Hydraulic (Piezometric) Grade Lines (HGL) and

Transcription

1 Hydraulic (Piezometric) Grade Lines (HGL) and Energy Grade Lines (EGL)

2 When the energy equation is written between two points it is expresses as in the form of: Each term has a name and all terms have dimension of length: hl maj H P / hl min T : Total : Total Major Minor head loss head loss (m) (m) : Head used by the Pump/ Head generated by the Turbine (m)

3 The total energy of a real fluid decreases as the fluid moves through the pipe from one point to another along the direction of the flow due to major and minor losses.

4 Engineers find it useful to use - the "energy grade line" (EGL) and - the "hydraulic grade line" (HGL) in working with the pipe systems. These imaginary lines help the engineers in finding the trouble spots in the system (usually points of low pressure).

5 5. 1 Hydraulic Grade Line (HGL) If a hole is open in any pressured liquid carrying pipe and a colourless pipe is installed on top of this hole, the pressurized liquid within that pipe will rise within that colourless pipe to some level. This level is referred as Hydraulic Grade point and can be determined by the algebraic sum of the pressure head (p/γ) and the elevation head (z) which was measured from preselected datum at that point. HG point in any pressurized piping system is the algebraic summation of the height between the centerline of the pressurized pipe to the liquid level within the colourless pipe and the topographic elevation of that pipes centerline from any datum. H.G.P.= Combination of these points form the Hydraulic Grade Line (HGL).

6 5.2 Energy Grade Line (EGL) The total energy at any point along the pressurized liquid carrying pipe is the algebraic sum of the pressure head (p/γ), the elevation head (z) which was measured from preselected datum and the average velocity head of that point. The energy grade line (EGL) forms by the connection these points.

7 Figure 5.1: A close look of HGL and EGL at sudden contraction (EXAGERATED)

8 A jump occurs in the HGL and the EGL whenever useful energy is added to the system as in pumps. A sudden drop occurs if useful energy is extracted from the system as in turbines. H P H T

9 Important notes related with the HGL and the EGL. As the velocity at any point approaches to zero within the system, the HGL and the EGL approach each other. Thus, in an open reservoir the free water surface level and these two lines become identical and lie on top of each other at free water surface level. The EGL and HGL slope downward in the direction of the flow due to the head loss within the pipe. The greater the loss per unit length, the greater the slope ratio; as the average flow velocity within the pipe increases the loss per unit length increases. A sudden change upwards or downwards occur in the HGL and EGL whenever a loss occurs due to minor losses like sudden geometry changes and/or valves etc

10 Considering only the major loss effects...

11 Sudden enlargement loss reservoir reservoir Pipe major (friction) and minor losses (exaggerated) including change in pipe diameter

12 Hp reservoir reservoir Effect of pump with pipe major (due friction) and minor losses

13 Hydraulic Grade Line (HGL) Energy Grade Line (EGL) EGL and HGL due major and minor losses (exaggerated)

14 Scaled drawn pipeline system where the HGL cuts the pipeline, the pressure at that point is atmospheric (gage=0). For the pipeline portion above the HGL, the pressure is below atmospheric (negative gage or vacuum pressure). An intensive care should be taken so as to avoid these regions. 0 atm. Negative Pressure (below atm) HGL Pressure=0 atm. Positive Pressure (above atm) Positive Pressure (above atm)

15 Question 5.1 Water at 20 o C flows between two reservoirs at the rate of 0.06 m 3 /sec. For water, the cavitation occurs when the vapor pressure within any pipe is: P 20 C < 2338 Pa (abs). Take atmospheric pressure as 1 atm = kpa (abs)= 0 gage. a) Sketch the HGL and EGL of the system by considering all the critical locations. b) Obtain the water level elevation of reservoir II. c) Solve the occurrence of cavitation (if occurs) by: i- only changing the problematic pipe diameter (same material); D? ii- by adding a single pump of η=0.83; Power? I 20 m z I =20 m K v ava L A = 30 m D A = 20 cm Galvanized iron z K =17.85 m Square-edged Enterance L v avb L B = 20 m D B = 15 cm Cast iron z L =16.92 m M v avc L C = 35 m D C = 20 cm PVC z M =27.27 m N II z N =15.75 m DATUM z II =?

16 Q av =0.06 m 3 /s γ w = kn/m 3 ν 20 C = 1.004x10-6 m 2 /s v av1 = 1.91 m/s v av2 = m/s v av3 = 1.91 m/s k s1 /D 1 =0.015/20= Re 1 =3.81*10 5 f 1 = k s2 /D 2 =0.025/15= Re 2 =5.07*10 5 f 2 = k s3 /D 3 = SMOOTH Re 3 =3.81*10 5 f 3 = Atm. pressure head P atm /γ=101310/9789= m absolute The pressure head for NO cavitation P K+ /γ > 2338/9789 = m (absolute) Cavitation occurs = m with respect to atm. pressure

17 Atm. pressure head P atm /γ=101310/9789= m absolute The pressure head for NO cavitation P/γ > 2338/9789 = m (absolute) Cavitation starts = m with respect to atm. pressure

18 TOTAL ENERGY HEAD at critical points Total energy Head at reservoir I surface : E I = z I = m Enterance loss coefficient at point K (from reservoir I) k=0.50 Enterance loss at point K + = m Total energy Head at point K + = m Major loss along pipe A = m Total energy Head at point L - = E L- = m

19 Sudden Contaction k=0.24 (help of graphics) Sudden Contaction loss at point L + = 0.24*(3.395) 2 /19.62=0.141 m Total energy Head at point L + = E L+ = m Major loss along pipe B = m Total energy Head at point M - = E M- = m Sudden Enlargment k= 0.19 (help of graphics) Sudden Enlargment loss at point M + = 0.19*(3.395) 2 /19.62=0.112 m Total energy Head at point M + = E M+ = m

20 Major loss along pipe C = m Total energy Head at point N - = E N- = m Enterance to reservoir II k =1.0 Enterance loss N + = 1.0*(1.910) 2 /19.62=0.186 m Total energy Head at point N + = E N+ = m Total energy Head at reservoir II surface E II =z ıı = m

21 HYDRAULIC ENERGY at critical points Hydraulic energy head at reservoir I surface : E I = z I = H I = m Hydraulic energy head at point K + : H K+ = m Hydraulic energy head at point L - : H L- = m Hydraulic energy head at point L + : H L+ = m Hydraulic energy head at point M - : H M- = m Hydraulic energy head at point M + : H M+ = m Hydraulic energy head at point N - : H N- = m Hydraulic energy head at point N + : E N+ = m Hydraulic energy head at reservoir II surface : E II =z ıı =H II = m

22 PRESSURE HEAD at critical points Pressure Head at reservoir I surface : 0 m (atmospheric ) Pressure Head at point K + : P K+ /γ = m Pressure Head at point L - : P L- /γ = m Lowest value Pressure Head at point L + : P L+ /γ = 1.707m Pressure Head at point M - : P M- /γ = m < m *CAVITATION Pressure Head at point M + : P M+ /γ = m < m *CAVITATION Pressure Head at point N - : P N- /γ = m Pressure Head at point N + : P N+ /γ = m Pressure Head at reservoir II surface : 0 m (atmospheric)

23 The EGL and the HGL sketched on the figure includes local losses (entrance, sudden contraction & enlargement and the exit). Note for the large velocity head due smaller pipe diameters causes the difference between the EGL and the HGL to be more apart from each others. HGL

24 M Z M =27.27m I EGL K+ =19.907m HGL K+ =19.721m HGL L- =19.169m EGL L- =19.355m EGL L+ =19.214m Z I =20m K HGL L+ =18.627m Z K =17.85m L O EGL M- =17.365m EGL M+ =17.253m EGL N- =16.827m HGL N- =16.641m P/γ K+ =1.871m Z L =16.92m P/γ L- =2.249m HGL M+ =17.067m HGL M- =16.778m P II P/γ L+ =1.707m Z N =15.75m N Z II =16.641m P/γ N- =0.891m DATUM P/γ M- = m P/γ M+ = m

25 c) Hence, along pipe B the pressure head drops lower than the acceptable cavitation pressure head which is not recommended. i- Solving the problem by changing the pipe diameter size: implies that the pressure head at the end of the pipe B at point M - should be at least m. So writing energy equation along pipe B (between points L + and M - ) P L+ 2 γ + v avl+ 2g x D = D f 12.1D =f D 5 + z L+ = P M γ + v 2 avm 2g (type 3 problem) + z M + f L D Trial 1: Let D = 20 cm so f= Not OK Trial 2: Let D = 25 cm so f= Not OK Trial 3: Let D = 18 cm so f= Not OK Trial 4: Let D = 17 cm so f= Not OK Trial 5: Let D = 16 cm so f= Not OK Trial 6: Let D = 15.8 cm so f= OK 2 v avb 2g For no cavitation the required minimum head measured fom atmospheric pressure.

26 ii- Solving the problem by adding pump to the system (without changing pipe B diameter): So the pressure head at the end of the pipe B at point M - should be at least m. By writing energy equation along pipe B (between points L + and M - ) and adding a single pump along pipe B : To determine the required head for the pump no where cavitation should occur so select the most lowest value causing cavitation which is: Hpump= Hp= m Pp= γqhpump η Pp= 9789x0.06x0.382 = Watt 0.36 hp 0.83 (1 hp= W)

27 Question 5.2: If the water is at 15 C, ignoring the effect of minor losses and taking the atmospheric pressure to be kpa (absolute), a) determine the discharge passing from the system pipe, b) for the point having topographic elevation 62.3 m and is 850 m away from reservoir S along the pipe 1 at ( ) determine its: i- total energy, ii- hydraulic energy, iii- gauge pressure head and absolute pressure value iv- for no cavitation occurance at point S, the probable maximum topographic head value.

28 a) Discharge passing from the sistem: f 1 =0.025 ; f 2 = Q 1 = m 3 /s, Q 2 = m 3 /s Existing atm. Pressure head = /9.7965= m b) For point S ( ): i- total energy: H S = P S ii- hydraulic energy: iii- gauge pressure head: + v 2 S γ 2g + z S = m P S γ + z S = m P S γ = m (vacuum) m (absolute) [absolute pressure P S = kpa (abs)]. iv- evaporation pressure head causing cavitation = kpa (abs) m (abs) m (gage) so z S = m For no cavitation occurance at point S, the probable maximum topographic head value should be z S = m.