Chapter (4) Motion of Fluid Particles and Streams

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Chapter (4) Motion of Fluid Particles and Streams

Read all Theoretical subjects from (slides Dr.K.AlASTAL) Patterns of Flow Reynolds Number (R e ): A dimensionless number used to identify the type of flow. R = Inertia Forces Viscous Forces = ρ V D = V D μ ν V = mean velocity m s, D = pipe diameter (m) ρ = fluid density (Kg/m ) μ = Dynamic viscosity (Pa. s), ν = kinematic viscosity (m /s) For flow in pipe: If (R 2000) The flow is laminar If (2000 < R 4000) The flow is transitional If (R > 2000) The flow is turbulent Read the two Examples (in slides) Discharge and Mean Velocity: Discharge: The total quantity of fluid flowing in unit time past any cross section of a stream is called the discharge or flow at that section (flow rate). Discharge can be measured by one of the following two methods: 1. In terms of mass (Mass Flow Rate,m, ): Mass of fluid m = time taken to collect the fluid = dm dt = ρ Q (Kg/s). 2. In terms of volume (Volume Flow Rate or discharge,q,): Volume of Fluid Q = = V Time t (m /s). This method is the most commonly used method to represents discharge. There is another important way to represents Q: Q = = = Area Speed Q = A v (m /s). Page (89) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Continuity of Flow Matter cannot be created or destroyed (principle of conservation of mass) Mass entering per = Mass leaving + Increasing of mass in the control unit time per unit time volume per unit time If the flow is steady, no increase in the mass within the control volume. So, Mass entering per unit time = Mass leaving per unit time Continuity Equation for Steady Flow and Incompressible Flow: A v = A v = Q = constant This equation is a very powerful tool in fluid mechanics and will be used repeatedly throughout the rest of this course. The following problems clarify the concept of continuity of flow Problems 1. Water flows through pipe AB 1.3m diameter at speed of 3m/s and passes through a pipe BC of 1.6m diameter. At C the pipe branches. Branch CD is 0.7m in diameter and carries one third of the flow in AB. The velocity in branch CE is 2.7m/s. Find the flow rate in AB, the velocity in BC, the velocity in CD and the diameter of CE. Page (90) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Q = A v = π 4 1.3 3 = 3.982 m /s. Q = Q = 3.982 m /s Q = A v v = Q = 3.982 A π = 1.98m/s. 1.6 4 Q = 1 3 Q ( as given) Q = 1 3 3.982 = 1.327m /s Q = A v v = Q = 1.327 A π = 3.45 m/s. 0.7 4 Q = 2 3 Q = 2 3 3.982 = 2.65 m /s Q = A v A = Q v π 4 D = 2.65 2.7 D = 1.11 m. Page (91) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

2. Pipe flow steadily through the piping junction (as shown in the figure) entering section (1) at a flow rate of 4.5m 3 /hr. The average velocity at section (2) is 2.5 m/s. A portion of the flow is diverted through the showerhead 100 holes of 1-mm diameter. Assuming uniform shower flow, estimate the exit velocity from the showerhead holes. By applying continuity equation for the system given: Q = Q + Q Q = 4.5m /hr = 4.5 60 60 = 1.25 10 m /s Q = A v A = π 4 (0.02) = 3.14 10 m, v = 2.5m/s (given) Q = 3.14 10 2.5 = 7.854 10 m /s Q = Q Q Q = 1.25 10 7.854 10 = 4.646 10 m /s Q = Q in the 100 hole Q = Q 4.646 10 = = 4.646 10 m /s 100 100 Q = A v A = π 4 (0.001) = 7.854 10 m v = Q A = 4.646 10 7.854 10 = 5.68 m/s. Page (92) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

3. For the shown tank, if the inlet pipe is 40 cm in diameter and the outlet pipes are 20 cm in diameter. A. Is the tank filling or emptying and at what rate? B. What should be the velocity of water at the inlet pipe to keep the water level in the tank constant? A. Q = A v = π 4 0.4 3 = 0.377 m /s Q = A v = π 4 0.2 1.5 = 0.047 m /s Q = A v = π 4 0.2 2 = 0.063 m /s Q (Q + Q ) = 0.377 (0.047 + 0.063) = +0.267m /s Since the inlet flow rate is larger than outlet flow rate, the tank is filling and with flow rate of 0.267 m 3 /s. B. The water level in the tank will be constant if the inlet flow equals the outlet flow. Q = (Q + Q ) π 4 0.4 v = (0.047 + 0.063) v = 0.875 m/s. Page (93) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

4. In the figure shown is the tank filling or emptying? At what velocity is the water level rising or falling in this circular tank? 1 3 2 Q = Q = π 4 0.1 3 = 0.0235 m /s Q = π 4 0.15 2.1 = 0.0371 m /s Q = π 4 0.075 1.2 = 0.0053 m /s Q = Q + Q = 0.0371 + 0.0053 = 0.0424 m /s Q Q = 0.0235 0.0424 = 0.0189 Q < Q Emptying. The rate of emptying is Q = 0.0189 m /s Q = A v 0.0189 = π 4 1.8 v v = 0.0074 m/s. Page (94) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Fluid Mechanics 5. In the figure shown below, the diameter of pipes (1) and (2) is 3cm and the diameter of pipe (3) is 4cm. Alcohol (SG = 0.8) enters section (1) at 6m/s while water enters section (2) at 10 m/s. Assuming ideal mixing of incompressible fluids, compute the exist velocity and the density of the mixture in section (3). Calculation of velocity in section (3): Q + Q = Q Q = π 4 0.03 6 = 0.00424 m /s Q = π 4 0.03 10 = 0.00706 m /s 0.00424 + 0.00706 = π 4 0.04 v v = 9m/s. Calculation of density of mixture (in section 3): m + m = m We know that m = ρ Q ρ Q + ρ Q = ρ Q 800 0.00424 + 1000 0.00706 = ρ (0.00424 + 0.00706) ρ = 925 Kg/m. Page (95) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

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