Solutons Manual

1.3 Classify the following substances (maintained at constant temperature):

(a)

du/dy, rad/s 0 3 4 6 5 4

τ,1b/ft2 2 4 6 8 6 4

(b)

du/dy, rad/s 0 0.5 1.1 1.8

τ,N/m2 0 2 4 6

(c)

du/dy, rad/s 0 0.3 0.6 0.9 1.2

τ,1b/ft2 0 2 4 6 8

Solution

(a) Non Newtonian.

Hysteresis suggests thixotropy.

(b) Nay Newtonian

(c) Newtonian (linear relationship between τ and )

The constant slope of the line gives the absolute viscosity μ：

1.4 A Newtonian liquid flows down an inclined plane in a thin sheet of thickness

t(Fig. 1.8). The upper surface is in contact with air, which offers almost no resistance

to the flow. Using Newton’s law of viscosity, decide what the value of du/dy, y

measured normal to the inclined plane, must be at the upper surface. Would a linear

variation of u with y be expected?

Solution

Since air offers no resistance to the flow, τ=0.

At free surface : τ=0

At the plane : τ≠0

Therefore, since we have change in (slope) the relationship

between μ and y is non-linear.

1.11 Does the weight of a 20 N bag of flour at sea level denote a

force or the mass of the flour? What is the mass of the flour in

kilograms? What are the mass and weight of the flour at a location

where the gravitational acceleration is one-seventh that of the

earth’s standard?

Solution

The weight always denotes force.

The mass is the same under any conditions.

1.12 A Newtonian fluid is in the clearance between a shaft and a concentric sleeve.

When a force of 600 N is applied to the sleeve parallel to the shaft, the sleeve attains a

speed of l m/s. If a 1500 N force is applied, what speed will the sleeve attain? The

temperature of the sleeve remains constant.

Solution

Therefore : ￠．u ； ￠=constant

￠． and ￠

and :

1.16 Determine the viscosity of fluid between the shaft and sleeve in Fig. 19.

Solution

A=Π．D．L

Therefore:

1.17 A flywheel weighing 600 N has a radius of gyration of 300 mm. When it is

rotating at 600 rpm, its speed reduces 1 rpm/min owing to fluid viscosity between the

sleeve and shaft. The sleeve length is 50 mm; shaft diameter is 20 mm; and radial

clearance is 0.05 mm. Determine the fluid viscosity.

Solution

I = mass moment of inertia

K = radius of gyration

T = torque

and :

I = mk2 ; ;

A = 2ΠrL ; ;

Therefore :

1.18 A 25-mm-diameter steel cylinder 300 mm long falls, because of its own weight,

at a uniform rate of 0.1 m/s inside a tube of slightly larger diameter. A castor-oil film

of constant thickness is between the cylinder and the tube. Determine the clearance

between the tube and the cylinder. The temperature is 38℃. The specific gravity of

steel = 7.85.

Solution

; A =Π．D．L

So :

Therefore :

so : t = 5.82 × 10-5m = 0.06mm

Note : for T = 38℃ from figure C.1 we find that : μ= 2.8 × 10-1 N．S/m2

1.20 A 12 kg cube slides down an inclined plane making an angle of 30° with the

horizontal. A fluid film 0.1 mm thick separates solid and surface. The

fluid viscosity is 0.04 N．s/m2. Assuming the velocity distribution in

the film is linear, find the terminal velocity of the block. The area of

the cube in contact with the film is 0.25 m2.

Solution

Assume that the cube has reached

its terminal velocity (u) and

that the temperature (T) is constant.

; ; F = W．sin 30°

Therefore :

so :

1.24 A body weighing 120 lb with a flat surface area of 2 ft2 slides down a lubricated

inclined plane making a 30° angle with the horizontal. For viscosity of

0.002 lb．s/ft2 and body speed of 3 ft/s, determine the lubricant film

thickness.

Solution

W = 120 lb

A = 2ft2

μ= 0.002 lb．s/ft2

u = 3 ft/s

; ; F = W sin 30°

Therefore :

so :

or t = 0.0024 in.

1.25 What is the viscosity of gasoline at 25℃ in poises? Use Fig. C.1,

Appendix C.

Solution

From figure C.1 : ◎T = 25℃ μ= 6 × 10-6 lb．s/ft2

so :

1.26 A liquid has a specific weight of 48 lb/ft3 and a dynamic viscosity of 3.05

lb．s/ft2. What is the kinematic viscosity?

Solution

;

;

so :

1.27 What is the specific volume in cubic feet per pound mass and cubic feet per slug

of a substance of specific gravity 0.75?

Solution

and

1.29 The density of a substance is 2.94 g/cm3. In SI units, what is its (a) specific

gravity, (b) specific volume, and (c) specific weight?

Solution

(a)

(b)

(c)

1.42 Three wastewater samples were taken from the same site at the same time, and

after analyzing them in the laboratory the following results were obtained

Sample Volume (mL) Mass suspended solids (g)

1 75 23.0

2 83.2 35.6

3 80 Glass container broken

The three samples are of the same density (ρ), and the first one contains

solids of specific gravity 1.93. Find the density ρ and the

concentration of the suspended solids in the three samples. Take the

mass fraction of solids in Sample 3 to be the average of the first

two.

Solution

Sample 1 :

Now

Therefore :

All three samples have the same density ρ.

Now :

Therefore :

and :

Finally :

1.47 A gas with molecular weight 28 has a volume of 4.0 ft3 and a

pressure and temperature of 2000 psfa (lb/ft2 abs) and 600°R,

respectively. What are its specific volume and specific weight?

Solution

1.53 An open tank contains water at 22℃ with 32.7-percent (by

weight) suspended solids of specific gravity 2.32. If the volume of

the mixture is 1.2 m3, calculate the concentration of the solids in

1bm/ft3 and kg/m3.

Solution

From equation (1.5.8) for T=22℃ we find that :

Therefore:

(mixture)

(solids)

Therefore :

Now :

and :

Or

so:

1.58 A steel container expands in volume 1-percent when the

pressure within it is increased by 10,000 psi. At standard pressure,

14.7 psi absolute, it holds 0.5 slug of water; ρ= 1.94 slug/ft3. For K

= 300,000 psi, when it is filled, how many slugs of water must be

added to increase the pressure to 10,000 psi ? What is the weight of

the water added?

Solution

Now :

and

Therefore 0.0233 slug of water must be added.

△W = g．△m = (32.174) (0.0233) → △W = 0.75 lb

1.62 What is the isothermal bulk modulus for air at 0.4 MPa abs?

Solution

so :

Therefore : K = P = 0.4MPa abs. The relation K = P holds for any

ideal gas, at any pressure.

1.66 A small circular jet of mercury 0.1 mm in diameter issues from

an opening. What is the pressure difference between the inside and

outside of the jet when at 20℃ ?

Solution

at T = 20℃ from table

we find : σ= 0.51 N/m

Now : Fσ = 2σL (1)

FP = P．D．L (2)

H should be FP = Fσ

So : (3)

Therefore :

Or P = 10.2 KPa

1.69 Using the data given in Fig. 1.6, estimate the capillary rise of

tap water between two parallel glass plates 0.20 in. apart.

Solution

For the two parallel plates (figure) :

(1)

and P = H (2)

From (1) and (2) : (3)

For a vertical tube :

or : (4)

where D is the tube diameter. From (3) and (4) we find :

(5)

Now letting a=θ and D=W, equation (5)

gives :

From figure 1.6, for we find that h = 0.09in (T = 68℉)

So :

1.70 A method of determining the surface tension of a liquid is to find the force

needed to pull a platinum wire ring from the surface (Fig. 1.10). estimate the force

necessary to remove a 20-mm-diameter ring from the surface of water at 20℃.

Solution

Let D be the ring diameter.

The wire diameter is not important.

Assume that the surface is vertical at the

breaking point A.

From table C.1 :

1.71 Calculate the capillary rise h in the tube of Fig. 1.11 in terms of

θ,σ,γ, and r.

Solution

If is the

Component of surface tension down,

then a force Balance gives :

But Δp = γ．h

So :

1.73 What vertical force due to surface tension would be required to

hold the tube of Fig. 1.11? Consider the tube wall thickness to be

very small.

Solution

0

The total force required

to hold the tube is :

Therefore :

1.75 Develop a formula for capillary rise h between two concentric

glass tubes of radii R and r and contact angle θ.

Soution

The pressure force is :

Where :

Therefore : (1)

The total force due to surface tension down is :

(2)

Neglecting the weight W of the fluid between the two tubes, a force

balance gives:

or :

2.3 What is the pressure at a point 10 m below the free surface in a fluid that has a

variable density in kilograms per cubic meter given by σ= 450+ah, in which

a=12 kg/m4 and h is the distance in meters measured from te free

surface ?

Solution

(1)

From (1) by integration we obtain:

(2)

Substitute the values for g, a, b and h, and find :

2.10 Express a pressure of 50 kPa in (a) millimeters of mercury, (b)

meters of water, (c) meters of acetylene tetrabromide, S = 2.94.

Solution

,

(a) S=13.57

So :

(b)

So :

(c)

P = 50kPa = 1.734m acetylene tetrabromide

2.11 A Bourdon gage reads 2 psi suction, and the barometer is 29.5

in. Hg, Express the pressure in six other customary ways.

Solution

For absolute pressures :

2.12 Express 4 atm in meters of water gage with a barometer

reading of 750mm Hg.

Solution

So : Pabs = 135.63ft = 41.34m

2.15 The container of Fig. 2.41 holds water and air as shown. What

is the pressure at A, B, C, and D in pounds per square foot and in

pascals.

2.23 In Fig.2.11a S1 = 1.0, S2 = 0.95, S3 = 1.0, h1 = h2 = 280 mm,

and h3 = 1 m.

Compute PA – PB in millimeters of water.

Solution

Therefore :

2.24 In Prob. 2.23 find the gage difference h2 for pA – pB = -350 mm

H2O.

Solution

From problem 2.23

Then

The manometer equation is :

Thus :

2.30 In Fig. 2.12 determine R, the gage difference, for a difference in

gas pressure of 9 mm

Solution

From equation 2.4.1

Since

(that is, )

We find that :

Therefore :

2.31 The inclined manometer of Fig. 2.13 reads zero when A and B

are the same pressure. The reservoir diameter is 2.0 in. and that of

the inclined tube is 1/4 in. For θ=30° and gage fluid S = 0.832, find

pA – pB in pounds per square inch as a function of gage reading R in

feet.

Solution

Combination of the above equations yields:

Where :

Therefore :

2.35 The container shown in Fig. 2.48 has a circular cross section.

Determine the upward force on the surface of the cone frustum

ABCD. What is the downward force on the plane EF ? Is this force

equal to the weight of the fluid? Explain.

Solution

The force up on the come frustum ABCD is:

The weight of the water in the container is :

2.38 A vertical right-angled triangular surface has a vertex in the

free surface of a liquid (Fig. 2.50). Find the force on one side (a) by

integration and (b) by formula.

Solution

(a)

Therefore :

(b)

2.39 Determine the magnitude of the force acting on one side of the

vertical triangle ABC of Fig. 2.51 (a) by integration and (b) by

formula.

(a) The low of cosines gives:

and

From similar triangles: , so :

and :

(b)

2.40 Find the moment about AB of the force acting on one side of

the vertical surface ABC of Fig. 2.50. .

Solution

Now from Appendix A :

∴

For figure 2 then :

Then for our problem (figure 3) we have :

Finally :

Where MAB in N．m and b, h in m

(y axes in Figures 2 and 3 point to opposite directions)

2.42 Find the moment about AB of the force acting on one side of

the vertical surface ABC of Fig. 2.51.

Solution

From equation (2.5.10)

The law of cosines give :

Then :

Finally :

and :

2.43 Locate a horizontal line below AB of Fig. 2.51 such that the

magnitude of pressure force on the vertical surface ABC is equal

above and below the line.

Solution

Let F1 be the force on the area between AB and EF, and F2 on the

area below EF. We need to find y* such that F1 = F2.

Using the results from problem 2.34 we have :

H = 2.4 ft; (1)

From equation (1) :

Therefore for F1 = F2 we find that :

Solving the latest equation by trial error we find :

or that EF is 0.774 ft below AB

2.48 Locate the distance of the pressure center below the liquid

surface in the triangular area ABC of Fig. 2.51 (a) by integration and

(b) by formula.

Solution

The law of cosines gives :

So : H = 4 sin B = 2.4 ft

(a) From similar triangles

Now

(b)

Therefore :

2.49 By integration locate the pressure center horizontally in the

triangular area of ABC of Fig. 2.51.

Solution

From problem 2.46 ; H = 2.4 ft ; B = 36.87°

Now

Take moments about CD :

Then :

From problem 2.39 : F = 1914 lb, so :

2.51 Determine by integration the pressure center for Fig. 2.50.

Solution

From similar triangles :

and

and , so :

and , so :

2.52 Locate the pressure enter for the annular area of Fig. 2.54.

Solution

Therefore :

Finally :

or :

2.53 Locate the pressure center for the gate of Fig. 2.55.

Solution

where : b = 4ft ; h = 6ft

2.56 Locate the pressure center for the vertical area of Fig. 2.56.

Solution

Therefore :

2.59 Determine the pivot location y of the rectangular gate of Fig.

2.59 so that it will open when the liquid surface is as shown.

Solution

want y at pressure center

(critical location) measured

from the free surface.

= 1.5556 m

Therefore : y = 2-1.556 → y = 0.444 m

2.66 The gate of Fig. 2.62 weighs 300 lb/ft normal to the paper. Its

center of gravity is 1.5 ft from the left face and 2.0 ft above the

lower face. It is hinged at O. Determine the water-surface position

for the gate just to start to come up. (The water surface is below the

hinge.)

Solution

W = 300 lb/ft

and (for equilibrium)

Therefore :

Which yields : h = 2.116 ft

2.70 For linear stress variation over the base of the dam of Fig. 2.64

(a) locate where the resultant crosses the base and (b) compute the

maximum and minimum compressive stresses at the base. Neglect

hydrostatic uplift.

Solution

(a)

Take moments about A :

Which yields : x = 11.588m

(b) Taking moments about A :

which gives :

Also :

Therefore :

2.71 Solve Prob. 2.70 with the addition that the hydrostatic uplift

varies linearly from 20 m at A to zero at the toe of the dam.

Solution

Refere to problem 270

(a)

(b)

The above equation yields :

Also :

Therefore :

2.74 To what height h will the water on the right have to rise to open

the gate shown in Fig. 2.67? The gate is 5 ft wide and is constructed

of material with specific gravity 2.5. Use the pressure-prism method.

Solution

The line of action of F2 is at

For equilibrium it should be :

So :

Therefore : h1 = 0.615 ft and h2 = 4.406 ft

If then the gate will open to the right

then the gate will open to the left

For we choose h = 4.406 ft

2.78 (a) Determine the horizontal component of force acting on the

radial gate (Fig. 2.70) and its line of action. (b) Determine the

vertical component of force and its line of action. (c) What force F is

required to open the gate, neglecting its weight? (d) What is the

moment about an axis normal to the paper and through point O?

Solution

(a)

Therefore :

And :

(b)

(d)Since the pressure force is normal to the surface, the resultant of

FH and FV has to pass through O. Taking moments about O, we

see that F = O (For equilibrium ).

check：

(d) Answered in question (c)

Mo = 0

2.80 What is the force on the surface whose trace is OA of Fig. 2.53?

The length normal to the paper is 3 m and

Solution

Where

The resultant fore is :

Or : F = 28814 N = 28.814 KN

2.84 Calculate the force F required to hold the gate of Fig. 2.74 in a

closed position when R = 2 ft.

Solution

Taking moments about A

we see that Fx = 0

Hence F = horizontal

Component of liquid force

2.85 Calculate the force F required to open or hold closed the gate

of Fig. 2.74 when R = 1.5 ft.

Solution

From problem 2.84 :

Therefore :

2.89 A log holds back water and oil as shown in Fig. 2.76. Determine

(a) the force per meter pushing it against the dam, (b) the weight of

the cylinder per meter of length, and (c) its specific gravity.

Solution

(a) First observe that the horizontal forces

due to water cancel out. Therefore,

the horizontal force, Fx, is due to oil only.

Consider unit length perpendicular to paper.

(b) The vertical force up, Fy is :

on lower half due to water + oil head added

on upper tourth (left) due to oil

2

2

22

2(9806) (0.8 9806)(2 2 )

2

(2) (9806 0.8) (2)

4

117637.28yF N

(c)

Therefore :

2.91 The hemispherical dome in Fig. 2.78 is filled with water. The

dome arrangement as shown weighs 28 kN and is fastened to the

floor by bolts equally spaced around the circum-ference at the base.

Find the total force required to hold down the dome.

Solution

Let Fv be the vertical force (due to water) on the dome.

So : Fv = 242.47 kN (up)

The net force pulling up is :

Therefore the total force required to hold down the dome is : 214.47

kN

2.92 A cable and semicircular ring suspend a spherical container by

the small piezometer tube, Fig. 2.79. The top of the tube is open to

the atmosphere. Calculate (a) the force on the bottom half of the

sphere, (b) the force on the top half of the sphere, and (c) the total

tension in the cable. Neglect the weight of the container.

Solution

(a)

(b)

(assume that the diameter of the tube is very small)

(c)

So : T = 17.3 kN

2.99 The cylinder gate of Fig. 2.80 is made from a circular cylinder

and a plate hinged at the dam. The gate position is controlled by

pumping water into or out of the cylinder. The center of gravity of

the empty gate is on the line of symmetry 4 ft from the hinge. It is

at equilibrium when it empty in the position shown. How many cubic

feet of water must be added per foot of cylinder to hold the gate in

its position when the water surface is raised 3 ft?

Solution

The shaded area has volume, :

(per foot of length)

and :

Consider the moments about H “added” when the liquid level in the

reservoir is raised 3ft : +7

M (water in cylinder) =

M (add, head on projection AA’) =

M (shaded area, vertical)

M (shaded area, horizontal)

Now :

2.100 A sphere 250 mm in diameter, S = 1.4, is submerged in a

liquid having a density varying with the depth y below the surface

given by ρ = 1000+ 0.03y kg/m3. Determine the equilibrium position

of the sphere in the liquid.

Solution

In general, any homogeneous body completely submerged in a fluid

with a linear density gradient will float such that the fluid density at

the depth of the centroid is equal to the body density. Let the

subscripts B and L indicate body and liquid. Given that ;

and β = 0.03 we have,

or : . Let

. Centroid is such that : , hence

so : and it is independent of the body shape.

Complete submergence must be Confirmed before using this

approach. Then :

2.102 A cube 2 ft on an edge has its lower half of S = 1.4 and upper

half of S = 0.6. It is submerged into a two-layered fluid, the lower S

= 1.2 and the upper S = 0.9. determine the height of the top of the

cube above the interface.

Solution

drop and solve for x :

2.103 Determine the density, specific volume, and volume of an

object that weights 3 N in water and 4 N in oil, S = 0.83.

Solution

Subscripts O, A, W refer to oil, air and water respectively.

WA = actual weight (in air) ; ;

So :

2.104 Two cubes of the same size, 1 m3, one of S = 0.80 and the

other of S = 1.1, are connected by a short wire and placed in water.

What portion of the lighter cube is above the water surface, and

what is the tension in the wire?

Soution

From static equilibrium, the weight of the cubes (w) must equal the

buoyant force FB

, therefore

Since , then

So :

The tension in wire is :

2.113 A wooden cylinder 600 mm in diameter, S = 0.50, has a

concrete cylinder 600 mm long of the same diameter, S = 2.50,

attached to one end. Determine the length of the wooden cylinder

for the system to float in stable equilibrium with axis vertical.

Solution

To float it should be :

So :

Taking moments :

Weigth

hence :

Then :

Set to obtain the max. value of L :

Therefore : L = 4769.85 mm (maximum)

2.120 In Fig. 2.86, and ay = 0. Find the imaginary free

liquid surface and the pressure at B, C, D, and E.

Solution

The plane of zero pressure has slope :

2.121 In Fig. 2.86, ax = 0 and ay = -8.05 ft/s2. Find the pressure at B,

C, D, and E.

Solution

P = -37.43y

Where p(psf) and y(ft)

PB = -37.43 × (-1) = 37.43 psf = 0.26 psi

PC = -37.43 × (-1) = 37.43 psf = 0.26 psi

PD = -37.43 × (2) = -74.86 psf = -0.52 psi

PE = -37.43 × (2) = -74.86 psf = -0.52 psi

2.123 In Fig. 2.87, ax = 9.806 m/s2 and ay = 0. Find the pressure at

A, B, and C

Solution

ax = 9.806 m/s2 = g

ay = 0

Consider unit length normal to paper

(or Pc)

2.124 In Fig. 2.87, . Find the pressure

at A, B, and C.

Solution

So y = 0.1375 m

PA = 0

PB = (1.3-0.1375) × 19612 = 22,798.95 Pa

2.127 The tube of Fig. 2.88 is filled with liquid, S = 2.40. When it is

accelerated to the right 8.05 ft/s2, draw the imaginary free surface

and determine the pressure at A. For PA = 8 psi vacuum determine

ax.

Solution

a)

b)

Given PA = 8 psi vacuum =

and

2.133 The U tube of Fig. 2.88 is rotated about a vertical axis 6 in. to

the right of A at such a speed that the pressure at A is zero gage.

What is the rotational speed?

Solution

r1 = 0.5 ft

r2 = 2.5 ft

h = 1 ft

Method is the same as for problem 2.132

rad/sec

or

w = 31.27 rpm.

2.134 Locate the vertical axis of rotation and the speed of rotation

of the U tube of Fig. 2.88 so that the pressure of liquid at the

midpoint of the U tube and at A are both zero.

Solution

Due to symmetry dist.

from A to axis equals dist.

from mid-point to axis

thus r1 = 1.5 ft and r2 = 0.5 ft

Now equate PB and PO , Pc and Po

hence

and

or w = 54.16 rpm

2.144 The U tbue of Fig. 2.88 is rotated about a vertical axis through

A at such a speed that the water in the tube begins to vaporize at

the closed end above A, which is at 70℉. What is the angular

velocity? What would happen if the angular velocity were increased?

Solution

P = 14.7 psi = 2116.8 psf

then :

Thus w = 23.09 rad/sec = 220.5 rpm

If w increases then a vapor pocket will form.