The equivalent diameter for fluid flow through a channel of constant non-circular cross section of area 'A' is given by (where, P = perimeter of the channel in contact with the fluid)

The equivalent diameter for fluid flow through square cross section channel of side 'x', for pressure drop calculation purpose is given by

A

4x

B

2x

C

x

D

$$\sqrt {\text{x}} $$

The most economical channel section for the fluid flow is the one for which the discharge is maximum for a given cross-sectional area. Vertical velocity distribution in an open channel for laminar flow can be assumed to be

A

Parabolic

B

Hyperbolic

C

Straight line

D

None of these

For annular flow of a fluid, the ratio of the equivalent diameter for pressure drop calculation to the equivalent diameter for heat transfer calculation is

A

$$\frac{{{{\text{d}}_1}}}{{{{\text{d}}_1} + {{\text{d}}_2}}}$$

B

$$\frac{{{{\text{d}}_2}}}{{{{\text{d}}_1} + {{\text{d}}_2}}}$$

C

$$\frac{{{{\text{d}}_1} + {{\text{d}}_2}}}{{{{\text{d}}_1}}}$$

D

$$\frac{{{{\text{d}}_1} + {{\text{d}}_2}}}{{{{\text{d}}_2}}}$$

Pharmacists recently conducted a study with respect to the reasons their customers purchased eye drops to soothe eye dryness. Dry eyes were more frequently experienced by customers who wore contact lenses than by custorners who did not wear contact lenses. The pharmacists concluded that wearing contact lenses, by itself, can cause contact wearers to have dry eyes.

A

An inherited condition can cause both weak eyesight and dry eyes

B

Physical exertion causes dry eyes in many people who wear contact lenses.

C

Most people who have dry eyes do not wear contact lenses.

D

Most people who wear contact lenses do not have dry eyes.

The ratio of the hydraulic radius to the diameter of the channel, for maximum mean velocity of flow in a circular channel, in open channel flow is

A

0.3

B

0.9

C

0.03

D

0.66

The ratio of the depth of flow to the diameter of the channel for maximum discharge in a circular channel in open channel flow is

A

0.1

B

0.55

C

0.95

D

1.85

Two circular dials of exactly the same size are mounted on a wall side by side in such a way that their perimeters touch at one point. Dial 1, which is on the left , spins clockwise around its center , and dial 2 , which is on the right spins counter clockwise around its center . (Assume that there is no friction at the point of contact between the dials.) Each dial is marked on its perimeter at three points that are at equale distances around the points arked ondial 1 are N, O, and P and the points marked on dial 2 are X, Y, and Z. If points O and Z are just meeting at the point of contact between the dials , and if dial 1 spins at the same speed as dial 2, what is the smallest number of revolutions of each dial that will bring O and Z together again?

A

1

B

2

C

3

D

4

The area of a rectangle R with width 4 feet is equal to the area of a square S, which has a perimeter of 24 feet. The perimeter of the rectangle R, in feet, is :

A

9

B

16

C

24

D

26

The area of a rectangle R with width 4 feet is equal to the area of a square S which has a perimeter of 24 feet. The perimeter of the rectangle R, in feet is :

A

9

B

16

C

24

D

26

If the area of rectangle R with length 4 feet is equal to the area of square S, which has a perimeter of 24 feet , then the perimeter of rectangle R equals -

A

9 Feet

B

16 Feet

C

24 Feet

D

26 Feet

The equivalent diameter for fluid flow through a channel of constant non-circular cross section of area 'A' is given by (where, P = perimeter of the channel in contact with the fluid)

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