1. R. Shanthini 05 Feb 2010

2. Whole System Design: An integrated Approach to Sustainable Engineering Units 6 – 10: Worked Examples July 2007 Citation: Stasinopoulos, P., Smith, M., Hargroves, K. and Desha, C. (2009) Whole System Design: An Integrated Approach to Sustainable Engineering, Earthscan, London, and The Natural Edge Project, Australia

3. R. Shanthini 05 Feb 2010 Whole System Design: An integrated Approach to Sustainable Engineering Unit 6: Industrial Pumping Systems

4. R. Shanthini 05 Feb 2010 Pumps move liquid from one place to another against pipe friction and against changes in height and direction. Industrial pumping systems account for 20% of the world’s industrial electrical energy demand, and 12% of world’s electricity. (Motors use 60% of the world’s electricity.) Improving the efficiency of industrial pumping systems can reduce industrial energy consumption and hence greenhouse gas emissions. Significance of Pumping Systems and Design

5. R. Shanthini 05 Feb 2010 Power plant loss 70% Transmission & distribution loss 9% Motor loss 10% Pump loss 25% Throttle loss 33% Pipe loss 20% 9.5 units of energy output 100 units of fuel energy input Drivetrain loss 2% Saving a single unit of pumping energy can save more than ten times that energy in fuel.

6. R. Shanthini 05 Feb 2010 Large pumping system (in the order of kW and MW) design is disciplined considering minimum velocities, thermal expansion, pipe work and maintenance. Small pumping system design is not disciplined since it accounts for a small fraction of the total cost of an industrial operation. However, it is likely that there are a lot more small pumping systems than large pumping systems. Both small and large pumping systems design could benefit from Whole System Design. Significance of Pumping Systems and Design (continued)

7. R. Shanthini 05 Feb 2010 1.Ask the right questions 2.Benchmark against the optimal system 3.Design and optimise the whole system 4.Account for all measurable impacts 5.Design and optimise subsystems in the right sequence 6.Design and optimise subsystems to achieve compounding resource savings 7.Review the system for potential improvements 8.Model the system 9.Track technology innovation 10.Design to create future options 10 Elements of applying a Whole System Design (WSD) approach

8. R. Shanthini 05 Feb 2010 Worked example overview A typical production plant scenario Design Challenge: Consider water at 20ºC flowing from reservoir A, through the system in the figure above to a tap with a target exit volumetric flow rate of Q = 0.001 m 3 /s. Select suitable pipes based on pipe diameter, D, and a suitable pump based on pump power, P, and calculate the cost of the system. Window (fixed into wall) Machine press (movable) (2) Elevation (Z 2 = 10 m) Elevation (Z 1 = 0 m) (1) A Q

9. R. Shanthini 05 Feb 2010 Conventional Design solution: Conventional system with limited application of the Elements of Whole System Design Whole System Design solution: Improved system using the Elements of Whole System Design Performance comparison: Comparison of the economic and environmental costs and benefit Design Process

10. R. Shanthini 05 Feb 2010 Conventional Design Solution

11. R. Shanthini 05 Feb 2010  The system accommodates the pre-existing floor plan (window) and equipment (machine press) in the plant.  Reservoir A exit is very well rounded.  The diameter of every pipe is D.  A globe valve, which acts as an emergency cut off and stops the flow for maintenance purposes, is fully open during operation.  The existing tap is replaced by a tap with an exit diameter of D. Conventional Design Solution (continued)

12. R. Shanthini 05 Feb 2010 The energy balance between point 1 and point 2 in the system is given by Bernoulli’s Equation: Σ P i /ρgA i V i = (p 2 /ρg - p 1 /ρg) + (α 2 V 2 2 /2g - α 1 V 1 2 /2g) + (z 2 - z 1 ) + Σ f i (L i /D i )(V i 2 /2g) + Σ K Li V i 2 /2g Friction head losses Head losses in pipe contractions, expansions, bends, joints and valves Pumping gains Step 8 of WSD: Model the system Pressure, kinetic energy and potential energy changes Ref: Appendix 6A

13. R. Shanthini 05 Feb 2010 Assumptions: p 1 = p 2 = atmospheric pressure; V 1 = 0; z 1 = 0 The diameter of every pipe is D, and therefore the cross sectional area of every pipe is A. The average velocity of the fluid in the downstream of the pump is constant and equal to V 2. The pipes are considered to be a single pipe of length L. Assume that head losses through reservoir A exit (well rounded), pump connectors and tap connectors are negligible. Step 8 of WSD: Model the system (continued)

14. R. Shanthini 05 Feb 2010 P/ρgAV 2 = α 2 V 2 2 /2g + z 2 + f (L/D)(V 2 2 /2g) + V 2 2 /2g (Σ K Li ) V 2 = Q / A = Q / (πD 2 /4) P = (8ρQ 3 /π 2 D 4 )(α 2 + f (L/D) + Σ K Li ) + ρgQz 2 Design equation relating the pump power (P) and pipe diameter (D) Under the assumptions, energy balance reduces to Velocity is related to volumetric flow rate by Combining them, we get Step 8 of WSD: Model the system (continued)

15. R. Shanthini 05 Feb 2010 The known variables are: ρ = 998.2 at 20 o C (Table 6A.3 in Appendix 6A) Q = 0.001 m 3 /s (design criterion) α 2 = 1 for uniform velocity profile > 1 for non-uniform velocity profile f depends on the Reynolds number of the flow L = 30 m (from system plan) K Li for 90 o threaded elbows = 1.5 (Table 6A.2) K LV for fully open glove valve = 10 (Table 6A.2) K LT for tab = 2 g = 9.81 m/s 2 z 2 = 10 m (from system plan) P = (8ρQ 3 /π 2 D 4 )(α 2 + f (L/D) + Σ K Li ) + ρgQz 2 Conventional Design Solution (continued)

16. R. Shanthini 05 Feb 2010 Re = ρV 2 D/μ = 4ρQ/ πDμ = 4(998.2 kg/m 3 )(0.001 m 3 /s)/ πD(1.002 x 10 -3 Ns/m 2 ) (μ is obtained from Table 6A.3 of Appendix 6A) Re = 1268.411 / D ReType of fluid flowf Re < 2100Laminar64 / Re 2100 < Re < 4000Transitional Re > 4000Turbulentfunction of Re and ε/D,where ε is the equivalent roughness. The friction factor, f, is dependent on the Reynolds number, Re For Re 0.604 m = 60.4 cm For Re > 4000, D < 0.317 m = 31.7 cm Conventional Design Solution (continued)

17. R. Shanthini 05 Feb 2010 Conventional Design Solution (continued) A pipe of diameter D = 0.317m is much larger than what would even be suitable for the system. Therefore, Re > 4000 and the flow is turbulent. Turbulent velocity profile can be assumed to be uniform, therefore α 2 = 1. P = (8.0911x10 -7 /D 4 )[f (30/D) + 19] + 97.923 Design equation therefore becomes For Re 0.604 m = 60.4 cm For Re > 4000, D < 0.317 m = 31.7 cm

18. R. Shanthini 05 Feb 2010 Conventional Design Solution (continued) P = (8.0911x10 -7 /D 4 )[f (30/D) + 19] + 97.923 Suppose drawn copper tubing of diameter D = 0.015 m was selected for the pipes. Then, Re = 1268.411/(0.015 m) = 84561 For drawn tubing, ε = 0.0015 mm (Table 6A.1 of Appendix 6A) Thus,ε/D = 0.0015/15 = 0.0001 Therefore, f = 0.0195 (from the Moody chart in Figure 6A.1 of Appendix 6A at Re = 84533 and ε/D = 0.0001)

19. R. Shanthini 05 Feb 2010 Conventional Design Solution (continued) P = (8.0911x10 -7 /D 4 )[f (30/D) + 19] + 97.923 Substituting D = 0.015 m and f = 0.0195 into the above equation, we get P = (8.0911x10 -7 /(0.015 m) 4 )[0.0195 (30/(0.015 m)) + 19] + 97.923 = 1025 W To generate an exit volumetric flow rate of Q = 0.001 m 3 /s, drawn copper tubing of 0.015 m diameter (D) pipe and 1025 W power pump (P) are required.

20. R. Shanthini 05 Feb 2010 From ‘Water pumps pricelist’ (Appendix 6B) we can select pump model: Waterco Hydrostorm Plus 150 at P = 1119 W (1.5 hp) Pump cost = $616 From ‘Hard drawn copper tube (6M length)’ in ‘Kirby copper pricelist’ (Appendix 6C) we can select pipe: T24937 at D = 15 mm (5/8 in) cost is $57.12 per 6m Pipe cost = ($57.12 per 6m)(30 m)/6 = $285.60 Total Cost of the Conventional Design Solution

21. R. Shanthini 05 Feb 2010 Total Cost of the Conventional Design Solution Elbow cost = ($2.34)(4) = $9.36 (For standard radius 90º elbows of 15mm (5/8 in) diameter J00231 ‘Copper fittings’ in Appendix 6C gives a cost of $2.34 each.) Estimated globe valve cost = $13 (US$10) (For a globe valve of diameter 15mm (5/8 in), by interpolating a ‘Components pricelist’, Appendix 6D) Tap cost = $6.70 (For a tap of exit diameter 0.015 m in ‘Components pricelist’, Appendix 6D) Installation costs = ($65/hr)(8 hrs) = $520 (Installation costs for 8hrs at $65/hr) Total capital cost of the system = $616 + $285.60 + $9.36 + $13 + $6.70 + $520 = $1451

22. R. Shanthini 05 Feb 2010 Total Cost of the Conventional Design Solution For the pump running at output power P = 1025 W, the monthly pump running costs for 12 hrs/day, 26 days/month are: Running cost = ($0.1/kWh)(1.025 kW)(12 hrs/day)(26 day/mth)/(0.47) = $68/month cost of electricity (2006 price for large energy users) pump efficiency for an electrical pump

23. R. Shanthini 05 Feb 2010 Whole System Design Solution Step 1 of WSD: Ask the right questions Could we change the pipe configuration so as to reduce the head losses? Did the selection procedure for pipe diameter, D, and pump power, P, address the whole system?

24. R. Shanthini 05 Feb 2010 Whole System Design Solution (continued) Step 7 of WSD: Review the system for potential improvements From Bernoulli’s equation, we can see that increasing diameter dramatically reduces the power required Can the system be designed with less bends? Can the system be designed with more-shallow bends? Is it worthwhile moving the plant equipment (machine press) ? Is an alternative pipe material more suitable? Is there a more suitable valve? Do we even need a valve? P = (8ρQ 3 /π 2 D 4 )(α 2 + f (L/D) + Σ K Li ) + ρgQz 2

25. R. Shanthini 05 Feb 2010 Whole System Design Solution (continued)

26. R. Shanthini 05 Feb 2010 Assumptions are the same as for the conventional system: p 1 = p 2 = atmospheric pressure; V 1 = 0; z 1 = 0 The diameter of every pipe is D, and therefore the cross sectional area of every pipe is A. The average velocity of the fluid in the downstream of the pump is constant and equal to V 2. The pipes are considered to be a single pipe of length L. Assume that head losses through reservoir A exit (well rounded), pump connectors and tap connectors are negligible. Whole System Design Solution (continued) P = (8ρQ 3 /π 2 D 4 )(α 2 + f (L/D) + Σ K Li ) + ρgQz 2 Same design equation as before

27. R. Shanthini 05 Feb 2010 The known variables are: ρ = 998.2 at 20 o C Q = 0.001 m 3 /s α 2 = 1 (assuming turbulent velocity profile) f depends on the Reynolds number of the flow L = 30 24 m K Li for 90 o 45 o threaded elbows = 1.5 0.4 (Table 6A.2) K LV for fully open glove gate valve = 10 0.15 (Table 6A.2) K LT for tab = 2 g = 9.81 m/s 2 z 2 = 10 m (from system plan) P = (8ρQ 3 /π 2 D 4 )(α 2 + f (L/D) + Σ K Li ) + ρgQz 2 Whole System Design Solution (continued) P = (8.0911x10 -7 /D 4 )[f (24/D) + 3.95] + 97.923

28. R. Shanthini 05 Feb 2010 P = (8.0911x10 -7 /D 4 )[f (24/D) + 3.95] + 97.923 Suppose drawn copper tubing of diameter D = 0.015 0.03 m was selected for the pipes. Then, Re = 1268/(0.03 m) = 42280 For drawn tubing, ε = 0.0015 mm (Table 6A.1 of Appendix 6A) Thus,ε/D = 0.0015/30 = 0.00005 Therefore, f = 0.0195 0.0215 (from the Moody chart in Figure 6A.1 of Appendix 6A at Re = 42267 and ε/D = 0.00005) Whole System Design Solution (continued)

29. R. Shanthini 05 Feb 2010 Substituting D = 0.03 m and f = 0.0215 into the above equation, we get P = (8.0911x10 -7 /(0.03 m) 4 )[0.0215 (24/(0.03 m)) + 3.95] + 97.923 = 1025 119 W P = (8.0911x10 -7 /D 4 )[f (24/D) + 3.95] + 97.923 Whole System Design Solution (continued) To generate an exit volumetric flow rate of Q = 0.001 m 3 /s, drawn copper tubing of 0.03 m diameter (D) pipe and 119 W power pump (P) are required.

30. R. Shanthini 05 Feb 2010 From ‘Water pumps pricelist’ (Appendix 6B) we can select pump model: Monarch ESPA Whisper 500 at P = 370 W (0.5 hp) From ‘Hard drawn copper tube (6M length)’ in ‘Kirby copper pricelist’ (Appendix 6C) we can select pipe: T22039 at D = 31.75 mm (1¼ in) Whole System Design Solution (continued) Is this a optimal solution for the whole system?

31. R. Shanthini 05 Feb 2010 Whole System Design Solution (continued) Step 3 of WSD. Design and optimize the whole system D (m)Reε/DfP (W) 0.015845610.00010.0195660 0.02634210.0000750.0205242 0.025507360.000060.0210148 0.03422800.000050.0215119 0.04317100.00003750.0230104 Pump power calculated for a spectrum of pipe diameters The capital and running costs for each pipe and pump combination are calculated in a similar way as for the conventional solution, and is given in the next slide.

32. R. Shanthini 05 Feb 2010 Whole System Design Solution (continued) D (m) Pipes and compone nts cost P (W) Pump selected Pump cost Total capital cost Running cost Life cycle cost (-NPV) 0.015$602660 Monarch ESPA Whisper 1000 $357$959$49/mth$10,227 0.02$745242 Monarch ESPA Whisper 500 $331$1076$19/mth$4,670 0.025$827148$331$1158$12/mth$3,428 0.03$914119$331$1245$9/mth$2947 0.04$1126104$331$1457$8/mth$2970

33. R. Shanthini 05 Feb 2010 Whole System Design Solution (continued) The efficiency of Monarch ESPA Whisper Model 1000 = 42% Model 500 = 40% The life cycle economic cost of each solution is estimated as the net present value (NPV) calculated over a life of 50 years and at a discount rate of 6%. Assumptions used for the cost calculation on the previous slide:

34. R. Shanthini 05 Feb 2010 Comparing the cost of the two solutions SolutionD (m) Pipes and componen ts cost P (W) Pump cost Total capital cost Running cost Life cycle cost (-NPV) Conven- tional 0.015$835 1025 $616$1451$61/mth$12,989 WSD0.03$914 119 $331$1245$9/mth$2947