f’(P<sub>n-1</sub>) |
| D. | P |
| Answer» D. P | |
| 6. |
Given_ε>0._Then$# |
| A. | f(a)f(b)>0 |
| B. | f(a)f(b)=0 |
| C. | f(a)f(b)=1 |
| D. | f(a)f(b)<0 |
| Answer» E. | |
| 7. |
Newton’s method is a fixed point iteration method, As Pn |
| A. | g(P<sub>n-1</sub>) |
| B. | (P<sub>n-1</sub>) |
| C. | g |
| D. | P<sub>n</sub> |
| Answer» B. (P<sub>n-1</sub>) | |
| 8. |
The θj factor is a$ |
| A. | Subtracted |
| B. | Additive |
| C. | Multiplier |
| D. | Divider |
| Answer» D. Divider | |
| 9. |
The initial assumption to Tomich method is |
| A. | Set initial Temperature |
| B. | Set total stages |
| C. | Final Pressure |
| D. | Initial Pressure |
| Answer» B. Set total stages | |
| 10. |
The stages flow rate and the vapor flow rates are expressed, as |
| A. | [E<sub>1</sub> E<sub>2</sub> E<sub>3</sub> …. E<sub>N-1</sub> E<sub>N</sub>, V<sub>1</sub> V<sub>2</sub> K<sub>3</sub> …. V<sub>N-1</sub> V<sub>N</sub>]. |
| B. | [T<sub>1</sub> T<sub>2</sub> T<sub>3</sub> …. T<sub>N-1</sub> T<sub>N</sub>, V<sub>1</sub> V<sub>2</sub> V<sub>3</sub> …. V<sub>N-1</sub> V<sub>N</sub>]. |
| C. | [S<sub>1</sub> S<sub>2</sub> S<sub>3</sub> …. S<sub>N-1</sub> S<sub>N</sub>, E<sub>1</sub> E<sub>2</sub> E<sub>3</sub> …. E<sub>N-1</sub> E<sub>N</sub>]. |
| D. | [K<sub></sub>1 K<sub>2</sub> K<sub>3</sub> …. K<sub>N-1</sub> K<sub>N</sub>, T<sub>1</sub> T<sub>2</sub> T<sub>3</sub> …. T<sub>N-1</sub> T<sub>N</sub>]. |
| Answer» C. [S<sub>1</sub> S<sub>2</sub> S<sub>3</sub> ‚Äö√Ñ√∂‚àö√묨‚àÇ. S<sub>N-1</sub> S<sub>N</sub>, E<sub>1</sub> E<sub>2</sub> E<sub>3</sub> ‚Äö√Ñ√∂‚àö√묨‚àÇ. E<sub>N-1</sub> E<sub>N</sub>]. | |
| 11. |
The independent variables of independent functions are used as, Fi = |
| A. | [E<sub>1</sub> E<sub>2</sub> E<sub>3</sub> …. E<sub>N-1</sub> E<sub>N</sub>, K<sub>1</sub> K<sub>2</sub> K<sub>3</sub> …. K<sub>N-1</sub> K<sub>N</sub>]. |
| B. | [T<sub>1</sub> T<sub>2</sub> T<sub>3</sub> …. T<sub>N-1</sub> T<sub>N</sub>, K<sub>1</sub> K<sub>2</sub> K<sub>3</sub> …. K<sub>N-1</sub> K<sub>N</sub>]. |
| C. | [S<sub>1</sub> S<sub>2</sub> S<sub>3</sub> …. S<sub>N-1</sub> S<sub>N</sub>, E<sub>1</sub> E<sub>2</sub> E<sub>3</sub> …. E<sub>N-1</sub> E<sub>N</sub>]. |
| D. | [K<sub>1</sub> K<sub>2</sub> K<sub>3</sub> …. K<sub>N-1</sub> K<sub>N</sub>, T<sub>1</sub> T<sub>2</sub> T<sub>3</sub> …. T<sub>N-1</sub> T<sub>N</sub>]. |
| Answer» D. [K<sub>1</sub> K<sub>2</sub> K<sub>3</sub> ‚Äö√Ñ√∂‚àö√묨‚àÇ. K<sub>N-1</sub> K<sub>N</sub>, T<sub>1</sub> T<sub>2</sub> T<sub>3</sub> ‚Äö√Ñ√∂‚àö√묨‚àÇ. T<sub>N-1</sub> T<sub>N</sub>]. | |
| 12. |
Which method uses the summation of the equations for vapor and liquid components? |
| A. | NR Method |
| B. | RF Method |
| C. | Tomich Method |
| D. | 2N Method |
| Answer» E. | |
| 13. |
XIJ and Yij are the component balances of |
| A. | Equilibrium relation |
| B. | Constant relation |
| C. | Final relation |
| D. | Direct relation |
| Answer» B. Constant relation | |
| 14. |
The 2N Newton method is based on algorithm of? |
| A. | N X N |
| B. | 2 X N |
| C. | N X M |
| D. | 2 X M |
| Answer» C. N X M | |