1. Hydrostatics describes pressure variations in static fluids. Pressure increases with depth due to the weight of the fluid above. 2. The Bernoulli equation relates pressure, velocity, and elevation in fluid flow. It states that the sum of pressure, kinetic energy, and potential energy remains constant along a streamline. 3. Calculations using hydrostatics and the Bernoulli equation allow determining pressures, flows, and forces in applications like pipes, tanks, dams and aircraft wings.

2. 1 Hydrostatics
For all relevant problems Kkg/J287R = , kg/N81.9g =
1/1 [ ]Pa?pp 0A =−
1/2 [ ]Pa?pp 21 =−
1/3 Section 1-2: 3
12 m/kg3.1=ρ
Section 3-4: 3
34 m/kg1.1=ρ
[ ]Pa?pp 14 =−
1/4 Pa10p 5
0 ≈ (for the calculation of ρ )
Outside (air): C0T1
o
=
In chimney (smoke):



=
≈
C250T
mmHg760p
2
2
o
[ ]Pa?pp 21 =−
Hydrostatics 4
1/5 The figure shows a vertical section of a gas pipe. At the lower tap
there is an overpressure of 500 Pa. How big is the overpressure at the
upper tap?
There is no flow in the pipe.
3
gas
3
air
m/kg7.0
m/kg2.1
=ρ
=ρ
1/6
kgK/J288R
air
m/kg2.1
m/N10p
0z 3
0
25
0
=







=ρ
=
=
a.) [ ]K?T0 =
b.) [ ]Pa?pA = ,
if the temperature is constant for m2000z0 <≤ .
1/7 25
A m/N105.0p ⋅=
air
m/kg25.1
m/N10p
0z 3
0
25
0








=ρ
=
=
[ ]m?zA = if the temperature is constant for Azz0 <≤ .
1/8 The vehicle is filled with oil.
[ ]Pa?pp
s/m3a
m/kg950
0A
2
3
oil
=−
=
=ρ

3. Hydrostatics 5
1/9 The vehicle is filled with oil.
[ ]2
0A
3
oil
s/m?a
Pa0pp
m/kg950
=
=−
=ρ
1/10 The tank wagon shown in the figure is taking a
curve with a centripetal acceleration of 2
s/m3a = .
The tank is filled with water.
a.) How high will climb the water surface on the
A-B side?
b.) How big force will affect the A-B side,
when the vehicle is 1.6 m long?
1/11 Where are the both surfaces of the liquid
situated if the pipe accelerates to the left
with an acceleration of
2
g
a = ?
1/12 min/11000n =
[ ]Pa?pp
m/kg1000
0A
3
water
=−
=ρ
Hydrostatics 6
1/13 The pipe is filled with water.
Pa10p 5
0 =
How high angular velocity is needed to
a.) reach Pa108.0p 5
A ⋅= ?
b.) empty the A-B section and have pressure of
Pa108.0 5
⋅ in it?
1/14 Effect of gravity is negligible.
?pp
min/16000n
m/kg800
0A
3
=−
=
=ρ
1/15 Effect of gravity is negligible.
[ ]Pa?pp
m/kg800
m/kg1000
s/1100
0A
3
oil
3
water
=−
=ρ
=ρ
=ω
1/16 What area does an ice-floe have, which can carry a person weighing 736 N? The thickness
of the ice-floe is 10 cm and its density is 900 kg/m3
?
1/17 The rope is weightless.
N200G
mm300r
m/kg1000
m/kg2300
Sphere
Sphere
3
Water
3
Cube
=
=
=ρ
=ρ
[ ]2
m/s?a =
surface at
standstill

4. Kinematics 9
2/5 Calculate the circulation along the dashed line.
2
r
2
v =
[ ]s/m? 2
=Γ
2/6
[ ] [ ]2
Akonv
1
s/m?a
.const
s/m20v
=
=ρ
=
3 Bernoulli Equation
3/1 Pa103p 5
t ⋅=
[ ]s/m?v
Pa10p 5
0
=
=
3/2 s/m10v =
[ ]Pa?pp
m/kg10
s/m4u
0A
33
=−
=ρ
=
3/3 Friction losses are negligible.
[ ]s/m?v
m/kg2.1
2
3
=
=ρ
3/4 Steady flow with
min/m1.0q 3
V = .
[ ]m?h =

5. Bernoulli Equation 11
3/5 Pa106.1p 5
1 ⋅=
[ ]s/m?q
Pa102.1p
3
V
5
2
=
⋅=
3/6
2
sm12a =
[ ]s/m?q
Pa105.0p
Pa10p
3
V
5
t
5
0
=
⋅=
=
3/7 s/125=ω
[ ]s/m?w =
(w: relative velocity)
3/8 s/m3w =
[ ]s/1?=ω
(w: relative velocity)
Bernoulli Equation 12
3/9 25
0 m/N10p =
[ ]2
A
A
A
s/m?a
s/m4v
0p
=
=
=
3/10 Pa10p 5
0 =
Pa109.0p 5
1 ⋅=
Friction losses are
negligible.
a.) How big is the starting
acceleration ’a’ when
opening the tap?
b.) [ ]m?H = in case of steady flow?
3/11 How big is the starting acceleration
in point B when opening the tap?
3/12 How big is the starting acceleration at the
end of the pipe?
0v
)reoverpressu(m/N102p 24
t
=
⋅=

6. Bernoulli Equation 13
3/13 s/m1v =
2
s/m1a =
Friction is negligible. How big
force is needed to push the
piston?
3/14 h/km72u =
s/m4v =
Friction is negligible.
a.) [ ]s/m?q 3
V =
b.) How big power is needed to move the pipe?
3/15 3
alc m/kg800=ρ
[ ]s/m?v
m/kg2.1 3
air
=
=ρ
3/16 The inner diameter of an orifice flowmeter is mm200d = . Flow coefficient 7.0=α
Compressibility factor 1=ε . The measured difference pressure is 2
m/N600p =∆ .
3
m/kg3.1=ρ .
[ ]s/m?q 3
V =
3/17 Width of the flow is 1 m.
a.) Construct the velocity distribution diagram
along the vertical line over the outlet.
b.) Calculate the flow rate [ ]s/mq 3
V !
Bernoulli Equation 14
3/18 Irrotational, horizontal, two-dimensional flow.
s/m5v
m8.0r
m5.0r
0
2
1
=
=
=
a.) What kind of velocity distribution has
developed in the arc?
b.) [ ]Pa?pp BA =−
c.) ?
r
r
f
2
v
pp
1
2
2
0
BA






=
ρ
−
(Draw a diagram!)

7. RESULTS
1 Hydrostatics
1/1 2
0 6200 m/NppA =−
1/2 2
21 m/N12360pp =−
1/3 2
14 m/N392pp =−
1/4 2
21 m/N486pp =−
1/5 The overpressure at the upper
tap is 600 Pa.
1/6
a.) K
R
p
T 290
0
0
0 =
ρ
=
b.) g
p
p
g
dz
dp
0
0
ρ−=ρ−=
A
p
p
z
p
g
p
dpA
0
0
0
ρ
−=∫
A
A
z
p
g
p
p
ln
0
0
0
ρ
−=
25
107880 m/N.pA ⋅=
1/7 m5650h =
Results 28
1/8 23
0 10237 m/N.ppA ⋅=−
1/9 2
452 s/m.a =
1/10 a) m422.0h =
b) NF 1400=
1/11 The surface at the left side is situated at the left lower corner, the other surface in the right
vertical section at a height of 100 mm.
1/12 Volumes are the same in standstill and rotation:
1
2
0
2
2
1
zrzR π=π
Points of equivalent potential:
2
12
22
1
2
0
2 ω
==
ω
−⋅
gz
r;
r
zg
After substitution:
m.
g
z
Rzz
gz
zR 2360
2
2
1 0
112
1
0
2
=ω=
ω
=
2
22
0 14300
2
m/N
R
zgpp AA =




 ω
−ρ−=−
1/13 After writing the equation
const
r
gzp +




 ω
−ρ−=
2
22
.
for the both known points (surfaces in the left and the right section), the angular velocity can
be calculated.
a.) s/. 1421=ω
b.) s/. 1324=ω
1/14 Equation const
r
gzp +




 ω
−ρ−=
2
22
written for the surface of the fluid:
2
22
0
0
ω
ρ−=
r
p.const
[ ] 252
0
2
A
2
0A m/N107.19rr
2
pp ⋅=−
ω
ρ=−

8. Results 31
[ ] 2
5
42
Aconvective
5
4
1
2
1
convective
3
2
11
s/m132
8.0
05.0
075.0
05.0202
a
x
r
r
rv2
x
v
va
x
r
r
2
rv
x
r
r
v
x
v
−=
⋅
−=
∆
∆
−=
∂
∂
=
∆
∆




−=
∂
∂
∂
∂
=
∂
∂
3 Bernoulli Equation
3/1 hg
p
2
vp 0
2
t
⋅+
ρ
+=
ρ
s/m8.19v =
3/2 ( ) Pa108.1uv
2
pp 42
0A ⋅=−
ρ
=−
3/3 s/m4.7v1
50
100
v
2
hg
4
2
water =⇒








−




ρ
=⋅⋅ρ
3/4
m141.0
g2
A
q
h
V
==
3/5 s/m793.0q 3
V =
3/6 ( )
2
vp
hag
p 2
0t
+
ρ
=⋅++
ρ
s/m00589.0q 3
v =
3/7 Observing in an absolute co-ordinate system, the flow is irrotational ( 0vrot = ). In a co-
ordinate system rotating with the pipe, ω= 2wrot , so the term sdwrotw∫ × is equal to
sdw2∫ ω× , the Coriolis force term. ( w – relative velocity) The Bernoulli equation can be
written after simplifying the terms above:
Results 32
( )
2
r
hg
2
v
2
r
2
r 22
2
2
2
22
1
2
1 ω
−⋅+=
ω
−
ω−
Point 1 is situated on the water surface on an arbitrary radius r1 , point 2 at the upper end of
the pipe.
s/m8.10v2 =
3/8 s/124=ω
3/9 ∫ ∂
∂
+⋅+=
ρ
A
0
2
A0
sd
t
v
hg
2
vp
2
A
AA
A
0
s/m1.24a
m3alasd
t
v
=
⋅=⋅=
∂
∂
∫
3/10 a.) [ ] s/m55.6a 0t ==
b.) m52.1H =
3/11 BB
B
A
a5.75
20
5
10asd
t
v
=





+=
∂
∂
∫
[ ] 2
0tB s/m31.1a ==
3/12 [ ] 2
0t2 s/m94.7a ==
3/13 N451F =
3/14 a) The Bernoulli-Equation has to be written between the surface point (1) and the pipe’s
outlet point (2), in a co-ordinate system moving with the pipe. It means that s/m24v1 = .
From the Bernoulli-equation:
s/m116.0qs/m4.23v 3
V2 =⇒=
b) the power is necessary to lift the water and to increase its kinetic energy. The change
of the kinetic energy must be calculated with the absolute velocity ’v’.
kW85.8
2
vv
hgqP
2
1
2
2
V =




 −
+⋅⋅ρ= .
3/15 s/m36
p2
v
air
=
ρ
∆⋅
=

9. Results 33
3/16 s/m67.0
p2
4
d
q
2
V =
ρ
∆π
ε⋅α=
3/17 Because the stream lines leaving the outlet are straight and
parallel, there is only a hydrostatic pressure variation along
the vertical axis. It follows that the outlet velocity is
constant.
s/m15.3q 3
V = .
3/18 a) in the arc
r
K
v = , because 0vrot = .
b)
1
2
12
r
r12
mean
r
r
ln
rr
K
dr
r
K
rr
1
v
2
1
−
=
−
= ∫ Because of continuity: 0mean vv =
( )
2.3
r
r
ln
rrv
K
1
2
12mean
=
−
=⇒
s/m4.6
r
K
v,s/m4
r
K
v
1
B
2
A ====⇒
From the Bernoulli-equation:
( ) Pa1025.1vv
2
pp 42
A
2
BBA ⋅=−
ρ
=−
c.)
( )
22
3
2
0
A
2
0
B
2
0
BA
n
1n
nln
1n
...
...
v
v
v
v
v
2
pp
+−
=
=





−





=
ρ
−
with
1
2
r
r
n =
Results 34
4 Integral Momentum Equation
4/1 N12100Fx =
4/2 After writing the Bernoulli equation for points situated upstream and downstream the blade
we get the result:
12 vv =
4/3 N510F = , direction 45° from the horizontal plane (’Northeast’)
4/4 N109F =
4/5 N57F =
4/6 N14G =
4/7 The integral momentum equation written for a control surface including only the plate and
the upper end of the jet:
vvAvAG 00
2
⋅⋅⋅ρ=⋅⋅ρ=
with v, the speed at the lower surface of the control surface.
According to the Bernoulli equation:
hg2vv
2
0 ⋅⋅−=
s/m55.4v0 =
4/8 Write the integral momentum equation for both
directions x and y:
a) N636F =
b) 8.5A/A 21 =
Solution with constructing the momentum rate
vectors:
(It has to be considered that
2
2
2
1
2
0 vAvAvA ⋅⋅ρ+⋅⋅ρ=⋅⋅ρ )

10. 1
1 FLUID STATICS, RELATIVE EQUILIBRIUM OF MOVING FLUIDS
1.1 Determine the gravity force W that can be sustained by the force F acting on
the piston of the Figure.
F= 1 MN
E
0
"U
E
E
0
~
II
"U
1
0= 240 mm diam
• • I
w
.. · ~ ,.... , ..'-~
. ' ._, .. .. .... "' . ". ,- .._ • . . ' ' • ;.. I
,.. ~ . . ' ' ..
• • ' • • . ._ I
" .. ,,... .. . . "' .- - . • • • -, - • • • I
.. I " • - _,,,,,..,, ... :'J· .,.,, ..... ··
' - ·. . •' -- 0 ·1 . . . ' . ' ._,• , -. • ·, ' . . I ~ ' '.. . , . ,... - • ,,
.- ..·,.. '..<.·-::.-:• :.:.''_-...._-: :·. __.._:: ,'
Hydraulic Jack
1.2 Neglecting the mass of the container find the force F tending to lift the circular
top AB.
...
"" f.h= l m
•.
'
· .
...,.
.... 3 3
:~ 9 =-10 kg Im
.·J--;...-" •
,.
'..::
.....·.·...
....
:7
·'.,,
·.~
1.3 For isothermal air at O°C, determine the pressure and density at H = 3000 m
when the pressure is p0
= 0.1 MPa abs at sea level (gas constant
R =287 Nm/(kgK) ).
1.4 In the Figure the liquids at A and B are water (p) and the manometer liquid is
oil (pJ.
h1 = 300 mm; h2 = 200 mm; h3 = 600 mm
Find pressure difference PA - Ps .

11. ..
...~·~
.,
~ --........
1.5 h1 =1.3 m; h2 = 1.5 m; h3 =1 m
Find the gauge pressure in point A
i; h1
.. h2
..~
·;
2
p
0
Mercury
(q =13600kg lm'>
m
1.6 The ratio of cistern diameter to tube diameter is 10. When the air in the tank is
at atmospheric pressure, the free surface in the tube is at position 1. When the
cistern is pressurized, the liquid in the tube moves 20 mm Y=200 mm} up the
tube from position 1 to position 2. What is the cistern pressure that causes
this deflection? The density of the liquid is p =800 kg/mJ .

12. 3
Cistern
1.7 Find the gauge difference N111 in the U-tube if the water level in the container is
raised by ,1/-1 =1.5 m (see the Figure).
- - - - - - - - - - - - - - Water
=-=-=-=-=--=--=-=-=-=-=-=-=-=-lqw=-1000 kg/~)_-_- _-_ - _-_- ~w- _- _-_-_- _- -
- - -- --- ------,____------- - - -- -- --- ---
A
A'
E
c.C>
If)
0
0
1.8 An aquarium at Marineland has a window as shown.
Find:
Mercury 3
{qm= 13600 kg/m ) t.h
11
(a) The resultant force from seawater (p =1015 kg/m3) on the window, F
(b) The line of action of the resultant force, in metre below the water surface, k.

13. 45
SOLUTIONS TO CHAPTER 1
1.1 The pressure p can be considered to be constant in the container:
Solving for W, yields
1.2 The gauge pressure (pAB - pc) at the inner side of the top is
1.3 Equation of hydrostatics for this barotropic fluid:
gz+P =canst
where for isothermal change of state P = Po
P Po
pd
P= J..!£.= Poln.E._
p0
P Po Po
Hence
gz+ PoIn .E._ =const
Po Po
Boundary condition: p =p0
at z =0, so canst= 0
Solving the equation for p yields the variation of pressure with elevation
p=Po exp(-_JJ_z)RTO

14. Then
46
P1 =p(z =H) =p0 exp(-_J/_H)=0.687kPa
RTO
From the perfect gas law
Hence
P1 = p(z = H) =A= 0.8764 kg/ m3
RT0
1.4 PA - Pw9h, - Po9h2 +Pw9h3 = Pa
PA -Pa =-1373.4 Pa
1.5 PA - Po9h1 +Pw9h2- Pm9h3 =Po
PA -Po= Q (p0 h1 +pmh3-pwh2)=1.3018 X 105
Pa
1.6. Continuity: eAt= NIAC. Then .6.h =eA, =2 mm
4
Pcistem =pg(f sin a+NI) =261.24 Pa
1.7 PA=Pe
Po +pgH = Po+Pm9'1h1
where H is the height of the water column above point A.
When the level of the water in the container is raised by .6.H: pi =p9
.
(2)
Combining equations (1) and (2), yields
(1)

15. 1.8
1.9
47
p +gz =const =Po +gd
p p
Hence the gauge pressure: p• =p - p0
=pg(d - z)
(a) F =p/A =pg(d-zc)ab;
where C is the centroid of area of the window,
Zc =d - c- a/ 2
F =Fx =5.3769 kN
(b) The moment of the resultant force is equated to the moment of
distributed forces about the axis A-A
whence
k =0.65 m
wh3
I }2 h 13
y =H - k= -r-+y = + H - -= - m
P A Ye c wh(H-~) 2 6
k = H-yp =5 16 m.
1.10 The moment at A have to be equal in magnitude to the moment of the
resultant of individual forces acting on the gate.
The gauge pressure distribution in the fluid: p - p0 =pg(H- z)

16. 17
3 APPUCATON OF BERNOULLI'S EQUATION
3.1 Air flows steadily and at low speed through a horizontal nozzle, discharging to
the atmosphere. At the nozzle inlet, the area is 0.1 m2
• At the nozzle exit the
area is 0.02 m2
. The flow is essentially incompressible, and frictional effects
are negligible. Determine the gauge pressure required at the nozzle inlet to
produce an outlet speed of 50 m/s. (p=1 .23 kg/m3
)
ICD
I
%,
At =- 0.1m
3.2 Neglecting losses determine the discharge Q in the Figure.
(p1 =750 kg/m3
; p2 =1 OOO kg/m3
; h1
=1 m; h2
=1.5 m; d=0.1 m)
Large
t ank
lnrniscible
fluids
z - - - - - - h 2 - - -
-- 9 - -- --- · -2 - - - --
_- - - x -- - - - - - - - - - -
v
--- - - - - --
3.3 A glass tube with a 90° bend is open at both ends. It is inserted into a flowing
stream of oil so that one opening is directed upstream and the other is
directed upward. Oil inside the tube is 50 mm higher than the surface of the
flowing oil. Determine the velocity measured by the tube.

17. p
atm
18
- - - - -
3.4 Water may be considered to flow without friction through the siphon. The
water flow rate is 0.03 m3
/sec, its temperature is 20 °C, and the pipe diameter
is 75 mm. Compute the maximum allowable height h, so that the pressure at
point A is above the vapour pressure of the water. (Pv = 2.5kPa;
p = 998.2 kg/m3
; Patm =100.5 kPa)
A
•
h
0
D=75mrn
- _ - _ - - - - q -_ - - -
Large tank
3.5 A smoothly contoured nozzle is connected to the end of a garden nose. At the
nozzle inlet where the velocity is negligible, the water pressure is 160 kPa
(gauge). Pressure at the nozzle exit is atmospheric. Assuming that the water
remains in a single stream that has negligible aerodynamic drag, estimate the
maximum height above the nozzle outlet that the stream could reach.

18. 19
h
3.6 Two probes are often combined, as in the pitot-static tube in the Figure. The
inner tube is used to measure the the stagnation pressure at point B while the
static pressure at C is sensed by the small holes in the outer tube. In flow
fields where the static pressure variation in the streamwise direction is small,
the pitot-static tube may be used to infer the velocity at point A in the flow, by
assuring PA =Pc- (Note that when PA:.i: Pc· this procedure will give erroneous
results).
(p=l000kg/m3
; Pm =13600kg/m3
; .1h=300mm)
Static pressure holes
Flow ~·--0 - - - - - - --
A ~ > v
E-

19. 20
3.7 Water flows in a circular pipe. At one section the diameter is 0.3 m, the static
pressure is 260 kPa (gauge). the velocity is 3 m/s, and the elevation is 10 m
above ground level. The elevation at a section downstream is O m, and the
pipe diameter is 0.15 m. Find the gauge pressure at the downstream section if
frictional effects may be neglected.
3.8 An airspeed indicator used frequently on World War II aircraft consisted of a
converging-diverging duct as shown in the Figure. The diameter of the duct at
2 is three-quarters of the entrance diameter at 1. The differential pressure
gauge records a pressure of 4000 Pa, and the density of the air is 1 kg/m3
.
The air flow is steady, incompressible, inviscid and irrotational. Determine the
airspeed U.
u
3.9 A large tank contains air, gasoline ( PG =680 kg / m3
}, light oil
( p0 =800 kg / m3
) and water. The pressure p of the air is 150 kPa gauge. If we
neglect friction, what is the mass flow mof oil from a 20-mm-diameter jet?
( .. 2m
1
.. .. " . . .·. .. ..
- - Gasoline- - -
~.""··~~:·.·.·-·:;
... ~· ' .0 1l , .. . ._ .
I ' ' •
- - - - qG -- - - _:::
' . . .. ..... .
t - ',;' :' '
. ' .
'. ·.~ :,"' ·' . . . .
-_-_-_-_- _s--=-- -_-_- _c -=--_---=-
-_-_- ---water-- - -
-- --- - - -- -----------

20. 21
3.10 For the venturi meter and manometer installation shown in the Figure
determine the volume rate of flow for the manometer reading .1h.
Data: 01 =200mm; 02 =150mm; z1 = 1m; z2 =1.3m; .1h = 0.2m;
p=1000kg/m3
; Pm=13600kg/m
3
......__ __ A ~s -~L---z= O
'Sm
3.11 Neglecting losses and surface tension effects, derive an equation for the
water surface r of the jet of the Figure in terms of y/H.
- ---- 1--- ---- - --- -------- - - - - ------
H
..11. 2r
3.12 Neglecting losses, calculate R in terms of H for the Figure.

21. 22
p I
a.m
E
- - - - - vlarge tank- -- - - - - -- - -- - - - - - - - -,...
H
>----
0
3
q =-1000 kg /m !'--.
t-A I B
•
""' - 3
~ -3000 kg/m
m
3.13 The tank shown in the Figure has a well-rounded orifice with area Aj = 7cm2
•
At the time t=O, the water level is at height H=2 m. Develop an expression for
the water height, z1, at any later time, t. The cross-sectional area of the tank is
A= 0.7 m2. You can neglect frictional effects, and the quasi-steady form of
the Bernoulli's equation can be used.
H
- Tank area - -
3.14 The tank with hemisphere shape has a well rounded orifice with area
Ai= 0.01 m2
• At time t=O the water level is at height R=2 m. Develop an

22. 23
expression for the water height, z1 , at any later time, t. Determine time T,
belonging to z1 = R/2. You can neglect the unsteady term in the Bernoulli's
equation.

23. 3.1 Basic equations
Bernoulli's equation:
Equation of continuity:
69
SOLUTIONS TO CHAPTER 3
P v 2 p v2
_1 +gz +-1-=_g_+gz +-2-
p 1 2 p 2 2
Since z1 = z2
, the combination of equations (1) and (2) yields
3.2 Basic equation:
P v2
- +gz +- =canst
p 2
Application of equation (1) between points 1 and 2:
P v 2 p v2
_1 +gh2+ -1- = atm +-
P2 2 P2 2
Application of equation (1) between points 0 and 1:
P v 2 p 2
atm +g(fli +h2) +-o- =-1 +gh2 +~
~ 2 ~ 2
v1 =V0 =0 here.
Equation of continuity:
d21r
Q=-V
4
The combination of equations (2)-(4), yields
(1)
(2)
(1)
(2)
(3)
(4)

24. 3.3
3.4
70
Apply Bernoulli's equation between points 1 and 2:
From the hydrostatic pressure variation:
P1 =Patm +pgh and P2 =Pa1m +pg(h +'111)
Hence
Combining equations (1)and (2), yields
Bernoulli's equation between points Oand A:
From the equation of continuity:
v2 8Q2
2- 04 rr2
The combination of equations (i) and (2), yields
8Q2
h =Po-Pv - =7.604m
p D4rr2g
(1)
(2)
(1)
(2)

25. 3.5
3.6
3.7
3.8
71
Neglecting friction write the Bernoulli's equation between points 1 and 2:
whence
P1=Potm +gH
p p
H = p1 -Patm =16.310m
pg
Apply Bernoulli's equation between points A and B:
Apply hydrostatic equation for the two limbs of the U-tube:
Taking into account that PA= Pc and combining equations (1) and (2). yield
Application of the Bernoulli's equation and the equation of continuity, yields
p2 =p, +pg{z2 -z,)+ ~[1-(~J]=2906kPa
The combination of the Bernoulli's equation and the equation of continuity
written between points 1 and 2, yields
(1)
(2)

26. 3.9
72
Bernoulli's equation between points C-D:
Pc= Po+~ +g/2
Po Po 2
Equation of hydrostatics between points A and B:
Equation of continuity:
. d27r
m=pv-o 4
Taking into account that p8 =Pc and the combination of equations (1) -(3),
yields
3.10
Bernoulli's equation:
Equation of continuity:
P v 2 p v 2
_1 +gz +-1-=2 +gz +-2-
p 1 2 p 2 2
Equation of hydrostatics for the two limbs of the U-tube:
(1)
(2)
(3)
(1)
(2)
(3)

27. 73
The combination of equations (1)-(3), yields
3.11
0 = D12te
4
& _7
(£)'-1
2gLlh = 0.02653 m' /s
Application of the Bernoulli's equation between points 1 and 2, and 1 and 3,
yield
V3 =.J2g(H+y)
Equation of continuity between points 2 and 3 (neglecting jet contraction):
Combination of equations (1)-(3) gives
r=--2.. 1+l...
d ( )--0.25
2 H
3.12
Bernoulli's equation between points 0 and E:
Equation of hydrostatics for the two limbs of the U-tube (PA =p8 ):
(1)
(2)
(3)
Patm+pgH =Po +(Pm-p)gR (2)
Combination of equations (1) and (2), yields
R =0 ( for an arbitrary value of H )

28. 74
3.13
Bernoulli's equation for points 1 and 2, yields
Equation of continuity:
The definition of v1:
dz1
v =--
1 dt
Combination of equations (1) and (2) gives
2g r:;-
2 .yz,
(~) -1
Combination of equations (3) and (4) can be integrated finally to give
3.14
Jii-~
2
2
2
~ t =(1.414213562 - 0.002214725t)2
(~) -1
(1)
(2)
(3)
(4)
The method of solution is the same as in Problem 3.13. The only difference is
that the cross-sectional area of the tank varies with z1
A1 =(2Rz1 -z/ )ir
The relation between z1 and t is obtained as follows
Ti =213.70 sec