Sourabh V. Apte 308 Rogers Hall sva@engr.orst.edu 1
Topics Quick overview of Fluid properties, units Hydrostatic forces Conservation laws (mass, momentum, energy) Flow through pipes (friction loss, Moody Chart) Examples and Problem Solving Procedure Sample examples are from past FE exams Morning session (short answer) problems 2
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- Pressure in a continuously distributed uniform static fluid varies only with the vertical distance and is independent of the shape of the container - Pressure is same at all points on a given horizontal plane in a static continuous fluid. Pressure increases with depth in the fluid. 5
Projected area 6
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Problem 1 In a static fluid, pressure at a given level is same. 3 1 2 4 5 8
In a static fluid, pressure at a given level is same. Start from A and write equations at intermediate points 1,2,3,4 up to B for pressure using the hydrostatic formula. 3 1 2 4 5 Choose proper and consistent units Evaluate each equation and intermediate pressures (OR) add the equations to directly evaluate the pressure difference between points A and B. 9
Problem 2 Take the projected area, and compute the horizontal force on the projected area. Magnitude of horizontal force A h=10 4 It acts at y cp distance from the centroid γh B A y Take moments about hinge (A) to find force at B F A x B x Do a force balance on the gate. Take all forces acting on the gate In the x-direction, add them and equate to zero. 10
Take the projected area, and compute the horizontal force on the projected area. γ(h-4) γh A B 4 γ(h-4) h=10 = + = γ(4) F 1 F 2 A B A y A x B x B x width width - Now that we know the forces and where they act, take moments about the hinge and equate to zero. This will give B x - Sum up all forces in x-directions and equate to zero. This will give A x - Note that you should end up with the same answer, except here you don t have to remember the formula for y p 11
Problem 3 Force diagram (FBD) Weight of fluid In the region gives the vertical force => b=(2)/2 + 0.083 F v = weight of fluid in the shaded region shown = F v1 + F v2 = ρg(vol 1 + Vol 2 ) =1000x9.8x[(2)(2)(3)+(2)π/4(2) 2 N=117.5 kn+ 61.5kN=179 kn F v1 acts at the centroid of the rectangular section so @ x = 1m from O F v2 acts at the centroid of the quarter circle so @ x= 4r/3π = 0.848 m from O Take moments around O to determine F: -F (2)+F H (1+0.083) - F v1 (1) - F v2 (0.848) = 0 => F = 0 12
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wall wall 14
Forces on CV may include - forces due to gage pressure - weight of fluid within CV - viscous shear stresses - any other reaction forces or external forces Vector equation. Must take proper components of velocity. 15
Work input to pump Head: H 16
Some Morning Session Questions 17
ρ= S (ρ_water) = 1.59 (1000) 18
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P_gage = ρgh = S (ρ_water) h =(1.025)(1000)(9.8)(152.4) Pa 20
P 1 /γ + V 12 /2g + z 1 = P 2 /γ + V 22 /2g + z 2 A 1 V 1 = A 2 V 2 21
F h /w = ρ g h cg A = (0.8)(1000)(9.8)(0.91+1.22)/2 (0.91+1.22) F v /w= ρ g V_displaced = (0.8)(1000)(9.8)[1.22x(0.91+1.22)-π(1.22) 2 /4] rectangle Quarter circle 22
Re = ρvd/µ = ρ(q/a)d/µ 23
inlet outlet P 1 /γ + V 12 /2g + z 1 = P 2 /γ + V 22 /2g + z 2 -H_pump +h L V 1 = V 2 = V; P 1 = P 2 = P atm ; z 1 = 100m; z 2 = 145.7 Pump Power = γ Q H_pump 24
h f = (4τ w /γ) (L/D) If not known: Linear momentum for the pipe: P 1 πr 2 - P 2 πr 2 + τ w (2πRL) = m(v 2 -V 1 ) = 0 Energy Equation: P 1 /γ + V 12 /2g + z 1 = P 2 /γ + V 22 /2g + z 2 + h f Z 1 = Z 2 ; V 1 = V 2 25
h f = (4τ w /γ) (L/D) If not known: Linear momentum for the pipe: P 1 πr 2 - P 2 πr 2 + τ w (2πRL) = m(v 2 -V 1 ) = 0 Energy Equation: P 1 /γ + V 12 /2g + z 1 = P 2 /γ + V 22 /2g + z 2 + h f Z 1 = Z 2 ; V 1 = V 2 26
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Pipes in Series f 1, L 1,d 1,V 1 f 2, L 2,d 2,V 2 f 3, L 3,d 3,V 3 V 1 d 1 2 = V 2 d 2 2 = V 3 d 3 2 Δh A B = V 2 1 2g + V 2 2 2g f 1 L 1 + K 1 d 1 f 2 L 2 + K 2 d 2 + V 2 3 2g f 3 L 3 + K 3 d 3 1. If flow rate and pipe dimensions are given, one can easily evaluate the total head loss in pipes in series and power required to pump fluid through this pipe 2. If head loss is given and we need flow rate, itera?ve procedure is required. Assume rough turbulent flow, guess fs, and use Colebrook rela?on to arrive at solu?on
Pipes in Parallel f 1, L 1,d 1,V 1 f 2, L 2,d 2,V 2 f 3, L 3,d 3,V 3 Conserva?on of Mass m 1 + m 2 + m 3 = m = ρ Q Q = Q 1 + Q 2 + Q 3 Q = π ( 4 V 1d 2 1 + V 2 d 2 2 2 + V 3 d ) 3 Total Head loss for parallel pipes Δh A B = Δh 1 = Δh 2 = Δh 3
Problem 4 Free stream P s P 33
Problem 5 1 Steady State energy equation 2 major minor In units of ft 2 elbows Major loss Valve entrance exit - Losses dependent on Reynolds number Re = VD/ν - But, the velocity V is unknown, so we cannot find the Reynolds number, and thus we cannot find f, the friction factor We have to do some iterative procedure. Assume that the flow is fully turbulent. In this case, the friction factor is function of the surface roughness of the pipe only (independent of Re. Check Moody Diagram). For steel pipe ε=0.00015 ft => ε/d=0.0003. From Moody chart, choose the line corresponding to this surface roughness, choose the friction factor for fully turbulent flow. 34
0.016 35
Guess 1: Fully turbulent flow => f 0.016. Substitute this f above. Find the velocity V = 12.27 ft/s. Now verify whether this velocity really gives a fully turbulent flow. Re = ρvd/µ = Guess 2: With this Reynolds number, and the surface roughness, check Moody diagram and get a better value for f => f 0.02. Substitute this f above. Find the velocity V = 11.2 ft/s. Recalculate Re. Re = 4 x 10 5 Guess 3: Use this Re and the surface roughness, find f. f=0.02 Same as above. Thus, we have reached the end of the iteration. V = 11.2 ft/s 36
Problem 6 V 1 and V 2 and P 1 and P 2 are not known. Mass conservation between sections 1 and 2 Bernoulli gives us another equation Rearranging and using result from mass-conservation From manometer reading we can find the pressure drop => Then calculate V 1, and the ideal discharge is given as Q = V 1 A 1 and mass flux is = ρv 1 A 1 37
In reality, there is a boundary layer that develops inside the venturi, the actual flow rate is slightly less than the one calculated above. This is expressed in terms of Q_actual = Cd V1 A1 The discharge coefficient depends on the Reynolds number and the area ratio of the venturi. Find Re. and knowing the area ratios, find the discharge coefficient. Use it to find Q_actual C_d ~ 0.985 Q_actual = 0.175 m3/s 38
Problem 7 P 2 is unknown, V 2 is unknown Conservation of Mass: ρv 1 A 1 = ρv 2 A 2 Bernoulli: => Consider forces on the fluid => F x = 7740 lb to the left => F y = 1910 lb upward 39