1 / 61

Tie Rod Case Study

Tie Rod Case Study. Water Handling Operations. Unit 4A – Water and Hose Hydraulics. Objectives. Identify the elements that should be considered when calculating pump discharge pressure.

gudrun
Download Presentation

Tie Rod Case Study

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Tie Rod Case Study

  2. Water Handling Operations Unit 4A – Water and Hose Hydraulics

  3. Objectives • Identify the elements that should be considered when calculating pump discharge pressure. • Perform friction loss calculations using Principles of Hydraulics – Rules of Thumb and friction loss calculators. • Identify ways to reduce friction loss.

  4. Objectives • Discuss how hydraulics affect drafting procedures used to refill an engine. • Describe the effects of ejector use during pumping operations.

  5. Pump Discharge Pressure (PDP) Pump discharge pressure is the amount of pressure in pounds per square inch (PSI) as measured at the pump discharge.

  6. Friction Loss - (F) Friction loss is pressure loss caused by the turbulent movement of water or solution against the interior surface of fire hose, pipe, or fittings • Normally measured in pressure loss (in PSI) per length of hose or pipe

  7. Appliances - (A) Appliances are items (gated wyes, in-line tees, and check valves) that connect or interconnect the hose and pump together to form a hose lay. • Appliances have an insignificant effect on friction loss calculations at flows used in wildland suppression and have been omitted from calculations in this lesson.

  8. Elevation/Head Pressure - (E) Elevation/head pressure is the weight of a given height (depth) of water column at its base. • Alsoknown as lift, back pressure, gravity loss or gain • Measured in terms of feet of water • One foot of water exerts a pressure of .5 PSIat the base of a column of water.

  9. Nozzle/Working Pressure - (N) Nozzle/working pressure is pressure that must be delivered to the nozzle to produce an effective stream. • Optimum operating pressures: • Straight stream and Forester nozzles is 50 PSI • Variable pattern or fog nozzle is 100 PSI

  10. Friction Loss Calculations

  11. Why do we need to perform friction loss calculations? Engine operators are more efficient and safe in their duties when they know the capabilities and limitations of their equipment.

  12. References and Tools for Making Calculations • Wildland Fire Hose Guide • Water Handling Equipment Guide • Incident Response Pocket Guide • Fire stream/friction loss calculator • Principles of Hydraulics—Rules of Thumb

  13. “Rule of Thumb” • A rule based on experience or practice rather than on scientific knowledge. • Any method of estimating that is practical though not precise.

  14. Friction Loss Formula Pump Discharge Pressure (PDP) = F+/- E+ N = Friction Loss +/- Elevation (Head Pressure) + Nozzle Pressure

  15. Considerations • ENOPs must know the gallons per minute (GPM) that each nozzle will discharge. • For simple hose lays, always start at the nozzle and work towards the pump. • For progressive hose lays, always start from the most distant nozzle and work towards the pump. • Always pump to the highest pump discharge pressure required. • Gate down any laterals that require substantially less pressure.

  16. Friction Loss Calculation Exercises(Pre-course Work)

  17. Exercise 1Pre-course Work 20 GPM You are pumping a 1½” hose lay 500’ long with a 5/16” tip. What is the GPM? What is the PDP? + F = +/-E = + N = PDP = 5 friction loss in 500’ (1 PSI/100’) of 1½” hose flowing 20 GPM 0 no elevation change 50 straight stream tip 55 PSI

  18. You are pumping 600’ of 1½” hose 50’ above the pump with a ⅜” tip. What is the GPM? What is the PDP? Exercise 2 Pre-course Work 30GPM + F = +/- E = + N = PDP = 18 friction loss in 600’ (3 PSI/100’) of 1½” hose flowing at 30 GPM +25 head pressure for 50’ of elevation gain (50 ÷ 2) 50 straight stream tip 93 PSI

  19. Exercise 3 Pre-course Work 13 GPM 26 GPM 13 GPM You are pumping 300’ of 1½” hose through a wye to 2 sections of 1” hose, each 100’ long with ¼” tips (remember tips are 50 PSI). What are the GPMs? What is the PDP? + F = + F = +/- E= + N = PDP = 3 friction loss in 100’ of 1” hose flowing 13 GPM 6 friction loss in 300’ (2 PSI/100’) of 1½” hose flowing 26 GPM 0 no elevation change 50 straight stream tip 59PSI

  20. Exercise 4Pre-course Work You are pumping to a hose lay with the first lateral at 1,500’. You have another 100’ of 1½” hose to another lateral. Both laterals have ⅜”tips. What are the GPMs? What is the PDP? 30GPM 60GPM 30GPM + F = + F = + F = +/-E = + N = PDP = 14 friction loss in 100’ of 1” hose flowing 30 GPM 3 friction loss in 100’ of 1½” hose flowing 30 GPM 135 friction loss in 1,500’ (9 PSI/100’) of 1½” hose flowing 60 GPM 0 no elevation change 50 straight stream tip 202PSI

  21. Friction Loss Calculation Exercises(In-class)

  22. Exercise 5 In-class 60 GPM 90 GPM You have 600’ of 1½” parallel hose to a Siamese valve. Attached to the wye is 100’ of 1½-inch hose to another wye and the first lateral, another 200’ of 1½” hose to the second lateral, and another 100’ of 1½” hose to last lateral. Each lateral is 100’ of 1” hose with a variable flow nozzle flowing 30 GPM. Consider this flat ground. What are the GPMs? What is the PDP? + F = + F = + F = + F = + F = +/-E = + N = PDP = 14 friction loss in 100’ of 1” hose flowing 30 GPM 3 friction loss in 100’ if 1½’’ hose flowing 30 GPM 18 friction loss in 200’ (9 PSI/100’) of 1½” hose flowing 60 GPM 20 friction loss in 100’ of 1½” hose flowing 90 GPM 30 friction loss in 600’ (5 PSI/100’) of 1½” hose with 2 Siamesed lines 0no elevation change 100 variable flow nozzle 185 PSI

  23. Exercise 6 In-class 60 GPM You have 400’ of 1½” parallel hose to a Siamese valve. Attached to the Siamese is 100’ of 1½” hose to a wye and the first lateral, another 200’ of 1½” hose to the second lateral. Each lateral is 100’ of 1” hose with a variable flow nozzle flowing 30 GPM. There is a 60’ rise in elevation. What are the GPMs? What is the PDP? + F = + F = + F = + F = +/- E = + N = PDP = 14 friction loss per 100’ of 1” hose flowing 30 GPM 6friction loss for 200’ (3 PSI/100’) of 1½” hose flowing 30 GPM 9 friction loss in 100’ of 1½” hose flowing 60 GPM 12 friction loss in 400’ (3 PSI/100’) of 1½” flowing 60 GPM with 2 Siamesed lines +30 head pressure for 60’ of elevation gain (60 ÷ 2) 100 variable flow nozzle 171 PSI

  24. Exercise 7 In-class 50 GPM You have 500’ of 1½” parallel hose to a Siamese valve with a 30’ drop in elevation. Attached to the Siamese valve is 300’ of 1½” hose to a wye and the first lateral has a gain of 10’ in elevation. There is another 100’ of 1½” hose to the second and last lateral. Each lateral is 100’ of 1” hose with a variable flow nozzle flowing 25 GPM. What is the PDP? + F = + F = + F = + F = +/-E + N = PDP = 10 friction loss in 100’ of 1” hose flowing 25 GPM 2 friction loss in 100’ of 1½” hose flowing 25 GPM 18 friction loss in 300’ (6 PSI/100’) of 1½” hose flowing 50 GPM 10 friction loss in 500’ (2 PSI/100’) of 1½” flowing 50 GPM with 2 Siamesed lines -10 head pressure for 20’ of elevation loss (20 ÷ 2) 100 fog nozzle 130 PSI

  25. Friction Loss Calculations (“Rules of Thumb” and IRPG)

  26. Exercise 1 “Rules of Thumb” 25 5 PSI/100’ x 5 5 @ appx 20 GPM, 1 PSI/100’ x 5 0 no change 0 no change 50 smooth bore tip 50 smooth bore tip 75 PSI 55 PSI 20 GPM You are pumping a 1½” hose lay 500’ long with a 5/16” tip. What is the GPM? What is the PDP? Cheat Sheet IRPG + F = +/-E = + N = PDP = + F = +/_E = + N = PDP =

  27. You are pumping 600’ of 1½” hose 50’ above the pump with a 3/8” tip. What is the GPM? What is the PDP? Exercise 2 “Rules of Thumb” 30 5 PSI/100’ x 6 18 @ appx30 GPM, 3 PSI/100’ x 6 +25 50’ gain +25 50’ gain 50 smooth bore tip 50 smooth bore tip 93 PSI 105 PSI 30 GPM Cheat Sheet IRPG + F = +/-E = + N = PDP = + F = +/-E = + N = PDP =

  28. Exercise 3 “Rules of Thumb” 10 10 PSI/100’ x 1 4 @ appx 15 GPM x 1 9 @ appx 30 GPM, 3 PSI/100’ x 3 15 5 PSI/100’ x 3 0 no change 0 no change 50 smooth bore tip 50 smooth bore tip 63 PSI 75 PSI 15 GPM 30 GPM 15 GPM You are pumping 300’ of 1½” hose through a wye to 2 sections of 1” hose, each 100’ long with ¼” tips (remember tips are 50 PSI). What are the GPMs? What is the PDP? Cheat Sheet IRPG + F = + F = +/-E = + N = PDP = + F = + F = +/-E = + N = PDP =

  29. Exercise 4 “Rules of Thumb” 10 PSI/100’ x 1 15 @ 30 GPM x 1 3 @ 30 GPM, 3 PSI/100’ x 1 5 5 PSI/100’ x 1 150 @ 60 GPM, 10 PSI/100’ x 15 150 10 PSI/100’ x 15 0 no change 0 no change 50 smooth bore tip 50 smooth bore tip 218 PSI 215 PSI 30 GPM 60 GPM 30 GPM You are pumping to a hose lay with the first lateral at 1,500’. You have another 100’ of 1½” hose to another lateral. Both laterals have ⅜”tips. What are the GPMs? What is the PDP? Cheat Sheet* IRPG + F = + F = + F = +/-E = + N = PDP = + F = + F = + F = +/-E = + N = PDP = * At high flows greater than >60 GPM and hose lengths greater than 10 PSI/100’ friction loss for 1½” hose.

  30. Exercise 5 “Rules of Thumb” 10 10 PSI/100’ x 1 15 @ 30 GPM x 1 3 @ 30 GPM, 3 PSI/100’ x 1 5 5 PSI/100’ x 1 20 @ 60 GPM, 10 PSI/100’ x 2 20 10 PSI/100’ x 2 20 @ 90 GPM x 1 10 10 PSI/100’ x 1 30 @ 90 GPM, 20 PSI/100’ x 6 ÷ 4(Siamese) 30 5 PSI/100’ x 6 0 no change 0 no change 100 variable/fog nozzle 100 variable/fog nozzle 188 PSI 175 PSI 90 GPM 60 GPM You have 600’ of 1½” parallel hose to a Siamese valve. Attached to the wye is 100’ of 1½” hose to another wye and the first lateral, another 200’ of 1½” hose to the second lateral, and another 100’ of 1½” hose to last lateral. Each lateral is 100’ of 1” hose with a variable flow nozzle flowing 30 GPM. Consider this flat ground. What are the GPMs? What is the PDP? Cheat Sheet IRPG + F = + F = + F = + F = + F = +/-E = + N= PDP = + F = + F = + F = + F = + F = +/-E = + N= PDP =

  31. Exercise 6 “Rules of Thumb” 10 10 PSI/100’ x 1 15 @ 30 GPM x 1 6 @ 30 GPM, 3 PSI/100’ x 2 10 5 PSI/100’ x 2 10 @ 60 GPM, 10 PSI/100’ x 1 10 10 PSI/100’ x 1 10 @ 60 GPM, 10 PSI/100’ x 4 ÷ 4(Siamese) 20 5 PSI/100’ x 4 +30 60’ gain +30 60’ gain 100 variable/fog nozzle 100 variable/fog nozzle 180 PSI 171 PSI 60 GPM You have 400’ of 1½” parallel hose to a Siamese valve. Attached to the Siamese valve is 100’ of 1½” hose to a wye and the first lateral, another 200’ of 1½” hose to the second lateral. Each lateral is 100’ of 1” hose with a variable flow nozzle flowing 30 GPM. There is a 60’ rise in elevation. What is the GPM? What is the PDP? Cheat Sheet IRPG + F = + F = + F = + F = +/- E = + N = PDP = + F = + F = + F = + F = +/-E = + N = PDP =

  32. Exercise 7“Rules of Thumb” 10 10’ x 1 10 @ 25 GPM x 1 2 @ 25 GPM, 2 PSI/100’ x 1 5 5 PSI/100’ x 1 24 @ 50 GPM, appx 8 PSI/100’ x 3 15 5 PSI/100’ x 3 10 @ 50 GPM, appx 8 PSI/100’ x 5 ÷ 4 (Siamese) 25 5 PSI/100’ x 5 10 20’ total drop 10 20’ total drop 100 variable/fog nozzle 100 variable/fog nozzle 136 PSI 145 PSI 50 GPM You have 500’ of 1½” parallel hose to a Siamese valve with a 30’ drop in elevation. Attached to the Siamese valve is 300’ of 1½” hose to a wye and the first lateral has a gain of 10’ in elevation. There is another 100’ of 1½” hose to the second and last lateral. Each lateral is 100’ of 1” hose with a variable flow nozzle flowing 25 GPM. What is the GPM? What is the PDP? Cheat Sheet IRPG + F = + F = + F = + F = +/-E = + N = PDP = + F = + F = + F = + F = +/-E = + N = PDP =

  33. Why are there differences? • Friction Loss Calculator • Derived from a mathematical equation • “Rule of Thumb “ • An estimation

  34. Are the differences significant enough . . .? NO! • The goal is to supply an adequate amount of water and pressure to the nozzle operator(s). • Calculations for neither method is exact. • Both methods provide a starting point to meet the goal.

  35. Friction Loss Calculator Use • The calculator provides consistency and validation (the theory behind hydraulics). • Field use may be slow and tedious.

  36. Reasons for using the Principles of Hydraulics—Rules of Thumb • Based on theories. • Provide quick solutions for fire situations. • Make fast and practical decisions during hose lay construction.

  37. Reducing Friction Loss • Reduce pump pressure. • Reduce the nozzle tip size. • Lay parallel hose. • Calculate friction loss for one hose, then divide by 4. • Eliminate unnecessary plumbing parts. • Increase hose diameters where possible.

  38. Pump Cavitation • Pump cavitation occurs when more water is being discharged from the pump than is being supplied creating a low-pressure area within the center (or eye) of the pump.

  39. Effects of Cavitation • When water enters the low pressure area at the eye of the impeller, water vaporizes or boils more easily. • When vapor bubbles reach the pressure or discharge side of the pump, they implode or forcefully collapse resulting in an intense shock wave which creates dings and pits of the pump walls and impeller. • Damage is cumulative and progressive. • Excessive cavitation will eventually lead to impeller failure.

More Related