EXAMPLES OF WRITTEN TEST 1 PROBLEMS

Properties of fluids

1. A 1 m2

metal plate glides on top of an 1 mm thick oil layer with constant velocity along a sloping plane. The

weight of the metal plate 100 N and the dynamic viscosity of the oil is 0.1 Pas. Determine the velocity of the

metal plate if the sloping angle of the plane is 5o. Disregard all end effects around the plate. Assume laminar

flow and that the velocity profile in the oil is linear.

2. A lubricant between a piston and cylinder according to below figure has a kinematic viscosity of 2.8∙10-5 m2/s

and a relative density of 0.92. If the piston has a mean velocity of 6 m/s what approximate power loss will

develop due to friction. The piston has a diameter of 150 mm and length 300 mm. The cylinder has a diameter

of 150.2 mm.

3. A block weighing 1 kN, has a bottom area of 200 mm by 200 mm, and is sliding down a slope on top of an

oil film with thickness 0.005 mm according to below figure. Assuming linear velocity profile in the oil and no

edge effects, what is the final velocity of the block? The dynamic viscosity of the oil is 710-3 kg /(sm).

150 mm 150.2 mm

300 mm

Lubricant

300

0.005mm

1kN

4. In the figure below a tool for determining the surface tension of a liquid is shown. If a force, F, is needed to

lift the thin ring with diameter, D, what is the surface tension, , expressed in terms of F and D?

5. Derive an expression for capillary rise h of a liquid between two pipes with radius ro and ri, respectively, and

contact angle θ according to below.

ro Liquid

Air

h

ri

θ

Seen from above

Seen from the side

Liquid

Hydrostatics

6. The tube in below figure is filled with oil of relative density 0.9. Calculate pressure head in meters of water at

M and N.

7. What is the water level difference between the two open water containers in the below figure?

8. A manometer is connected to a water tank according to the below figure. Determine the water depth, H, in the

tank.

Oil

Oil

r.d = 0.9

M

N Air

3m

1m Patm , relative pressure = 0

9. The half sphere container in the figure is filled with water and fixed to the floor by two bolts (one on each

side of the sphere). How large is the force in each bolt that is required to hold the sphere down if its weight is 25

kN?

10. A circular plate with a diameter of 4 m has a central circular hole of diameter 1.5 m and is submerged in

water at angle according to below figure. Determine the resultant force from water pressure on one side of the

plate and the vertical location of the resultant force from water surface (tips second moment area for a circular

area with a centric hole is IG = (R4-r4)/4, where R = radius of circular area and r = radius of hole).

1

m

3

m

1.5m

θ

Water

4

m

Water surface

11. A 1 m long pole is fixed to a circular gate. A floating cylinder is connected to the pole´s other end through a

chain (see figure below). The cylinder is 25 cm in diameter and has weight of 200 N. The liquid is water. The

weight of the pole and chain can be disregarded. Determine the length of the chain so that the gate just starts to

open when the water depth above the pivot is 10 m.

12. The linear pressure distribution over the base of a concrete (wconcrete = 24 kN/m3

) dam creates a lifting force.

Determine the maximum water depth, H, on the left-hand side in order for the dam not to fall over (w=wH2O = 10

kN/m3

; w=γ in below figure).

13. The rigid gate OAB in below figure is hinged at O and rests against a rigid support at B. What minimum

horizontal force P is required to hold the gate closed if its width is 5 m? Neglect the weight of the gate and

friction in the hinge. To the right and under the gate is atmosphere.

14. Determine the force P needed to keep the gate in its position according to the figure below. The quarter

cylindrical gate has width (in to the paper) of 5 m.

Wate

r

P

4 m

3 m

2 m

Hing

e

A B

O

Atmosphere

15. What force P is necessary to keep the 5 m wide (in to the paper) quarter cylinder gate below in its closed

position.

Basic equations

16. A flow field is periodic in such a way that the streamline pattern is repeated at fixed intervals. During the

first second the fluid is moving upwards to the right at a 45o angle and during the next second the fluid is

moving downwards to the right at a 45o angle etc according to Fig. a). The flow velocity is constant = 10 m/s.

After 2.5 s the particle track for a particle that is released at point A at time zero is shown in Fig. b). If colour is

injected continuously at point A from time 0 how will the resulting streakline look like after 2.5 s?

Water Air

Hinge

6 m

Concrete floor

Radius = 3 m

P

Air

17. The system below is at equilibrium at a temperature of 20oC. If the pressure at A is equal to 100 kPa,

determine the pressure at B.

18. Determine the flow in the pipe according to the below figure assuming that the pitot tube measures the mean

velocity.

19. Water flows in the horizontal pipe below. If cavitation is observed at the constriction B, what is the flow?

Disregard all energy losses. Pvapor = 2 kPa, Patmosphere = 100 kPa.

6 m

3 m

4 m

Water

Air Air

Air

A

Pipe

Water

Q

6 m

Hg; relative density

13.55

3 m

Water

B

10 m

Q

20. The glass tube in the below figure is used to measure the pressure p1 in the water tank. The tube´s diameter

is 1 mm and the water has a temperature of 20oC. After correction for surface tension effects, what is the real

piezometric height in the glass tube? The contact angle between glass and water is 0o.

21. The space between two quadratic plates is filled by oil. The length of each plate´s side is 720 mm. The

thickness of the oil film is 15 mm. The upper plate that is moving with velocity of 3 m/s needs a pulling force of

120 N to keep a constant velocity. Determine 1) the dynamic viscosity for the oil and 2) the kinematic viscosity

for the oil if the relative density for the oil is 0.95.

22. Determine relative pressure Pa and Pb and absolute pressure at Q according to below figure. The barometer

air pressure is 720 mm Hg. Disregard density of air.

Air

Air Air

Water

Oil

SOLUTIONS

1. The component of the plate´s weight along the plane, F = 100sin5. Put the plate velocity to V, oil film

thickness to h, area of plate to A. Use Newton’s viscosity law.

Answer: The velocity of the plate is 8.7 cm/s.

2. ρ = r.d. ∙ 1000 = 920 kg/m3; μ = υ∙ρ = 2.8∙10-5∙920 = 0.0258 Pa∙s; τ = μ(V/h) = 0.028∙(6/0.0001) = 1548 Pa.

The friction force F on the piston is F = τ∙A= 1548∙π∙0.15∙0.3 = 219 N. The power loss P due to friction is P =

F∙V = 219∙6 = Answer: 1.313 kW.

3. Assume a linear velocity profile in the oil and vT the velocity:

Friction forces will balance gravity forces and 1000*sin300=56.0vT, therefore, Answer: vT = 8.93 m/s.

4. A force balance gives:

5. Force balance between liquid weight and capillary force σ gives:

h · w · (π · ro2 – π · ri

2) = σ · (2 · π · ri + 2 · π · ro) · cos θ →

h = (2 · σ · (ri + ro) · cos θ)/ (w· (ro2 – ri

2)) = (2 · σ · cos θ)/ (w· (ro – ri))

Answer: h = (2 · σ · cos θ)/ (w · (ro – ri))

6. For M; hw = - (1+3) 0.9 = -3.6 m H20.

For N; hw = - 1 0.9 = -0.9 m H20.

7. Draw a horizontal line through the dividing area between oil and water at the manometer left side. Denote the

dividing point between the line and the left of the manometer by 1 and the dividing point between the line and

manometer right point by 2. Denote the difference in level between the reservoirs by x, and the vertical

distance between the left reservoir surface and point 1 by l. Manometry gives:

Answer: The level difference is 38 mm.

Answer:

8. Draw a horizontal line through the dividing area between mercury and water on the manometer left side.

Denote the dividing point between the line and the left of the manometer by 1 and the dividing point between

the line and manometer right point by 2. Manometry gives:

Answer: The water depth in the tank is 2.87 m.

9. (The solution is given during Lecture 10). Answer: The force in each bolt is 178.1 kN.

10. Total force from pressure F = w h A = (9.81 (42 – 1.52) 2) / 4 = 212 kN. sin =1/2

LF = (IG/ALG) + LG = ((R4-r4)/4)/(A LG) + LG = ((24 – 0.754))/(4(22-0.752) 4) + 4 = 4.285 m. hF = 4.285/2 =>

Answer: 2.14 m.

11. In below w=γ.

12.

13. Fh = pressure force on gate side OA; Fv = pressure force on gate side AB; lh = lever arm for Fh about hinge;

lv = lever arm for force Fv about hinge; yc,OA, yc,AB = are vertical distances from water surface to center of gate

side OA and AB, respectively; yp,OA = vertical distance from water surface to center of pressure on gate side

OA; Ic,OA = second moment of area for gate side OA about its center; AOA, AAB = surface areas for gate side OA

and AB, respectively.

yc,OA = 4+1.5 = 5.5 m

yc,AB = 4+3 = 7 m

AOA = 3∙5 = 15 m2

AAB = 2∙5 = 10 m2

Ic,OA = 5∙33/12 = 11.25 m4

yp,OA = yc,OA + Ic,OA/( yc,OA∙ AOA) = 5.636 m

Fh = ρH20∙ g ∙yc,OA ∙ AOA = 809.3 kN

Fv = ρH20∙ g ∙yc,AB ∙ AAB = 686.7 kN

lh = (yp,OA – 4) = 1.636 m

lv = 1 m

The moment about the hinge then becomes:

Pmin∙3 = Fh ∙ lh + Fv ∙ lv => Answer: Pmin = 670 kN.

14. (The solution is given during lecture 10).

15. Select the quarter cylinder as a control volume and

expose forces.

Forces:

Fx = 7.5 · w · 3 · 5 = 1104 kN

The total vertical force will be due buoyancy

of a rectangular body plus a quarter cylinder

body according to:

Fyr = 5 · 6 · 3 · w = 882.9 kN

Fyt = ((5 · 32)/4) w = 346.5 kN

Levers:

lx = hP – 6 = (5·33)/(12·15·7.5)+7.5-6 = 1.6 m

lyr = 1.5 m

lyt = 3 – (4·3)/(3) = 1.73 m

Moment balance gives → Answer: P = 1.23 MN

16. (The solution is given during Lecture 6).

Answer: The streakline will be:

17. Denote PB the pressure at B, PL pressure at the left-hand side surface, and PR pressure at the right-hand side

surface between water and Hg in the manometer. This gives:

PL = PR

PL = 100 + 3 wHg = 100 + 3 13550 9.81 = 498.8 kPa

PR = 3 wH20 + PB = 3 9810 + PB = 29.4 + PB kPa

Answer: PB = 469.4 kPa

18. The difference in level between pitot tube and piezometer in the figure corresponds to the velocity head:

19. (The solution is given during Lecture 10).

P

Fy

Fx

20. Table 1 in Handouts gives σ20oC = 0.0728 N/m. A force balance for capillary rise in a tube gives:

h = σ·2·cosθ/(w·r) = 2·0.0728·cos0o/(9810·0.0005) = 3.0 cm

This gives the real piezometric height as 17.0 -3.0 = 14.0 cm.

Answer: Real piezometric height is 14.0 cm.

21. Given F=120N; S=0.95; u=3m/s; dy=15mm; length of the square plate= 720 mm; Area of the plate, A=

(0.72)2=0.5184 m2.

We have from Newton’s law of viscosity,

To find Kinematic viscosity,

{Mass density of oil, ρo=S*ρw=0.95*1000=950 kg/m3}

Answer: Dynamic viscosity = 1.16 Ns/m2; Kinematic viscosity = 0.00122 m2/s.

22.

Neglecting the air pressure, we have,

p = w h = 13.555 9.82 720/1000 = 98.99 kPa