# Fluid mechanics. Pressure at depth In a cylindrical column of water, as in any cylinder the volume is the height x cross sectional area The density of.

## Presentation on theme: "Fluid mechanics. Pressure at depth In a cylindrical column of water, as in any cylinder the volume is the height x cross sectional area The density of."— Presentation transcript:

Fluid mechanics

Pressure at depth In a cylindrical column of water, as in any cylinder the volume is the height x cross sectional area The density of the water = mass ÷volume

Pressure at depth The mass of water = density x volume = density x height x area The weight of water = mass x gravity = density x height x area x gravity ρhAg ( ρ = density) ρ

Pressure at depth Pressure = force ÷ area (weight ÷ area) =(density x height x area x gravity) ÷ area = Density x height x gravity (area cancels out) Pressure = pgh

Pressure at depth h pgh p A Pressure = pgh Since the increases uniformly with depth The average pressure = pgh/2 or pgh (h = mean depth)

Pressure at depth h pgh p A Pressure = force ÷ area Force is pressure x area F = pgAh This is the hydrostatic thrust acting on the side of the cylinder The point at which it acts Is 2/3 the depth from the surface of the water (1/3 from the bottom) FF

Archimedes F1F1 m mg If a mass is suspended By a cable in air the tension force in the cable is equal to the weight of the mass F 1 = mg = (ρVg) body where ρ is the density of the body V is the volume of the body g = gravity

Archimedes If a body is totally or partially submersed in water (or other liquid) it will displace some of the water (the water level will rise). The volume of water displaced is the same as the volume of the water F1F1 m F mg F up

Archimedes The body will experience an apparent loss in weight which is equal to the weight of water (or other liquid displaced) This apparent loss in weight is equal to the up-thrust force of the liquid on the body F2F2 m F mg F up

Archimedes F2F2 m F mg F up F 2 = F mg – F up F 2 = (ρVg) body – (ρVg) water where ρ is the density of the body V is the volume of the body g = gravity and ρ is the density of the water (liquid) V is the volume of the water(liquid)

Flow through a tapered pipe Flow Volume flow = Volume/time (m 3 /s) Mass flow = Mass /time (kg/s) Volumetric flow symbol V V = Av (m 3 /s) m = ρ V A = area v= velocity Ρ = density Av (kg/s)

Flow through a tapered pipe Flow Volume flow = Volume/time (m 3 /s) Mass flow = Mass /time (kg/s) Volumetric flow symbol V V = Av (m 3 /s) m = ρ V A = area v= velocity Ρ = density Av (kg/s)

Flow through a tapered pipe Flow Rearranging for v v = ρA A = area v= velocity Ρ = density V A1A1 A2A2 Since V is the same at inlet area A1 and area A2 and ρ is the same we can calculate the velocity of flow at each by using the two different areas in the equation

Flow through a tapered pipe Flow Rearranging for v v = A A = area v= velocity Ρ = density V A1A1 A2A2 V V v 1 = v 2 = A1A1 A2A2 Area = πr 2

Download ppt "Fluid mechanics. Pressure at depth In a cylindrical column of water, as in any cylinder the volume is the height x cross sectional area The density of."

Similar presentations