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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
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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.
6 Projected area
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In a static fluid, pressure at a given level is same.
1 2
3
4 5
Problem 1
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1 2
3
4 5
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.
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.
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Take the projected area, and compute the horizontal force on the projected area.
Magnitude of horizontal force
It acts at ycp distance from the centroid
F
Bx
Ax
Ay
Take moments about hinge (A) to find force at B
Do a force balance on the gate. Take all forces acting on the gate In the x-direction, add them and equate to zero.
h=10’
γh
4’
A
B
Problem 2
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Take the projected area, and compute the horizontal force on the projected area.
Bx
γ(4)
γ(h-4)
+
h=10’
γh
4’
γ(h-4) A
B = =
Ax
Ay
F1
F2
A
B Bx
- Now that we know the forces and where they act, take moments about the hinge and equate to zero. This will give Bx
- Sum up all forces in x-directions and equate to zero. This will give Ax - Note that you should end up with the same answer, except here you don’t have to remember the formula for yp
width
width
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Force diagram (FBD) Weight of fluid In the region gives the vertical force
Fv = weight of fluid in the shaded region shown = Fv1 + Fv2 = ρg(Vol1 + Vol2) =1000x9.8x[(2)(2)(3)+(2)π/4(2)2 N=117.5 kN+ 61.5kN=179 kN
Fv1 acts at the centroid of the rectangular section so @ x = 1m from O Fv2 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)+FH(1+0.083) - Fv1 (1) - Fv2(0.848) = 0 => F = 0
=> b=(2)/2 + 0.083
Problem 3
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wall
wall
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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.
16 Work input to pump Head: H
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Some Morning Session Questions
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ρ= S (ρ_water) = 1.59 (1000)
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P_gage = ρgh = S (ρ_water) h =(1.025)(1000)(9.8)(152.4) Pa
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P1/γ + V12/2g + z1 = P2/γ + V2
2/2g + z2
A1V1 = A2V2
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Fh/w = ρ g hcg A = (0.8)(1000)(9.8)(0.91+1.22)/2 (0.91+1.22)
Fv/w= ρ g V_displaced = (0.8)(1000)(9.8)[1.22x(0.91+1.22)-π(1.22)2/4] rectangle Quarter circle
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Re = ρVD/µ = ρ(Q/A)D/µ
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P1/γ + V12/2g + z1 = P2/γ + V2
2/2g + z2 -H_pump +hL
V1 = V2 = V; P1 = P2 = Patm; z1 = 100m; z2 = 145.7
Pump Power = γ Q H_pump
inlet outlet
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hf = (4τw/γ) (L/D)
If not known:
Linear momentum for the pipe: P1 πR2 - P2πR2 + τw(2πRL) = m(V2-V1) = 0
Energy Equation: P1/γ + V12/2g + z1 = P2/γ + V2
2/2g + z2 + hf
Z1 = Z2; V1 = V2
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hf = (4τw/γ) (L/D)
If not known:
Linear momentum for the pipe: P1 πR2 - P2πR2 + τw(2πRL) = m(V2-V1) = 0
Energy Equation: P1/γ + V12/2g + z1 = P2/γ + V2
2/2g + z2 + hf
Z1 = Z2; V1 = V2
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Pipes in Series
f1,L1,d1,V1 f2 ,L2 ,d2 ,V2 f3,L3,d3,V3
V1d12 = V2d2
2 = V3d32
ΔhA→B =V12
2gf1L1d1
+ K1∑
+V22
2gf2L2d2
+ K2∑
+V32
2gf3L3d3
+ K3∑
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
Conserva?on of Mass m1 + m2 + m3 = m = ρ Q
∴ Q = Q1 + Q2 + Q3 ⇒ Q =π4V1d1
2 +V2d22 +V3 d3
2( )
ΔhA→B = Δh1 = Δh2 = Δh3
Total Head loss for parallel pipes
f1,L1,d1,V1
f2 ,L2 ,d2 ,V2
f3,L3,d3,V3
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P
Free stream Ps
Problem 4
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Steady State energy equation
1
2
major minor
In units of ft
entrance exit
Valve 2 elbows Major loss
- 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
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.
We have to do some iterative procedure.
Problem 5
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0.016
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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
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 105
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V1 and V2 and P1 and P2 are not known.
Bernoulli gives us another equation
Rearranging and using result from mass-conservation
From manometer reading we can find the pressure drop
=>
Then calculate V1, and the ideal discharge is given as Q = V1 A1 and mass flux is = ρV1 A1
Mass conservation between sections 1 and 2
Problem 6
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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
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
Q_actual = Cd V1 A1 The discharge coefficient depends on the Reynolds number and the area ratio of the venturi.
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P2 is unknown, V2 is unknown
Bernoulli: =>
Consider forces “on” the fluid
=> Fx = 7740 lb to the left
=> Fy = 1910 lb upward
Conservation of Mass: ρV1 A1 = ρV2 A2
Problem 7
