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2011 chevy chevrolet silverado 1500 owners manual

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2011 chevy chevrolet silverado 1500 owners manualInsurance Newest Full li thuy?t 4 chuong - nvjdfnvj HD-d?ch - good Translation - good Edtech Hosodangkythamdu- Final Full li thuy?t 4 chuong - vdfjnvfk SB Unit Guide Test exam- chap 3 FI 3- accounting for FA - Impairment FI-5- Accounting for counpound instrument Chapter 4 - Power Distribution System Configuration Chapter 3 - Economics analysis Chapter 1 - Introduction to Power Supply Systems Chapter 16 van hoa kinh doanh- van hoa Tu. Mass flow requires the presence of two regions at different chemical compositions, and it refers to the movement of a chemical species from a high concentration region towards a lower concentration one relative to the other chemical species present in the medium. Mass transfer cannot occur in a homogeneous medium. The concentration of a commodity is defined as the amount of that commodity per unit volume. The concentration gradient is defined as the change in the concentration C of a commodity per unit length in the direction of flow x. The diffusion rate of the commodity is expressed as dC kdiff A dx where A is the area normal to the direction of flow and kdiff is the diffusion coefficient of the medium, which is a measure of how fast a commodity diffuses in the medium. Examples of different kinds of diffusion processes: (a) A gallon of gasoline left in an open area will eventually evaporate and diffuse into air. (b) A spoon of sugar in a cup of tea will eventually dissolve and move up. (c) gas: A moth ball left in a closet will sublimate and diffuse into the air. (d) Air dissolves in water. Although heat and mass can be converted to each other, there is no such a thing as and mass transfer cannot be studied using the laws of radiation transfer. Mass transfer is analogous to conduction, but it is not analogous to radiation. (a) Temperature difference is the driving force for heat transfer, (b) voltage difference is the driving force for electric current flow, and (c) concentration difference is the driving force for mass transfer.http://flexi-cms.com/uploads/dvmatte-pro-manual.xml

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(a) Homogenous reactions in mass transfer represent the generation of a species within the medium. Such reactions are analogous to internal heat generation in heat transfer. (b) Heterogeneous reactions in mass transfer represent the generation of a species at the surface as a result of chemical reactions occurring at the surface. Such reactions are analogous to specified surface heat flux in heat transfer. PROPRIETARY MATERIAL. 2007 The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Mass Diffusion In the relation dx), the quantities, k, A, and T represent the following in heat conduction and mass diffusion: Rate of heat transfer in heat conduction, and rate of mass transfer in mass diffusion.A Area normal to the direction of flow in both heat and mass transfer. T Temperature in heat conduction, and concentration in mass diffusion. (a) T (b) F (c) F (d) T (e) F (a) T (b) F (c) F (d) T (e) T In the law of diffusion relations expressed as diff, A dwA and dx dy diff, A AB A, the diffusion coefficients DAB are the same.In a binary ideal gas mixture of species A and B, the diffusion coefficient of A in B is equal to the diffusion coefficient of B in A. Therefore, the mass diffusivity of air in water vapor will be equal to the mass diffusivity of water vapor in air since the air and water vapor mixture can be treated as ideal gases. Solids, in general, have different diffusivities in each other. At a given temperature and pressure, the mass diffusivity of copper in aluminum will not be the equal to the mass diffusivity of aluminum in copper. We would carry out the hardening process of steel carbon at high temperature since mass diffusivity increases with temperature, and thus the hardening process will be completed in a short time. The molecular weights of CO2 and N2O gases are the same (both are 44).http://www.kurashi-kyoiku.com/cms/dat/upimg/dvmax-user-manual.xml Therefore, the mass and mole fractions of each of these two gases in a gas mixture will be the same. PROPRIETARY MATERIAL. 2007 The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. The masses of the constituents of a gas mixture are given. The mass fractions, mole fractions, and the molar mass of the mixture are to be determined. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. The mole fractions of the constituents of a gas mixture are given. The mass of each gas and apparent gas constant of the mixture are to be determined. Assumptions None. Properties The molar masses of H2 and N2 are 2.0 and 28.0 respectively (Table Analysis The mass of each gas is H2: mH 2 N H 2 M H 2 (8 kmol) (2 16 kg N2: m N 2 N N 2 M N 2 2 kmol) (28 56 kg The molar mass of the mixture and its apparent gas constant are determined to be m m 16 56 kg 7.2 N m 8 2 kmol 8 kmol H2 2 kmol N2 Ru 8.314 K 1.15 K M 7.2 The mole numbers of the constituents of a gas mixture at a specified pressure and temperature are given. The mass fractions and the partial pressures of the constituents are to be determined. Assumptions The gases behave as ideal gases. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. The error involved in assuming the density of air to remain constant during a humidification process is to be determined. Properties The density of moist air before and after the humidification process is determined from the psychrometric chart to be T1 0.0727 3 air,1 and T1 0.07117 3 air, 2 Analysis The error involved as a result of assuming constant air density is then determined to be air air,1 100 0.0727 0.0712 3 0.0727 3 100 Air 14.7 psia which is acceptable for most engineering purposes. The diffusion coefficient of hydrogen in steel is given as a function of temperature. The diffusion coefficients at various temperatures are to be determined. Analysis The diffusion coefficient of hydrogen in steel between 200 K and 1200 K is given as D AB 1.65 T ) m 2 Using this relation, the diffusion coefficients at various temperatures are determined to be 200 K: D AB 1.65 200) 1.46 m 2 500 K: D AB 1.65 500) 1.57 m 2 K: D AB 1.65 m 2 1500 K: D AB 1.65 1500) 7.53 10 m 2 PROPRIETARY MATERIAL. 2007 The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. EES Prob. is reconsidered. The diffusion coefficient as a function of the temperature is to be plotted. Analysis The problem is solved using EES, and the solution is given below.Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. The molar concentration Ci of the gas species i in the solid at the interface Ci, solid side (0) is proportional to the partial pressure of the species i in the gas Pi, gas side(0) on the gas side of the interface, and is determined from C i, solid side (0) S Pi, gas side (0) where S is the solubility of the gas in that solid at the specified temperature. Using constant data for a gas dissolved in a liquid, the mole fraction of the gas dissolved in the liquid at the interface at a specified temperature can be determined from law expressed as yi, liquid side (0) Pi, gas side (0) H where H is constant and Pi, gas side(0) is the partial pressure of the gas i at the gas side of the interface. This relation is applicable for dilute solutions (gases that are weakly soluble in liquids). The permeability is a measure of the ability of a gas to penetrate a solid. The permeability of a gas in a solid, P, is related to the solubility of the gas P SDAB where DAB is the diffusivity of the gas in the solid. The mole fraction of CO2 dissolved in water at the surface of water at 300 K is to be determined. Assumptions 1 Both the CO2 and water vapor are ideal gases. 2 Air at the lake surface is saturated. Properties The saturation pressure of water at 300 K is 3.60 kPa (Table The constant for CO2 in water at 300 K is 1710 bar (Table Analysis The air at the water surface will be saturated. Therefore, the partial pressure of water vapor in the air at the lake surface will simply be the saturation pressure of water at Pvapor 3.60 kPa Assuming both the air and vapor to be ideal gases, the partial pressure and mole fraction of dry air in the air at the surface of the lake are determined to be Pdry air P Pvapor 100 3.60 96.4 kPa The partial pressure of CO2 is PCO2 y CO2 Pdry air (0.005)(96.4) 0.482 kPa 0.00482 bar y CO2 PCO2 0.00482 bar 10 1710 bar H PROPRIETARY MATERIAL. 2007 The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. The mole fraction of the water vapor at the surface of a lake and the mole fraction of water in the lake are to be determined and compared. Assumptions 1 Both the air and water vapor are ideal gases. 2 Air is weakly soluble in water and thus law is applicable. Properties The saturation pressure of water at is 0.3632 psia (Table constant for air dissolved in water at (294 K) is given in Table to be H 66,800 bar. Analysis The air at the water surface will be saturated. Therefore, the partial pressure of water vapor in the air at the lake surface will simply be the saturation pressure of water at Pvapor 0.3632 psia Saturated air 13.8 psia Assuming both the air and vapor to be ideal gases, the mole fraction of water vapor in the air at the surface of the lake is determined from Eq.Therefore, the mole fraction of water in the lake near the surface is y water, liquid side 1 y dry air, liquid side 1 1.39 Discussion The concentration of air in water just below the interface is 1.39 moles per moles. The amount of air dissolved in water will decrease with increasing depth. The mole fraction of the water vapor at the surface of a lake at a specified temperature is to be determined. Assumptions 1 Both the air and water vapor are ideal gases. 2 Air at the lake surface is saturated. Properties The saturation pressure of water at is 1.705 kPa (Table Analysis The air at the water surface will be saturated. Therefore, the partial pressure of water vapor in the air at the lake surface will simply be the saturation pressure of water at Pvapor 1.7051 kPa Saturated air 13.8 psia yH2O, air side Assuming both the air and vapor to be ideal gases, the partial pressure and mole fraction of dry air in the air at the surface of the lake are determined to be Pdry air P Pvapor 100 1.7051 98.295 kPa y dry air Pdry air P yH2O, liquid side 1.0 Lake, 98.295kPa 0.983 (or 100 kPa Therefore, the mole fraction of dry air is 98.3 percent just above the interface. PROPRIETARY MATERIAL. 2007 The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. A rubber plate is exposed to nitrogen. The molar and mass density of nitrogen in the rubber at the interface is to be determined. Assumptions Rubber and nitrogen are in thermodynamic equilibrium at the interface. Properties The molar mass of nitrogen is M 28.0 (Table The solubility of nitrogen in rubber at 298 K is 0.00156 (Table Rubber plate Analysis Noting that 250 kPa 2.5 bar, the molar density of nitrogen in the rubber at the interface is determined from Eq.A rubber wall separates O2 and N2 gases. The molar concentrations of O2 and N2 in the wall are to be determined. Assumptions The O2 and N2 gases are in phase equilibrium with the rubber wall. Properties The molar mass of oxygen and nitrogen are 32.0 and 28.0 respectively (Table The solubility of oxygen and nitrogen in rubber at 298 K are 0.00312 and respectively (Table Analysis Noting that 750 kPa 7.5 bar, the molar densities of oxygen and nitrogen in the rubber wall are determined from Eq.PROPRIETARY MATERIAL. 2007 The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. A glass of water is left in a room. The mole fraction of the water vapor in the air and the mole fraction of air in the water are to be determined when the water and the air are in thermal and phase equilibrium. Assumptions 1 Both the air and water vapor are ideal gases. 2 Air is saturated since the humidity is 100 percent. 3 Air is weakly soluble in water and thus law is applicable. Properties The saturation pressure of water at is 2.339 kPa (Table constant for air dissolved in water at (293 K) is given in Table to be H 65,600 bar. Molar masses of dry air and water are 29 and 18 respectively (Table Analysis (a) Noting that air is saturated, the partial pressure of water vapor in the air will simply be the saturation pressure of water at Pvapor Psat 20 C 2.339 kPa Assuming both the air and vapor to be ideal gases, the mole fraction of water vapor in the air is determined to be y vapor Pvapor P 2.339 kPa 0.0241 97 kPa Air 97 kPa Evaporation (b) Noting that the total pressure is 97 kPa, the partial pressure of dry air is Pdry air P Pvapor 97 2.339 94.7 kPa 0.947 bar From law, the mole fraction of air in the water is determined to be y dry air,liquid side Pdry air,gas side H 0.947 bar 1.44 65,600bar Water Discussion The amount of air dissolved in water is very small, as expected. PROPRIETARY MATERIAL. 2007 The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. A carbonated drink in a bottle is considered. Assuming the gas space above the liquid consists of a saturated mixture of CO2 and water vapor and treating the drink as a water, determine the mole fraction of the water vapor in the CO2 gas and the mass of dissolved CO2 in a 200 ml drink are to be determined when the water and the CO2 gas are in thermal and phase equilibrium. Assumptions 1 The liquid drink can be treated as water. 2 Both the CO2 and the water vapor are ideal gases. 3 The CO2 gas and water vapor in the bottle from a saturated mixture. 4 The CO2 is weakly soluble in water and thus law is applicable. Properties The saturation pressure of water at is 6.33 kPa (Table constant for CO2 issolved in water at (310 K) is given in Table to be H 2170 bar. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. During mass diffusion of species A through a plane wall, the species A content of the wall will remain constant during steady mass diffusion, but will change during transient mass diffusion. PROPRIETARY MATERIAL. 2007 The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. A thin plastic membrane separates hydrogen from air. The diffusion rate of hydrogen diffusion through the membrane under steady conditions is to be determined. Assumptions 1 Mass diffusion is steady and since the hydrogen concentrations on both sides of the membrane are maintained constant. Also, there is symmetry about the center plane of the membrane. 2 There are no chemical reactions in the membrane that results in the generation or depletion of hydrogen. Properties The binary diffusion coefficient of hydrogen in the plastic membrane at the operation temperature is given to be The molar mass of hydrogen is M 2 (Table Analysis (a) We can consider the total molar concentration to be constant (C CA CB CB constant), and the plastic membrane to be a stationary medium since there is no diffusion of plastic molecules ( B 0 ) and the concentration of the hydrogen in the membrane is extremely low (CA 1). Then the molar flow rate of hydrogen through the membrane diffusion per unit area is determined from C A,1 C A, 2 j diff diff D AB A L (0.045 0.002) 3 Plastic m 2 m membrane B 2.s H2 B Air The mass flow rate is determined multiplying the molar flow rate the molar mass of hydrogen, mdiff diff M j diff (2 10.s) 2 L 2.s (b) Repeating the calculations for a thick membrane gives C A,1 C A, 2 j diff diff D AB A L ( 0.045 0.002) 3 m 2 m 2.s and diff M j diff (2 2.s) 2.s The mass flow rate through the entire membrane can be determined multiplying the mass flux value above the membrane area. PROPRIETARY MATERIAL. 2007 The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Natural gas with hydrogen content is transported in an above ground pipeline. The highest rate of hydrogen loss through the pipe at steady conditions is to be determined. Assumptions 1 Mass diffusion is steady and since the hydrogen concentrations inside the pipe is constant, and in the atmosphere it is negligible. Also, there is symmetry about the centerline of the pipe. 2 There are no chemical reactions in the pipe that results in the generation or depletion of hydrogen. 3 Both H2 and CH4 are ideal gases. Properties The binary diffusion coefficient of hydrogen in the steel pipe at the operation temperature is given to be The molar masses of H2 and CH4 are 2 and 16 respectively (Table A1). The solubility of hydrogen gas in steel is given as w H 2 2.09 10 exp( T ) PH0.25. The density of steel pipe is 7854 (Table Analysis We can consider the total molar concentration to be constant (C CA CB CB constant), and the steel pipe to be a stationary medium since there is no diffusion of steel molecules ( B 0 ) and the concentration of the hydrogen in the steel pipe is extremely low (CA 1). The molar mass of the H2 and CH4 mixture in the pipe is B M M i i (0.08)(2) (0.92)(16) 14.88 Noting that the mole fraction of hydrogen is 0.08, the partial pressure of hydrogen is y H2 PH 2 P B PH 2 (0.08)(500 kPa ) 40 kPa 0.4 bar Then the mass fraction of hydrogen becomes wH 2 T ) PH 2 0.5 4 293)(0.4) 0.5 Steel pipe 293 K Natural gas H2, 500 kPa H2 diffusion The hydrogen concentration in the atmosphere is practically zero, and thus in the limiting case the hydrogen concentration at the outer surface of pipe can be taken to be zero. Then the highest rate of hydrogen loss through a 100 m long section of the pipe at steady conditions is determined to be diff,A,cyl D AB w A,1 w A, 2 ln(r2 r1 ) (100m)(7854 3 ) 0 PROPRIETARY MATERIAL. 2007 The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Books You don't have any books yet. Studylists You don't have any Studylists yet. Recent Documents You haven't viewed any documents yet. Students also viewed Cengel Cimbala Solutions Chap02 Cengel Cimbala Solutions Chap04 Cengel Cimbala Solutions Chap10 Cengel Cimbala Solutions Chap11 Cengel Cimbala Solutions Chap15 Azer Other related documents Psikolojiye Giris I - Ders Notlar? - 2013 MATH210-HW1 MAK 307- Fluid Mech. -I Syllabus(2018-2019 ) 344550115 Plato and Paul Socrates and Solomon on issues of Justice Just man True Happiness MAK307 FINAL EXAM. 09.01.2018-SOL MAK307 Midterm exam 13 Preview text Chapter 14 Turbomachinery Solutions Manual for Fluid Mechanics: Fundamentals and Applications Cimbala CHAPTER 14 TURBOMACHINERY PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The Companies, Inc.No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or any means, electronic or otherwise, without the prior written permission of PROPRIETARY MATERIAL. 2006 The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 14 Turbomachinery General Problems Solution We are to discuss energy producing and energy absorbing devices. Analysis A more common term for an energy producing turbomachine is a turbine. Turbines extract energy from the moving fluid, and convert that energy into useful mechanical energy in the surroundings, usually in the form of a rotating shaft. Thus, the phrase is from a frame of reference of the fluid the fluid loses energy as it drives the turbine, producing energy to the surroundings. On the other hand, a more common term for an energy absorbing turbomachine is a pump. Pumps absorb mechanical energy from the surroundings, usually in the form of a rotating shaft, and increase the energy of the moving fluid. Thus, the phrase is from a frame of reference of the fluid the fluid gains or absorbs energy as it flows through the pump. Discussion From the frame of reference of the surroundings, a pump absorbs energy from the surroundings, while a turbine produces energy to the surroundings. Thus, you may argue that the terminology also holds for the frame of reference of the surroundings. This alternative explanation is also acceptable. Solution We are to discuss the differences between fans, blowers, and compressors. Analysis A fan is a gas pump with relatively low pressure rise and high flow rate. A blower is a gas pump with relatively moderate to high pressure rise and moderate to high flow rate. A compressor is a gas pump designed to deliver a very high pressure rise, typically at low to moderate flow rates. Discussion The boundaries between these three types of pump are not always clearly defined. Solution We are to list examples of fans, blowers, and compressors. Analysis Common examples of fans are window fans, ceiling fans, fans in computers and other electronics equipment, radiator fans in cars, etc. Common examples of blowers are leaf blowers, hair dryers, air blowers in furnaces and automobile ventilation systems. Common examples of compressors are tire pumps, refrigerator and air conditioner compressors. Discussion Students should come up with a diverse variety of examples. Solution We are to discuss the difference between a turbomachine and a dynamic turbomachine. Analysis A turbomachine is a device that contains a closed energy is transferred to the fluid (pump) or from the fluid (turbine) via movement of the boundaries of the closed volume. On the other hand, a dynamic turbomachine has no closed instead, energy is transferred to the fluid (pump) or from the fluid (turbine) via rotating blades. Examples of pumps include well pumps, hearts, some aquarium pumps, and pumps designed to release precise volumes of medicine. Examples of turbines include water meters and gas meters in the home. Examples of dynamic pumps include fans, centrifugal blowers, airplane propellers, centrifugal water pumps (like in a car engine), etc. Examples of dynamic turbines include windmills, wind turbines, turbine flow meters, etc. Discussion Students should come up with a diverse variety of examples. PROPRIETARY MATERIAL. 2006 The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 14 Turbomachinery Solution water pump. We are to determine how the average speed at the outlet compares to the average speed at the inlet of a Assumptions 1 The flow is steady (in the mean). 2 The water is incompressible. Analysis Conservation of mass requires that the mass flow rate in equals the mass flow rate out. Thus, Conservation of mass: in Ain out outVout Aout or, since the area is proportional to the square of diameter, 2 Vout D Vin in in Vin in Dout Dout 2 (1) (a) For the case where Dout Din, Vout must be greater than Vin. (b) For the case where Dout Din, Vout must be equal to Vin. (c) For the case where Dout Din, Vout must be less than Vin. Discussion A pump does not necessarily increase the speed of the fluid passing through it. In fact, the average speed through the pump can actually decrease, as it does here in part (c). Solution For an air compressor with equal inlet and outlet areas, and with both density and pressure increasing, we are to determine how the average speed at the outlet compares to the average speed at the inlet. Assumptions 1 The flow is steady. Analysis Conservation of mass requires that the mass flow rate in equals the mass flow rate out. The areas of the inlet and outlet are the same. Thus, Conservation of mass: in Ain out outVout Aout or Vout Vin (1) Since Vout must be less than Vin. Discussion A compressor does not necessarily increase the speed of the fluid passing through it. In fact, the average speed through the pump can actually decrease, as it does here. PROPRIETARY MATERIAL. 2006 The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 14 Turbomachinery Pumps Solution We are to list and define the three categories of dynamic pumps. Analysis The three categories are: Centrifugal flow pump fluid enters axially (in the same direction as the axis of the rotating shaft) in the center of the pump, but is discharged radially (or tangentially) along the outer radius of the pump casing.Discussion There are also some dynamic pumps, such as jet pumps and electromagnetic pumps, that are not discussed in this text. Solution (a) False: Actually, blades yield the highest efficiency. (b) True: The pressure rise is higher, but at the cost of less efficiency. (c) True: In fact, this is the primary reason for choosing blades. (d) False: Actually, the opposite is true a pump with blades usually has more blades, but they are usually smaller. Solution We are to choose which pump location is better and explain why. Analysis The two systems are identical except for the location of the pump (and some minor differences in pipe layout). The overall length of pipe, number of elbows, elevation difference between the two reservoir free surfaces, etc.In addition, the length of pipe from the lower reservoir to the pump inlet is smaller in Option (a), and there is one less elbow between the lower reservoir and the pump inlet, there decreasing the head loss upstream of the pump both of which also increase NPSH, and reduce the likelihood of cavitation. Discussion Another point is that if the pump is not Option (b) may run into problems if the free surface of the lower reservoir falls below the elevation of the pump inlet. Since the pump in Option (a) is below the reservoir, is not an issue. Solution We are to define and discuss NPSH and NPSHrequired. Analysis Net positive suction head (NPSH) is defined as the difference between the inlet stagnation pressure head and the vapor pressure head, P V2 P NPSH v g g g 2 pump inlet We may think of NPSH as the actual or available net positive suction head. On the other hand, required net positive suction head (NPSHrequired) is defined as the minimum NPSH necessary to avoid cavitation in the pump. As long as the actual NPSH is greater than NPSHrequired, there should be no cavitation in the pump. Discussion Although NPSH and NPSHrequired are measured at the pump inlet, cavitation (if present) happens somewhere inside the pump, typically on the suction surface of the rotating pump impeller blades. PROPRIETARY MATERIAL. 2006 The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 14 Turbomachinery Solution (a) False: Since the pumps are in series, the volume flow rate through each pump must be the same:. (b) True: The net head increases H1 through the first pump, and then H2 through the second pump. The overall rise in net head is thus the sum of the two. (c) True: Since the pumps are in parallel, the total volume flow rate is the sum of the individual volume flow rates. (d) False: For pumps in parallel, the change in pressure from the upstream junction to the downstream junction is the same regardless of which parallel branch is under consideration. Thus, even though the volume flow rate may not be the same in each branch, the net head must be the same: H H1 H2. Solution Shutoff head Pump performance curve We are to label several items on the provided plot. Analysis The figure is here, and the requested items are labeled. Havailable H System curve Discussion Also labeled are the available net head, corresponding to the pump performance curve, and the required net head, corresponding to the system curve. The intersection of these two curves is the operating point of the pump. Solution equation.