University of Hartford Department of Civil, Environmental and Biomedical Engineering CE 320 Water Resources Engineering Spring Semester 2017 Lab #5 - Pump Performance Characteristics Report Outline (1) Title Page (2) Answers to Centrifugal Pump Questions from Book 1: Understanding Centrifugal Pumping - Classroom Basics (see Pages 2-5) - 60 points (neatness counts!!) (3) Results - See Pages 6-7 for Required 5 Tables and 2 Figures - 70 points (4) Findings - 15 points a. Operating Point for Pump at 1800 rpm b. Operating Point for Pump at 1600 rpm c. Net Positive Suction Head required (NPSHr) (5) Appendix a. Hand Calculations (Equations are given on Page 8) - 20 points b. Excel Spreadsheet - 5 points 1 BOOK 1: UNDERSTANDING CENTRIFUGAL PUMPING - CLASSROOM BASICS Lesson 1: Define Centrifugal Pumping Focus 1: How Gravity and Velocity Relate to Centrifugal Pumping? 1. Calculate the velocity of a marble at impact if it is dropped from an \"X\" story building (each story is 12 ft. high). Disregard any air resistance. 2. At what speed would we need to launch the marble from the ground level to reach the top of the building? Group A: Group B: Group C: Group D: X = 5 Story Building X = 6 Story Building X = 7Story Building X = 8 Story Building Focus 2: Sizing and Impeller 1. For a centrifugal pump spinning at 3600 RPM, what is the impeller diameter that would be needed to fling water up \"X\" feet (a.k.a. develop a head of \"X\" feet)? Group A: Group B: Group C: Group D: X = 100 feet X = 125 feet X = 150 feet X = 175 feet Certification Quiz Groups A and B: What are the two main components of a centrifugal pump? Groups C and D: What typically drives a centrifugal pump? Lesson 2: Specific Gravity of a Fluid (Why head is expressed in feet?) 1. You have \"X\" gallons of mercury with a Specific Gravity of 13.5. When you weigh a gallon of water at STP, the scale reads 8.3 lbs. What will the scale read when you weigh \"X\" gallons of mercury? Group A: Group B: Group C: Group D: X = 1.5 gallons X = 2.0 gallons X = 2.5 gallons X = 3.0 gallons 2. A train leaves Chicago carrying 500 gallons of diesel fuel (specific gravity of 0.82) for its engines, and travels 50 mph which is 400 miles away. The 2 engines combine to burn \"X\" gallons of fuel per hour. How much less will the train weigh when it gets to Topeka? Group A: Group B: Group C: Group D: X = 40 gallons/hour X = 45 gallons/hour X = 50 gallons/hour X = 55 gallons/hour Lesson 3: Viscosity of a Fluid 1. What is viscosity? 2. How do you measure viscosity? 3. Using the chart in this section: What happens to Pump Efficiency and Total Dynamic Head it can produce as the viscosity of the fluid it pumps get larger? What about the required power to drive the pump? Why? Lesson 4: Friction and Friction Head 1. Based on the supplied charge on Page 14 (Figure 4.2), what would be the added friction head to your system calculation for a 17 year old, \"X\" inch diameter, 200 foot long pipe that is used to pump 120 gpm of water? Group A: Group B: Group C: Group D: X = 1 \" pipe diameter X = 2\" pipe diameter X = 2 \" pipe diameter X = 3\" pipe diameter 2. A brand new 2 inch pipe is installed right next to a pipe that has been in service for 25 years. Each pipe has to deliver 100 gpm of output at the end of a 100 foot run. How much friction does each pump have to overcome for each pipe? 3. If you are pumping 10 gpm through a 1 inch pipe diameter and also pumping the same amount through a 2 inch diameter pipe, which pipe offers the most friction drag? Why? 3 Lesson 5: Introduction to Suction Conditions Focus 1: Pressure 1. If you have a soccer ball with a pressure indication gauge on it, and pumped it up to 10 psig, what is this pressure called? How does that relate to atmospheric pressure? How does that relate to absolute pressure? 2. Which requires a higher suction pressure when drinking through a straw; cola or a thick milk shake? Why? 3. What is the absolute vapor pressure of water if feet at \"X\" F? Group A: Group B: Group C: Group D: X = 120F X = 150F X = 180F X = 210F Focus 2: Capacity and Suction Lift 1. Using Figure 5.3 (page 21), What happens to the suction lift and head of a centrifugal pump when you increase the flow from 220 gpm to 290 gpm? Certification Quiz 1. You are drinking a glass of water with a straw inside a building that is pressurized above atmospheric pressure. If you go outside, will it be easier, harder, or make no difference on how hard you have to suck on the straw to drink the water? Why? Lesson 6: Introduction to Suction Conditions Focus 1: Net Positive Suction Head Available (NPSHa) 1. Define what NPSHa means for a centrifugal pump? Focus 2: Net Positive Suction Head Required (NPSHr) 1. Define what NPSHr means for a centrifugal pump? 2. What is the best way to determine NPSHr? 3. What is the phenomena we are trying to avoid by maintaining desired NPSH values? 4 Certification Quiz 1. Can water boil at virtually any temperature? Why? 2. True or False; centrifugal pumps create a partial vacuum in the inlet piping during operation. Lesson 7: Cavitation - Another Word for Trouble 1. Give 3 problems cavitation creates for centrifugal pump operator. 2. Does cavitation occur on the discharge side or intake side of the pump? Certification Quiz 1. Cavitation bubbles collapse in the system, rather than burst like those in boiling pot of water. Which transmits more energy by their action? 2. Does cavitation occur when the NPSHa (available) at the pump suction inlet falls below or above the minimum NPSHr (required) by a particular pump? Why? 3. You are working in a dairy processing plant. Your boss points to 2 pumps which are transferring milk from the individual holding tank to a process area. He tells you pump B does not seem to be getting the milk to the processing area as well as pump A. What might be the problem and what could you do to change the situation? 5 Results: Excel Tables and Figures (see page 8 for equations) - Flow Rate in gpm and Head in feet a. Table 1 - Pump Characteristics at 1800 rpm from Closing Discharge Valve (Focus 1 - Page 21) i. Flow Rate ii. Discharge Pressure Head iii. Discharge Velocity Head iv. Discharge Static Head v. Total Dynamic Discharge Head vi. Suction Pressure Head vii. Suction Velocity Head viii. Suction Static Head ix. Total Dynamic Discharge Head x. Total Dynamic Head b. Table 2 - Pump Characteristics at 1600 rpm from Closing Discharge Valve (Focus 1 - Page 26) i. Flow Rate ii. Discharge Pressure Head iii. Discharge Velocity Head iv. Discharge Static Head v. Total Dynamic Discharge Head vi. Suction Pressure Head vii. Suction Velocity Head viii. Suction Static Head ix. Total Dynamic Discharge Head x. Total Dynamic Head c. Table 3 - System Operating Curve - (Focus 2 - Page 27) i. Flow Rate ii. Discharge Pressure Head iii. Discharge Velocity Head iv. Discharge Static Head v. Total Dynamic Discharge Head vi. Suction Pressure Head vii. Suction Velocity Head viii. Suction Static Head ix. Total Dynamic Discharge Head x. Total Dynamic Head 6 d. Table 4 - Pump Characteristics at 1800 rpm from Closing Suction Valve (Focus 3 - Page 30) i. Flow Rate ii. Discharge Pressure Head iii. Discharge Velocity Head iv. Discharge Static Head v. Total Dynamic Discharge Head vi. Suction Pressure Head vii. Suction Velocity Head viii. Suction Static Head ix. Total Dynamic Discharge Head x. Total Dynamic Head e. Table 5 - Net Positive Suction Head at 1800 rpm (Focus 3- Page 30) i. Flow Rate ii. Suction Pressure Head iii. Suction Velocity Head iv. Vapor Pressure Head at 70F v. Net Positive Suction Head Available f. Figure 1 - Pump Characteristics and System Operating Curve (results from Tables 1, 2, and 3 all on same graph) i. Pump Characteristics at 1800 rpm (TDH vs Flow) ii. Pump Characteristics at 1600 rpm (TDH vs Flow) iii. System Operating Curve (TDH vs Flow) iv. Show operating point for the pump operating at 1600 rpm and 1800 rpm g. Figure 2 - Net Positive Suction Head Required (results from Tables 1, 4, and 5 all on same graph) - See Page 54 i. Pump Characteristics at 1800 rpm from closing discharge valve (TDH vs Flow from Table 1) ii. Pump Characteristics at 1800 rpm from closing suction valve (TDH vs Flow from Table 4) iii. Net Positive Suction Head Available vs Flow from Table 5 iv. Show Net Positive Suction Head Required on Graph where TDH decreases from 3% of its normal value due to cavitation 7 Equations: Total Dynamic Head (TDH) - Calculations for Table 1 (Pump at 1800 rpm), Table 2 (Pump at 1600 rpm), and Table 3 (System Operating Curve) 1) Total Dynamic Head (TDH) Total Dynamic Discharge Head (TDDH) - Total Dynamic Suction Head (TDSH) a. TDDH = Pressure Head + Velocity Head + Static Head i. Pressure Head = Pdischarge/ where = 62.4 lbf/ft3 ii. Velocity Head = v2/2g where discharge pipe diameter = 1.6 in. iii. Static Head = z (We also call this elevation head) - See system diagram below b. TDSH = Pressure Head + Velocity Head + Static Head i. Pressure Head = Psuction/ where = 62.4 lbf/ft3 ii. Velocity Head = v2/2g where suction pipe diameter = 1.85 in. iii. Static Head = z (We also call this elevation head) - See system diagram below Net Positive Suction Head Available (NPSHa) 1) NPSHa = Suction Pressure Head + Suction Velocity Head - hvp(T) a. Suction Pressure Head = Psuction/ where = 62.4 lbf/ft3 b. Suction Velocity Head = v2/2g where suction pipe diameter = 1.85 in. c. Water Vapor Pressure at 70F in feet 8 Example for Figure 2 9 Book Classroom Basics Understanding Centrifugal Pumping Turbine Technologies, Ltd., 410 Phillips Street, Chetek, WI, 54728 USA. Phone: 715-924-4876, Fax: 715-924-2436, www.turbinetechnologies.com BOOK 1: Understanding Centrifugal Pumping - Classroom Basics INTRODUCTION Upon completion of this book, the student will have a good understanding of important issues related to centrifugal pumping, including; the relationship of gravity and velocity, pressure head, pump impeller sizing, fluid specific gravity and viscosity, system friction head, suction head/lift, Net Positive Suction Head (NPSH) and Cavitation. This knowledge will serve as the foundation for learning to effectively operate a live centrifugal pump system. Turbine Technologies, Ltd., LabVIEW, National Instruments are trademarks of respective systems and equipment used in equipment manufactured by Turbine Technologies, Ltd. Copyright 2012 by Turbine Technologies, Ltd. All rights reserved. No part of this lesson series may be reproduced, translated or transmitted in any form or by any means without the permission of Turbine Technologies, Ltd. EVALUATION COPY Turbine Technologies, Ltd., 410 Phillips Street, Chetek, WI, 54728 USA. Phone: 715-924-4876, Fax: 715-924-2436, www.turbinetechnologies.com 2 Understanding Centrifugal Pumping Copyright 2012 by Turbine Technologies, Ltd. Book 1: Understanding Centrifugal Pumping - Classroom Basics TABLE OF CONTENTS Book 1: Introduction to Centrifugal Pumping - Classroom Basics Lesson 1: Define Centrifugal Pumping/Introduction of Primary Components.......................................4 Focus 1: How Gravity and Velocity Relate to Centrifugal Pumping Focus 2: Sizing an Impeller Knowledge Certification Quiz: Lesson 1 Lesson 2: Specific Gravity of a Fluid (Why head is expressed in feet)................................................. 10 Knowledge Certification Quiz: Lesson 2 Lesson 3: Viscosity of a Fluid....................................................................................................................... 12 Knowledge Certification Quiz: Lesson 3 Lesson 4: Friction and Friction Head.......................................................................................................... 14 Knowledge Certification Quiz: Lesson 4 Lesson 5: Introduction to Suction Conditions........................................................................................... 17 Focus 1: Suction Lift Focus 2: Capacity and Suction Lift Knowledge Certification Quiz: Lesson 5 Lesson 6: Net Positive Suction Head (NPSH) & Introduction to Cavitation......................................... 23 Focus 1: Net Positive Suction Head Available (NPSHa) Focus 2: Net Positive Suction Head Required (NPSHr) Knowledge Certification Quiz: Lesson 6 Lesson 7: Cavitation - Another Word for Trouble..................................................................................... 29 Knowledge Certification Quiz: Lesson 7 EVALUATION COPY Understanding Centrifugal Pumping Copyright 2012 by Turbine Technologies, Ltd. 3 BOOK 1: Understanding Centrifugal Pumping - Classroom Basics Lesson 1: Define Centrifugal Pumping Approximate Lesson Duration: 25 min. Student will understand what a centrifugal pump is, what the major components of a centrifugal pump are, what pumping head is, and how to size a pump impeller. Centrifugal pumps are used in many applications including water, sewage, petroleum and petrochemical pumping. They have become a primary component of many industries. A centrifugal pump is a machine that gives energy to a fluid. This energy infusion can cause a liquid to flow, rise to a higher level, or both. The centrifugal pump is a very simple rotary machine which consists of two major parts: 1) impeller (rotating component) and 2) volute (the stationary structure or casing). The figure shows these two parts (along with the eye, which is the fluid inlet into the volute). The centrifugal pump's function is as simple as its design. It is filled with liquid and the impeller is rotated (usually with an electric motor). The impeller vanes grab the liquid and \"fling\" it, giving it energy, which causes it to exit the impeller's vanes at a greater velocity than it possessed when it entered. IMPELLER EYE VOLUTE Figure 1.1: Pump Primary Components EVALUATION COPY 4 Understanding Centrifugal Pumping Copyright 2012 by Turbine Technologies, Ltd. Book 1: Understanding Centrifugal Pumping - Classroom Basics DISCHARGE This outward flow (flinging) reduces the pressure at the impeller eye (the center where the inlet piping allows the fluid to enter the casing), allowing more liquid to enter. The liquid exits the impeller at its edge (peripheral velocity) and is collected in the casing (volute) where its velocity is converted to pressure before it leaves the pump's discharge. Figure 1.2: Impeller Outward Flow EYE What does a real centrifugal pump look like? Figure 1.3: Commercial Centrifugal Pump What does a real centrifugal impeller look like? Figure 1.4: Centrifugal Pump Impeller EVALUATION COPY Understanding Centrifugal Pumping Copyright 2012 by Turbine Technologies, Ltd. 5 BOOK 1: Understanding Centrifugal Pumping - Classroom Basics Focus 1: Let's Fling Stones! (How gravity and velocity relate to centrifugal pumping) Gravity is one of the more important forces that a centrifugal pump must overcome. You will find that the relationship between final velocity, due to gravity, and initial velocity, due to impeller speed, is a very useful one. Let's use an example that's easy to visualize to better understand what we're talking about. If a stone is dropped from the top of a building its speed (velocity) will increase at a known rate of 32.2 feet per second for each second that it falls. This increase in velocity is known as acceleration due to gravity. If we ignore the effect of air resistance on the falling stone, we can predict the velocity it will hit the ground based upon its starting height off the ground and the effect of acceleration due to gravity (that 32.2 feet per second PER second). Now, with help of mathematics and even Sir Isaac Newton himself (you remember that he discovered this thing called gravity), a convenient equation was developed which describes the relationship of velocity, height, and gravity as it applies to a falling body, like that stone as: v2 = 2gh Where: v = the velocity of the body in ft/sec g = the acceleration due to gravity @ 32.2 ft/sec/sec (or ft/sec2) h = the distance through which the body falls For example, if a stone is dropped from a building 100 feet high: v2 = 2 x 32.2 ft/sec2 x 100 ft, which gives us... v2 = 6440 ft2/sec2 v = square root of 6440 ft2/sec2 = 80.3 ft/sec The stone, therefore, will strike the ground at a velocity of 80.3 feet per second (about 55 miles per hour). EVALUATION COPY 6 Understanding Centrifugal Pumping Copyright 2012 by Turbine Technologies, Ltd. Book 1: Understanding Centrifugal Pumping - Classroom Basics Now, let's reverse this scenario. That same equation allows us to figure out how fast we would have to throw that stone (initial velocity) upward from ground level for it to reach the top of the building (100 feet). This is true because the final velocity of a falling body happens to be equal to the initial velocity required to launch it to get it to that same height from which it fell (got that?). Using the example above, the initial velocity required to throw the stone to a height of 100 feet is 80.3 feet per second, the same as its final velocity. So what does dropping and throwing stones have to do with pumping? Here's the cool part: the same equation we used for the stone applies when pumping water with a centrifugal pump. The velocity of the water as it leaves the impeller determines the head developed. In other words the water is \"thrown\" to a certain height. To reach this height it must start with the same velocity it would attain if it fell from that height. So, if we rearrange the falling body equation we get: h = v2/2g Now we can determine the height to which a body (or water) will rise given a particular initial velocity. For example, at 10 ft per sec: h = (10 ft/sec x 10 ft/sec) / (2 x 32.2 ft/sec2) h = 100 ft2/sec2 / 64.4 ft/sec2 h = 1.55 ft Lesson 1, Focus 1 1. Calculate the velocity of a marble at impact if it is dropped from a 6 story building (each story is 12 ft. high). Disregard any air resistance. 2. At what speed would we need to launch the marble from the ground level to reach the top of the building? 3. If you were to try this with several different initial velocities, you would find out that there is an interesting relationship between the height achieved by a body and its initial velocity. This relationship is one of the Affinity Laws of centrifugal pumps which can be studied independently. EVALUATION COPY Understanding Centrifugal Pumping Copyright 2012 by Turbine Technologies, Ltd. 7 BOOK 1: Understanding Centrifugal Pumping - Classroom Basics Focus 2: Sizing an Impeller As a finale to this section, let's apply what we have learned to a practical application. Follow each step carefully and you will gain a good understanding of how this works (you'll feel so smart). Problem: We have a centrifugal pump that is spinning at 1800 RPM. We need to figure out what impeller diameter would be needed to fling the water up 200 feet (a.k.a. develop a head of 200 feet). First we must calculate the initial velocity required to develop a head of 200 feet. Using our falling body equation from before: v2 = 2gh v2= 2 x 32.2 ft/sec2 x 200 ft v2= 12,880 ft2/sec2 v = square root of 12,880 ft2/sec2 = 113 ft/sec We also need to know the number of rotations the impeller undergoes each second: 1800 RPM / 60 sec = 30 RPS (Revolutions per Second). Now we can compute the number of feet a point on the impellers rim travels in a single rotation: 113 ft/sec divided by 30 rotations/sec = 3.77 ft/rotation Since distance traveled per rotation is the same as the circumference of the impeller we can compute the diameter using the circular relation formula as follows: Diameter = Circumference / Pi Diameter = 3.77 Ft / 3.1416 Diameter = 1.2 Feet or 14.4 Inches Therefore an impeller with a diameter of approximately 14.4\" turning at 1800 RPM will produce a head of (fling water up) 200 Feet. Lesson 1, Focus 2 We have a centrifugal pump that is spinning at 1800 RPM. What impeller diameter would be needed to fling the water up 122 feet (a.k.a. develop a head of 122 feet). _________________ EVALUATION COPY 8 Understanding Centrifugal Pumping Copyright 2012 by Turbine Technologies, Ltd. Book 1: Understanding Centrifugal Pumping - Classroom Basics Knowledge Certification Quiz: Lesson 1 Please answer all of the following questions without referring to the text. If you can answer all questions, you are ready for Lesson 2. If not, please review Lesson 1 and retake certification quiz. 1. Name 3 areas centrifugal pumps are used in. 2. What are the 2 main components of a centrifugal pump? 3. What typically drives a centrifugal pump? 4. Calculate the velocity of a cue ball that was dropped from a 100 foot building as it hits the ground. 5. \u0007How fast would you need to throw that cue ball upward from ground level for it to reach the top of the 100 foot building? 6. What would the diameter of a pump impeller have to be to throw water up that high? I certify that I have answered all certification quiz questions correctly and am ready for the next lesson. ____________________________________\t_____________________________ Your Signature Date EVALUATION COPY Understanding Centrifugal Pumping Copyright 2012 by Turbine Technologies, Ltd. 9 BOOK 1: Understanding Centrifugal Pumping - Classroom Basics Approximate Lesson Duration: 10 min. Lesson 2: Specific Gravity of a Fluid (Why head is expressed in feet) Chances are pretty good that you will be pumping a liquid other than water in your field. The good news is that in pumping, each liquid is compared to water (at a standard temperature and pressure) to get a handle on how a pump will perform when pumping it. This comparison is known as Specific Gravity. This section will explain Specific Gravity and how it relates to head. The Specific Gravity of a substance is the ratio of the weight of a given volume of the substance to that of an equal volume of water at standard temperature and pressure (STP). Assuming the viscosity of a liquid (see Section 1, Part 3) is similar to that of water, the following statements will always be true regardless of the specific gravity: 1) A Centrifugal pump will always develop the same head in feet regardless of a liquid's specific gravity. 2) Pressure will increase or decrease in direct proportion to a liquid's specific gravity. 3) Brake HP required will vary directly with a liquid's specific gravity. Figure 2.1 illustrates the relationship between pressure (in psi) and head (in ft). If that same pump requires 10 HP when pumping water, it will require 11 HP when pumping ethylene glycol and only 8.2 HP when pumping jet fuel (based on the specific gravity ratios). The preceding discussion of Specific Gravity illustrates why centrifugal pump head (or pressure) is expressed in feet. Since pump specialists work with many liquids of varying specific gravity, head in feet is the most convenient system of designating head. When selecting a pump, always remember that factory tests and curves are based on water at STP. If you are working with other liquids always correct the HP required for the specific gravity of the liquid being pumped. Specific Gravity Pressure Variations at Constant Head We can see that the level in each of the three tanks is 100 feet. The resulting pressure at the bottom of each varies substantially as a result of the varying specific gravity. Now, if we keep pressure constant as measured at the bottom of each tank, then it would stand to reason that the fluid levels would vary in a similar way. Notice that the specific gravity (remember definition) of Jet Fuel is only 82% that of water, while ethylene glycol will be 110% of water. It is very interesting to look up a chart of the specific gravity of a number of liquids. This really gives you a feel for how things compare and may even surprise you. For instance, olive oil has a specific gravity of 91% while glucose is around 140%. One element we'll talk about later is the Viscosity of the liquid. A centrifugal pump can also develop 100' of head when pumping water, jet fuel, and ethylene glycol. The resulting pressures, however, will vary just as those seen in the figure. 100 Ft. 35.5 PSI 100 Ft. 47.6 PSI 43.3 PSI Jet Fuel SP. GR. = 0.82 100 Ft. Water SP. GR. = 1.0 Ethylene Glycol SP. GR. = 1.1 Figure 2.1: Specific Gravity Comparisons (pressure related to head) EVALUATION COPY 10 Understanding Centrifugal Pumping Copyright 2012 by Turbine Technologies, Ltd. Book 1: Understanding Centrifugal Pumping - Classroom Basics Lesson 2 What is the basic definition of Specific Gravity? You have a gallon of mercury with a Specific Gravity of 13.5. When you weigh a gallon of water at STP, the scale reads 8.3 lbs (something to keep in mind). What will the scale read when you weigh a gallon of mercury? A train (yes, a train) leaves Chicago carrying 500 gallons of diesel fuel (specific gravity of 0.82) for its engines, and travels at 50 MPH toward Topeka, which is 400 miles away. The engines combine to burn 50 gallons of fuel per hour. How much less will the train weigh when it gets to Topeka? Knowledge Certification Quiz: Lesson 2 1. \u0007What is the specific gravity of water at standard temperature and pressure? 2. \u0007If 3 identical tanks are filled with 3 mystery fluids to a height each of 50 feet and the pressure readings at the bottom of the tanks all read the same pressure, what does that tell us about the specific gravity of each fluid. 3. \u0007The 3 tanks from question 2 are drained. One is filled with water to 100 feet. The second is filled with a Mystery Fluid A to 100 feet. The third is filled with a Mystery Fluid B to 100 feet. If the volume in each tank is 1,000 gallons and Mystery Fluid 1 has a Specific Gravity of 0.8 and Mystery Fluid B has a Specific Gravity of 1.2, how much does each tank of mystery fluid weigh if the water weighs 8.3 lbs/gallon. I certify that I have answered all certification quiz questions correctly and am ready for the next lesson. ____________________________________\t_____________________________ Your Signature Date EVALUATION COPY Understanding Centrifugal Pumping Copyright 2012 by Turbine Technologies, Ltd. 11 BOOK 1: Understanding Centrifugal Pumping - Classroom Basics Approximate Lesson Duration: 5 min. Lesson 3: Viscosity of a Fluid This lesson will explain the importance of viscosity in fluid pumping and how it changes pump performance. Viscosity is a fluid property that is independent of specific gravity. Just as resistivity is the inherent resistance of a particular conductor, viscosity is the internal friction of a fluid. The coefficient of viscosity of a fluid is the measure of its resistance to flow. Fluids having a high viscosity are sluggish in flow. Examples include molasses and heavy oil. Viscosity usually varies greatly with temperature with viscosity decreasing as temperature rises. The instrument used to measure viscosity is the viscometer. Although there are many, the Saybolt Universal is the most common. It measures the time in seconds required for a given quantity of fluid to pass through a standard orifice under STP. The unit of measurement is the SSU or Seconds Saybolt Universal. Normally, small and medium sized centrifugal pumps can be used to handle liquids with viscosities up to 2000 SSU. Below 50 SSU the Characteristic Curves remain about the same as those of water; however, there is an immediate decrease in efficiency when viscosity increases over that of water. Viscosities over 2000 SSU are usually better suited for positive displacement pumps. This is good to know, especially if you're hired to work in a molasses factory! SSU 2000 1000 SSU HE AD 500 SSU 200 SSU .5 SSU ATER) 31 B.H.P. (W SU 0S U S S 50 00 U 10 SS 00 20 CA PA CI TY 20 (W 0 S AT SU ER )3 1.5 SS U U S S CY EN I C FI EF 5 31. R) E T WA %( 200 S SU 500 SS U 1000 SSU 2000 SSU Figure 3.1: Pump Performance at Various Viscosities EVALUATION COPY 12 Understanding Centrifugal Pumping Copyright 2012 by Turbine Technologies, Ltd. Book 1: Understanding Centrifugal Pumping - Classroom Basics Lesson 3 1. What is viscosity? 2. How do you measure viscosity? 3. Using the chart shown on page 12 (Figure 3.1), answer this question: If a fluid has a viscosity of 1000 SSU and is being pumped at 90 gallons per minute, how does the efficiency compare if you were pumping water (SSU of 315)? Knowledge Certification Quiz: Lesson 3 1. What is viscosity? 2. If you were interested in how much a gallon of fluid weighed, would finding its viscosity be the best way to do this? 3. Name an industry device which measures viscosity. 4. Using the chart in this section: What happens to Pump Efficiency and Total Dynamic head it can produce as the viscosity of the fluids it pumps get larger? What about the required power to drive the pump? Why? I certify that I have answered all certification quiz questions correctly and am ready for the next lesson. ____________________________________\t_____________________________ Your Signature Date EVALUATION COPY Understanding Centrifugal Pumping Copyright 2012 by Turbine Technologies, Ltd. 13 BOOK 1: Understanding Centrifugal Pumping - Classroom Basics Approximate Lesson Duration: 10 min. Lesson 4: Friction and Friction Head Friction occurs when a fluid flows in or around a stationary object or when an object moves through a fluid. An automobile and an aircraft are subjected to the effects of friction as they move through our atmosphere. Boats create friction as they move through the water. It so happens that liquids create friction as they move through a closed pipe. This lesson will discuss how friction head contributes to centrifugal pumping requirements. Designers and engineers spend a great deal of money on the design and redesign of aircraft, automobiles and boats to reduce friction (also called drag). Why? Because friction produces heat and where there is heat there is energy, wasted energy that is. Making these vehicles more aerodynamic, or \"slippery\