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

2. 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?

4. 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

5. Control Volume

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

7. Outputs
Inputs
ΣρQin = ΣρQout
Conservation of Mass

8. Vout
Aout
Vin
Ain
Outputs
Inputs
ΣρVinAin = ΣρVoutAout
Continuity

9. Vout
Aout
Vin
Ain
Outputs
Inputs
ΣVinAin = ΣVoutAout
Continuity

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

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

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

13. Example 1 – Solve for Q3 and V3 below.
Q1 = m3/s
Q2 = 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 = m3/s m3/s = m3/s

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

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

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

17. 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

18. 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

19. Bernoulli RelationshipC
Bernoulli Relationship

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

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

22. 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)

23. 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)

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

25. 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
= m2(s2) = 0.27 m
s2(2)(9.81 m)

26. 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
= ft2(s2) = 0.87 ft
s2(2)(32.2 ft)

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

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

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

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

31. 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

32. 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

33. 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 x 103 ft·lb(min)gal = 218 ft
(min)500 gal(8.34 lb)

34. 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

35. 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 x 103 ft·lb(min)gal = 159 ft
(min)500 gal(8.34 lb)

36. ρΣQinVin
ρΣQoutVout
ΣFx =
ρΣQoutVout -
ρΣQinVin
Linear Momentum

37. 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

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

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

40. 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

41. Angular Momentum

42. vt2 r2 r1
vt1

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

44. 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?

45. = 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

46. 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)

47. 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