1. In the bottom of a vessel with mercury... m. At 

1. In the bottom of a vessel with mercury there is a round hole of diameter 70 m. At
what maximum thickness of the mercury layer will the liquid still not flow out
through this hole? The surface tension of mercury is 0.5 N/m.
The excess hydrostatic pressure at the bottom of the vessel is p   gh . The surface
tension provides the excess pressure within the hole as given by the Laplace formula
2
,
p
R
where R is the curvature radius of the surface within the hole. The equilibrium condition:
 gh 
h
2
.
R
2
4
4  0.5


 0.2 m
3
 gR  gd 13.6 10 10  70 106
2.
Find the difference in height of mercury columns in two communicating vertical
capillaries whose diameters are d1=0.5 mm and d2=1 mm, if the contact angle the
surface tension is 0.5 N/m.
The curvature radius of the meniscus is
R
d
d
d


o
2sin(  90 ) 2  0.74 1.5
The pressure equilibrium condition:
 gh 
h
2 2

R2 R1
2 1.5  1 1 
   =1.1 cm
 g  d1 d 2 
3. A wide vessel with a small hole in the bottom is filled with water and kerosene.
Find the velocity of the water flow, if the thickness of the water layer is equal to
h1=30 cm and that of the kerosene layer to h2 = 20 cm. Density of kerosene is
0.8 g/cm3.
The pressure at the kerosene-water boundary is p  patm  k gh2 . Bernoulli eq. for water flow:
p
w
 gh1 
patm
w
 12 v2
 p  patm



v  2
 gh1   2 g  k h2  h1 
 w

 w

4. Through a pipe of diameter D, water is pumped from a lake to the water-tower.
(a) If the rate of water supply (volume per unit time) is Q, what is the speed of
the water in the pipe? (b) What pressure is necessary to deliver this flow? (c)
What is the power of the pump? The height of the water-tower is H.
v=
Q 4Q

S  D2
p   gh  12  v2
A mass of water dm acquires the energy dE  dm
N
dE dm

dt
dt

1
2


1
2
v2  gh  Q

v2  gh . The required power is

1
2
v2  gh

.‫ כמתואר באיור‬d ‫ משתמשים בצינור גינה פלסטי עם קוטר פנימי‬,‫ על מנת לרוקן בריכת שכשוך‬.5
.‫ אפשר להזניח את הצמיגות‬.A ‫מצא\י את קצב הזרימה (נפח ליחידת זמן) בצינור ולחץ בנקודה‬
The Bernoulli equation for a flow line connecting the surface and the hose nozzle:
patm   gH  patm   gh  12  v 2
v= 2 g ( H  h)
The flow rate Q  vS  4 d 2 v  4 d 2 2 g ( H  h)
In order to find the pressure at the point A, write the Bernoulli equation for the surface
and the point A
patm   gH  pA   g ( H  a)  12  v2
The continuity equation, Sv=const, implies that the velocity within the hose does not
vary because the cross section is constant. Then
pA  patm   ga  12  v2  patm   ga  12   2 g ( H  h)  patm   g (a  h  H )
Note that pA<patm.
‫ ואילו הקוטר של הצינור עצמו הוא‬d=4cm ‫ הקוטר של זרבובית של הצינור לכיבוי אש הוא‬.6
‫? אפשר‬q=20litr/s ‫ איזו עודף לחץ דרוש בתוך הצינור על מנת לספק קצב זרימה‬.D=8cm
.‫להזניח את הצמיגות‬
q=Sv.
v1 
p
p  p atm 

4q
4q
; v2 
2
D
d2
 v12
2
 patm 
 v22
2
2
2
1  4q   4q   8q 2
  2    2    2 2
2  d   D    d
8  2  10 4

 2 34

2
  d 2 
1    
  D  
1  8  4  10 8

 4  10 6  400kPa
1  4  
10  81
 2 