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Fluid Mechanics
Exercise sheet 2 – Statics
last edited April 27, 2015
These lecture notes are based on textbooks by White [4], Çengel & al.[6], and Munson & al.[8].
Except otherwise indicated, we assume that fluids are Newtonian, and that:
ρwater = 1 000 kg m −3 ; p atm. = 1 bar; ρatm. = 1,225 kg m −3 ; µatm. = 1,5 · 10 −5 N s m −2 ;
д = 9,81 m s −2 . Air is modeled as a perfect gas (R air = 287 J K −1 kg −1 ; γair = 1,4).
2.1
Pressure in a fluid
A small water container whose geometry is described in fig. 2.7 is filled with water. What
is the pressure at the bottom of the container?
Figure 2.7: A small water container.
figure CC-0 o.c.
2.2
Pressure measurement
A tube is connected to a pressurized vessel as shown in fig. 2.8. The U-tube is filled with
water. What is the pressure in the vessel?
What would be the height difference shown if mercury (ρmercury = 13 600 kg m−3 ) was
used instead of water?
31
Figure 2.8: Working principle of a simple liquid tube manometer.
figure CC-0 o.c.
2.3
Water lock
A lock is setup to allow boats to travel up and down along a 20 m height (fig. 2.9). They
have a length of 50 m and a width of 10 m; the canal has a depth of 3 m.
Figure 2.9: Schematic layout of a simple lock (top: side view; bottom: view from the top). In
practice, in order to counter the moment generated about the doors hinges by the water, the
doors are often set-up in a diagonal position, as described in exercise 2.8.
figure CC-0 Olivier Cleynen
1. Sketch the distribution of pressure on the downstream doors when the lock is full.
2. When the lock is full, what is the force resulting from the water pressure on each
of the doors?
32
3. What is the moment about each hinge?
4. What is the moment about each hinge when a barge with a mass of 500 t transits
through the lock?
2.4
Buoyancy force
~ = ρ~
1. Starting from eq. (2/14): ∇p
д , show that when the temperature Tcst. is assumed
to be uniform, the atmospheric pressure p at two points 1 and 2 separated by a
height difference ∆z is such that:
"
#
д∆z
p2
= exp
(2/20)
p1
RTcst.
A student contemplates a tin can of height 10 cm and diameter 7 cm. The room temperature is 20 ◦C.
2. What is the buoyancy force generated by the atmosphere on the can when it is
positioned vertically?
3. What is the force generated when the can is immersed in water at a depth of 20 cm?
At a depth of 10 m?
4. What is the buoyancy force generated when the can is immersed in the water in a
horizontal position?
2.5
Buoyancy of a barge
A barge of very simple geometry is moored in a water reservoir (fig. 2.10).
Figure 2.10: Basic layout of a barge floating in water.
figure CC-0 o.c.
1. Sketch the distribution of pressure on each of the immersed walls of the barge.
2. What is the force resulting from pressure efforts on each of these walls?
3. What is the weight of the barge?
33
2.6
Atmospheric buoyancy force
Estimate the buoyancy force exerted on an Airbus A380, both on the ground and in cruise
flight (ρcruise = 0,4 kg m−3 , Tcruise = −40 ◦C).
Figure 2.11: Airbus A380-800
Drawing CC-by-sa by Julien Scavini
Length overall
72,73 m
Wingspan
79,75 m
Height
24,45 m
Wing area
845 m2
Aspect ratio
7,5
Wing sweep
33,5°
Maximum take-off weight
560 000 kg
Typical operating empty weight
276 800 kg
Table 2.1: Characteristics of the Airbus A380-800.
2.7
Standard atmosphere
The integration we carried out in §2.3.4 to model the pressure distribution in the atmosphere was based on the hypothesis that the temperature was uniform and constant.
This may however not always be so.
1. If the temperature decreases with altitude at a constant rate of −6 K km−1 , what is
the pressure distribution?
A successful fluid dynamics lecturer purchases an apartment at the top of the Burj Khalifa
tower (800 m above the ground). The ground temperature is 30 ◦C.
2. What atmospheric pressures in the apartment are predicted by the constanttemperature and constant-temperature-gradient models?
34
2.8
Lock with diagonally-mounted doors
The doors of the lock studied in exercise 2.3 are replaced with diagonally-mounted
doors at a 20° angle, mounted such that no bending moment is sustained by the hinges
(fig. 2.12).
What are the efforts exerted on the doors?
Figure 2.12: Lock doors mounted with an angle relative one to another. The angle relative to the
case in exercise 2.3 is 20°.
figure CC-0 o.c.
2.9
Reservoir door
Munson & al. [8] 2.87
A water reservoir has a door of width 3 m which is held in place with a horizontal cable,
as shown in fig. 2.13.
Figure 2.13: A sealed, hinged door in a water reservoir. The width across the drawing (towards
the reader) is 3 m
figure CC-0 o.c.
The door has a mass of 200 kg and the hinge exerts negligible bending moment. What is
the force in the cable?
35
Answers
2.1
p A = p atm. + 0,039 bar ≈ 1,039 bar.
2.2
1) p inside = p atm. + 0,0157 bar ≈ 1,0157 bar;
2) ∆z 2 = 1,1765 cm.
h i 23 m
= 12,753 MN per door;
2) F net = ρдL 21 z 2
3m
L
3) M = F net 2 = 31,88 MN m per hinge. 4) It has no effect at all on the previous
calculation!
2.3
2.4
1) See §2.3.4;
3) F vertical = 3,775 N;
2.5
2) F rear = 0,2453 MN, F side = 2,1714 MN, F bottom = 13,734 MN, F front = 0,3468 MN;
3) F buoyancy = 13,979 MN (1 425 t).
2.6
Assuming a volume of approx. 2 600 m3 , we obtain approx. 30,9 kN on the ground,
10,4 kN during cruise.
д
p
2 kR
;
2) Top of tower pressure: p 2 = 0,9131p 1 .
1) p21 = 1 + kz
T1
2.7
2.8
2.9
36
2) F vertical = 4,487 mN upwards;
4) There is no change.
The moment is brought to zero by an inter-door force F sideways = 19,84 MN (perpendicular to the canal).
h
i
dp sin θ 3 R
H 2
= 0,4186 MN m; so, F cable = 81,13 kN.
M water = L dz
2r − 3 r
water
2