1. PUMP HYDRAULICS: Brief Basics Rupp Carriveau University of Windsor Student Lecture PUMP HYDRAULICS

2. lecture overview 1. INTRODUCTION A. The importance of pumps B. The importance of this lecture C. An exercise in humility 2. THEORETICAL FOUNDATION A. The hydraulic system B. The pump capacity C. Finding your operating point 3. APPLIED EXAMPLE A. Problem layout B. Develop the system curve C. Develop pump curve D. Find operating point 4. KNOWLEDGE EXTENSION A. What we assumed B. Things to think about C. Resources to learn more PUMP HYDRAULICS

3. introduction the importance of pumps before pumps after pumps pump applications PUMP HYDRAULICS

4. introduction the importance of this lecture A B C U of W water source insert pump here pumpsystem PUMP HYDRAULICS

5. pump hydraulics introduction an exercise in humility A B C pump system

6. pump hydraulics theoretical foundation the hydraulic system H system = H static + h L but, h L = Σ friction losses + Σ minor losses We know: [A] [B] k1k1 k2k2 k3k3 k4k4 k5k5 k6k6 k7k7 k8k8 ( L s dsds e s ) ( L dd e d ) El. 50 m El. 55 m El. 70 m How much head is required to move the water from A to B for a given flow rate? Required: A B

7. pump hydraulics theoretical foundation the system equation Develop expression for system head, H system = H system (f,L,d,k,Q) Required: let’s work Q into [1]: v = Q/A = Q/(  (d/2) 2 ) [2]H system = H static + Σ fLv 2 /d2g + Σ kv 2 /2g We know: [1] so [1] into [2]: H system = H static + Σ fL/d2g (Q 2 /(  2 d 4 /16)) + Σ k/2g (Q 2 /(  2 d 4 /16)) [3] collecting terms: H system = H static + 8/gd 4  2 (Σ fL/d + Σ k) Q 2 [4] ‘C’‘C’ or simply: H system = H static + C Q 2 [5]

8. pump hydraulics theoretical foundation the system equation plot this simple relationship: H system = H static + C Q 2 H static = static head

9. pump hydraulics theoretical foundation the pump capacity Develop expression for head, H pump that can be delivered by the pump for variable Q Required: Pumps will supply declining head with increasing flow We know: Manufacturers may supply pump curve or data points which may be fit to an expression similar to [6] H pump = H o - A Q 2 [6] H o = shutoff head

10. pump hydraulics theoretical foundation operating point for the entire system Develop a relationship between the pump response and system response i.e., find an operating point for the pump and system working together Required: If we can equate the system curve and the pump curve we can find the operating point for the entire system We know: This can be done algebraically, two equations [5] and [6], and two unknowns, or graphically H operation Q operation

11. applied example problem layout Required: Find the operational head and flow for the given pump and system configuration k1k1 pump hydraulics Given: El. 50 m k2k2 k3k3 k4k4 k5k5 k6k6 k7k7 k8k8 ( L s dsds e s ) ( L dd e d ) El. 55 m El. 70 m Solution Steps: 1. Build expression for system curve 2. Build pump curve 3. Match curves to find operation point

12. pump hydraulics applied example solution step 1a: get system curve We know: H system = H static + (C s + C d ) Q 2 20 m and: C s = 8/(g  2 d s 4 ) [f s L s /d s + Σk s ] = C s = 8/(g  2 d s 4 ) [f s L s /d s + k 1 +k 2 +k 3 +k 4 ] C d = 8/(g  2 d d 4 ) [f d L d /d d + Σk d ] so: C s = 8/(9.81m/s 2 *  2 (0.20m) 4 ) [(0.015*30m/0.2m)+ (0.5 +2.0+2.5+0.9)] = 421 C d = 7573 but: assume fully turbulent flow, i.e., Re>3000, then use Swamee and Jain (1976), for ‘ f ’ f = 0.25/[log(e/3.7d) + 5.74Re 0.9 )] 2 f s = 0.015 f d = 0.016

13. pump hydraulics applied example Now, if we collect terms and plot: H system = 20 + 7994 Q 2 solution step 1b: plot system curve

14. pump hydraulics applied example We have been given 4 data points, we can try a quadratic fit to these points That is, we know the curve will look like: H pump = H o – AQ 2 and, we already know the shutoff head, H o = 48 m thus, find ‘A’ try a simple fit routine:A = AVG [(48-45)/0.02 2, (48-37)/0.04 2, (48-20)/0.06 2 ] A = 7384 H pump = 48 - 7384 Q 2 now we can write and plot our pump curve: solution step 2: get pump curve

15. H pump = 48 - 7384 Q 2 H system = 20 + 7994 Q 2 Solution: pump hydraulics applied example Q operation = 0.043 m 3 /s H operation = 34.54 m solution step 3: find operation point

16. pump hydraulics knowledge extension assumptions We must understand the major assumptions made in our solution process, so that we might appreciate when this technique is not applicable A1. Computation of friction factor f - are we within the valid range for the empirical expression we used? - if pumping conditions change will computed ‘f’ still be valid? A2. Pump curve fitting constant - did we choose the best method for fitting a trend line to the pump data?

17. pump hydraulics knowledge extension things to think about 1 It is interesting to consider what the equations have taught us about how a pumped hydraulic system behaves, let’s look at the system curve: Look at how sensitive ‘C’ is to changes in the system’s pipe diameter ‘d’ H system = H static + 8/g d 4  2 (Σ fL/d + Σ k) Q 2 [T1] ‘C’‘C’ - if changing the system was an option, then we might be able to move the operation point to the right by increasing the pipe diameter

18. pump hydraulics knowledge extension things to think about 2 Pumps operate at different efficiencies under different loads, we want to choose a pump that operates closest to its peak efficiency

19. pump hydraulics knowledge extension resources to learn more For those who wish to explore the topic further: “Pumps : selection, systems, and applications” R.H. Warring, Golf Publishing Company, 1984 “Pumps and pumping operations” Nicholas P. Cheremisinoff, Paul N. Cheremisinoff, Prentice Hall, 1992 “Hydraulics” Andrew L. Simon, Scott F. Korom., Prentice Hall, 1997