The Hoover - USA

The Three Gorges - China

Thrust on an immersed plane surface P5calculate the resultant thrust and overturning moment on a vertical rectangular retaining surface with one edge in the free surface of a liquid Definition of terms: Resultant thrust = overall pushing force Newton Overturning moment = product of force and distance from the pivot Nm Retaining surface in a liquid = something like a dam or sluice gate. Hoover dam

Thrust on an immersed plane surface To calculate thrust (force) on the surface it is crucial to find the pressure acting due to the weight of the fluid. From previous work – pressure at depth = ρ x g x h When: ρ = density (kg/m 3 ) g = 9.81 (ms -2 ) h = fluid depth (m) The previous equation can be used to obtain a value of pressure acting at the depth given.

Thrust on an immersed plane surface Try: 1.A diver swims in sea-water of relative density at a depth of 20m below surface. Calculate the pressure on the diver at this depth. (201 kPa) 2.A tanker carries litres of petrol of relative density Calculate the total weight of the petrol. (218 kN) Note:1m 3 or 1000 dm 3 is approx. equal to 1000 litres

Thrust on an immersed plane surface Pressure at depth is converted to thrust (force) using F = PA (Force = Pressure x area) eg1. Calculate the pressure and thrust acting on the base of a tank 2m wide x 4m x 750mm deep holding oil of density 800 kgm -3 Solution. Pressure = ρgh when ρ = 800 kgm -3, g = 9.81, h = 0.75m Pressure = 800 x 9.81 x 0.75 = 5886 Pa (5.886 kN/m 2 ) Area on which the pressure acts (base of tank) = 2m x 4m = 8m 2 Thrust (Force) = Pressure x area = 5886 Pa x 8m 2 = N Thrust = kN

Thrust on an immersed plane surface Try: 1. A tank contains fuel oil of relative density 0.72 to a depth of 5m. Calculate the load on a valve plate of area 500 mm 2 in the base of the tank due to the weight of the oil. (17.6N) 2. A hatch cover for a submarine has a surface area of 0.4 m 2. What is the total force on it due to the weight of water at a depth of 100m ? Assume water density is 1025 kg m -3 (402 kN)

Thrust on an immersed plane surface If the pressure at depth is converted to thrust (force) using F = PA then it is important to consider where on the surface the force can be assumed to act before taking moments. eg1. Calculate the pressure and thrust acting on a flat surface 5m wide holding back pure water to a depth of 4 metres. Solution. Pressure = ρgh when ρ = 1000 kgm -3, g = 9.81, h = 4 m Pressure = 1000 x 9.81 x 4 = Pa (39.24 kN/m 2 ) at the bottom !

Thrust on an immersed plane surface If the surface suffering pressure runs down into the liquid then:- At the surface where h = 0 metres the pressure due to water = 0Pa It is most sensible to use the average pressure value between 0 and 4 m of water depth. ie.divide max. pressure (at bottom) by 2 Average pressure = 39240/2 = Pa (19.62 kN/m 2 ) To calculate the thrust then Force = pressure x area is used. when the ‘wet’ surface area = 5m x 4m = 20 m 2 Thus:- thrust = Pa x 20 m 2 = N (392.4 kN) Note:- average pressure of each fluid must be used when dealing with layered fluids – oil on water etc.

Thrust on an immersed plane surface Try: Find the max. pressure, average pressure and thrust on a flat surface 750 mm wide holding back oil of relative density 0.8 to a depth of 500mm. When:- ρ = 0.8 x 1000 kgm -3, g = 9.81, h = 0.5m Max. Pressure (P max ) = 3924 Pa Average pressure (P av ) = 3924/2 = 1962 Pa (1962 N/m 2 ) Thrust = Average Pressure x ‘wet’ Area When rectangular wet area = 0.5m x 0.75m = m 2 Thrust = 1962 N/m 2 x m 2 = N

Thrust on an immersed plane surface The previous work using average pressure assumes a depth of ½ of the wet depth. This is acceptable for flat surfaces and the position is known as the centroid of the surface. The total thrust is seen to be the pressure at the centroid x the wet surface area. To find the actual position of the force acting it is important to consider how the pressure will vary with depth.

Thrust on an immersed plane surface To find the actual position of the force acting it is important to consider how the pressure will vary with depth. Pressure = 0 Pressure = Max. Fluid Flat surface Pressure gradient

Thrust on an immersed plane surface On a flat surface the pressure gradient is found to be a triangle The thrust can be assumed to act at a point known as the centre of pressure. The centre of pressure is the centroid of the pressure gradient. This is 1/3 of the height from the base of the triangle Fluid Flat surface

Thrust on an immersed plane surface If overturning moments are to be calculated it must be remembered that the total thrust will act at the centre of pressure 1/3 of the height of the wetted area from the base up. Consider eg.1 to calculate the overturning moment on the surface assuming it to be hinged at a point 1m above the water (ie 5m up from the base) From before:- Average pressure = Pa (19.62 kN/m 2 ) thrust = N Centre of pressure from the base = 1/3 x 4m = 4/3m (1.33m) BUT from the pivot point the distance = 5m – 1.33m = 3.67m overturning moment = N x 3.67m = Nm

Thrust on an immersed plane surface Try An oil storage tank of width 2m, length 4m and depth 1m is filled to a depth of 900mm with oil of density 800 kg/m 3. Calculate the horizontal thrust on the side and on the end of the tank due to the oil. If one end panel of the tank is hinged at its top calculate the overturning moment on the panel due to the weight of the oil. Hinge Fluid level

Thrust on an immersed plane surface Average pressure P av = (ρgh) ÷ 2 when:- ρ = 800 kgm -3, g = 9.81, h = 0.9m P av = 800 x 9.81 x 0.9 = ÷ 2 = Pa Area of side = 4m x 0.9m = 3.6 m 2 Area of end = 2m x 0.9m = 1.8 m 2 Thrust = Average pressure x area Thrust on side = Pa x 3.6 m 2 = N Thrust on end = Pa x 1.8 m 2 = N Centre of pressure = 1/3 from base 0.9m x 1/3 = 0.3m from base Distance from hinge = 1m – 0.3m = 0.7m turning moment on end plate = thrust x distance = x 0.7 Overturning Moment = Nm