Hydrostatics Hydrodynamics Pressure p
F F p A A
Pressure = force / area SI – unit 1 N/m2 = 1 Pascal = 1 Pa
Other units: 1 bar = 100 000 Pa 1 atm = 1.013 bar 1 mmHg = 1/760 atm Hydrostatic pressure
Example: Calculate the pressure caused by the 10 m high brick wall onto the lowest brick layer. The density of bricks is 7.0 kg/m3.
Let the base area of the brick wall be A and the height h. Then the weight of the wall is
h mg =Ahg
The pressure p = mg/A = hg = 7.0*10*9.81Pa = 700 Pa pressure p
The same calculation applies also to the pressure under water or under some other liquid or gas. At depth h the hydrostatic pressure due to the weight of the water (or liquid) above is p = hg Archimedes Law Buoyancy F = Vg
In liquids and gases the pressure = density of the liquid spreads h1 h2 uniformly to all directions p1=h1g A h=h1-h2 p2=h2g
Net force upwards is called buoyancy and it is F =p*A =(p2-p1)A = (h2-h1)Ag = Ahg = Vg
“Archimedes law”: The net force acting on a body in a liquid is the same as the weight of the equal volume of liquid. Determination of the density of a solid “sinking” body
Measurements: • Weigh the body in the air F1 • Weigh the body in the water F2
Calculations:
F1= xVg
F2= xVg - Vg = (x- )Vg F1-F2 = Vg
x/ = F1/(F1-F2)
x = F1/(F1-F2) * Determination of the density of a solid “floating” body
Measurements: 1) Weigh the body in the air F1 2) Fill a tube with water. Put the body carefully in the tube and take the spilling water and weigh it F2 Calculations:
F1= xVg and F2= Vg
=> x / = F1/F2
=> x =F1/F2* Determination of the density of an unknown liquid
Needs: A body heavier than water Measurements: 1) Weigh the body in the air F1 1) Weigh the body in water F2 2) Weigh the body in the liquid F3
Calculations: , w and x = F1= Vg , F2= (- w)Vg, F3= (- x)Vg are the densities of the F1-F2 = wVg and F1-F3= xVg body, water and x/ w = (F1-F2)/(F1-F3) unknown liquid
x = (F1-F2)/(F1-F3) * w U -tube barometer
The pressure of the air is equal to gh , where = 13600 kg/m3 is the density of mercury
Example: Calculate the difference of Hg-levels , when the air pressure is 1.01 bar. Solution: From p = gh we get h = p/g =101000Pa/ (13600kg/m3 *9.81 m/s2) = 0.76 m = 760 mm Hydrodynamics Continuum equation
Liquid flow in a tube.
v v1 2
A A1 2
In the time interval t the same volume of liquid must pass through both the areas A1 and A2:
A1v1 t =A2v2 t Cancelling t we get the continuum equation
A1v1 =A2v2 Bernoulli equation
Bernoulli equation is the energy principle applied to liquid flows or gas flows = density of the liquid p1 m = mass of the liquid h1 between the ends of the tube
v1 liquid flow
h2 v2 p2
2 2 mgh1 + ½ mv1 + W = mgh1 + ½ mv1 W=Fs=F/A*As 2 2 mgh1 + ½ mv1 + (p1 - p2)V = mgh1 + ½ mv1 =>W=pV Dividing by volume:
2 2 gh1 + ½ v1 + p1 = gh2 + ½ v1 + p2 Bernoulli formula
At every point of the liquid or gas flow
p + gh + ½ v2 = constant
p = pressure at the point gh = the hydrostatic pressure ½ v2 = the dynamic pressure Applications of the Bernoulli law Torricelli law
How to calculate the velocity of a fluid flowing out through a hole beneath the surface
p + gh + ½ v2 = constant p A 0 applied to points A and B gives
p + gh = p + ½ v2 h 0 0 B v gh = ½ v2 Torricelli law
p0 Pitot tube
The tube measures the speed of the liquid.
2 pA + ½ v = pB
½ v2 = p = gh
v = (2gh)
* simple Pitot tube
Pitot tube have been used in submarines, ships and aeroplanes measuring speed. Another type of Pitot tube The wing
B The velocity difference causes a v1 pressure difference and a net force A upwards. v2
2 2 p1 + ½ v1 = p2 + ½ v2
Pressure difference:
2 2 p = ½ v1 - ½ v2
Force:
2 2 F = (½ v1 - ½ v2 )A Venture tube measures the the volume flow dV/dt in pipe systems
h2 h1
A1 v1 A2 v2
2p 2gh V A1 2 A1 2 2 A21 A1 p1 + ½ v1 = p2 + ½ (1v2 2 ) (1 2 ) A2 A2
From continuum equation A1v1 = A2v2 we get v2 = A1/A2 v1
The volume flow Suction effect on a roof in storm
v1 = storm velocity
v2 = 0 under the roof
Pressure difference: 2 2 2 p = ½ v1 - ½ v2 = ½ v1
SUction force on the roof:
2 F = ½ v1 A