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HYDROENERGY

Problems

Fundamentals of fluid flows

1. The siphon of Figure 1 is filled with water and discharging 150 l/s. Find the losses

from point 1 to point 3 in terms of velocity head푣 2푔⁄ (keep in mind that points 1

and 3 are at atmospheric pressure and that at point 1 the velocity is zero while at

point 3 the velocity is 푣 2푔⁄ ). Find the pressure at point 2.

Figure 1

2. For the water shooting out of the pipe and nozzle under the conditions shown in

Figure 2, find the height, h, above the nozzle, that the jet attains. Assume

negligible the head loss along the siphon.

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Figure 2

3. Water flows from section 1 to section 2 in the pipe shown in Figure 3. Determine

the velocity of the flow and the fluid pressure at section 2. Assume the total head

loss between sections 1and 2 is 3.00 m.

Figure 3

4. A nozzle is attached to a pipe as shown in Figure 4. The inside diameter of the

pipe is 100 mm, while the water jet exiting from the nozzle has diameter of 50 mm.

If the pressure at section 1 is 500 kPa, determine the water jet velocity. Assume

the head loss in the jet is negligible.

Figure 4

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5. A horizontal pipe transports a liquid with density of 0.85 and includes a

venturimeter, according to what is presented in the Figure 5.

Figure 5

Considering that the difference of elevations of the free surfaces of the mercury of

the pressure meter is ∆y=0.06 m, determine the discharge flowing in the pipe.

Consider that α=1.15 and that the head loss between the two cross sections

containing the axis of the pressure meter intakes is 5% of the kinetic head in the

first of the cross sections.

Momentum conservation. Forces exerted by the flow

6. In the conditions of Problem4, determine the component of the forces exerted by

the flow in the pipe (consider the bend and the fire hose).

7. Consider a vertical jet of water with a diameter of 50 mm and a velocity of 10 m/s

impinging on a horizontal surface. Determine the force exerted by the jet on the

surface.

Bernoulli equation for real fluids. Pumps, turbines and head losses

8. In a pumping system like the one represented in Figure 6, a discharge of 60 l/s is

pumped from a reservoir with water at elevation 20.00 to another one with water at

elevation 100.00. The pump inlet and outlet axis are at elevations 15.00 and 16.00

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and the respective diameters are 0.25 and 0.20 m. The pipes upstream and

downstream of the pump are 500 and 1000 m long and the respective unit head

losses are 0.004 and 0.010.

Considering that the local head losses are negligible, and considering α=1.1,

determine:

a) the grade line elevations at the pump inlet and outlet cross sections;

b) the pressure heads at the axis in the same sections;

c) the head at which the pump is operating and the power of the pump (η=0.80).

Figure 6

9. Consider the concrete pipe with a nozzle discharging water from the reservoir to

the atmosphere represented in Figure 7. Determine the area of the nozzle B, with

axis at elevation 20.00 (disregarding the contraction of the jet), necessary to obtain

a discharge of 0.8 m3/s. Consider an entry heal loss coefficient of 0.5. Solve the

problem considering the continuous head losses calculated according the

Colebrook-White formula (ks=0.04 mm), considering the Hazen-William formula

(C=120) and considering the Manning-Strickler formula (K=85 m1/3/s).

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Figure 7

10. Determine:

a) the net head of a turbine with efficiency η=0.82 so that 1 m3 of water produces

1 kWh of energy;

b) the head at which a pump is operating so that with an efficiency η=0.75 that

pumping of 1 m3 requires 1 kWh of energy.

11. Consider the cast iron pipe with D=450 mm (C=110) connecting two reservoirs like

shown in Figure 8. The pipe is 40 m long and the discharge is 550 l/s. Determine

the difference in elevation between the water surfaces in the two reservoirs.

Considerer the local head losses at the entry and at the outlet of the conduit.

Figure 8

12. Figure 9 shows two reservoirs with water at 20º connected by a concrete pipe

(k=0.03 mm). The water in the upper reservoir is to be drained to other reservoir at

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a lower level. The total length of the pipe is 35 m and its diameter is 120 mm.

Consider that the following local head loss coefficients:

• entrance – Kv=0.50;

• partially closed gate valve – Kv=17.00;

• 90º elbow – Kv=0.65;

• outlet – Kv=1.00.

What will be the discharge through the pipe for the water surface elevations shown

in Figure 9 and what will be the continuous head loss and the total of local head

losses?

Figure 9

13. Water flows at a rate of 0.020 m3/s from reservoir A to reservoir B through three

concrete pipes connected in series, as shown in Figure 10. Find the difference H in

water-surface elevations in the reservoirs. Neglect local head losses and consider

the Hazen-Williams formula with C1=100; C2=110; C3=105 to compute the

continuous head losses.

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Figure 10

14. Consider reservoirs A and B connected through steel pipes, as shown in Figure

10. Supposing that the water-surface elevations are 125 and 100 m, and that the

diameter of the middle conduit is unknown, find the diameter necessary to obtain a

discharge of 0.025 m3/s. Neglect local losses and consider the lengths, the other

diameters shown in Figure 10 and the Hazen-Williams coefficients indicated in

Problem 13.

15. Water flows from an upper reservoir to a lower one while passing through a

turbine, as shown in Figure 11. Find the power generated by the turbine for a

turbine efficiency of 0.90. Neglect local losses and use the Colebrook-White

formula for the continuous head losses (ks=0.02 mm).

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Figure 11

16. Water from a reservoir is being used for hydropower generation, as shown in

Figure 12. What is the water surface elevation in the upper reservoir necessary for

the turbine to generate 700 kW, for a turbine efficiency of 0.85 ? Neglect local

losses.

Figure 12

17. The turbine shown in Figure 13 is operating with a discharge of 0.2 m3/s and the

pressures at points 1 and 2 are 150 and -35 kPa, respectively. Determine the

power generated by the turbine, for a turbine efficiency of 0.85.

Figure 13

Q= 1.0 m3/s

Elev = 100 m D = 500 mm L = 850 m C = 110

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SOLUTIONS

Problem Solution

1 p2 = -31 kPa

2 h = 4.82 m

3 V2 = 8.00 m/s

p2 = 260 kPa

4 V1 = 0.250 V2

V2 = 32.7 m/s

5 Q = 0.285 m3/s

6 F = 2351 N

7 F = 196 N

8 a) Hpm= 18 m

Hpj= 110 m

b) pm/γ= 2.92 m

pj/γ= 93.85 m

c) Hp= 92 m

Pp= 67.62 kW

9 Colebrook-White - A=0.0261 m2

Hazen-Williams - A=0.0262 m2

Manning-Strickler - A=0.0261 m2

10 a) Hu = 448.00 m

b) Ht = 275.50 m

11 ∆H = 2.06 m

12 Q = 34 l/s

∆Hc = 2.07 m

∆Hl = 8.93 m

13 H = 21.44 m

14 D = 375 mm

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15 P = 191 kW

16 Elevation = 228.00 m

17 P = 34.4 kW

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Table – Dynamic viscosity and kinematic viscosity of water

Temperature Dynamic viscosity

Kinematic viscosity

t µ υ

(oC) (Pa s, N s/m2) x 10-3 (m2/s) x 10-6

0 1.787 1.787 5 1.519 1.519

10 1.307 1.307 20 1.002 1.004 30 0.798 0.801 40 0.653 0.658 50 0.547 0.553 60 0.467 0.475 70 0.404 0.413 80 0.355 0.365 90 0.315 0.326 100 0.282 0.29

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Table - Hazen–Williams roughness coefficients

Material Hazen-Williams coefficient

C

ABS - Acrylonite Butadiene Styrene 130

Aluminum 130 - 150

Asbestos Cement 140

Asphalt Lining 130 - 140

Brick sewer 90 - 100

Cast-Iron - new unlined (CIP) 130

Cast-Iron 10 years old 107 - 113

Cast-Iron 20 years old 89 - 100

Cast-Iron 30 years old 75 - 90

Cast-Iron 40 years old 64-83

Cast-Iron, cement lined 140

Cement lining 130 - 140

Concrete 100 - 140

Concrete lined, steel forms 140

Concrete lined, wooden forms 120

Concrete, old 100 - 110

Copper 130 - 140

Corrugated Metal 60

Ductile Iron Pipe (DIP) 140

Ductile Iron, cement lined 120

Fiber 140

Fiber Glass Pipe - FRP 150

Galvanized iron 120

Glass 130

Lead 130 - 140

Plastic 130 - 150

Polyethylene, PE, PEH 140

Polyvinyl chloride, PVC, CPVC 150

Steel new unlined 140 - 150

Steel, corrugated 60

Steel, welded and seamless 100

Steel, projecting girth and horizontal rivets 100

Steel, vitrified, spiral-riveted 90 - 110

Steel, welded and seamless 100

Tin 130

Vitrified Clay 110