Equation of continuity

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1. Presentation Submitted To Mr. Zahid Iqbal Presented By 2016-BT-Mech-703 2o16-BT-Mech-726 Department Mechanical M.N.S U.E.T MULTAN

2. Continuity Equation in Fluid Mechanics The product of cross sectional area of the pipe and the fluid speed at any point along the pipe is constant. This product is equal to the volume flow per second or simply flow rate. Mathematically it is represented as Av = Constant Continuity equation derivation Consider a fluid flowing through a pipe of non uniform size.The particles in the fluid move along the same lines in a steady flow.

3. M If we consider the flow for a short interval of time Δt,the fluid at the lower end of the pipe covers a distance Δx 1  with a velocity v 1  ,then: Distance covered by the fluid =  Δx 1  = v 1 Δt Let A 1  be the area of cross section of the lower end then volume of the fluid that flows into the pipe at the lower end =V= A 1  Δx 1  = A 1 v 1   Δt

4. If ρ is the density of the fluid,then the mass of the fluid contained in the shaded region of lower end of the pipe is: Δm 1 =Density × volume Δm 1  = ρ 1 A 1 v 1 Δt ——–( 1 ) Now the mass flux defined as the mass of the fluid per unit time passing through any cross section at lower end is: Δm 1/Δt = ρ 1 A 1 v 1 Mass flux at lower end =  ρ 1 A 1 v 1 ———————(2) If the fluid moves with velocity v 2  through the upper end of pipe having cross sectional area A 2  in time  Δt,then the mass flux at the upper end is given by: Δm 2/Δt  = ρ 2 A 2 v 2  Mass flux at upper end = ρ 2 A 2 v 2  ———————–(3)

5. Since the flow is steady,so the density of the fluid between the lower and upper end of the pipe does not change with time.Thus the mass flux at the lower end must be equal to the mass flux at the upper end so: ρ 1 A 1 v 1 =  ρ 2 A 2 v 2  ———————-(4) In more general form we can write : ρ A v =constant This relation describes the law of conservation of mass in fluid dynamics.If the fluid is in compressible ,then density is constant for steady flow of in compressible fluid so   ρ 1  =ρ 2 Now equation (4) can be written as: A 1  v 1 =  A 2  v 2 In general: A v = constant

6. Question: Calculate the velocity if 10 m3/h of water flows through a 100 mm inside diameter pipe.If the pipe is reduced to 80 mm inside diameter. Solutions (1) Velocity of 100 mm pipe ( 10 m3/h)(1/3600 h/s)=v100(3.14(0.1 m)2/4) Or v100 =(10 m3/h)(1/3600 h/s)(3.14(0.1)2/4) V100 = 0.35   m/s Answer

7. (2) Velocity of 80 mm pipe According To Formula (10 m3/h)(1/3600 h/s)=v80(3.14(0.08 m)2/4) Or v80=(10 m3/h)(1/3600 h/s)(3.14(0.08 m)2/4) V80 = 0.55  m/s Answer :

8. Question :- Water runs through a water main of cross-sectional area 0.4 m 2 with a velocity of 6 m/s. Calculate the velocity of the water in the pipe when the pipe tapers down to a cross-sectional area of 0.3 m 2 . Solution using formula

9. Question:- Water enters a typical garden hose of diameter 1.6 cm with a velocity of 3 m/s. Calculate the exit velocity of water from the garden hose when a nozzle of diameter 0.5 cm is attached to the end of the hose. Solutions First, find the cross-sectional areas of the entry (A 1 ) and exit (A 2 ) sides of the hose .

10. Next, apply the continuity equation for fluids to solve for the water velocity as it exits the hose (v 2 ).