MCQOPTIONS

Explore topic-wise MCQs in Distillation Design.

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‚Äö√Ñ√∂‚àö√ë‚àö¬•(P<sub>n+1</sub>)
B.	f‚Äö√Ñ√∂‚àö√ë‚àö¬•(P)
C.	f‚Äö√Ñ√∂‚àö√ë‚àö¬•(P<sub>n-1</sub>)
D.	P
Answer» D. P

6.	Given_‚âà√≠¬¨¬µ>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‚Äö√Ñ√∂‚àö√ë‚àö¬•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 ‚âà√≠‚Äö√†√®j 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> ‚Äö√Ñ√∂‚àö√ë¬¨‚àÇ. E<sub>N-1</sub> E<sub>N</sub>, V<sub>1</sub> V<sub>2</sub> K<sub>3</sub> ‚Äö√Ñ√∂‚àö√ë¬¨‚àÇ. V<sub>N-1</sub> V<sub>N</sub>].
B.	[T<sub>1</sub> T<sub>2</sub> T<sub>3</sub> ‚Äö√Ñ√∂‚àö√ë¬¨‚àÇ. T<sub>N-1</sub> T<sub>N</sub>, V<sub>1</sub> V<sub>2</sub> V<sub>3</sub> ‚Äö√Ñ√∂‚àö√ë¬¨‚àÇ. V<sub>N-1</sub> V<sub>N</sub>].
C.	[S<sub>1</sub> S<sub>2</sub> S<sub>3</sub> ‚Äö√Ñ√∂‚àö√ë¬¨‚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√ë¬¨‚àÇ. E<sub>N-1</sub> E<sub>N</sub>].
D.	[K<sub></sub>1 K<sub>2</sub> K<sub>3</sub> ‚Äö√Ñ√∂‚àö√ë¬¨‚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√ë¬¨‚àÇ. 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√ë¬¨‚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√ë¬¨‚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> ‚Äö√Ñ√∂‚àö√ë¬¨‚àÇ. E<sub>N-1</sub> E<sub>N</sub>, K<sub>1</sub> K<sub>2</sub> K<sub>3</sub> ‚Äö√Ñ√∂‚àö√ë¬¨‚àÇ. K<sub>N-1</sub> K<sub>N</sub>].
B.	[T<sub>1</sub> T<sub>2</sub> T<sub>3</sub> ‚Äö√Ñ√∂‚àö√ë¬¨‚àÇ. T<sub>N-1</sub> T<sub>N</sub>, K<sub>1</sub> K<sub>2</sub> K<sub>3</sub> ‚Äö√Ñ√∂‚àö√ë¬¨‚àÇ. K<sub>N-1</sub> K<sub>N</sub>].
C.	[S<sub>1</sub> S<sub>2</sub> S<sub>3</sub> ‚Äö√Ñ√∂‚àö√ë¬¨‚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√ë¬¨‚àÇ. E<sub>N-1</sub> E<sub>N</sub>].
D.	[K<sub>1</sub> K<sub>2</sub> K<sub>3</sub> ‚Äö√Ñ√∂‚àö√ë¬¨‚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√ë¬¨‚àÇ. 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√ë¬¨‚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√ë¬¨‚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