Whole System Design: An integrated Approach to Sustainable Engineering Units 6 – 10: Worked Examples July 2007

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. Whole System Design: An integrated Approach to Sustainable Engineering Unit 6: Industrial Pumping Systems

4. Significance of Pumping Systems and Design 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.

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

6. Significance of Pumping Systems and Design (continued) 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.

7. 10 Elements of applying a Whole System Design (WSD) approach • Ask the right questions • Benchmark against the optimal system • Design and optimise the whole system • Account for all measurable impacts • Design and optimise subsystems in the right sequence • Design and optimise subsystems to achieve compounding resource savings • Review the system for potential improvements • Model the system • Track technology innovation • Design to create future options

8. (2) Elevation (Z2 = 10 m) Window (fixed into wall) Elevation (Z1 = 0 m) Q (1) Machine press (movable) A 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 m3/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.

9. Design Process 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

10. Conventional Design Solution

11. Conventional Design Solution (continued) • 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.

12. Step 8 of WSD: Model the system The energy balance between point 1 and point 2 in the system is given by Bernoulli’s Equation: Σ Pi/ρgAiVi = (p2/ρg - p1/ρg) + (α2V22/2g - α1V12/2g) + (z2 - z1) + Σ fi (Li/Di)(Vi2/2g) + Σ KLiVi2/2g Pumping gains Pressure, kinetic energy and potential energy changes Friction head losses Head losses in pipe contractions, expansions, bends, joints and valves Ref: Appendix 6A

13. Step 8 of WSD: Model the system (continued) • Assumptions: • p1 = p2 = atmospheric pressure; V1 = 0; z1 = 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 V2. • 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.

14. Step 8 of WSD: Model the system (continued) Under the assumptions, energy balance reduces to P/ρgAV2 = α2V22/2g + z2 + f (L/D)(V22/2g) + V22/2g (Σ KLi) Velocity is related to volumetric flow rate by V2 = Q / A = Q / (πD2/4) Combining them, we get P = (8ρQ3/π2D4)(α2 + f (L/D) + Σ KLi) + ρgQz2 Design equation relating the pump power (P) and pipe diameter (D)

15. Conventional Design Solution (continued) P = (8ρQ3/π2D4)(α2 + f (L/D) + Σ KLi) + ρgQz2 • The known variables are: • ρ = 998.2 at 20oC (Table 6A.3 in Appendix 6A) • Q = 0.001 m3/s(design criterion) • α2 = 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) • KLi for 90o threaded elbows = 1.5 (Table 6A.2) • KLV for fully open glove valve = 10 (Table 6A.2) • KLT for tab = 2 • g = 9.81 m/s2 • z2 = 10 m (from system plan)

16. Conventional Design Solution (continued) The friction factor, f, is dependent on the Reynolds number, Re Re = ρV2D/μ = 4ρQ/ πDμ = 4(998.2 kg/m3)(0.001 m3/s)/ πD(1.002 x 10-3 Ns/m2) (μ is obtained from Table 6A.3 of Appendix 6A) Re = 1268.411 / D For Re < 2100, D > 0.604 m = 60.4 cm For Re > 4000, D < 0.317 m = 31.7 cm

17. Conventional Design Solution (continued) For Re < 2100, D > 0.604 m = 60.4 cm For Re > 4000, D < 0.317 m = 31.7 cm 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 α2 = 1. Design equation therefore becomes P = (8.0911x10-7/D4)[f (30/D) + 19] + 97.923

18. Conventional Design Solution (continued) P = (8.0911x10-7/D4)[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. Conventional Design Solution (continued) P = (8.0911x10-7/D4)[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 m3/s, drawn copper tubing of 0.015 mdiameter (D) pipe and 1025 W power pump (P) are required.

20. Total Cost of the Conventional Design Solution 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

21. 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. 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. 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. 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 = (8ρQ3/π2D4)(α2 + f (L/D) + Σ KLi) + ρgQz2

25. Whole System Design Solution (continued)

26. Whole System Design Solution (continued) • Assumptions are the same as for the conventional system: • p1 = p2 = atmospheric pressure; V1 = 0; z1 = 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 V2. • 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. P = (8ρQ3/π2D4)(α2 + f (L/D) + Σ KLi) + ρgQz2 Same design equation as before

27. Whole System Design Solution (continued) P = (8ρQ3/π2D4)(α2 + f (L/D) + Σ KLi) + ρgQz2 • The known variables are: • ρ = 998.2 at 20oC • Q = 0.001 m3/s • α2 = 1 (assuming turbulent velocity profile) • f depends on the Reynolds number of the flow • L = 30 24 m • KLi for 90o 45o threaded elbows = 1.5 0.4(Table 6A.2) • KLV for fully open glove gate valve = 10 0.15(Table 6A.2) • KLT for tab = 2 • g = 9.81 m/s2 • z2 = 10 m (from system plan) P = (8.0911x10-7/D4)[f (24/D) + 3.95] + 97.923

28. Whole System Design Solution (continued) P = (8.0911x10-7/D4)[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)

29. Whole System Design Solution (continued) P = (8.0911x10-7/D4)[f (24/D) + 3.95] + 97.923 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 To generate an exit volumetric flow rate of Q = 0.001 m3/s, drawn copper tubing of 0.03 mdiameter (D) pipe and 119 W power pump (P) are required.

30. Whole System Design Solution (continued) 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) Is this a optimal solution for the whole system?

31. Whole System Design Solution (continued) Step 3 of WSD. Design and optimize the whole system 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. Whole System Design Solution (continued)

33. Whole System Design Solution (continued) Assumptions used for the cost calculation on the previous slide: 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%.

34. Comparing the cost of the two solutions