Hydraulics and hydrology

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1 Hydraulics and hydrology - project exercises - Class 4 and 5 Pipe flow Discharge (Q) (called also as the volume flow rate) is the volume of fluid that passes through an area per unit time. The discharge in a pipe is related to the flow velocity and cross-sectional area. The volume flow rate through a differential area of the section is v da, and the total volume flow rate is obtained by integration the velocity distribution over the entire cross-section. Q = v da A Flow mean velocity ( v ) (average velocity) is defined as the discharge divided by the cross-sectional area: Q v = A Hence, the volume flow rate can be calculated as: Continuity equation for flow in a pipe Q = v A Pipe flow governing equations Q 1 = Q or v 1 A1 = v A The Bernoulli equation p1 α v1 p α v z = z γ g γ g where: Total head loss (or combined head loss) h = Σh + Σh T P C h T Minor head loss (or component head loss) is associated with flow through devices such as valves, bends, elbows, contractions, expansions etc. v h C = ζ g ζ is head loss coefficient

2 Example 1 Water flows though an orifice inside a pipe of diameter d = 0 cm as shown in figure below. Estimate the minor head loss coefficient (ζ) for the orifice if the discharge in the pipe is Q = 6,8 l/s, and the water surface levels in the piezometers, attached just upstream and downstream the orifice, are at h 1 = 55 cm and h = 35 cm respectively. Neglect major losses. h 1 h d Q solution: orifice Example Find the water flow discharge (Q) through the Venturi s meter shown in figure below. Diameters are: d = 10 [cm] and D = 0 [cm]. Piezometric head difference H is 0 cm. Both minor and major loses can be neglected. H D Q d solution:

3 Major head loss (or pipe head loss) is associated with fully developed flow in conduits, and is caused by shear stresses that act on the flowing fluid. Major head loss can be determined using the Darcy-Weisbach formula: h L v = λ P d g Friction factor (resistance coefficient) λ = f ( R, ε ) Reynolds number Relative roughness coefficient v d R e = ν k ε = d k is absolute roughness (or equivalent sand-grain roughness) Equivalent sand-grain roughness for various pipe materials Material Glass, plastic e Equivalent sand-grain roughness k [mm] smooth Cooper or brass tubing 0,0015 Wrought iron, steel 0,046 Galvanized iron 0,15 Cast iron 0,45 Concrete 0,3 3,0 Riveted steel 0,9 9,0 Rubber pipe (straight) 0,05 Moody diagram Effects of wall roughness Depending on the parameter range the friction factor (λ) may be influenced either by the Reynolds number (R e ) or the relative roughness (ε) or by both these parameters as shown in the table below: Type of Flow Parameter Ranges Influence of Parameters on λ Laminar Flow R e < 000 NA Turbulent Flow, Smooth Tube R e > 3000 ε R e < 10 Transitional Turbulent Flow R e > < ε R e < 1000 Fully Rough Turbulent Flow R e > 3000 ε R e > 1000 λ depends on R e λ is independent of ε λ depends on R e λ is independent of ε λ depends on R e λ depends on ε λ is independent of R e λ depends on ε

4 0,130 Moody Diagram (PN-76/M-34034) 0,100 Laminar flow Transitional zone Fully rough turbulent flow 0,090 0, ,070 0, ,050 Resistance coefficient λ λ a a r c i t n i k ó ł c z y p W s 0,040 0,030 0,05 0,00 0, λ = R e Smooth pipes a - n 8 d ę 6 g l z 4 w o ś ć a t w 10 o - 3 p o 6 h r 4 c Relative roughness ε 0,010 0,009 0,008 0,007 0, Reynolds L i c z b a number R e y n o l d Rs e a vd R e = ν construction is co-financed by the European Union within the confines of the European Social Fund

5 Example 3 A horizontal pipe of diameter d = 50 cm carries water at the temperature T = 10 o C. Determine the piezometric head difference if the distance between the piezometers is L = 800 m, the volume flow rate is Q = 1000 m 3 /h, and absolute roughness k = 0,3 mm. H=? d Q L solution: Example 4 Three pipes of equal length (L) but different diameters (d) and absolute roughness carry water at different temperatures (T) and velocities (v). Indicate the pipe which causes the highest total head loss. Justify your answer by calculations. pipe I: L = 1000 m, d = 10 cm, v = cm/s, T = 10 o C, k = 0,1 mm, pipe II: L = 1000 m, d = 0 cm, v = 50 cm/s, T = 10 o C, k = 0,4 mm, pipe III: L = 1000 m, d = 40 cm, v = 60 cm/s, T = 0 o C, k = 1 mm, solution: Energy and Hydraulic Grade Line The Hydraulic Grade Line (HGL) and the Energy Grade Line (EGL) are graphical presentations of the Bernoulli equation that show head in a system. These imaginary lines help the engineers locate the trouble spots in the system (usually points of low pressure). The EGL is a line that indicates the total head at each location along a pipe and is related to terms in the Bernoulli equation by: p v EGL = z + + γ g The HGL is a line that indicates the piezometric head at each point along a pipe and therefore is coincided with the liquid free surface in a piezometer. p HGL = z + γ

6 where: z elevation head, γ p pressure head, v g velocity head. The piezometric head represents the potential energy of flowing liquid whereas the total head represents the total energy (potential and cinematic) of the liquid. Example 5 For a system shown in figure below the EGL is given. Determine all minor and major losses. Draw the HGL for the system and describe all drops on it. Assume that all pipes have the same absolute roughness. Neglect velocities in the tanks. Where is the potential trouble spot in the system (point of the lowest pressure)? gas p g > p atm L 1 L 1 L 1 water d d 1 L d 1 L 1 solution:.....

7 Pipe Entrance squared Minor head losses coefficients (PN-76/M-34034) chamfered rounded ζ = 0,5 re-entrant rounded ζ = 0,5 re-entrant sharp edged ζ = 0,10 0,06 angled ζ = 0,56 ζ = 1,30 ζ = 0,5 + 0,3 sinϕ + 0, sin ϕ Valves butterfly valve ϕ ζ 0,4 0,5 0,9 1,54,51 3,91 6, 10,8 18, ball valve ϕ ζ 0,05 0,9 1,56 5,17 17,3 31, 5, sluice-valve D s s/d 1/8 1/4 3/8 1/ 5/8 3/4 7/8 ζ 0,07 0,6 0,81,06 5,5 17,0 97,8

8 Change of direction elbow ζ = 0,946 sin +,05 sin 4 0 o 40 o 60 o 80 o 90 o 100 o 10 o 140 o 160 o ζ 0,04 0,14 0,36 0,74 0,98 1,6 1,86,43,85 bend r R ζ = 0, ,847 r R 3,5 o 90 o r/r 0 o 40 o 60 o 80 o 90 o 100 o 10 o 140 o 160 o 180 o 0,1 0,09 0,058 0,088 0,117 0,13 0,146 0,175 0,05 0,34 0,63 0, 0,031 0,061 0,09 0,1 0,138 0,153 0,183 0,14 0,45 0,75 0,3 0,035 0,070 0,106 0,141 0,158 0,176 0,11 0,46 0,81 0,317 0,4 0,046 0,091 0,137 0,183 0,06 0,9 0,74 0,30 0,366 0,41 0,5 0,065 0,131 0,196 0,6 0,94 0,37 0,39 0,458 0,53 0,589 0,6 0,098 0,196 0,93 0,391 0,440 0,489 0,587 0,684 0,78 0,880 0,7 0,147 0,94 0,441 0,588 0,661 0,734 0,881 1,08 1,175 1,3 0,8 0,17 0,434 0,651 0,868 0,977 1,085 1,30 1,50 1,737 1,954 0,9 0,313 0,66 0,939 1,5 1,408 1,565 1,878,191,504,817 1,0 0,440 0,879 1,319 1,758 1,978,198,637 3,077 3,516 3,956

9 expansion (sudden enlargement) d D D ζ = d 1 D /d 1, 1,4 1,6 1,8,0,5 3,0 3,5 4,0 4,5 5,0 6,0 ζ 0,04 0,16 0,36 0,64 1,00,5 4,00 6,5 9,00 1,5 16,5 5,00 Example 6 Both pipes shown in figure below have an equivalent sand roughness (k) of 0,1 mm and a discharge (Q) of 0,1 m 3 /s. Also, D 1 = 15 cm, L 1 = 50 m, D = 30 cm, and L = 160 m. Determine the difference (H) in the water-surface elevation between the two reservoirs. Draw the EGL and HGL. water at T = 0 o C H =? squared entrence D 1 butterfly valve =5 o D L 1 Solution: L

10 Example 7 Two open reservoirs are connected by a 0 m long pipe as shown in figure below. Find the diameter (d) of the pipe if the difference in the water-surface elevation between the two reservoirs H = m, the discharge Q = 6 l/s, the bend radiuses are R 1 = 100 mm, and R = 50 mm, the absolute roughness k = 0,08 mm, and the kinematic viscosityν = 0,1 cm /s, Assume the re-entrant sharp edged pipe entrance. H R 1 R Solution:

11 Example 8 A pipe of length L = 35 m, and diameter d = 40 mm carries water from a closed pressurized tank to a point at height of H = 15 m above the water surface in the tank as shown in figure below. Find the discharge (Q) through the pipe if the gas in the tank is under pressure of p gas = 300 kpa. The bend radius is R = 00 mm, the absolute roughness k = 0,15 mm, and the kinematic viscosity ν = 0,1 cm /s. Assume the squared pipe entrance. Use the atmospheric pressure p atm = 100 kpa and the specific weight of water γ = 9,81 [kn/m 3 ]. H R ball valve = 45 o Q =? gas water sluice-valve s/d = 0,5 Solution:

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