CEE 410 Hydraulic Engineering 1- Foundational Concepts Michael D. Doran, P.E. DEE Professor of Practice

What is Hydraulic Engineering? Incompressible fluids Drinking water, wastewater, stormwater, surface water Other fluids (be very careful) Analysis and design of infrastructure What is Hydraulic Engineering?

Outcomes for today Understand application of ‘Continuity Equation’ Understand application of ‘Bernoulli Principle’ Understand volume, mass and energy balances for a ‘Control Volume’ Understand concepts of fluid momentum Outcomes for today

Control Volume

Outputs Inputs Σ 𝑚 in= Σ 𝑚 out Conservation of Mass

Outputs Inputs ΣρQin = ΣρQout Conservation of Mass

Vout Aout Vin Ain Outputs Inputs ΣρVinAin = ΣρVoutAout Continuity

Vout Aout Vin Ain Outputs Inputs ΣVinAin = ΣVoutAout Continuity

Example 1 – Solve for Q3 and V3 below. Q1 = 0.069 m3/s Q2 = 0.044 m3/s 0.21 m Dia 0.21 m Dia 0.15 m Dia Q3? V3?

Example 1 – Solve for Q3 and V3 below. Q1 = 0.069 m3/s Q2 = 0.044 m3/s 0.21 m Dia 0.21 m Dia 0.15 m Dia Q3? V3? Q1 = Q2 + Q3 (ΣQin = ΣQout)

Example 1 – Solve for Q3 and V3 below. Q1 = 0.069 m3/s Q2 = 0.044 m3/s 0.21 m Dia 0.21 m Dia 0.15 m Dia Q3? V3? Q1 = Q2 + Q3 (ΣQin = ΣQout) Q3 = Q1 – Q2

Example 1 – Solve for Q3 and V3 below. Q1 = 0.069 m3/s Q2 = 0.044 m3/s 0.21 m Dia 0.21 m Dia 0.15 m Dia Q3? V3? Q1 = Q2 + Q3 (ΣQin = ΣQout) Q3 = Q1 – Q2 Q3 = 0.069 m3/s - 0.044 m3/s = 0.025 m3/s

Example 1 – Solve for Q3 and V3 below. Q1 = 0.069 m3/s Q2 = 0.044 m3/s 0.21 m Dia 0.21 m Dia 0.15 m Dia Q3? V3? Q3 = A3(V3) = 0.025 m3/s

Example 1 – Solve for Q3 and V3 below. Q1 = 0.069 m3/s Q2 = 0.044 m3/s 0.21 m Dia 0.21 m Dia 0.15 m Dia Q3? V3? Q3 = A3(V3) = 0.025 m3/s V3 = Q3/A3

Example 1 – Solve for Q3 and V3 below. Q1 = 0.069 m3/s Q2 = 0.044 m3/s 0.21 m Dia 0.21 m Dia 0.15 m Dia Q3? V3? Q3 = A3(V3) = 0.025 m3/s V3 = Q3/A3 = 0.025 m3 = 1.4 m/s s[π(0.15)2/4]m2

Bernoulli Relationship Total Head (Potential to do Work) = (Head due to elevation + Head due to pressure + Head due to fluid motion) = Constant for a given system Bernoulli Relationship

Bernoulli Relationship Total Head (Potential to do Work) = Head due to elevation + Head due to pressure + Head due to fluid motion Note – Head has units of length (usually m or ft) Bernoulli Relationship

Bernoulli Relationship C Bernoulli Relationship

Bernoulli Relationship zA B zB Head due to elevation (C = 0): ElevHeadA = zA ElevHeadB = zB C Bernoulli Relationship

Bernoulli Relationship H = p/ρg = p/γ B Head due to pressure: PressHeadB = pB/γ PressHeadC = pC/γ C Bernoulli Relationship

Example 2 – Calculation of Pressure Head (SI) for point in Control Vol where p = 2.7 atm γ = 9.6 kN/m3 101,300 Pa/atm 1 Pa = 1 N/m2 p = 2.7 atm (101,300 Pa/atm)(1 N/m2/Pa) = 274 x 103 N/m2 H = p/γ = 274 x 103 N(m3) = 28.5 m m2 (9.6 kN)

Example 2A – Calculation of Pressure Head (US) for point in Control Vol where p = 40 psi γ = 62.4/ft3 144 in2/ft2 p = 40 lb/in2 (144 in2/ft2) = 5.76 x 103 lb/ft2 H = p/γ = 5.76 x 103 lb(ft3) = 92.3 ft ft2 (62.4 lb)

Bernoulli Relationship v = (2gh)0.5 v2 = 2gh H = v2/2g B C Bernoulli Relationship

Example 3 – Compute Velocity Head (SI) for point in Control Vol where v = 2.3 m/s g = 9.81 m/s2 H = v2/2g = 2.32 m2(s2) = 0.27 m s2(2)(9.81 m)

Example 3A – Compute Velocity Head (US) for point in Control Vol where v = 7.5 ft/s g = 32.2 ft/s2 H = v2/2g = 7.52 ft2(s2) = 0.87 ft s2(2)(32.2 ft)

Total Hydraulic Head A B C = HTA = HzA + HpA + HvA = HTB = HzB + HpB + HvB HTC = HzC + HpC + HvC B C Total Hydraulic Head

Bernoulli Relationship Bernoulli Hydraulic Gradeline A Observed Hydraulic Gradeline B Hydraulic Gradeline: Plot of Total Head along length of Control Volume C Bernoulli Relationship

Conservation of Energy Work In or Out Outputs Inputs Σ 𝐸 in + Σ 𝐸 w = Σ 𝐸 out + Σ 𝐸 L Conservation of Energy

Conservation of Energy Work In or Out Outputs Inputs ΣQinHin + ΣQinEw = ΣQoutHout + ΣQoutHL Conservation of Energy

Conservation of Energy Work In or Out Outputs H = p/γ + z +V2/2g Inputs H = p/γ + z +V2/2g ΣQinHin + Σ 𝐸 w = ΣQoutHout + ΣQoutHL Conservation of Energy

Example 4 (US) - If no headlosses are considered, compute the discharge head on a 25 hp centrifugal pump discharging 500 gpm when the total head at the pump inlet is 20 ft. Ignore difference in elevation of pump nozzles. 500 gpm; HT = 20 ft 500 gpm; HT = ? ft 25 hp

25 hp (33,000 ft·lb)(1/min·hp) = 825 x 103 ft·lb/min 𝐸 W: 25 hp (33,000 ft·lb)(1/min·hp) = 825 x 103 ft·lb/min Hout: Hout = Hin + 𝐸 W /γQ Hout = 20 ft + 825 x 103 ft·lb(min)gal = 218 ft (min)500 gal(8.34 lb)

Example 4A (US) - If no headlosses are considered, compute the discharge head on a 25 hp centrifugal pump discharging 500 gpm when the total head at the pump inlet is 20 ft. Ignore difference in elevation of pump nozzles. Pump has hydraulic efficiency η of 70%. 500 gpm; HT = 20 ft 500 gpm; HT = ? ft 25 hp

25 hp (33,000 ft·lb)(1/min·hp)(0.70) = 578 x 103 ft·lb/min 𝐸 W = 𝐸𝜂 25 hp (33,000 ft·lb)(1/min·hp)(0.70) = 578 x 103 ft·lb/min Hout: Hout = Hin + 𝐸 W /γQ Hout = 20 ft + 578 x 103 ft·lb(min)gal = 159 ft (min)500 gal(8.34 lb)

ρΣQinVin ρΣQoutVout ΣFx = ρΣQoutVout - ρΣQinVin Linear Momentum

Example 5 (US) – What is the force developed by a jet ski pumping 300 gpm when Vin = 40 ft/s and Vout = 60 ft/s? Ignore losses. Q = 300 gpm Vin = 40 ft/s Q = 300 gpm Vout = 60 ft/s

ρ = 8.34 lbm/gal = 62.4 lbm/ft3 gc = 32.2 lbm·ft lbf·s2

ρ = 8.34 lbm/gal = 62.4 lbm/ft3 gc = 32.2 lbm·ft lbf·s2 ΣFx = ρQ(Vout – Vin)

ρ = 8.34 lbm/gal = 62.4 lbm/ft3 gc = 32.2 lbm·ft lbf·s2 ΣFx = ρQ(Vout – Vin) = 8.34 lbm(300 gal)(min)(60 – 40)ft(lbf·s2) gal(min)(60 s)(s)(32.2 lbm·ft) = 26 lbf

Angular Momentum

vt2 r2 r1 vt1

T = torque vt = tangential velocity P = power ω = rotational speed (rad/s) T = ρQ(vt2r2 – vt1r1) P = Tω = γQEw

Example 6 (US) – At a flow of 100 gal/min a pump produces a tangential velocity of 100 ft/s at the edge of a 12 in diameter impeller (r = 6 in). The tangential velocity is near zero at the eye of the pump. If the pump is rotating at 1,800 rpm what is the torque and power produced?

= 8.34 lbm(100 gal)(min)(100 – 0)ft(0.50 ft)lbf·s2 gc = 32.2 lbm·ft lbf·s2 T = ρQ(vt2r2 – vt1r1) = 8.34 lbm(100 gal)(min)(100 – 0)ft(0.50 ft)lbf·s2 gal(min)(60 s)(s)(32.2 lbm·ft) = 22 ft·lb

gc = 32.2 lbm·ft lbf·s2 ω = 1,800 rev(2π rad) min(rev) = 11 x 103 rad/min T = ρQ(vt2r2 – vt1r1) = 8.34 lbm(100 gal)(min)(100 – 0)ft(0.50 ft)lbf·s2 gal(min)(60 s)(s)(32.2 lbm·ft) = 22 ft·lbf P = Tω = 22 ft·lbf(11 x 103)(min·hp) = 7.3 hp min(33,000 ft·lbf)

Outcomes for today Understand application of ‘Continuity Equation’ Understand application of ‘Bernoulli Principle’ Understand volume, mass and energy balances for a ‘Control Volume’ Understand concepts of fluid momentum Outcomes for today