This

**preview**shows pages 1-3. to view the full**11 pages of the document.** 1

Solutions to CE2200 Exam #1

1. Choose the correct answer from the following options (20 pt: 4 points each):

(1) At boundaries, fluid particles adhere to the walls, and so their velocities are

zero relative to the wall. This characteristic is referred to as

(A) surface tension; (B) bulk modulus of elasticity; (C) no-slip condition.

(2) How many meters of water are equivalent to 760 mm Hg (in terms of

pressure)? SHg = 13.6.

(A) 13.60; (B) 10.34; (C) 0.76.

wh = Hg 0.76 h = 0.76(Hg/w) = 0.7613.6 (m)

(3) The center of pressure is always the centroid.

(A) below; (B) above; (C) coincident with.

(4) The dimension of force F is .

(A) ML/T2; (B) lbf; (C) N.

(5) Which of the following is the formula for the gage pressure within a very

small spherical droplet of water?

(A) p = /(2r); (B) p = 2/r; (C) p = 4/r.

2r = pr2 p = 2/r

Only pages 1-3 are available for preview. Some parts have been intentionally blurred.

2

2. (20 pt): Students are given a simple device called a rotational cylindrical viscometer,

as shown in Fig. 1, and asked to determine the viscosity of an unknown liquid. The

outer cylinder is fixed while the inner cylinder is rotating at a constant angular speed

of by applying a torque T. (A) Derive an equation for the viscosity in terms of

angular velocity, , torque, T, submerged inner cylinder height, L, inner cylinder

radius, Ri, and outer cylinder radius, Ro; (B) Calculate the viscosity value when =

55 rev/min, T = 0.9 N m, L = 0.3 m, Ri = 0.12 m and Ro = 0.13 m.

SOLUTION:

Step-1: Find the dynamic viscosity of the fluid: = ?

Step-2: Schematic as shown in the figure.

Velocity Profile of the Fluid between the Gap of the Cylinders

Step-3: Assumptions: (1) The liquid is a Newtonian fluid; (2) The gap between the

cylinders is small and the velocity profile is linear, i.e., du/dr u/r=U/(Ro - Ri); (3)

Neglect the end effects of the cylinders; and (4) the force (F) is perpendicular to its

moment arm (r), hence the torque is given by T = Fr.

Step-4: Equations

The torque is given by multiplying the force with its moment arm. That

is,

T = Fr

The force can be expressed in terms of the shear stress and area as

follows:

F = (2rL) = /r

Rearrange terms to yield

Only pages 1-3 are available for preview. Some parts have been intentionally blurred.

3

= / 2r2L

According to Newton's law of viscosity for Newtonian fluids,

= du/dr

The negative sign is inserted to indicate that u decreases as r increases. Note that the

velocity profile is assumed to be linear, as shown in the figure. The end conditions are

known and are given by: u = Ri at r = Ri and u = 0 at r = Ro.

Equating the above two equations and integrating from

r = Ri to Ro:

Step-5: Known parameters

= 55 rev/min, T = 0.9 N m, L = 0.3 m, Ri = 0.12m and Ro = 0.13 m.

Step-6: Calculations

Substitution of the known values into the derived equation for dynamic viscosity yields

the viscosity of the liquid

Step-7: Discussion

Based on the assumptions, the dynamic viscosity can also be approximated as:

2

3s/mN 480.0

2

)(

i

io

RL

RRT

The thinner the gap layer is, the more accurate the solution is.

###### You're Reading a Preview

Unlock to view full version