CIVIL , ENVIRONMENTAL & GEOMATIC ENGINEERING DEPARTMENT
1st YEAR FLUID MECHANICS
CEGE1009 MECHANISMS
SOLUTIONS for TUTORIAL SHEET 3
3
3
Density of water ρ = 1000 kg / m ; for air, R = 287 J/(kgK); ρmercury= 13560 kg/m ; Patm = 101.4 kPa
1.
A hollow cylinder with closed ends is 300 mm diameter and 450 mm high, has a mass of 27 kg and has
a small hole in the bottom. It is slowly lowered into the water, its axis remaining vertical. Calculate the
depth to which it will sink, the height to which the water will rise in it, and the gauge pressure of the air
inside it. Disregard the effect of the thickness of the wall but assume that it is uniform and that the
compression of the air is isothermal.
[Answer: 398 mm; 16.04 mm; 3.747 kPa]
Cylinder mass M; p, pa are absolute pressures
What force actually keeps the body afloat, and where does it act? In fact, the only place where a hydrostatic
force can act vertically to balance the weight is on the underneath of the TOP of the cylinder. (Note that, if the
cylinder walls are of negligible thickness, the hydrostatic pressures on the inside and outside of the bottom of
the cylinder are the same - because of the hole). Hence:
(p - p a )
 d2
27 x 9.81 x 4 kg m s-2
= M g ; (p - p a ) =
[
] = 3747 [kg m-1 s- 2 ] = 3747 Pa
2
2
4
 x 0.3
m
Taking the compression of the air within the cylinder to be isothermal, we have P1 V1 = P2 V2 and hence:
pa
p - pa
 d2 H
 d 2 (H - h )
3747
=p
or h =
H =
x 0.45 m = 0.01604 metres
4
4
p
105147
Now equating pressure at level C - C' inside and outside the cylinder:
p +  g h = p a +  g x or x =
p - pa
3747
+h =
+ 0.016 = 0.398 metres
g
1000 x 9.81
2.
A rigid spherical balloon of diameter 800 mm is to be filled with helium and is intended to carry
meteorological instruments to a height of 6000 m above sea level. The mass of the balloon is 0.1 kg,
and the mass of the instruments, whose volume is negligible, is 0.06 kg. Assume that the balloon does
not expand, that the atmospheric temperature decreases with increasing altitude at a rate of 0.0065 K/m,
0
and that the sea level temperature is 15 C. Determine from first principles the mass of helium required.
[Answer: 16.9 g]
When in equilibrium at 6000m, the buoyancy force on the balloon must equal its total weight.
FB   airVg  mg, where V is the volume, and m is the mass of the balloon, instruments and helium.
(weight of air displaced)
Hence, the total mass m =  airV
-Equation 1
Taking p   air RT and assuming: T  T0  z at any elevation z above the Earth’s surface, we can now
express the vertical pressure gradient in the air as:
p
pg
pg
  g  

;
z
RT
R (T0  z )

p
 gz
g
log e (T0  z )  const

; this integrates to give: log e p 
R
p
R(T0  z )
Applying boundary conditions: p = p0 at z=0: const  log e p 0 
 T  z 
(T  z )
p
g
Hence: log e
or p  p0  0

log e 0

p0 R
T0
 T0 
g
g
log e (T0 )
R
R
-Equation 2
To find the required mass of Helium, we need to solve Equation 1 for  air :
But we know from above (Gas Law etc.) that: p   air RT and T  T0  z , so  
p
R (T0  z )
Now substituting Equation 2:
 T0  z 
p0
p




R(T0  z ) R(T0  z )  T0 
g
R
;
101400
 288  .0065 x6000 
And, at z=6000m,  

287(288  .0065 x6000) 
288
Hence m = 0.66
 (0.8) 3
6
9.81
.0065x 287
 0.66
kg
m3
 0.1769 kg, or 176.9g; and mass of helium = 176.9-100-60 = 16.9g
3.
A rectangular gate of height 5a and width a seals a hole of the same size and shape in a vertical partition
in a tank. The gate is pivoted about a horizontal axis through O, leaving 3a of the gate above O and 2a below it.
The tank contains liquid of density ρ; on the left hand side its free surface is at a height c above the pivot O (c >
3a ), on the right hand side the free surface is level with the pivot. Both free surfaces are exposed to
atmospheric pressure. Working from first principles, derive expressions for the force on the gate and the
moment exerted about the pivot.
Answer: F = ρga2[5c-9a/2]; M = ρ g a3 [9a - 5c/2]
pa
Adopt origin at the gate pivot O;
Forces on element dy are,
- from the left: dFl   pa  g (c  y)a dy
- from the right: dFr   p a  gy a dy
c
Hence total resultant force (+ve from the left) is:
3a
pa
O
y
2a
dy
F
2a
2a
3 a
0
 ga(c  y)dy   gaydy

F  ga cy  12 y 2

2a
3a
 
 ga 12 y 2
F  ga(2ac  2a 2  (3ac) 
F  ga 2 (5c 
9a 2
 2a 2 )
2
9a
)
2
Taking moments about the gate pivot O (anticlockwise moment +ve);
Moments on element dy are,
- from the left: dM l   p a  g (c  y)ya dy
- from the right: dM r   p a  gy ya dy
Hence total resultant moment (+ve anticlockwise) is:
 ga(c  y)y dy   gay dy
2a
M
2a
2
3 a
0

M  ga 12 cy 2  13 y 3
M  ga(2ca 2 

2a
3a
 
 ga 13 y 3
2a
0
2a
0
8a 3 9ca 2  27a 3 8a 3
9c
5c



)  ga 3 (9a  2c  )  ga 3 (9a  )
3
2
3
3
2
2
4.
The hinged gate formed from two rectangular elements which are rigidly attached to each other as
shown below is to open automatically when the water reaches a given level above the hinge. Show that this
occurs when h > b3
Consider unit width of gate into the paper.
pa
Adopt the origin at O
dx
dF
dy
Vertical force on element dy of horizontal
part of the gate is:
dFv   pa  gh  pa dy  gh dy
h
x
Anticlockwise moment on element dy of
horizontal part of gate is:
pa
O
y
dM   pa  gh  pa  y dy  ghy dy
And total anticlockwise
horizontal part of gate is:
b
b
M   ghy dy  gh
0
b
2
moment
on
2
Now considering clockwise moment about O on element dx of vertical part of the gate:
dM  g (h  x) x dx
And total moment (clockwise) about O on vertical part of gate is:
h
 hx 2 x 3 
h3
M   g (h  x) x dx  g 
   g
3 0
6
0
2
h
Gate opens when the clockwise moment on the vertical part of the gate equals or exceeds the anticlockwise
moment on the horizontal part of the gate:
Hence gate opens when: g
or h 2  3b 2 ,
or h  b 3
2
h3
b
 gh ,
6
2
5.
A circular gate of radius a in the vertical wall of a tank is hinged about a horizontal axis through its
centre. The free surface of the liquid in the tank is a distance s above the gate centre (s > a) and both the free
surface and the other side of the gate are exposed to the atmospheric pressure. Derive expressions for the force
on the gate and the moment about the hinge, and show that the moment is independent of the liquid depth.
[Answer: F = ρgπa2 s; M = ρgπa4 / 4]
pa
y=a
s
dF
x
dy
y
x
θ
y
O
pa
O
a
y = -a
Adopt the origin at the hinge O;
Force on horizontal element of width 2x: dF   p a  g ( s  y)  pa 2 x dy  2 g ( s  y) x dy
Moment on element about O: dM   pa  g ( s  y)  pa 2 xy dy  2 g ( s  y) xy dy
a
Therefore, total force on gate F 
 2g (s  y) x dy
a
a
And total moment M 
 2g (s  y) xy dy ; but x = a sinθ
and y = a cosθ , so dy = - a sinθ dθ
a
 0
Hence: F 

 2g (s  a sin  )a sin  (a sin  ) d  2ga  (s sin
 
2

  a sin 2  cos  )d
2
0



 sin 3  
s (1  cos 2 )
sin 2 
2 
d  2 ga 3 
 ga 2 s
And F  2 ga 
  2 ga s  

2
4 0
2
 3 0
0
2
 0
Also: M 
 2g (s  a sin  )a
 
2
sin  cos  (a sin  ) d  2 ga



3
 (s sin
 cos   a sin 2  cos 2  )d
2
0


sin 2
1
1  cos 4
1
sin 4 
1

)d  ga 4 
d  ga 4  
 ga 4
And M  2 ga  (

4
2
2
4
4 0 4

0
0
4
2