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Module 3A: Flow in Closed Conduits

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Module 3A: Flow in Closed Conduits Module 3a: Flow in Closed Equation of Continuity Conduits At any given location (assuming incompressible fluid): Continuity, Momentum, and Energy (Bernoulli) Flow In = Flow Out Qin =Qout Robert Pitt Since Q = (Velocity)(Cross-Sectional Area of Flow) = VA University of Alabama Where V = average (mean) velocity across the and profile. Shirley Clark (VA) = (VA) Penn State - Harrisburg in out Continuity Equation Continuity Equation Example: Example: Water flows in a 10-cm diameter pipe at a mean velocity of Water flows in a 10-cm diameter pipe at a mean velocity of 1.5 m/sec. What is the discharge rate of flow at a 1.5 m/sec. What is the discharge rate of flow at a temperature of 5oC? temperature of 5oC? Had the example asked for the mass rate of flow, the mass Using the continuity equation, rate of flow is equal to the flow Q multiplied by the density Q = VA of the fluid at the temperature of interest. Substituting: Mass rate of flow = (0.012 m3/sec)(1000 kg/m3 ) ⎛ π ⎞ 2 Q = ()1.5m /sec ⎜ ⎟()0.10m ⎝ 4 ⎠ Mass rate of flow = 12 kg/sec Q = 0.012m3 /sec 1 Continuity Equation Continuity Equation Example: Example: Water is flowing in a 2-inch diameter pipe at a velocity of 16 Find the cross-sectional area of flow at points 1 and 2 (assume ft/sec. The pipe expands to a 4-inch diameter pipe. Find the that the pipe is flowing full). velocity in the 4-inch diameter pipe. 2 πD1 2 π (2in)(1 ft /12in) 2 By the Continuity Equation: A1 = = = 0.022 ft 4 4 2 V1A1 = V2A2 πD 2 π ()(4in 1 ft /12in ) A = 2 = = 0.086 ft 2 2 4 4 Substituting: V1 A1 = V2 A2 2 2 ()16 ft /sec ()()0.022 ft = V2 0.086 ft V2 = 4.09 ft /sec Momentum Equation Momentum Equation • This is a vector relationship, i.e., the force equation may act in more than one direction (x-component, y-component, and Example: possible z-component). Determine the force exerted by the nozzle on the pipe shown 3 • The Law of Conservation of Momentum: when the flow rate is 0.01 m /sec. Neglect all losses. The time rate of change in momentum (defined as the mass rate of flow ρAV multiplied by the velocity V) along the path of flow will result in a force called the impulse force. • Net force on a fluid caused by the change in momentum: F = M(V2 –V1) = ρQ(V2 –V1) Where F = net force M = mass flow rate = ρQ Assume the fluid is water. Need to find velocities using V = velocity continuity equation. Need cross-sectional area of flow for Q = flow rate continuity equation. ρ = density 2 Momentum Equation Momentum Equation Solution: Solution: Area at point 1: Velocity at point 1: 3 2 2 2 Q1 = Q = A1V1 = 0.01m /sec = (0.007854m )V1 πD1 π (0.1m) 2 A1 = = = 0.007854m 4 4 V1 = 1.273m/sec Area at point 2: Velocity at point 2: 2 2 3 πD2 π (0.025m) 2 Q 0.01m /sec A2 = = = 0.000491m V2 = = 2 4 4 A2 0.000491m V2 = 20.37m/sec Momentum Equation Momentum Equation Solution: • Momentum equation usually applied to determining Calculating the net force caused by a change in forces on a pipe in a bend. momentum: F = ρQ(V2 −V1) F = (998.2kg / m3 )(0.01m3 /sec)(20.37 −1.273m /sec) F = 190.6kg − m /sec2 = 190.6N After looking at significant figures: F = 190 N From: Metcalf & Eddy, Inc. and George Tchobanoglous. Wastewater Engineering: Collection and Pumping of Wastewater. McGraw-Hill, Inc. 1981. 3 Momentum Equation Momentum Equation Equation for the force in the x-direction: Example: F = p A cosθ -pA + ρQ(V cosθ -V) • Determine the magnitude and direction of the force needed to x 2 2 1 1 2 1 counteract the force resulting from the change in momentum in a horizontal 90o bend in a 200-mm force main. The rate of flow 3 Equation for the force in the y-direction: through the force main is 0.1 m /sec. F = p A sinθ + ρQV sinθ Note: The x-y plane y 2 2 2 is horizontal in this example with equal gravitation forces on If the y-direction is vertical, the weight of the water and all sections of pipe. the pipe will need to be added to the right side of the Fy equation. Momentum Equation Momentum Equation Find the cross-sectional area of flow: Solution: • By continuity, the flow rate does not change. Therefore, Q = ⎛ π ⎞ 2 ⎛ π ⎞ 2 V A = V A . The problem indicates that at both points 1 and 2, A = ⎜ ⎟D = ⎜ ⎟()0.20m 1 1 2 2 ⎝ 4 ⎠ ⎝ 4 ⎠ the diameter is 0.20 m. A = 0.0314m2 Therefore, A1 = A2, and by continuity V1 = V2. Also given: D = 0.20 m Find the velocity of flow: θ = 90o Q 0.10m3 /sec 3 o V = = Looking up: ρ = 998.2 kg/m at 20 C (assume T) A 0.0314m2 γ = 9789 kg/m2-sec2 at 20oC V = 3.18m/sec 4 Momentum Equation Momentum Equation Since the pressure in the system is based only on a Find the pressure (convert velocity into an energy head term velocity component at both points 1 and 2, p1 = p2. which equals the pressure head term): Since V1 = V2 and A1 = A2 by continuity (and same diameter pipe on both sides of the bend) and since p By Bernoulli’s equation: 1 = p2, simplify the force equations: p v 2 1 = 1 F = pAcosθ -pA + ρQ(Vcosθ -V) γ 2g x F = pA(cosθ -1) + ρQV(cosθ -1) Substituting : x Fx = (pA + ρQV)(cosθ -1) p (3.18m / sec)2 1 = 2 2 2 9789kg / m − sec 2(9.81m / sec ) Fy = pAsinθ + ρQVsinθ F = (pA + ρQV)sinθ 2 y p1 = 5045.4 kg/m-sec = 5045 Pa Momentum Equation Substituting: ()pA + ρQV = [(504.5kg / m − sec2 )(0.0314m2 )+ (998.2kg / m3 )(0.10m3/sec)(3.18m /sec)] ()pA + ρQV = 475.8kg − m /sec2 = 475.8N cos θ = cos(90o) = 0 sin θ = sin(90o) = 1 Fx = (475.8N)(0 −1) Fx = −475.8N Fy = (475.8N)(1) Fy = 475.8N Therefore, a thrust block capable of resisting 480 N (sig figs) must be placed against the pipe in both the x-direction and the y-direction and pipe hangers or appropriate bedding will be required to support the pipe from downward gravitational forces. 5 Figure 2.17 Chin 2000 Bernoulli’s Equation (aka Energy Equation) Bernoulli’s (Energy) Equation For Fluid Flow Between Two Points (in a pipe or channel): P V 2 P V 2 1 + 1 + z = 2 + 2 + z + h γ 2g 1 γ 2g 2 f1−2 Where P/γ = pressure head V2/2g = velocity head z = static head hf1-2 = head loss between two points(usually resulting from shear stress along walls of pipe, within fluid, and from momentum changes at entrances, exits, changes in cross-section or direction, and fittings) – also abbreviated HL or hL. Friction Slope = rate at which energy is lost along length Hydraulic Grade = pressure head + elevation head of flow (channel or pipe) Energy Grade = hydraulic grade + velocity head 6 Comparison of Bernoulli’s Equation for Pipe Bernoulli’s Equation Flow vs. Open-Channel Flow From: Terence McGhee. Water Supply and Sewerage, Sixth Edition. From: Metcalf & Eddy, Inc. and George Tchobanoglous. Wastewater McGraw-Hill, Inc., New York, NY. 1991. Engineering: Collection and Pumping of Wastewater. McGraw-Hill, Inc. 1981. Bernoulli’s Equation Bernoulli’s Equation Example: Example: • Water is flowing through a 2-inch pipe at a velocity of 16 ft/sec. Since the centerline does not change elevation z1 = z2, and z’s The pipe expands to a 4-inch pipe. Given that the pressure in the cancel out. 2-inch pipe is 40 psig. What is the pressure in the 4-inch pipe Since friction is negligible, h is negligible (set equal to zero). f1-2 just after expansion, assuming that friction is negligible? Substituting: Given: V = 16 ft/sec 40 lb /in 2 (12 in/ft)2 (16 ft/sec)2 P (12 in/ft)2 (4.1 ft/sec)2 1 f + = 2 + 2 3 2 3 2 g = 32.2 ft/sec 62.4 lbf /ft 2(32.2 ft/sec ) 62.4 lbf /ft 2(32.2 ft/sec ) 3 2 γ = 62.4 lbf/ft 92.31 ft + 3.98 ft = (2.31 in − ft/lbf )P2 + 0.26 ft 2 96.03 ft = (2.31 in - ft/lbf )P2 2 96.03 = (2.31 in /lbf )P2 Solving the Continuity Equation earlier, V2 = 4.1 ft/sec 2 P2 = 41.57 lbf /in = 41.6 psig 7 Bernoulli’s Equation Bernoulli’s Equation Example: Example: What is the pressure at a depth of 300 feet in fresh water? What is the theoretical velocity generated by a 10-foot hydraulic head? Elevation (Depth) Head = 300 feet From Bernoulli’s Equation, look at velocity term (expresses From Bernoulli’s Equation, look at pressure term (all energy is the kinetic energy in system): potential to do work as expressed by the pressure head term): Velocity Head = V2/2g Pressure Head = P/γ g = 32.2 ft/sec2 For water, γ = 62.4 lb /ft3 f Substituting: Substituting: V 2 P 10 ft = 300 ft = 3 2 62.4lbf / ft 2(32.2 ft /sec ) 2 2 2 V 2 = 10 ft(2)(32.2 ft /sec2 ) = 644 ft 2 /sec2 P = (18720lbf / ft )(1ft /144in ) 2 V = 25.4 ft /sec P = 130lbf /in = 130 psi Bernoulli’s Equation including Pumps in Bernoulli’s Equation the System •Example: 2 2 A 1200-mm diameter transmission pipe carries 126 L/sec p V p V from an elevated storage tank with a water surface 1 + 1 + z + E = 2 + 2 + z + h γ 2g 1 pump γ 2g 2 f1−2 elevation of 540 m.
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