1 K1.01 Distinguish between static pressure, dynamic pressure, and total pressure.K1.02 Define head loss. K1.03 Discuss operational considerations of viscosity as related to head loss. K1.04 Explain operational implications of water hammer. K1.05 Define or explain the following terms and concepts: Mass flow rate K1.06 Define or explain the following terms and concepts: Two-phase flow K1.07 Define or explain the following terms and concepts: Pressure spike K1.08 Define or explain the following terms and concepts: Gas binding K1.09 Define or explain the following terms and concepts: Recirculation ratio K1.10 Define or explain the following terms and concepts: Water hammer K1.11 Define or explain the following terms and concepts: Cavitation K1.12 Explain why flow measurements must be corrected for density changes. K1.13 Explain the relationship between pressure head and velocity head in a fluid system. K1.14 Discuss the velocity profiles for laminar flow and turbulent flow. K1.15 Describe the methods of controlling system flow rates. Operator Generic Fundamentals Thermodynamics – Fluid Statics and Dynamics

2 Terminal Learning ObjectivesAt the completion of this training session, the trainee will demonstrate mastery of this topic by passing a written exam with a grade of 80 percent or higher on the following Terminal Learning Objectives (TLOs): Explain the fundamental properties of fluids and closed fluid systems. Describe laminar and turbulent flows. Explain Bernoulli’s equation then use it to determine fluid properties at different points in a system. Explain the various changes in fluid flow, their effect on plant operation, and methods of control. TLOs

3 Fundamental Properties of FluidsTLO 1 – Explain the fundamental properties of fluids and closed fluid systems. 1.1 Explain the following terms: Buoyancy Static pressure Dynamic pressure Total pressure 1.2 Explain the following terms: Pascal’s Law Volumetric flow rate Mass flow rate Conservation of mass Continuity equation Density compensation TLO 1

4 Fundamental Properties of FluidsELO 1.1 – Explain the following terms: buoyancy, static pressure, dynamic pressure, and total pressure. A fluid is any substance that flows freely and its molecules lack rigid attachment within a crystalline structure. Liquids, gases, and materials normally considered solids, such as glass. Incompressible or compressible describe fluid properties. Maintain a constant density with changes in pressure; for example, liquids. Related KA – K1.01 Distinguish between static pressure, dynamic pressure, and total pressure ELO 1.1

5 Fundamental Properties of FluidsCompressible fluids like gases change density easily. Gases expand and completely fill their container. Properties of fluids affect the way in which the fluid behaves. Temperature Pressure Mass Specific volume Density Briefly review these terms with the class. ELO 1.1

6 Fundamental Properties of FluidsBuoyancy: Describes how an object responds when placed in a fluid. Our bodies float in water, as do wood and ice. A rock underwater is lighter than when removed from the water. Ships rely on buoyant force to float. The Greek philosopher Archimedes calculated buoyant forces. An object placed in a fluid is buoyed by a force equal to the weight of the fluid it displaces. An object weighing more than the liquid it displaces sinks, but loses an amount of weight equal to the displaced liquid. A floating object displaces its own weight in the fluid in which it floats. An object in a liquid loses weight, but retains its mass, mass is unnaffected by this phenomenon. ELO 1.1

7 Fundamental Properties of FluidsStatic Pressure: A confined, non-moving fluid is a static fluid. Static pressure exerted due to presence of a substance and the random movement of its molecules (PV Energy). Pressure exerted by a static fluid is the force exerted by the fluid on the walls of the container per unit area. Dynamic Pressure: The pressure due to flow is called the dynamic pressure (Kinetic Energy). Fluid flow is a dynamic event. Total Pressure: Total pressure is the combination of the static pressure and the dynamic pressure. Barometeric pressure is an indication of a static pressure, and wind creates dynamic pressure on top of that static pressure. ELO 1.1

8 Closed System Fluid PropertiesELO 1.2 – Explain the following terms: Pascal’s Law, mass flow rate, volumetric flow rate, conservation of mass, continuity equation, and density compensation. To understand fluid response in complex open boundary systems first requires a basic understanding of: Fluid characteristics within a closed system Terms describing quantities and rates of fluid flow Related KAs Define or explain the following terms and concepts: K1.05 Mass flow rate ; K1.12 Explain why flow measurements must be corrected for density changes ELO 1.2

9 Pascal’s Law Pressure applied to a confined fluid transmits undiminished throughout the confining vessel of the system. All pistons have same 1 square inch cross-sectional area Piston A exerting 50 lbf Piston B, at top has 50 lbf applied from force of piston A Piston C has 50 lbf plus 10 lbf force from weight of water above Pistons D and E have 50 lbf force plus weight of water based on location Review question: What is the approximate height of this tank of water? Since P = pgz/gc, and the conversion factor = 0.433 Solving the formula down using a density of 62.4 results in: 30/0.433 = 69.3 ft Figure: Pascal's Law ELO 1.2

10 Closed System Fluid PropertiesVolumetric Flow Rate: The volumetric flow rate of a fluid system is a measure of the volume of fluid passing a point in the system per unit of time. It is the product of the cross-sectional area (A) for flow and the average flow velocity (v). 𝑉 =𝐴v If cross-sectional area is in square feet and velocity in feet per second, the above equation results in volumetric flow rate of cubic feet per second. Other common units for volumetric flow rate include gallons per minute, cubic centimeters per second, liters per minute, and gallons per hour. ELO 1.2

11 Closed System Fluid PropertiesVolumetric Flow Rate Example: A pipe with an inner diameter of 4 inches contains water that flows at an average velocity of 14 feet per second. Calculate the volumetric flow rate of water in the pipe. Solution: 𝑉 = 𝜋 𝑟 2 𝑣 𝑉 = 𝑓𝑡 𝑓𝑡 𝑠𝑒𝑐 𝑉 =1.22 𝑓 𝑡 3 𝑠𝑒𝑐 ELO 1.2

12 Closed System Fluid PropertiesVolumetric Flow Rate and Pressure Relationship: Velocity is a function of the square root of the change in pressure drop Doubling velocity doubles the flow rate, which quadruples pressure drop This relationship holds well for water, but less accurately for gases 𝑉 ∝ 𝐷𝑃 or 𝑉 𝑓𝑖𝑛𝑎𝑙 𝑉 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝐷𝑃 𝑓𝑖𝑛𝑎𝑙 𝐷𝑃 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 Other factors, for example the friction factor, Reynolds number and viscosity, influence this relationship but not significantly for small changes in pressure drops and flow rates. More discussion on this later in this module. ELO 1.2

13 Closed System Fluid PropertiesExample: A 20 gpm leak to atmosphere has developed from a cooling water system that is operating at 200 psig. If pressure decreases to 50 psig, what is the new leak rate? Solution: The change in pressure is now a quarter of what it was. The square root of .25 = .5. Therefore, flow (and velocity) will decrease by half or to 10 gpm. Mathematically: 𝑉 𝑓𝑖𝑛𝑎𝑙 𝑉 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝐷𝑃 𝑓𝑖𝑛𝑎𝑙 𝐷𝑃 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 therefore 𝑉 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑉 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = = =10 ELO 1.2

14 Closed System Fluid PropertiesMass Flow Rate: Mass of fluid passing a point in the system per unit time Mass flow rate relates to the volumetric flow rate: 𝑚 =𝜌 𝑉 Where: ρ = density of the fluid ELO 1.2

15 Closed System Fluid PropertiesMass Flow Rate: If the volumetric flow rate is in cubic feet per second (ft3/sec) and the density is in pounds-mass per cubic foot (lbm/ft3) Mass flow rate measured in pounds-mass per second A common term used for mass flow rate is pounds-mass per hour (lbm/hr) 𝑚 =𝜌𝐴𝑣 Density of water is 𝑙𝑏𝑚 𝑓𝑡 3 ; (works for low temps for fluids) Substituting volumetric flow rate with velocity and area the formula can be rewritten. Density is the mass of a substance per unit volume. Typical units are lbm/ft3. ELO 1.2

16 Example: Mass Flow RateWith a volumetric flow rate of 1.22 ft3/sec, calculate the mass flow rate. Assume a density of lbm/ft3. Solution: 𝑚 =𝜌 𝑉 𝑚 = 𝑙𝑏𝑚 𝑓 𝑡 𝑓 𝑡 3 𝑠𝑒𝑐 𝑚 =76.2 𝑙𝑏𝑚 𝑠𝑒𝑐 ELO 1.2

17 Conservation of Mass The basic principles of fluid flow:Principle of momentum (leads to equations of fluid forces) Conservation of energy (leads to the First Law of Thermodynamics) Conservation of mass (leads to the continuity equation) The First Law of Thermodynamics is a version of the conservation of energy for thermodynamic systems principle or rule. Conservation of Energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but cannot be created or destroyed. The change in the internal energy of a closed system equals the amount of heat supplied to the system minus the amount of work done by the system on its surroundings. ELO 1.2

18 Continuity Equation Conservation of mass states that all mass flow rates into a control volume equal all mass flow rates out of the control volume plus the rate of change of mass within the control volume. 𝑚 𝑖𝑛 = 𝑚 𝑜𝑢𝑡 + ∆𝑚 ∆𝑡 Where: ∆𝑚 ∆𝑡 = increase or decrease of the mass within the control volume over a specified time period A control volume is a fixed region in space where the masses and energies cross the region’s boundaries. This concept of a control volume is useful in analyzing fluid flow problems and also used as the physical boundary through which the flow occurs. The control volume concept used in fluid dynamics applications uses principles of continuity, momentum, and energy. Once a control volume and its boundary are established, the various forms of energy crossing the boundary with the fluid can be used to solve fluid problems in equation form. The control volume approach is an open system analysis. ELO 1.2

19 Continuity Equation The continuity equation is a mathematical expression of the conservation of mass principle For a control volume that has a single inlet and a single outlet, the mass flow rate into the volume must equal the mass flow rate out 𝑚 𝑖𝑛𝑙𝑒𝑡 = 𝑚 𝑜𝑢𝑡𝑙𝑒𝑡 (𝜌𝐴𝑣) 𝑖𝑛𝑙𝑒𝑡 = (𝜌𝐴𝑣) 𝑜𝑢𝑡𝑙𝑒𝑡 ELO 1.2

20 Continuity Equation Knowledge CheckA reactor is operating normally at 100 percent power. Reactor coolant enters the reactor vessel at a temperature of 556°F and a total flow rate of 320,000 gpm. The reactor coolant leaves the reactor vessel at 612°F. What is the approximate flow rate of the reactor coolant leaving the reactor vessel? 320,000 gpm 330,000 to 339,000 gpm 340,000 to 349,000 gpm 350,000 to 359,000 gpm Correct Answer is D. This bank question is a review from – Steam Correct answer is D NRC Bank Question – P3874 Analysis: Equation is animated in on Mouse Click after answer is shown. Give students a chance to perform the math before showing formula. Note that in order to obtain the specific volumes at 556ºF and 612ºF, the subcooled liquid approximation was used. This states that the specific volume (or enthalpy, entropy, etc.) of a subcooled liquid at a given temperature is approximately equal to the value for specific volume of a saturated liquid at that given temperature. ELO 1.2

21 Continuity Equation – Piping ExpansionFor a control volume with multiple inlets and outlets, the principle of conservation of mass requires that the sum of the mass flow rates into the control volume equals the sum of the mass flow rates out of the control volume. 𝑚 𝑖𝑛𝑙𝑒𝑡𝑠 = 𝑚 𝑜𝑢𝑡𝑙𝑒𝑡𝑠 ELO 1.2

22 Continuity Equation – Piping ExpansionExample 1: Steady-state flow exists in a pipe that undergoes a gradual expansion from a diameter of 6 inches to a diameter of 8 inches The density of the fluid in the pipe is constant at 60.8 lbm/ft3 If the flow velocity is 22.4 ft/sec in the 6 inch section, what is the flow velocity in the 8 inch section? 6 inch pipe 8 inch pipe ELO 1.2

23 Continuity Equation – Piping ExpansionSolution: From the continuity equation, the mass flow rate in the 6 inch section must equal the mass flow rate in the 8 inch section. 𝑚 1 = 𝑚 2 𝜌 1 𝐴 1 𝑣 1 = 𝜌 2 𝐴 2 𝑣 2 𝑣 2 = 𝑣 1 𝐴 1 𝜌 1 𝐴 2 𝜌 2 𝑣 2 = 𝑣 1 𝐴 1 𝜌 1 𝐴 2 𝜌 𝑣 2 = 𝑣 1 𝜋 𝑟 𝜋 𝑟 2 2 𝑣 2 = 22.4 𝑓𝑡 𝑠𝑒𝑐 𝑖𝑛 𝑖𝑛 𝑣 2 =12.6 𝑓𝑡 𝑠𝑒𝑐 Subscript 1 represents the 6 inch pipe section and subscript 2 represents the 8 inch pipe section The increase in pipe diameter from 6 inches to 8 inches caused a decrease in flow velocity from 22.4 to 12.6 ft/sec ELO 1.2

24 Continuity Equation – Centrifugal PumpExample: The inlet diameter of the cooling pump shown in the figure below is 28 inches The outlet flow through the pump is 9,200 lbm/second. The density of the water is 49 lbm/ft3. What is the velocity at the pump inlet? Figure: Centrifugal Pump in Continuity Equation ELO 1.2

25 Continuity Equation – Centrifugal PumpSolution: 𝐴 𝑖𝑛𝑙𝑒𝑡 =𝜋 𝑟 2 = 𝑖𝑛 1 𝑓𝑡 12 𝑖𝑛 2 =4.28 𝑓 𝑡 2 𝑚 𝑖𝑛𝑙𝑒𝑡 = 𝑚 𝑜𝑢𝑡𝑙𝑒𝑡 =9,200 𝑙𝑏𝑚 𝑠𝑒𝑐 (𝜌𝐴𝑣) 𝑖𝑛𝑙𝑒𝑡 =9,200 𝑙𝑏𝑚 𝑠𝑒𝑐 𝑣 𝑖𝑛𝑙𝑒𝑡 = 9,200 𝑙𝑏𝑚 𝑠𝑒𝑐 𝐴𝜌 = 9,200 𝑙𝑏𝑚 𝑠𝑒𝑐 𝑓 𝑡 𝑙𝑏𝑚 𝑓 𝑡 3 𝑣 𝑖𝑛𝑙𝑒𝑡 =43.9 𝑓𝑡 𝑠𝑒𝑐 The inlet diameter of the cooling pump shown in the below figure is 28 inches. The outlet flow through the pump is 9,200 lbm/second. The density of the water is 49 lbm/ft3. What is the velocity at the pump inlet? ELO 1.2

26 Continuity Equation – Multiple OutletsExample: A piping system has a Y configuration for separating the flow, shown below in the figure. The diameter of the inlet leg is 12 in., and the diameters of the outlet legs are 8 inches and 10 inches. The density of water is lbm/ft3. The velocity in the 10 inch leg is 10 ft/sec. The flow through the main portion is 500 lbm/sec. What is the velocity out of the 8 inch pipe section? Figure: Y Configuration Example ELO 1.2

27 Continuity Equation – Multiple OutletsSolution: 𝐴 8 =𝜋 4 𝑖𝑛 1 𝑓𝑡 12 𝑖𝑛 2 =0.349 𝑓 𝑡 2 𝐴 10 =𝜋 5 𝑖𝑛 1 𝑓𝑡 12 𝑖𝑛 2 =0.545 𝑓 𝑡 2 𝑚 𝑖𝑛𝑙𝑒𝑡𝑠 = 𝑚 𝑜𝑢𝑡𝑙𝑒𝑡𝑠 𝑚 12 = 𝑚 𝑚 8 𝑚 8 = 𝑚 12 − 𝑚 10 𝜌𝐴𝑣 8 = 𝑚 12 − 𝜌𝐴𝑣 10 𝑣 8 = 𝑚 12 − 𝜌𝐴𝑣 10 𝜌𝐴 8 = 500 𝑙𝑏𝑚 𝑠𝑒𝑐 − 62.4 𝑙𝑏𝑚 𝑓 𝑡 𝑓 𝑡 2 10 𝑓𝑡 𝑠𝑒𝑐 62.4 𝑙𝑏𝑚 𝑓 𝑡 𝑓 𝑡 2 𝑣 8 =7.3 𝑓𝑡 𝑠𝑒𝑐 A piping system has a Y configuration separating the flow. The diameter of the inlet leg is 12 in., and the diameters of the outlet legs are 8 in. and 10 in. The density of water is 62.4 lbm/ft3. The velocity in the 10 in. leg is 10 ft/sec. The flow through the main portion is 500 lbm/sec. What is the velocity out of the 8 inch pipe section? ELO 1.2

28 Continuity Equation – Multiple Outlets (cont.)Solution: 𝑆𝑖𝑛𝑐𝑒: 𝑚 12 = 𝑚 𝑚 8 The ratio of mass flow rate going through each section is a function of the area (or diameter squared). 𝑚 10 = 500 𝑙𝑏𝑚/𝑠𝑒𝑐 𝑚 8 = 500 𝑙𝑏𝑚/𝑠𝑒𝑐 𝑚 10 = =305 𝑚 8 = =195 The ratio of volumetric or mass flow rates are the type of questions asked in the NRC bank on split piping arrangements. ELO 1.2

29 Density Compensation Flow detectors measure volumetric flow rateProportional to square root of pressure drop Temperature affects liquid and steam detectors Pressure only affects steam detectors Liquid is essentially incompressible Steam flow detectors, therefore require density compensation for pressure changes electronically corrected by a signal input from steam pressure Density compensation is required when measuring mass flow rates if the measured fluid has large changes in pressure or temperature affecting actual mass flow values. Mass flow rate calculations are density dependent. If the instrument measures volume, density compensation is unnecessary since volume is not density related. When compensation for density is required, it is usually performed electronically. ELO 1.2

30 Density Compensation The relationship of density to mass flow rate is directly proportional If the density increases, assuming a constant volumetric flow rate, the actual mass flow rate increases A density decrease results in lower actual flow However, temperature and density are inversely related An increase in temperature results in a decrease in density, and a decrease in actual mass flow rate The easiest way to understand density compensation of steam flow detectors is to relate m-dotactual to m-dotindicated. ρV-dotactual = ρV-dotindicated When steam pressure is constant, indicated equals actual. If actual steam pressure increases, density increases, increasing actual mass flow rate (assuming volumetric actual is constant). If the detector was NOT density compensated, all it would see is the constant volumetric flow rate. Actual mass flow rate would be greater than indicated mass flow rate. ELO 1.2

31 Density Compensation Knowledge CheckFlow instruments that measure the mass flow rate of steam often have a density compensation feature because, for a steam pressure increase at a constant volumetric flow rate, steam density will __________ and the actual mass flow rate will __________. decrease; increase increase; decrease increase; increase decrease; decrease Correct answer is C. Correct answer is C NRC Question P281 Analysis: An increase in steam pressure will result in more dense steam (more molecules per unit area). The result of an increase in steam density also increases actual mass flow rate (with a constant volumetric flow rate). If this detector did not have density compensation, when steam pressure increased, increasing actual mass flow rate, the detector would only see the constant volumetric flow rate (since that is what it is detecting). Therefore, indicated flow rate would be less that actual flow rate. ELO 1.2

32 Fluid Properties of a Closed SystemKnowledge Check – NRC Bank A heat exchanger has the following initial cooling water inlet temperature and differential pressure (ΔP) parameters: Inlet Temperature = 70 °F Heat Exchanger ΔP = 10 psi Six hours later, the current heat exchanger cooling water parameters are: Inlet Temperature = 85 °F In comparison to the initial cooling water mass flow rate, the current mass flow rate is... lower, because the density of the cooling water has decreased higher, because the velocity of the cooling water has increased the same, because the changes in cooling water velocity and density offset The same, because the heat exchanger cooling water ΔP is the same Correct Answer is A. Correct answer is A NRC Question P5342 Analysis: ELO 1.2

33 Laminar and Turbulent FlowsTLO 2 – Describe laminar and turbulent flows. 2.1 Describe the characteristics and flow velocity profiles of both laminar flow and turbulent flow. 2.2 Explain the terms viscosity and ideal fluid. TLO 2

34 Laminar and Turbulent FlowELO 2.1 – Describe the characteristics and flow velocity profiles of laminar flow and turbulent flow. Laminar and turbulent flow are different To understand the desirability of each of these flow types, it is necessary to understand their characteristics Related KA K1.14 Discuss the velocity profiles for laminar flow and turbulent flow. 1.8* 1.9* ELO 2.1

35 Flow Regimes Two categories or regimes classify fluid flow types.Laminar flow Turbulent flow The amount of fluid friction determines the amount of energy required to maintain the desired flow depends on the mode of flow Of concern are: Heat transfer to the fluid Energy required to create the desired flow Mixing within the fluid The flow regime selection, whether laminar or turbulent, is important in the design and operation of any fluid system. ELO 2.1

36 Laminar Flow Also known as streamline or viscous flowLayers of water flow over one another at different speeds virtually no mixing between layers Fluid particles move in definite and observable paths or streamlines. Great for fluid transport systems Lower flow, less headloss Laminar flow is characteristic of viscous (thick) fluid, or is one in which the fluid’s viscosity plays a significant role. ELO 2.1

37 Turbulent Flow Characterized by irregular movement of particles of the fluid No definite frequency as with wave motion Great for heat transfer systems Higher flow, better mixing of fluid Fluid particles travel in irregular paths with no observable pattern and no distinct layers ELO 2.1

38 Reynolds Number The regime of fluid flow may be determined from the Reynolds number Laminar flow exists where the Reynolds number is less than 2,000 Turbulent flow exists with Reynolds numbers greater than 3,500 Transition flow identifies with values between these two numbers Reynolds Number a function of: Average velocity (νav), pipe diameter (D), density (ρ), and dynamic viscosity (μ) ELO 2.1

39 Flow Velocity ProfilesFluid particles do not travel at the same velocity within a pipe The shape of the velocity profile depends on whether the flow is laminar or turbulent Higher velocities give a higher Reynolds number In a laminar flow, the velocity distribution at a cross-section has a parabolic shape with the maximum velocity at the center running about twice the average velocity. Turbulent flow has a flat velocity distribution across the pipe section, and the entire fluid flows at an approximately equal velocity. Actual velocity profile depends on the surface condition of the pipe wall. A smoother wall results in a more uniform velocity profile than a rough pipe wall. Figure: Laminar and Turbulent Flow Velocity Profiles ELO 2.1

40 Characteristics of FluidsELO 2.2 – Explain the terms viscosity and ideal fluid. Fluids include gases and liquids. Fluids have certain characteristics that are different from solids. One of these important characteristics is fluid viscosity Related KA – not related to any specific KAs, ACAD reference – properties of fluids. Certain characteristics of fluids help solve flow problems and assist in understanding the requirements for system design. ELO 2.2

41 Viscosity A measure of the resistance of a fluid to flowingThick oil has a high viscosity Water has a low viscosity The internal friction of fluids resists flow past a solid surface or other layers of the fluid Viscosity is a fluid property that measures the resistance of the fluid to deforming due to a shear force. ELO 2.2

42 Viscosity Depends on the temperature of the fluid LiquidsAs the temperature increases, viscosity decreases Gases As the temperature increases, viscosity increases Lubricating oil of engines is a good example of viscosity When oil is cold, the oil is viscous or thick At operating temperatures, viscosity of oil decreases significantly, and the oil flows more easily ELO 2.2

43 Ideal Fluid An ideal fluid is one that is incompressible and has no viscosity Ideal fluids do not exist, but are sometimes useful to simplify the problem’s fluid flow calculations Bernoulli’s (simplified) equation applies only to fluid flow treated as ideal, with no fluid friction More on Bernoulli in the next section. The extended Bernoulli’s equation considers fluid friction head losses ELO 2.2

44 Bernoulli’s Equation and EnergyTLO 3 – Explain Bernoulli’s equation and use it to determine fluid properties at different points in a system. 3.1 Using the simplified form of Bernoulli’s equation, explain the interrelationships of these energies or heads: Elevation head Velocity head Pressure head 3.2 Using the extended form of Bernoulli’s equation, explain both head loss and pump head. 3.3 Explain the following causes of head loss: Friction Viscosity Minor losses 3.4 Describe methods used to control fluid flow rates and the resulting pressure and flow rate changes. TLO 3

45 Simplified Bernoullis EquationELO 3.1 – Using the simplified form of Bernoulli’s equation, explain the interrelationships of these types of energy: elevation head, velocity head, and pressure head. Four forms of energy already discussed: Elevation Head Potential energy Velocity Head Kinetic energy Pressure Head Flow energy Internal energy Temperature Related KA K1.13 Explain the relationship between pressure head and velocity head in a fluid system Special case of the general energy equation Probably the most used tool for solving fluid flow problems. Easy way to relate the energies for elevation head, velocity head, and pressure head of a fluid. Bernoulli’s equation can be modified in a way that accounts for both head losses and pump work. ELO 3.1

46 General Energy EquationConservation of energy principle states that energy can be neither created nor destroyed. Equivalent to the First Law of Thermodynamics. Following equation is the general energy equation for an open system: 𝑄+ 𝑈+𝑃𝐸+𝐾𝐸+𝑃𝑉 𝑖𝑛 =𝑊+ 𝑈+𝑃𝐸+𝐾𝐸+𝑃𝑉 𝑜𝑢𝑡 + 𝑈+𝑃𝐸+𝐾𝐸+𝑃𝑉 𝑠𝑡𝑜𝑟𝑒𝑑 Where: Q = heat (BTU) U = internal energy (BTU) PE = potential energy (ft-lbf) KE = kinetic energy (ft-lbf) P = pressure (lbf/ft2) V = volume (ft3) W = work (ft-lbf) ELO 3.1

47 Simplified Bernoulli Equation𝑃𝐸+𝐾𝐸+ (𝑃𝑉) 1 =𝑃𝐸+𝐾𝐸+ (𝑃𝑉) 2 Substituting appropriate expressions for potential energy and kinetic energy, the above equation rewritten: 𝑚𝑔 𝑧 1 𝑔 𝑐 + 𝑚 v 𝑔 𝑐 + 𝑃 1 𝑉 1 = 𝑚𝑔 𝑧 2 𝑔 𝑐 + 𝑚 v 𝑔 𝑐 + 𝑃 2 𝑉 2 Where: m = mass (lbm) z = height above reference (ft) v = average velocity (ft/sec) g = acceleration due to gravity (32.17 ft/sec2) gc = gravitational constant, (32.17 ft-lbm/lbf-sec2) P = pressure (lbf/ft2) V = volume (ft3) Bernoulli’s equation results from the application of the general energy equation and the First Law of Thermodynamics to: a steady flow system in which no work is done on or by the fluid no heat is transferred to or from the fluid no change occurs in the internal energy of the fluid (for example, no temperature change) ELO 3.1

48 Simplified Bernoulli EquationThe gravitation constant (gc ) is only required when using the English System of Measurement and mass is measured in pounds mass (lbm). Essentially a unit conversion factor Not required if mass is in slugs or in the metric system Terms represent form of energy (potential, kinetic, and pressure related energies) Bernoulli’s equation represents a balance of the KE, PE, PV energies First Law of Thermodynamics For example, when pressure drops (PV), velocity increases (KE) ELO 3.1

49 Simplified Bernoulli EquationAnother form of Bernoulli’s equation: 𝑧 v 𝑔 + 𝑃 1 𝑣 1 = 𝑧 v 𝑔 + 𝑃 2 𝑣 2 𝑔 𝑐 𝑔 Where: z = height above reference level (ft) v = average velocity of fluid (ft/sec) P = pressure of fluid (lbf/ft2) 𝑣 = specific volume of fluid (ft3/lbm) g = acceleration due to gravity (ft/sec2) gc = gravitational constant, (32.17 ft-lbm/lbf-sec2) Bernoulli’s equation is not required to be committed to memory since a version of it is on the NRC Equation Sheet. Recall the proportionality that volumetric flow rate is proportional to the square root of the pressure drop. This is merely the KE and PV terms of Bernoulli's equation. In a fluid piping system the energy change in a REAL system as explained in Bernoulli’s equation is the drop in flow energy (pressure) and the increase in internal energy (temperature) due to friction. These losses due to friction must be made up by the head of the pump. This will be explained in the next ELO. ELO 3.1

50 Simplified Bernoulli Equation – EnergyThe different forms of energy in Bernoulli’s equation, referred to as heads, are: Elevation head (PE) Velocity head (KE) Pressure head (PV) Head can be expressed in ft or psi The total head is the sum of the elevation head, velocity head, and pressure head of a fluid Bernoulli’s equation states that the total head of the fluid is constant Engineers use the term head when referencing pressure. Head references height, typically in feet of a water column, equivalent to a supporting pressure and expresses the energy possessed by a fluid. Elevation head represents the potential energy of a fluid due to its elevation above a reference level. Velocity head represents the kinetic energy of the fluid due to its velocity; column height in feet a flowing fluid would rise if all of its kinetic energy was converted to potential energy. Pressure head is the flow energy of a column of fluid whose weight equals the pressure of the fluid. Recall the conversion factor (for low temperature liquids) or psi/ft ELO 3.1

51 Energy Conversions in Fluid SystemsAs diameter and flow area increase, the flow velocity must decrease to maintain the same mass flow rate With outlet velocity less than inlet velocity, the velocity head of the flow decreases If pipe is horizontal, no change in elevation head occurs Therefore, an increase in pressure head compensates for the decrease in velocity head For an incompressible ideal fluid, specific volume does not change Bernoulli’s equation allows examination of energy transfers among elevation head, velocity head, and pressure head. This equation makes it possible to examine individual components of piping systems and determine what fluid properties vary and the effect on energy balance. Only way that the pressure head for an incompressible fluid can increase is for the pressure to increase ELO 3.1

52 Energy Conversions in Fluid SystemsFor a constant diameter pipe containing an ideal fluid undergoing a decrease in elevation, the same net effect occurs. However: The flow velocity and the velocity head must be constant to satisfy the mass continuity equation Increase in pressure head compensates for decrease in elevation head Again, the fluid is incompressible so the increase in pressure head must result in an increase in pressure Although Bernoulli’s equation has several limitations, its use applies to many physical fluid problems. ELO 3.1

53 Example: Bernoulli’s EquationAssume frictionless flow in a long, horizontal, conical pipe. The diameter is 2.0 ft at one end and 4.0 ft at the other. The pressure head at the smaller end is 16 ft of water. If water flows through this cone at a rate of ft3/sec, find the velocities at the two ends and the pressure head at the larger end. Figure: Conical Pipe Shows Flow of Pressure Energy ELO 3.1

54 Example: Bernoulli’s Equation Solution𝑉 1 = 𝐴 1 v 1 v 1 = 𝑉 1 𝐴 1 v 2 = 𝑉 2 𝐴 2 v 1 = 𝑓 𝑡 3 𝑠𝑒𝑐 𝜋 1 𝑓𝑡 2 v 2 = 𝑓 𝑡 3 𝑠𝑒𝑐 𝜋 2 𝑓𝑡 2 v 1 =40 𝑓𝑡 𝑠𝑒𝑐 v 2 =10 𝑓𝑡 𝑠𝑒𝑐 𝑧 1 + v 1 2 2𝑔 + 𝑃 1 𝑣 1 𝑔 𝑐 𝑔 = 𝑧 2 + v 2 2 2𝑔 + 𝑃 2 𝑣 2 𝑔 𝑐 𝑔 𝑃 2 𝑣 2 𝑔 𝑐 𝑔 = 𝑃 1 𝑣 1 𝑔 𝑐 𝑔 + 𝑧 1 − 𝑧 2 + v 1 2 − v 2 2 2𝑔 =16 𝑓𝑡+0 𝑓𝑡+ 40 𝑓𝑡 𝑠𝑒𝑐 2 − 10 𝑓𝑡 𝑠𝑒𝑐 𝑓𝑡 𝑠𝑒 𝑐 2 =39.3 𝑓𝑡 Assume frictionless flow in a long, horizontal, conical pipe. The diameter is 2.0 ft at one end and 4.0 ft at the other. The pressure head at the smaller end is 16 ft of water. If water flows through this cone at a rate of ft3/sec, find the velocities at the two ends and the pressure head at the larger end. ELO 3.1

55 Extended Bernoulli's EquationELO 3.2 – Using the extended form of Bernoulli’s equation, explain both head loss and pump head. Modifying the simplified Bernoulli equation makes it possible to deal with both head losses and pump work Takes into account gains and losses of head Often used to solve most fluid flow problems To keep fluid moving pump work must overcome headloss due to friction Related KA – K1.02 Define head loss These limits are a problem, since real systems have: Fluid friction Pumps Practical applications of the simplified Bernoulli Equation to real piping systems are not possible: Calculating fluid friction applies only to ideal fluids. Calculating work on the fluid is not possible because no work is allowed on, or to be done by the fluid. Prevents two points in a fluid stream from being analyzed if a pump exists between the two points. ELO 3.2

56 Extended Bernoulli's EquationThe below equation is one form: 𝑧 v 𝑔 + 𝑃 1 𝑣 1 𝑔 𝑐 𝑔 + 𝐻 𝑝 = 𝑧 v 𝑔 + 𝑃 2 𝑣 2 𝑔 𝑐 𝑔 + 𝐻 𝑓 Where: z = height above reference level (ft) v = average velocity of fluid (ft/sec) P = pressure of fluid (lbf/ft2) 𝑣 = specific volume of fluid (ft3/lbm) Hp = head added by pump (ft) Hf = head loss due to fluid friction (ft) g = acceleration due to gravity (ft/sec2) The head loss due to fluid friction (Hf) is energy used in overcoming friction caused by every item the fluid must flow through or around. It represents a loss of energy from the standpoint of fluid flow, not normally a significant loss of the fluid’s total energy. Does not violate the law of conservation of energy since the head loss due to friction results in an equivalent increase in the internal energy (u) of the fluid. Losses are greatest as the fluid flows through entrances, exits, pumps, valves, fittings, and any other piping with rough inner surfaces. ELO 3.2

57 Example: Extended Bernoulli's EquationWater from a large reservoir is pumped to a point 65 feet higher than the reservoir. How many feet of head must the pump add if 8,000 lbm/hr flows through a 6 inch pipe and the head loss, due to friction, is 2 feet? The density of the fluid is 62.4 lbm/ft3, and the cross-sectional area of a 6-inch pipe is ft2. Before looking at the equation (which could be considered cumbersome), ask the students how much work does a pump have to do to lift water 65 feet with 2 feet of friction? Hopefully they will say 67 feet! ELO 3.2

58 Example: Extended Bernoulli's EquationUse reference points chosen at the surface of the reservoir (point 1) and the outlet of the pipe (point 2) The pressure at the surface of the reservoir is the same as the pressure at the exit of the pipe, such as, atmospheric pressure The velocity at point 1 is essentially zero (0) Using the equation for the mass flow rate, determine the velocity at point 2: Water from a large reservoir is pumped to a point 65 feet higher than the reservoir. How many feet of head must the pump add if 8,000 lbm/hr flows through a 6 inch pipe and the head loss, due to friction, is 2 feet? The density of the fluid is 62.4 lbm/ft3, and the cross-sectional area of a 6 inch pipe is ft2. ELO 3.2

59 Example: Extended Bernoulli's Equation𝑚 2 =𝜌 𝐴 2 𝑣 2 𝑣 2 = 𝑚 2 𝜌 𝐴 2 𝑣 2 = 8,000 𝑙𝑏𝑚 ℎ𝑟 62.4 𝑙𝑏𝑚 𝑓 𝑡 𝑓 𝑡 2 𝑣 2 =639 𝑓𝑡 ℎ𝑟 1 ℎ𝑟 3,600 𝑠𝑒𝑐 𝑣 2 =0.178 𝑓𝑡 𝑠𝑒𝑐 Water from a large reservoir is pumped to a point 65 feet higher than the reservoir. How many feet of head must the pump add if 8,000 lbm/hr flows through a 6 inch pipe and the head loss, due to friction, is 2 feet? The density of the fluid is 62.4 lbm/ft3, and the cross-sectional area of a 6 inch pipe is ft2. Use the Extended Bernoulli equation to solve for the required pump head – see next slide for the answer. ELO 3.2

60 Example: Extended Bernoulli's Equation𝑧 1 + v 1 2 2𝑔 + 𝑃 1 𝑣 1 𝑔 𝑐 𝑔 + 𝐻 𝑝 = 𝑧 2 + v 2 2 2𝑔 + 𝑃 2 𝑣 2 𝑔 𝑐 𝑔 + 𝐻 𝑓 𝐻 𝑝 = 𝑧 2 − 𝑧 1 + v 2 2 − v 1 2 2𝑔 + 𝑃 2 − 𝑃 1 𝑣 𝑔 𝑐 𝑔 + 𝐻 𝑓 =65 𝑓𝑡 𝑓𝑡 𝑠𝑒𝑐 2 − 0 𝑓𝑡 𝑠𝑒𝑐 𝑓𝑡 𝑠𝑒 𝑐 2 +0 𝑓𝑡+2 𝑓𝑡 𝐻 𝑝 = 𝑓𝑡 Extended Bernoulli equation used to solve for required pump head The solution of this example problem has a value that makes sense from the data given in the problem Total head increase of 67 ft is due primarily to the 65 ft evaluation increase and the 2 ft of friction head ELO 3.2

61 Bernoulli’s Equation ExtendedKnowledge Check – NRC Bank Refer to the drawing of a section of pipe that contains flowing subcooled water. Given: Pressure at P1 is 24 psig. Pressure at P2 is 16 psig. Pressure change due to change in velocity is 2 psig. Pressure change due to change in elevation is 10 psig. The pressure decrease due to friction head loss between P1 and P2 is _______; and the direction of flow is from ________. 2 psig; left to right 2 psig; right to left 4 psig; left to right 4 psig; right to left Correct Answer is D. Correct answer is D. NRC Question P5446 (actually of – Thermal Hydraulics, but applicable to this concept). Analysis: There are a few different ways to analyze this problem. Understanding how to utilize Bernoulli’s equation is imperative. However, using the simple “vector method” in Option 3 works every time and is simple to understand. 1. Bernoulli’s Equation (always best method) – Based on information provided: elevation head + velocity head + pressure head – headloss equals ZERO. Based on the actual equation and values provided: 10 psig + 2 psig +(-8 psig, based on P2 – P1) – HL = 0; HL = 4 psig; because headloss is a positive value, flow is from top to bottom (right to left) 2. Gage Pressures based on Total Pressure method - The bottom gage (P1) should be reading the pressure at P2 (16) plus the elevation pressure (10) and plus (or) minus the velocity pressure. Since the larger diameter pipe is at the bottom (where P1 is) the velocity is lower and pressure higher. Therefore, you ADD the velocity pressure (2). Therefore P1 should be reading P2(16) + elevation (10) + velocity (2), or it should be reading 28. But it is only reading 24. WHY? Because there is 4 psig of headloss. And, since it is reading less (24) than what it should be (28), it is on the downstream of the flow direction. Flow is right to left. 3. Simple vector analogy method – Pressure at P1 (24) is 8 psig higher than at P2 (16). Therefore there is an 8 psig “flow push” from left to right. Elevation is higher at P2 so there is a 10 psig “flow push” from right to left. Velocity is higher at P2 so there is a 2 psig “flow push” from right to left. Based on these values: 8 left t right+ 10 right to left+ 2 right to left, the vector difference is “4 right to left”. Figure: Pipe Section of Flowing Subcooled Water ELO 3.2

62 Applying Bernoulli’s Equation to a VenturiA Venturi is a flow measuring device consisting of a gradual contraction followed by a gradual expansion. By measuring the differential pressure between the inlet of the Venturi from point 1 and to the throat of the Venturi at point 2, the flow velocity and mass flow rate can be determined based on Bernoulli’s equation. Figure: Venturi Tube Flow Meter ELO 3.2

63 Applying Bernoulli’s Equation to a VenturiBernoulli’s equation states that the total head of the flow must be constant. Elevation does not change between points 1 and 2; the elevation head at the two points is the same and cancels each other out Bernoulli’s equation basically simplifies to the following: v 2 2 − v 𝑔 𝑐 = 𝑃 1 − 𝑃 2 𝑣 Or, the change in kinetic energy (KE) equals the change in flow energy (Pv) ELO 3.2

64 Applying Bernoulli’s Equation to a VenturiFor a liquid system around 100°F, specific volume is ≈ The only unit conversion required is 144 in2 = 1 ft2 Since 2gc = 64.4 The formula: v 2 2 − v 𝑔 𝑐 = 𝑃 1 − 𝑃 2 𝑣 Can be simplified to: v 2 2 − v 1 2 = 𝑃 1 − 𝑃 2 ( x 144 x 64.4) Or, v 2 2 − v = 𝐷𝑃 This simplification of Bernoulis equation will help answer bank questions – P4939 and – P3207. ELO 3.2

65 Venturi Sonic (Choke) FlowIf flow through a typical ∆P (change in pressure) Venturi becomes excessively large (due to a pipe break) Velocity through the nozzle can reach sonic speed in the throat In this state, if downstream pressure (P2) decreased further Flow does not increase above sonic flow in the throat Some Venturi designed as Sonic Chokes or Critical Flow Nozzles Used in control systems produce a desired fixed flow rate unaffected by downstream pressure Note: Some areas where sonic flow chokes are used are main steam line flow restrictors. These limit the rate of cooldown in the RCS by limiting the rate at which steam is released (in the event of a faulted SG). In a converging/diverging nozzle the maximum speed achieved in the throat is Mach 1(M1). If sonic conditions are reached in the throat than the flow will go supersonic in the diverging section. A further reduction in backpressure does not increase flow rate since max speed is achieved in the throat. Reducing backpressure further changes where the shock wave forms in the divergent section. It does not change the concept of sonic choking but is more technically correct to specify that speed in the throat cannot exceed M1 even though it will go faster in the divergent nozzle ELO 3.2

66 Venturi Knowledge Check – NRC BankRefer to the drawing of a venturi in a main steam line (see figure above). The venturi inlet and outlet pipe diameters are equal. A main steam line break downstream of the venturi causes the main steam mass flow rate through the venturi to increase. Soon, the steam reaches sonic velocity in the throat of the venturi. How will the main steam mass flow rate through the venturi be affected as the steam pressure downstream of the venturi continues to decrease? It will continue to increase at a rate that is dependent on the steam velocity in the throat of the venturi. It will continue to increase at a rate that is dependent on the differential pressure (P1 - P2) across the venturi. It will not continue to increase because the steam velocity cannot increase above sonic velocity in the throat of the venturi. It will not continue to increase because the differential pressure (P1 - P2) across the venturi cannot increase further once the steam reaches sonic velocity in the throat of the venturi. Correct answer is C. Correct answer is C. NRC Question 4243. Analysis: As pressure continues to be lowered downstream of the venturi, mass flow rate through the venturi will not increase. This is because the maximum speed a fluid can attain in a convergent nozzle is sonic speed. Once sonic speed is attained in the convergent nozzle (and likewise in the throat of the venturi), maximum mass flow rate is reached. Keep in mind that various NRC bank questions test variations of this concept. For example, in this case mass flow rate does not change. But what does downstream velocity do if pressure decreases further? It must increase since mass flow rate remains constant. Be careful as to which parameter the question asks for. Look at wording differences between P4243 and P6843 on venturis. ELO 3.2

67 Causes of Head Loss Head loss that occurs in pipes depends on:ELO 3.3 – Explain the following causes of head loss: friction loss, viscosity loss, and minor losses. Head loss that occurs in pipes depends on: Flow velocity Pipe length and diameter Friction factor based on the roughness of the pipe and the Reynolds number of the flow Head loss that occurs in the components of a flow path correlates to a length of pipe that causes an equivalent head loss ELO 3.3

68 Causes of Head Loss Head loss is a measure of the reduction in the total head of the fluid as it moves through a fluid system Pressure drop Caused by: Friction between the fluid and the walls of the pipe Friction between adjacent fluid particles as they move relative to one another Turbulence due to: flow directional changes piping entrances and exits pumps, valves, flow reducers, and fittings disrupting flow ELO 3.3

69 Friction Factor The friction factor depends on the Reynolds number for the flow and the degree of roughness of the pipe’s inner surface The quantity used to measure the roughness of the pipe is the relative roughness term average height of surface irregularities (ε) divided by the pipe diameter (D) 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑅𝑜𝑢𝑔ℎ𝑛𝑒𝑠𝑠= 𝜀 𝐷 This concept is not tested. ELO 3.3

70 Darcy’s Equation Combines pipe dimension, flow velocity, and the friction factor into an equation for headloss, due to friction, calculations. 𝐻 𝑓 =𝑓 𝐿 𝑣 2 𝐷2𝑔 Where: 𝑓 = friction factor (dimensionless) - Rougher piping walls or higher fluid viscosity, the greater the friction 𝐿 = length of pipe (ft) - Longer the pipe, the greater the friction 𝐷 = diameter of pipe (ft) - Greater the pipe diameter, the lower the friction 𝑣 = fluid velocity (ft/sec) - Greater the fluid velocity, the higher the friction. Exponentially, head loss varies with the square of fluid velocity. 𝑔 = gravitational acceleration (ft/sec2) - fixed The equality associated with Darcy’s Equation is not something tested by the NRC. However, understanding the “proportionality” example in a couple of slides is. ELO 3.3

71 Head Loss ProportionalityLength increases Headloss increases Velocity increases Headloss increases exponentially Diameter increases Headloss decreases NOTE: Opening a valve in a system increases velocity in the system, but it also increases the “effective pipe size” (diameter) Overall headloss decreases (due to increased diameter) 𝐻𝑒𝑎𝑑𝑙𝑜𝑠𝑠 ∝ 𝐿 𝑣 2 𝐷 Several bank questions on this topic! ELO 3.3

72 Viscosity Loss Recall:Viscosity decreases as the temperature increases Viscosity also affects head loss the more viscous a fluid is, the more resistant to flow it is Lower the temperature, higher the viscosity, higher the headloss Higher the temperature, lower the viscosity, lower the headloss This is referring to liquids, not gasses In gasses, viscosity follows temperature changes For example, if temperature increases, gas viscosity increases as well ELO 3.3

73 Minor Losses Losses in pipes due to bends, elbows, joints, valves, etc. Expressed by the loss coefficient (𝑘) The following equation calculates minor losses of individual fluid system components: 𝐻 𝑓 =𝑘 𝑣 2 2𝑔 Where: 𝑘 = loss coefficient 𝑣 2 = fluid velocity (ft/sec) These losses often are more important than the losses due to pipe friction. For all minor losses in turbulent flow, the head loss varies as the square of the velocity. Standard engineering and/or technical handbooks provide values of the loss coefficient (k) for typical situations and fittings. ELO 3.3

74 Causes of Energy Loss Knowledge CheckWhich one of the following will increase the head loss occurring in an operating cooling water system? Shifting two heat exchangers from parallel to series operation. Increasing the flow rate in the system by throttling open a flow control valve. Replacing a 20 foot section of 10-inch diameter pipe with a 10 foot section of 10-inch diameter pipe. Replacing a 20 foot section of 10-inch diameter pipe with a 20 foot section of 12-inch diameter pipe. Correct answer is A. Correct answer is A. NRC Bank Question – P7543 Analysis: Recall the HL proportionality equation: A. CORRECT. Shifting from parallel to series decreases effective the diameter of the pipe, which will INCREASE headloss. B. WRONG. This one is tricky! Even though you increase the velocity, resulting in higher velocity headloss, opening the valve results in an overall DECREASE in system headloss (due to increasing pipe diameter). C. WRONG. Decreasing piping length will DECREASE system headloss. D. WRONG. Raising pipe diameter will DECREASE system headloss. ELO 3.3

75 Flow Control Methods Multiple methods of controlling fluid flow existELO 3.4 – Describe methods used to control fluid flow rates as well as the resulting pressure and flow rate changes. Multiple methods of controlling fluid flow exist System application determines whether pump configuration changes or if throttle valves provide fluid flow control Related KA K1.15 Describe the methods of controlling system flow rates ELO 3.4

76 Flow Control Methods Changing the number of operating pumpsChanging pump speed control Positive displacement pumps - system flow rate varies with pump speed, but NOT pump head Raising pump speed increases flow rate, decreasing pump speed decreases flow rate Centrifugal pumps - system flow rate varies with pump speed and pump head Increasing head decreases flow rate, decreasing head increases flow rate The installed pump may be a positive displacement or centrifugal pump, each of these has different operational characteristics. Pump design characteristics and the fluid system itself determine the effect on system pressure and flow ELO 3.4

77 Flow Control Methods System designs often utilize throttle valves to control flow rate Opening the valve increases flow Closing the valve decreases the flow Changing throttle valve position changes the frictional head loss in the system (piping diameter changes) Closing the throttle valve increases Hf Opening the throttle valve decreases Hf Often, system design incorporates automatic throttle valve control for temperature, level, or pressure control. For example, steam generator water level control. As an example, if Hf decreases, flow should increase (all other factors held constant). Actual system flow and pressure changes depend on system design, pump capacity, etc. ELO 3.4

78 Flow Control Methods Varying pump speed or number of operating pumps affects the pump head (Hp) term If Hf remains constant, increasing pump head (Hp) increases system pressure and flow However, head loss varies with the square of velocity Increasing the system flow causes an exponential change in head loss, resulting in restrictions to flow gains For example, when starting the 2nd RCP you don’t get twice the flow Slightly less than twice the flow due to increased headloss If you had a four-loop plant with four 25% rated flow pumps, when you start the first pump you get 25% flow. When you start pump 2 you might only get 24% flow. Etc. In order to get 100% flow with all four pumps running, you would need four pumps rated higher than 25% flow each. The following numbers are used only as an example (27% flow pumps): When you start the first one you might get 27%, 2nd pump you get 26%, 3rd pump you get 24%, 4th pump you get 23% = 100 ELO 3.4

79 Flow Control Methods Knowledge Check – NRC BankA four-loop PWR nuclear power plant uses four identical reactor coolant pumps (RCPs) to supply reactor coolant flow through the reactor vessel. The plant is currently operating at 20% power with all RCPs in operation. Which one of the following describes the stable RCS flow rate through the reactor vessel following the trip of one RCP? (Assume that no operator actions are taken and the reactor does not trip.) Less than 75 percent of the original flow rate. Exactly 75 percent of the original flow rate. Greater than 75 percent of the original flow rate. Unpredictable without pump curves for the RCPs. Correct answer is C. Correct answer is C. NRC Question P3183 Analysis: When one of the Reactor Coolant Pumps (RCP’s) trips, flow will be reduced relative to the capacity of the pumps. Because all four pumps are identical, ONE MAY THINK THAT flow would tend to stabilize at 75% of original flow. However at a lower flow rate, less headloss will be present. Lower headloss will result in a flow slightly greater than 75% of original flow. ELO 3.4

80 Flow Control Methods Knowledge Check - NRC BankTwo identical centrifugal pumps (CPs) and two identical positive displacement pumps (PDPs) are able to take suction on a vented water storage tank and provide makeup water flow to a cooling water system. Pumps can be cross-connected to provide multiple configurations. In single pump alignment, each pump will supply 100 gpm at a system pressure of 1,000 psig. Given the following information: Centrifugal Pumps Shutoff head = 1,500 psig Maximum design pressure = 2,000 psig Flow rate with no backpressure = 180 gpm  Positive Displacement Pumps  Maximum design pressure = 2,000 psig Which one of the following pump configurations will supply the lowest makeup water flow rate to the system if system pressure is 1,700 psig? Two CPs in series Two CPs in parallel One PDP and one CP in series (CP supplying PDP) One PDP and one CP in parallel Correct answer is B. NOTE: Even though this question is in this bank, it will be more easily understood if – Pumps has already been taught. Correct answer is: B NRC Question: P1784 Analysis: Both the positive displacement pump and centrifugal pump will produce 100 gpm at 1,000 psig. One fundamental difference between a positive displacement pump vs. a centrifugal pump is that positive displacement pumps will produce a constant flow rate over varying discharge pressures; centrifugal pump flow rate varies inversely with discharge pressure. Note that shutoff head for the centrifugal pump is 1,500 psig; therefore, operating the centrifugal pump at a head of 1,700 psig will result in the pump operating above shutoff head (centrifugal pump provides no flow). In reality, this is a very bad condition which could overheat the pump and damage it! Placing two centrifugal pumps in parallel operating above shutoff head will result in no system flow. Any combination including a positive displacement pump will provide at least 100 gpm. Also, even though you are above the shutoff head of a CP, having two CPs in series might raise pressure enough to provide some flow. That is why Choice “A” is incorrect. Choices “C” and “D” will still provide about 100 gpm so they are incorrect choices. ELO 3.4

81 Fluid Flow Effects on Plant OperationTLO 4 – Explain the various changes in fluid flow, their effect on plant operation, and methods of control. 4.1 Describe the following terms: Gas Binding Cavitation SG Recirculation Ratio Two-phase flow Pressure spike 4.2 Describe water hammers, including severity and resulting pressure spikes. 4.3 List actions that operators can take to minimize the occurrence of water hammer. The steam generators utilize two-phase flow and conversion of subcooled feedwater to steam. Recirculation within the steam generators provides a process of preheating the feedwater and reducing thermal shock. TLO 4

82 Fluid Flow Terms Various things can impact fluid flow in a systemELO 4.1 Describe the following terms: Gas Binding, Cavitation, SG Recirculation Ratio, Two-phase flow, and Pressure spike. Various things can impact fluid flow in a system Certain things can have catastrophic effects Cavitation (pumps and valves) Gas binding Water hammer Related KAs Define or explain the following terms and concepts: K1.06 Two-phase flow , K1.07 Pressure spike , K1.08 Gas binding , K1.09 Recirculation ratio , K1.11 Cavitation ELO 4.1

83 Gas Binding Caused by gases coming out of solution in systemSystem not vented prior to starting pump Net effect: Pump can’t deliver designed flow and head In extreme cases, complete flow blockage can occur A pump that is air (gas) bound, it will not provide flow in a system. Some pumps are self priming and will fill themselves with the fluid they are intended to pump, but this is a very small minority of them. ELO 4.1

84 Cavitation Formation of vapor bubbles at the low pressure area of a pump Eye of the impeller Vapor bubbles collapse as pressure increases (Tsat – Psat relationship) Volute or casing Indications: Oscillating system flow and/or pressure Could sound like marbles or rocks in the pump Cavitation can also occur in a valve when throttled Net effect: Can cause damage to component The damage, due to cavitation, is caused by the collapse of the bubbles. This implosion causes a cyclical stress on the impeller, casing or piping and can lead to premature failure of a pump, usually due to bearing failure. It also causes pitting damage, the removal of material from pipes, casings or impellers. ELO 4.1

85 Steam Generator Recirculation FlowRecirculation is the mixture of incoming feedwater and moisture separator drainage Typical U-tube Steam Generator from Westinghouse Electric Company LLC Figure: Typical U-tube Steam Generator from Westinghouse Electric Company LLC ELO 4.1

86 Steam Generator Recirculation FlowFeedwater enters the SG subcooled Feedwater flows through the downcomer region to the bottom of the tube bundle Feedwater heating occurs from the primary coolant flowing through the U-tubes Feedwater flows upward through tube bundle heats to saturation, then to a steam-water vapor mixture Steam leaving tube bundle region enters moisture separators and removes moisture from the steam Liquid from the moisture separators flows down around the separator shroud, mixing with incoming feedwater Helps add sensible heat to preheat feedwater in SG Related KAs Define or explain the following terms and concepts: K1.05 Mass flow rate ; K1.09 Recirculation Ratio 1.9* 1.9* ELO 4.1

87 Recirculation Ratio (RCR)The recirculation ratio is the amount of recirculated moisture to the amount of steam produced 𝑅𝐶𝑅= 𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑙𝑒𝑎𝑣𝑖𝑛𝑔 𝑡ℎ𝑒 𝑡𝑢𝑏𝑒 𝑏𝑢𝑛𝑑𝑙𝑒 𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑠𝑡𝑒𝑎𝑚 𝑙𝑒𝑎𝑣𝑖𝑛𝑔 𝑡ℎ𝑒 𝑡𝑢𝑏𝑒 𝑏𝑢𝑛𝑑𝑙𝑒 Or 𝑅𝐶𝑅= 𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑟𝑒𝑡𝑢𝑟𝑛𝑖𝑛𝑔 𝑡𝑜 𝑡ℎ𝑒 𝑑𝑜𝑤𝑛𝑐𝑜𝑚𝑒𝑟 𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑠𝑡𝑒𝑎𝑚 𝑙𝑒𝑎𝑣𝑖𝑛𝑔 𝑡ℎ𝑒 𝑠𝑡𝑒𝑎𝑚 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 Mathematically, the recirculation ratio can be looked at a couple of ways. These expressions describe the mathematical relationship. ELO 4.1

88 Purpose of Steam Generator RecirculationTwo important functions: Preheats feedwater to near saturation prior to entering the bottom of the tube bundle Minimizes thermal stress on the nozzles and steam generator shell by reduction of their temperature differences Recirculation ratio decreases dramatically as power level increases, reaching a minimum value at 100% power Typical values: RCR of about 3% at 100% power 25% to 30% at 10% power These numbers vary for different plants and different steam generator models. The next figure graphically illustrates this. ELO 4.1

89 Steam Generator Recirculation RatioRCR decreases as power increases because: Steam flow increases as power increases Thermal driving head in the S/Gs maintains liquid recirculation flow constant at low power levels versus high power levels Liquid flow decreases proportionally to the increase in steam flow as power level increases; causing an decrease in RCR Result is more steam flow compared to recirculation flow Figure: Recirculation Ratio Versus Power Level ELO 4.1

90 Two-Phase Flow Flow in the core and in the steam generator tube bundle region are examples of two-phase flow Small amount of voiding at top of hotter channels at 100% pwr Two-phase fluid flow may cause operational problems Vapor formation and a subsequent collapse may produce severe water hammers Flow oscillations Level oscillations SG Shrink and swell Two phase flow is present in some places due to water (or other fluid) flashing to steam (or its vapor state). Other fluids can exhibit this behavior, for example, hydraulic fluid, oil and any of many liquid chemicals used at power plants. ELO 4.1

91 SG Shrink and Swell Shrink Caused by a turbine load rejectionsteam pressure and temperature increase less boiling in the secondary side of the steam generator resulting in a drop in SG level Swell Caused when the steam demand is increased pressure reduction increased boiling and expansion of the steam bubbles increase in the SG level New NRC Bank question – P7689 tests this concept. ELO 4.1

92 Two-Phase Flow Knowledge CheckReactor coolant system (RCS) hot leg temperature is constant at 568°F while RCS pressure is decreasing due to a small reactor coolant leak. Which one of the following RCS pressure ranges includes the pressure at which two-phase flow will first occur in the hot leg? 1,250 to 1,201 psig 1,200 to 1,151 psig 1,150 to 1,101 psig 1,100 to 1,051 psig Correct answer is B. Correct answer is B NRC Question P580 Analysis: BE SURE TO CONVERT PSIG TO PSIA! Two phase flow will first occur when the saturation pressure for the given reactor coolant system (RCS) saturation pressure is reached. Looking up the saturation pressure for 568ºF in the steam tables yields 1207 psia. Note that this pressure is absolute pressure. The answers/distractors are given in gauge pressure. Therefore, absolute pressure must be converted to gauge pressure: 1207 – 15 = 1192 psig ELO 4.1

93 Pressure Spike Sudden conversion of KE (velocity) to flow energy (pressure) Pressure spike depends on the velocity AND density of fluid Function of mass flow rate, not volumetric flow rate The following equation calculates the pressure peak: ∆𝑃= 𝜌𝑐∆v 𝑔 𝑐 Where: ∆P = Pressure spike (lbf/ft2) ρ = Density of the fluid (lbm/ft3) c = Velocity of the pressure wave (Speed of sound in the fluid)  (ft/sec) ∆v = Change in velocity of the fluid (ft/sec) gc = Gravitational constant (lbm-ft/lbf-sec2) Water hammers are discussed in the next section. ELO 4.1

94 Water Hammer Slugs of liquid move at high speeds (speed of sound)ELO 4.2 – Describe water hammer, including severity and resulting pressure spikes. Slugs of liquid move at high speeds (speed of sound) Liquid can slam into valves, pipe bends, tees, or other piping components Can break piping supports Deform piping bends Knock pumps out of alignment Breach system piping K1.04 Explain operational implications of water hammer K1.07 Pressure spike K1.10 Define or explain the following terms and concepts: Water hammer Water hammer, also known as steam hammer, is a physical shock to the components of a fluid system, liquid, or vapor, caused by impact of high-velocity liquids. These liquids tend to be dense and incompressible. ELO 4.2

95 Water Hammer All fluid hammers come from the momentum of a moving higher density liquid Results of a sudden change in the fluid’s momentum when the fluid’s velocity changes quickly Most water hammers result from fast operation of valves Opening or closing rapid warming of steam lines Hot or cold condensate sitting in steam lines operator misunderstanding or impatience Cold condensate sitting in a steam line – causes steam to collapse and create a vacuum void into which a mass of fluid travels at a high speed toward piping components. Hot condensate sitting in steam line – could be blown into collisions with piping components by the movement of steam. A confined mass of incompressible liquid moving at high speeds creates pressure when it strikes the piping components and is a common occurrence in the condensate system. ELO 4.2

96 Water Hammer The severity of water hammer caused by valve operation depends on: Initial system pressure Density of the fluid Speed of sound in the fluid Elasticity of the fluid and the pipe Change in the velocity of the fluid Diameter and thickness of the pipe Valve stroking time ELO 4.2

97 Water Hammer Closing a valve converts the kinetic energy to flow energy (pressure) The compressed liquid and stretched pipe walls start releasing the liquid in the pipe back to the source Forms another pressure wave to valve When shockwave reaches valve, pipe wall begins contracting The pressure waves travel back and forth several times until fluid friction dampens pressure waves The figure on this slide show visually the impact that this pressure surge has on the piping. The enlargement and constriction of the pipes during the water hammer dissipation are exaggerated to show how this pressure can potentially damage piping! Figure: Typical Water Hammer ELO 4.2

98 Water Hammer Knowledge CheckRefer to the drawing of two lengths of 6-inch diameter pipe, each containing an identical automatic isolation valve. The actual pipe lengths are proportional to their symbols in the drawing Water at 65°F is flowing at 1,000 gpm through each pipe. If the isolation valves instantly close, valve A piping will experience a pressure increase that is __________ the pressure increase experienced by valve B piping; and the pressure spike will dissipate quicker in the __________ length of pipe. equal to; shorter equal to; longer less than; shorter less than; longer Correct answer is A. Correct answer is A. NRC Question P4042 (there are a few variations of this question in the bank) Analysis: Because the flow rates and piping diameter are the same, when each valve is closed, the pressure increase will be equal. However, the pressure spike will dissipate quicker in the shorter length of pipe. Keep in mind that the pressure spike is a function of the mass flow rate, not the amount of mass in the pipe (pipe length). ELO 4.2

99 Preventing Water HammerELO 4.3 – List actions that operators can take to minimize the occurrence of water hammer. Water hammer is preventable Good operator practices include making slow flow changes in piping systems Some plants utilize interlocks Circ water pump/valve operation K1.04 Explain operational implications of water hammer K1.10 Define or explain the following terms and concepts: Water hammer ELO 4.3

100 Preventing Water HammerFor water hammer prevention, operators should follow these precautions: Ensure piping systems are properly filled and vented Operate manual system valves slowly Ensure compliance with all heatup rate limitations Drain and warm system steam lines prior to initiating flow When possible, initiate pump starts against a closed discharge valve Open the discharge valve slowly to initiate system flow Some systems designs include auto opening of discharge valves ELO 4.3

101 Preventing Water HammerStart smaller capacity pumps before larger capacity pumps, if possible Use warm-up valves around main steam stop valves Close pump discharge valves before stopping pumps, if possible Periodically verify proper function of moisture traps and steam traps during operation Use high point pump casing vents Primes the pump and removes non-condensable gases Some systems will have specific startup and shutdown procedures to prevent water hammer. For example, the circulating water system. ELO 4.3

102 Preventing Water HammerKnowledge Check The primary reason for slowly opening the discharge valve of a large motor-driven centrifugal cooling water pump after starting the pump is to minimize the... net positive suction head requirements. potential for a water hammer. motor running current requirements. potential for pump cavitation. Correct answer is B. Correct answer is B. NRC Question P2880 Analysis: A. WRONG. Net positive suction head (NPSH) requirements is a function of the suction of the pump, not the discharge. Having the SUCTION valve closed when starting would impact NPSH. This is explained in further detail in – Pumps. B. CORRECT. Slowly introducing a fluid in downstream piping by SLOWLY opening the discharge valve will minimize the possibility of water hammer. C. WRONG. Opening the pump discharge valve, regardless of the rate at which it is done, raises motor amps due to higher system flow. The discharge valve being closed when starting a centrifugal induction motor will minimize the duration of the STARTING current, not the RUNNING current. This concept is discussed in further detail in – Motors and Generators. D. WRONG. The potential for pump cavitation is more a function of the suction of the pump, not the discharge. Similar to Choice “A” above. ELO 4.3

103 Preventing Water HammerKnowledge Check Which one of the following will minimize the possibility of water hammer? Draining the discharge line of a centrifugal pump after shutdown. Draining condensate out of steam lines before and after initiating flow. Starting a centrifugal pump with its discharge valve fully open. Starting a positive displacement pump with its discharge valve partially closed. Correct answer is B. Correct answer is B: NRC Question P2079 Analysis: A. WRONG. Draining the discharge of a centrifugal pump after shutdown is not a common practice; if the pump were restarted with an empty discharge line, significant water hammer would occur due to the high initial kinetic energy of the fluid being pumped. B. CORRECT. See discussion above. C. WRONG. Starting a centrifugal pump with a fully open discharge valve fully open will rapidly introduce flow into a previously stagnant system, resulting in a conversion to kinetic energy and a resulting pressure wave. It will also lengthen the duration of the starting current which could impact the motor. D. WRONG. Starting a positive displacement pump with a closed discharge valve will result in a rapid pressure surge; positive displacement pumps are designed to provide a constant flow over varying discharge pressures. ELO 4.3

104 NRC KA to ELO Tie KA # KA Statement RO SRO ELO K1.01Distinguish between static pressure, dynamic pressure, and total pressure. 2.2 2.3 1.1 K1.02 Define head loss. 1.4 3.2 K1.03 Discuss operational considerations of viscosity as related to head loss. 1.7 1.8 3.3 K1.04 Explain operational implications of water hammer. 3.4 3.6 4.2, 4.3 K1.05 Define or explain the following terms and concepts: Mass flow rate 2.9 3.0 1.2 K1.06 Define or explain the following terms and concepts: Two-phase flow 2.8 4.1 K1.07 Define or explain the following terms and concepts: Pressure spike 2.7 K1.08 Define or explain the following terms and concepts: Gas binding K1.09 Define or explain the following terms and concepts: Recirculation ratio 1.9 K1.10 Define or explain the following terms and concepts: Water hammer K1.11 Define or explain the following terms and concepts: Cavitation 3.1 K1.12 Explain why flow measurements must be corrected for density changes. 2.5 2.6 K1.13 Explain the relationship between pressure head and velocity head in a fluid system. K1.14 Discuss the velocity profiles for laminar flow and turbulent flow. 2.1 K1.15 Describe the methods of controlling system flow rates.