1. Fluid Flow P7 - use the continuity of volume and mass flow for an incompressible fluid to determine the design characteristics of a gradually tapering pipe Definition of terms: Continuity of volume and mass flow = consider a fluid (eg water) moving thro’ a pipe, what goes in one end arrives out from the other. This can be applied to a volume of water such as 1 litre ( 1x10 -3 m 3 ) or to the mass of the water :– 1 litre = 1 kg Incompressible fluid = liquids such as water are deemed to be incompressible, they cannot be squashed ! unlike gases such as air that can be compressed, (squashed). Gradually tapering pipe = a normal hosepipe would be expected to have constant diameter throughout its length but may have a nozzle at the end to focus the flow, the nozzle will normally be in the form of a simple taper where the diameter reduces.

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3. Fluid Flow To calculate volumetric flow & mass flow the concept of fluid density is used. From previous work – Density = mass ÷ volume ρ= m/v When: ρ = density (kg/m 3 ) m = mass (kg) v = volume (m 3 ) Remember that the relative density of a substance is relative to water which has a mean value of 1000 kg/m 3

4. Fluid Flow The link between mass, volume and density can be used as follows:- eg 1.Calculate the mass of 500 litres of oil which has a relative density of 0.82 Soln. Density = mass/volumetherefore mass = density x volume ρ = m/v therefore m = ρ x v when density ρ = 0.82 x 1000 kg.m -3 = 820 kg.m -3, volume v = 500 litres = 0.5 m 3 mass = 820 kg.m -3 x 0.5 m 3 = 410 kg

5. Fluid Flow The previous equation can again be used even when a time factor is introduced. eg 2.Calculate the mass of oil of relative density 0.88 which is pumped through a heated pipeline in 2 hours if the pumping system is able to deliver 50 litres per minute. Soln. As before:- mass = density x volume m = ρ x v when density ρ = 0.88 x 1000 kg.m -3 = 880 kg.m -3, volume v = 50 litres = 0.05 m 3 per minute mass = 880 kg.m -3 x 0.05 m 3 = 44 kg per minute In 2 hours the mass delivered will be 44kg/min x 120 minutes Total mass in 2 hours = 5280 kg or 5.28 tonnes of oil

6. Fluid Flow An alternative method is: eg 2.Calculate the mass of oil of relative density 0.88 which is pumped through a heated pipeline in 2 hours if the pumping system is able to deliver 50 litres per minute. Soln. As before:- mass = density x volume m = ρ x v when density ρ = 0.88 x 1000 kg.m -3 = 880 kg.m -3, volume v = 50 litres = 0.05 m 3 per minute volume in 2 hours = 0.05 m 3 x 120 min = 6 m 3 total mass = 880 kg.m -3 x 6 m 3 = 5280 kg as before

7. Fluid Flow Try: 1.Calculate the volume of oil of relative density 0.8 held in a storage tank of total mass 3.26 tonnes if the tank has a mass of 860 kg when empty. 3000 litres 2.Calculate the time taken to drain the tank of oil if the drain tap allows the oil to flow out at a rate of 100 litres/minute. ½ hour 3.If pure water is pumped into the tank at a rate of 75 litres/minute find the total mass of tank and water after 1 hour of pumping. 5.36 tonne

8. Fluid Flow To calculate volumetric flow & mass flow the concept of quantity of flow is needed. Quantity of flow is usually given the symbol Q The quantity of flow or flowrate can be measured on a volumetric basis such as cubic metres per second (m 3 /s), this is to SI and is often termed the cumec. It may also be measured on a volumetric basis using units such as litres per minute (l.p.m.). This is not to SI but is very common and convenient in industry.

9. Fluid Flow If the density ρ of the fluid is known in kg/m 3 (kg.m -3 ) Then the mass flow rate can be found from ρ Q kg/s This is often more convenient as ρ. a. v kg/s The mass flowrate is sometimes given the symbol of m for mass but with a single dot above the m It is measured in kilograms per second kg/s or kg.s -1

10. Fluid Flow The quantity of liquid flowing Q through a pipe can be found from: the cross sectional area of the pipe a m 2 the velocity of liquid flowing thro’ the pipe v m/s It must be noted that as volume and velocity have a common symbol it is common to give fluid velocity the symbol c m/s The quantity of fluid flowing Q becomes: the cross sectional area of the pipe x the velocity of liquid flowing thus Q = a.v m 3 /s (rem. 1 m 3 = 1000 litres)

11. Fluid Flow The continuity equation for flow can now be established. For a simple horizontal converging nozzle as shown below:- The quantity fluid entering B must equal the quantity leaving at C The quantity fluid entering B = a 1 v 1 = Q in The quantity fluid leaving C = a 2 v 2 = Q out Exit CEntrance B Velocity v 1 area a 1 Velocity v 2 area a 2

12. Fluid Flow As the quantity fluid entering B must equal the quantity leaving at C Q in = Q out Therefore a 1 v 1 = a 2 v 2 This is called the equation of continuity. As what goes in = what comes out any reduction in the area of the outlet must cause an equivalent rise in the outlet velocity to maintain continuity. In most cases pipes/nozzles are circular giving an area of πd 2 /4 The equation can be written as:

13. Fluid Flow The equation can simplified as and eventually :- This shows that the flow velocity is inversely proportional to the square of the diameter. If the diameter is reduced by ½ the velocity will be increased by 4x the original value.

14. Fluid Flow Example. In the nozzle shown below water enters at B at a rate of 32 kg/s and is expected to leave C at 13 m/s. If the nozzle entrance diameter is 75mm find: a.the entrance velocity (v 1 m/s) b.the exit diameter (d 2 mm) Exit CEntrance B Velocity v 1 diameter d 1 Velocity v 2 diameter d 2

15. Fluid Flow Solution: (working to SI throughout) Part a. Entrance velocity v 1 As Q = av (flowrate = area x velocity then velocity = flowrate/area) Therefore v 1 = Q/a 1 Water enters at 32 kg/s which is 32/1000 m 3 /s (1 m 3 of water = 1000kg) Flowrate Q = 0.032 m 3 /s Area of entrance a 1 = πd 2 /4 = π x 0.075 2 /4 = 0.00442 m 2 Thus entrance velocity v 1 = Q/a 1 = 0.032 m 3 /s ÷ 0.00442 m 2 = 7.24 m/s

16. Fluid Flow Solution: (working to SI throughout) Part b. Exit diameter d 2 To find exit diameter first find exit area a 2 As Q = av (flowrate = area x velocity then area = flowrate/velocity) Therefore a 2 = Q/v 2 From previous:- Flowrate Q = 0.032 m 3 /s Exit velocity v 2 = 13 m/s (given) Area of exit a 2 = Q/v 2 = 0.032 m 3 /s ÷ 13 m/s = 0.00246 m 2 Area of exit a 2 = πd 2 /4 = 0.00246 m 2 Therefore exit diameter d 2 = = 0.056m = 56mm

17. Fluid Flow Try: 1. Water flows through a 50mm diameter pipe at a rate of 2.1 ms -1 Find:a.the volume rate of flow in m 3 /s b.the mass rate of flow in kg/s 2. Water flows through a 250mm diameter pipe at 570 l/min,Find the velocity of water in the pipe 3.Oil flows through a pipe which tapers from 200mm diameter to 100mm diameter at a rate of 3 m 3 /min. Find the velocity of the oil when a.the pipe diameter is 200mm b.the pipe diameter is 100mm