Ρευστά σε Κίνηση Γ΄ Λυκείου - Προβλήματα
28
Transcript of Ρευστά σε Κίνηση Γ΄ Λυκείου - Προβλήματα
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: 312 m/kg3.1=ρ
Section 3-4: 3
34 m/kg1.1=ρ
[ ]Pa?pp 14 =−
1/4 Pa10p 50 ≈ (for the calculation of ρ )
Outside (air): C0T1o=
In chimney (smoke):
=
≈
C250T
mmHg760p
2
2o
[ ]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.
3gas
3air
m/kg7.0
m/kg2.1
=ρ
=ρ
1/6
kgK/J288Rair
m/kg2.1
m/N10p0z
30
250
=
=ρ
==
a.) [ ]K?T0 =
b.) [ ]Pa?pA = ,
if the temperature is constant for m2000z0 <≤ .
1/7 25A m/N105.0p ⋅=
airm/kg25.1
m/N10p0z
30
250
=ρ
==
[ ]m?z A = if the temperature is constant for Azz0 <≤ .
1/8 The vehicle is filled with oil.
[ ]Pa?pps/m3a
m/kg950
0A
2
3oil
=−=
=ρ
Hydrostatics 5
1/9 The vehicle is filled with oil.
[ ]20A
3oil
s/m?a
Pa0ppm/kg950
=
=−=ρ
1/10 The tank wagon shown in the figure is taking a
curve with a centripetal acceleration of 2s/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 2ga = ?
1/12 min/11000n =
[ ]Pa?ppm/kg1000
0A
3water
=−=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 5A ⋅= ?
b.) empty the A-B section and have pressure of
Pa108.0 5⋅ in it?
1/14 Effect of gravity is negligible.
?ppmin/16000nm/kg800
0A
3
=−==ρ
1/15 Effect of gravity is negligible.
[ ]Pa?ppm/kg800
m/kg1000s/1100
0A
3oil
3water
=−=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
mm300rm/kg1000
m/kg2300
Sphere
Sphere
3Water
3Cube
=
==ρ
=ρ
[ ]2m/s? a =
surface at standstill
Kinematics 9
2/5 Calculate the circulation along the dashed line.
2r2v =
[ ]s/m? 2=Γ
2/6
[ ] [ ]2Akonv
1
s/m?a.const
s/m20v
=
=ρ=
3 Bernoulli Equation
3/1 Pa103p 5t ⋅=
[ ]s/m?vPa10p 5
0
==
3/2 s/m10v =
[ ]Pa?ppm/kg10
s/m4u
0A
33
=−=ρ
=
3/3 Friction losses are negligible.
[ ]s/m?vm/kg2.1
2
3
==ρ
3/4 Steady flow with
min/m1.0q 3V = .
[ ]m?h =
Bernoulli Equation 11
3/5 Pa106.1p 51 ⋅=
[ ]s/m?q
Pa102.1p3
V
52
=
⋅=
3/6 2sm12a =
[ ]s/m?q
Pa105.0p
Pa10p
3V
5t
50
=
⋅=
=
3/7 s/125=ω
[ ]s/m?w =
(w: relative velocity)
3/8 s/m3w =
[ ]s/1?=ω
(w: relative velocity)
Bernoulli Equation 12
3/9 250 m/N10p =
[ ]2A
A
A
s/m?a
s/m4v0p
=
==
3/10 Pa10p 50 =
Pa109.0p 51 ⋅=
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
=⋅=
Bernoulli Equation 13
3/13 s/m1v = 2s/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 3V =
b.) How big power is needed to move the pipe?
3/15 3alc m/kg800=ρ
[ ]s/m?vm/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 2m/N600p =∆ . 3m/kg3.1=ρ .
[ ]s/m?q 3V =
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 3V !
Bernoulli Equation 14
3/18 Irrotational, horizontal, two-dimensional flow.
s/m5vm8.0rm5.0r
0
2
1
===
a.) What kind of velocity distribution has developed in the arc?
b.) [ ]Pa?pp BA =−
c.) ?rrf
2v
pp
1
22
0
BA
=
ρ
−(Draw a diagram!)
RESULTS
1 Hydrostatics
1/1 20 6200 m/NppA =−
1/2 221 m/N12360pp =−
1/3 214 m/N392pp =−
1/4 221 m/N486pp =−
1/5 The overpressure at the upper tap is 600 Pa.
1/6
a.) KR
pT 290
0
00 =
ρ=
b.) gppg
dzdp
00
ρ−=ρ−=
A
p
p
zp
gp
dpA
0
0
0
ρ−=∫
AA z
pg
ppln
0
0
0
ρ−=
25107880 m/N.pA ⋅=
1/7 m5650h =
Results 28
1/8 230 10237 m/N.ppA ⋅=−
1/9 2452 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:
12
02
21 zrzR π=π
Points of equivalent potential:
212
22
12
02 ω
==ω
−⋅gz
r;rzg
After substitution:
m.gz
RzzgzzR 2360221 0
1121
02 =ω=
ω=
222
0 143002
m/NRzgpp AA =
ω−ρ−=−
1/13 After writing the equation
constrgzp +
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 constrgzp +
ω−ρ−=
2
22
written for the surface of the fluid:
2
220
0ω
ρ−=r
p.const
[ ] 2520
2A
2
0A m/N107.19rr2
pp ⋅=−ω
ρ=−
Results 31
[ ] 25
42
Aconvective
5
41
21
convective
32
11
s/m1328.0
05.0075.0
05.0202a
xr
rrv2
xvva
xr
r2rv
xr
rv
xv
−=⋅
−=
∆∆
−=∂∂
=
∆∆
−=
∂∂
∂∂
=∂∂
3 Bernoulli Equation
3/1 hgp2vp 0
2t ⋅+
ρ+=
ρ
s/m8.19v =
3/2 ( ) Pa108.1uv2
pp 420A ⋅=−
ρ=−
3/3 s/m4.7v150
100v2
hg4
2water =⇒
−
ρ
=⋅⋅ρ
3/4
m141.0g2
Aq
hV
==
3/5 s/m793.0q 3V =
3/6 ( )2vphagp 2
0t +ρ
=⋅++ρ
s/m00589.0q 3v =
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
rhg
2v
2r
2r 22
222
221
21 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
2A0 sd
tvhg
2vp
2A
AA
A
0
s/m1.24a
m3alasdtv
=
⋅=⋅=∂∂∫
3/10 a.) [ ] s/m55.6a 0t ==
b.) m52.1H =
3/11 BB
B
A
a5.7520510asd
tv
=
+=
∂∂∫
[ ] 20tB s/m31.1a ==
3/12 [ ] 20t2 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 3V2 =⇒=
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’.
kW85.82
vvhgqP
21
22
V =
−+⋅⋅ρ= .
3/15 s/m36p2vair
=ρ∆⋅
=
Results 33
3/16 s/m67.0p24
dq2
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 3V = .
3/18 a) in the arc rKv = , because 0vrot = .
b) 1
2
12
r
r12mean r
rln
rrKdr
rK
rr1v
2
1−
=−
= ∫ Because of continuity: 0mean vv =
( )2.3
rr
ln
rrvK
1
2
12mean =−
=⇒
s/m4.6rKv,s/m4
rKv
1B
2A ====⇒
From the Bernoulli-equation:
( ) Pa1025.1vv2
pp 42A
2BBA ⋅=−
ρ=−
c.)
( )22
3
2
0
A
2
0
B
20
BA
n1n
nln1n...
...vv
vv
v2
pp
+−=
=
−
=
ρ−
with 1
2
rrn =
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 002 ⋅⋅⋅ρ=⋅⋅ρ=
with v, the speed at the lower surface of the control surface. According to the Bernoulli equation:
hg2vv 20 ⋅⋅−=
s/m55.4v0 =
4/8 Write the integral momentum equation for both directions x and y:
a) N 636F = b) 8.5A/A 21 =
Solution with constructing the momentum rate vectors: (It has to be considered that
22
21
20 vAvAvA ⋅⋅ρ+⋅⋅ρ=⋅⋅ρ )
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 p
0 = 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. h 1 = 300 mm; h2 = 200 mm; h3 = 600 mm Find pressure difference PA - Ps .
.. ... ~· ~ .,
~ --........
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 .
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.h11
(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.
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 (p AB - 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 ..!£. = Po ln.E._
p0
P Po Po Hence
gz + Po In .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
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: e At= NIAC. Then .6.h = e A, = 2 mm 4
Pcistem = pg( f sin a+ NI) = 261. 2 4 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 = p 9
.
(2)
Combining equations (1) and (2), yields
(1)
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)
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 .1 m
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.
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. 5 kPa;
p = 998.2 kg/m3; Patm = 100.5 kPa)
A •
h
0
D=75mrn
- _ - _ - - - - q -_ - - -
Larg e 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.
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-
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 m of oil from a 20-mm-diameter jet?
( .. 2m 1
.. .. " . . . ·. .. ..
- - Gasoline- - -~.""··~~:·.·.·-·:; ... ~· ' . 0 1 l , .. . ._ . I ' '•
- - - - qG -- - - _::: ' . \. .. ..... . t - ',;' :' '\ . \ ' . '. ·.~ :,"' ·' . . . \ .
-_-_-_-_- _ s--=-- -_ -_- _c -=--_ ---=-
- _ -_ - ---water-- - -
-- --- - - -- -----------
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: 0 1 =200mm; 02 =150mm; z1 = 1m; z2 =1.3m; .1h = 0.2m;
p=1000kg/m3; Pm =13600kg/m3
......_ ___ 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
.. 1 1. 2r
3.12 Neglecting losses, calculate R in terms of H for the Figure.
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 = 7 cm2•
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
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.
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 = z 2 , 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)
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 O and A:
From the equation of continuity:
v2 8Q2
2- 0 4 rr2
The combination of equations (i) and (2), yields
8Q2 h =Po -Pv - =7.604m
p D4rr2g
(1)
(2)
(1)
(2)
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)
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)
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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 )
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
