Distribution Reservoirs. M5: Distribution and Storage Reservoirs, and Pumps. Types of Distribution and Service. Reservoirs

Size: px
Start display at page:

Download "Distribution Reservoirs. M5: Distribution and Storage Reservoirs, and Pumps. Types of Distribution and Service. Reservoirs"

Transcription

1 M5: Ditribution and Storage Reervoir, and Pu Robert Pitt Univerity of Alabaa and Shirley Clark Penn State - Harriburg Ditribution Reervoir Provide ervice torage to eet widely fluctuating deand ioed on yte. Accoodate fire-fighting and eergency requireent. Equalize oerating reure. Deign Period and Caacitie in Water-Suly Syte (Table. Chin 000) Tye of Ditribution and Service Reervoir Surface Reervoir at ground level large volue Standie cylindrical tank whoe torage volue include an uer ortion (ueful torage) uually le than 50 feet high Elevated Tank ued where there i not ufficient head fro a urface reervoir ut be ued to, but ued to allow gravity ditribution in ain yte

2 Location of Ditribution Reervoir Provide axiu benefit of head and reure (elevation high enough to develo adequate reure in yte) Near center of ue (decreae friction loe and therefore lo of head by reducing ditance to ue). Great enough elevation to develo adequate reure in yte May require ore than one in large etroolitan area Storage Tank Tye Deterining Required Storage Aount Function of caacity of ditribution network, location of ervice torage, and ue. To coute required equalizing or oerating torage, contruct a diagra of hourly rate of conution. Obtain hydrograh of hourly deand for axiu day (in Alabaa, thi would likely be either late ring or uer, and would include deand for lawn care, filling and/or aintaining wiing ool, outdoor recreation, etc.) Tabulate the hourly deand data for axiu day. Find required oerating torage uing a diagra, hydrograh, or calculated table.

3 Deterining Required Storage Aount: Exale Tie (hr) 0 Deand (g) 0 Tie (hr) 7 Deand (g),60 Tie (hr) Deand (g) 6,440 Tie (hr) 9 Deand (g) 9,,70 8 5,90 4 6,70 0 8,0,00 9 5,60 5 6,0 5,050,00 0 5, ,40,570 4,970 6, ,640,470 5,980 6,0 8 7,0 4,90 6,080 To calculate total volue required for torage Calculate hourly deand in gallon. Calculate cuulative hourly deand in gallon. Divide cuulative deand by 4 hour to get the average hourly uly needed. Calculate the urlu/deficit between the hourly uly and hourly deand for each hour. Su either the urlu or the deficit to deterine the required torage volue. REMEMBER: Thi reult ean that the tank i full when the urlu i greatet and ety when the deficit i greatet! It i filling when the loe of the u curve i greater than the loe of the deand curve and i etying when the loe of the deand curve i greater than the loe of the u curve. There i no exce for eergencie! Deterining Required Storage Aount: Exale Deterining Required Storage Aount: Exale Tie (hr) Deand (g),70 Deand (gal) 0,00 Cuulative Deand (gal) 0,00 Surlu/ Deficit* (gal) 56,008 Deficit (gal) urlu ,00 6,000 56,00 60,08 urlu ,00,00 77,400 65,008 urlu ,970 8,00 495,600 68,008 urlu ,980 8,800 64,400 67,408 urlu ,080 4,800 79,00 6,408 urlu * Surlu or deficit average hourly deand (86,08 gal) actual hourly deand

4 Deterining Required Storage Aount: Exale Tie (hr) Deand (g),60 Deand (gal) 7,800 Cuulative Deand (gal) 957,000 Surlu/ Deficit (gal) 68,408 Deficit (gal) urlu ,90,400,68,400-5,9-5, ,60 7,00,605,600-50,99-50, ,900 54,000,959,600-67,79-67, ,040 6,400,,000-76,9-76, ,0 79,00,70,00-9,99-9, ,440 86,400,087,600-00,9-00,9 Tie (hr) Deterining Required Storage Aount: Exale Deand (g) 5,050 Deand (gal) 0,000 Cuulative Deand (gal) 6,49,80 Surlu/ Deficit (gal) -6,79 Deficit (gal) -6, ,570 54,00 6,58,80,008 urlu 00-00,470 48,00 6,7,580 8,008 urlu ,90 7,400 6,868,980 48,808 urlu Average* Storage Required 86,08 gallon,465,75 gallon (u of all deficit) lu additional for eergencie Su of deficit,465,75 * Average cuulative hourly deand / 4 hour 6,868,980 gal/4 hr 86,08 gal/hr Deterining Required Storage Aount: Exale Tie (hr) Deand (g) 6,70 Deand (gal) 8,00 Cuulative Deand (gal),469,800 Surlu/ Deficit (gal) -95,99 Deficit (gal) -95, ,0 79,00,849,000-9,99-9, ,40 80,400 4,9,400-94,9-94, ,640 98,400 4,67,800 -,9 -, ,0 49,00 5,067,000-5,99-5, , 559,980 5,66,980-7,77-7, ,0 499,00 6,6,80 -,99 -,99 Surlu/Deficit Chin 000, Figure.4 4

5 Exale.8 (Chin 000) A ervice reervoir i to be deigned for a water uly erving 50,000 eole with an average deand of 600 L/day/caita and a fire flow of 7,000 L/in. The required torage i the u of: () volue to uly the deand in exce of the axiu daily deand, () fire torage, and () eergency torage () The volue to uly the eak deand can be taken a 5% of the axiu daily deand. The axiu daily deand factor i.8 tie the average deand. The axiu daily flowrate i therefore: 8 5 ( 600 L / d / caita )( 50,000 eole ).7 x 0 L / day.7 x 0 day.8 / The correonding volue i therefore: 5 ( 0.5)(.7 x 0 / day) 67,500 V eak Exale.9 (Chin 000) A water uly yte deign i for an area where the iniu allowable reure in the ditribution yte i 00 kpa. The head lo between the low reure ervice location (having a ieline elevation of 5.40 ) and the location of the elevated torage tank wa deterined to be 0 during average daily deand condition. Under axiu hourly deand condition, the head lo i increaed to. Deterine the noral oerating range for the water tored in the elevated tank. Under average deand condition, the elevation (z o ) of the hydraulic grade line (HGL) at the reervoir location i: where γ z h in in 9.79 L 0 : kn kpa / in z 0 z in γ h L Therefore, under average condition: 00kPa z kN / () The fire flow of 7,000 L/in (0.6 /ec) ut be aintained for at leat 9 hour. The volue to uly the fire deand i therefore: ( 0.6 /ec)( 9hour)(,600ec/ hr) 0,00 V fire () The eergency torage can be taken a the average daily deand: 6 ( 50,000 eole )( 600L / day / caita ) 50x 0 L 50,000 V eer The required volue of the ervice reervoir i therefore: V V eak V fire V eer 67,500 0,00 50,000 7,600 Mot of the torage for thi exale i aociated with the eergency volue. Of coure, the ecific factor ut be choen in accordance with local regulation and olicy. and under axiu deand condition, the elevation of the HGL at the ervice reervoir, z, i: 00kPa z kN / Therefore, the oerating range in the torage tank hould be between 46.0 and

6 PUMPING Two tye of u coonly ued in water and ewage work: Centrifugal Pu tyical ue i to tranort water and ewage Dilaceent Pu tyical ue i to handle ludge in a treatent facility Scheatic of hydraulic grade line for a ued yte (Walki, et al. 00 figure.6) Pu election guideline (Chin 000 Table.) Pu effect on flow in ieline (Chin 000 Figure.6) 6

7 Sizing Pu To deterine the ize of the u, ut know the total dynaic head that the u i exected to rovide. Total Dynaic Head (TDH) conit of: the difference between the center line of the u and the height to which water ut be raied the difference between the uction ool elevation and center line of the u friction loe in the u and fitting velocity head Static v. Dynaic Head HGL hydraulic grade line TDH total dynaic head TSH total tatic head DDH dynaic dicharge head SSH tatic uction head DSH dynaic uction head SSL tatic uction lift DSL dynaic uction lift SDH tatic dicharge head Head Added by Pu (Total Dynaic Head) If the u ha been elected, Bernoulli Equation can be rearranged to olve for the head added by a u: P P V V h γ g A Z Z h f where h a head added by u (total dynaic head) P atoheric reure γ ecific weight of fluid V velocity Z elevation h f head lo in attached ie and fitting 7

8 Intallation of Pu into Water Suly Syte Exale of Head Added by Pu Exale: A u i being ued to deliver 5 g of hot water fro a tank through 50 feet of -inch diaeter ooth ie, exiting through a ½- inch nozzle 0 feet above the level of the tank. The head lo fro friction in the ie i 6.7 feet. The ecific weight of the hot water i 60.6 lb f /ft. Exale for Head Added by Pu Solve: P P V V A Z Z h f h γ g Set reference oint. Let oint be the water level in the tank. Let oint be the outlet of the nozzle. 8

9 Exale for Head Added by Pu Since both the end of the nozzle and the to of the tank are oen to the atohere, let P P 0 (gage reure). In a tank, the velocity i o all a to be negligible, o V 0. Calculate V. ft / ec 5g 449g V 57.4 A π ( 0.5 in) ft ft 4 44in / ec Calculation of TDH fro Pu Tet Data TDH H L H F H V Where H L total tatic head (difference between elevation of uing ource and oint of delivery) H F friction loe in u, ie and fitting H V velocity head due to uing Exale for Head Added by Pu Subtituting into Bernoulli Equation: h h h A A A P γ P V 0 lbf / ft ft V g Z Z (57.4 ft / ec) (. ft / ec h 0 0 ft 0 ft 6.7 ft ) Calculation of TDH fro Pu Tet Data Subtituting ter fro Bernoulli Equation for Velocity Head: TDH H L H F V g Can lot yte head (total dynaic head) veru dicharge. TDH ay not be contant becaue of difference between elevation during uing (deleting uly and adding to torage). TDH alo ay not be contant becaue dicharge rate will affect friction loe (a well a the velocity head ter). 9 f

10 Pu efficiency curve (Net Poitive Suction Head) (Walki, et al. 00 figure.8) Effect of a change in deand on a contant-eed u in a cloed yte (Walki, et al. 00 figure 9. and table 9.) Syte oerating oint (Walki, et al. 00 figure.7) Exale. (Chin 000) Water i ued fro a lower reervoir to an uer reervoir through a ieline yte a hown below. The reervoir differ in elevation by 5. and the length of the teel ie (k 0.046) connecting the reervoir i.. The ie i 50 in diaeter and the erforance curve i given by: h where h (the u head) i in eter and i in L/ 0

11 Oerating oint in ieline yte (Exale.) (Chin 000 Figure.7) In general, f i a function of both the Reynold nuber and the relative roughne. However, if fully turbulent, then the friction factor deend on the relative roughne according to: log ε and the relative roughne i given by: f.7 k relative roughne ε D k equivalent and roughne for thi cae ε f f log Can alo ue the Moody diagra and read the friction factor by reading traight acro fro the roughne value if auing turbulent flow. Uing thi u, what flow do you exect in the ieline? If the otor on the u rotate at,400 r, calculate the ecific eed of the u in US Cutoary unit and tate the tye of u that hould be ued. Neglecting inor loe, the energy equation for the ieline yte i: fl h z 5. ga D fl ga D Where h i the head added by the u, f i the friction factor, L i the ie length, A i the cro-ectional area of the ie, and D i the diaeter of the ie. π 4 π 4 A D ( 0.05 ) Subtituting thi friction factor into the energy equation: h fl ga D in unit 5. of / ; ( 0.09)(. ) ( / )( ) ( 0.05 ) 9.8 for in L / ec Cobining the yte curve and the u characteritic curve lead to: L /ec The next te i to verify if the flow in the ieline wa coletely turbulent

12 V.09x 0 / A / ( / )( 0.05 ) VD Re.78x 0 6 ν.00 ( x 0 / ) 4 The friction factor can now be re-calculated uing the Jain equation: f and f log ε Re 0.9 log (.78x 0 ) Therefore the original aroxiation of the friction factor (0.09) wa in error and the flow calculation need to be reeated, leading to:.09l / Table. Chin 000 Therefore, with a ecific eed of 59 and the low flowrate, the bet tye of u for thee condition i a centrifugal u. ( L / ) 5. h In U.S. Cutoary unit: h.09l / 7. g ft ω,400r The ecific eed i given by: N ω h ( 400 )( 7. ) 0.75 ( 50. ) Cavitation If the abolute reure on the uction ide of a u fall below the aturation vaor reure of the fluid, the water will begin to vaorize, thi i called cavitation. Thi caue localized high velocity jet than can caue daage to the u through itting of the etal caing and ieller, reducing u efficiency and cauing exceive vibration. Thi ound like gravel going through a centrifugal u. Exale. (Chin 000) A u with a erforance curve of h -0.0 (h in, in L/) i above a water reervoir and u water at 4.5 L/ through a 0 diaeter ductile iron ie (k 0.6 ). If the length of the ieline i.5, calculate the cavitation araeter of the u. If the ecific eed of the u i 0.94, etiate the critical value of the cavitation araeter and the axiu height above the water urface of the reervoir that the u can be located. The cavitation araeter i defined by: σ Net head Potive added Suction by the Head u h NPSH Where NPSH (net oitive uction head) i: NPSH z h L γ o γ v

13 where : γ z o v atoheric reure, 0kPa ecific weight uction lift, of water, 9.79kN / aturated vaor reure of water at 0 o C,.4kPa The head lo can be etiated uing the Darcy-Weibach and inor lo coefficient. The flow rate and Reynold nuber are: / V.0 / A π ( 0.0 ) 4 ( / )( 0.0 ) 5 VD Re x ν.00x 0 / The Colebrook equation give a friction factor, f, of uing thee value. If the ecific eed of the u i 0.94, then the critical cavitation araeter i etiated to be 0. (uing the to cale for SI unit). Since the calculated cavitation araeter wa.04, cavitation hould not be a roble. Chin 000, Figure.8 The head lo, h L in the ieline between the reervoir and the ection ide of the u i: ( )(.5 ) ( 0.0 ) ( ) h fl V.0 0. L D g ( / ) The NPSH can now be calculated: v 0kPa.4kPa NPSH o z h L kN / 9.79kN / γ γ The head added by the u, h, can be calculated fro the u erforance curve and the given flow rate: ( L / ) h The cavitation araeter of the u i therefore: σ NPSH h When cavitation i iinent, the cavitation araeter i equal to 0.0 and the reultant NPSH i: NPSH σ h ( 6 ). c 0.0. z h L γ o γ v The head loe between the u and reervoir i etiated a: f D h L ( z 0.5 ) V g ( 0.057)( z 0.5 ) ( ) h L z ( / )

14 Cobining the equation reult in:. γ o 0kPa 9.79kN / z z h L γ v ( z ).4kPa 9.79kN / z and olving for z : z Therefore, the u hould be located no higher than 7.49 (about 5 ft) above the water urface in the reervoir to revent cavitation (thi ut alway be le than at, or about ft). Calculation for Pu Efficiency η P P BHP *00 where η P u efficiency P fluid ower BHP brake horeower actually delivered to the u Calculation of the Theoretical Required Power of a Pu Power (h) γ(tdh)/550 Where dicharge (ft /ec) γ ecific weight of water (at ea level, 6.4 lb f /ft ) TDH total dynaic head (ft) 550 converion factor fro ft-lb f /ec to horeower Each u ha it own characteritic relative to ower requireent, efficient and head develoed a a function of flow rate. Thee are uually given on u curve for each u. In general, the efficient for centrifugal u i between 50 and 85%, with u efficiency generally increaing with the ize and caacity of the u. Pu Power and Efficiency Exale: A u i being ued to deliver 5 g of hot water fro a tank through 50 feet of -inch diaeter ooth ie, exiting through a ½-inch nozzle 0 feet above the level of the tank. The head lo fro friction in the ie i 6.7 feet. The ecific weight of the hot water i 60.6 lb f /ft. What i the ower delivered to the water by the u? If the efficiency i 60%, calculate the ower delivered to the u. 4

15 Pu Power and Efficiency γ ( TDH ) Power ( h ) 550 Convert the dicharge to cf. 449 g cf 5g ft Pu Power and Efficiency Efficiency calculation: BHP BHP BHP P *00 η P (0.755h ) * h /ec Pu Power and Efficiency Subtituting back into Power equation: TDH wa calculated to be 87.9 ft. Power Power Power ( ( ( h h h ) ) ) γ ( TDH 550 ) ( ft /ec)( 6.4 lb / ft )( 87.9 ft) 550 ft lb 0.755h / h Generating Pu Curve Exale: The characteritic of a centrifugal u oerating at two different eed are lited below. Grah thee curve and connect the bet efficiency oint. Calculate the headdicharge value for an oerating eed of 450 r and lot the curve. Sketch the u oerating enveloe between 60 and 0 ercent of the bet efficiency oint. 5 f f

16 Generating Pu Curve Dicharge (g) 0 Seed 750 r Head (ft) 0 Efficiency (%) NA Dicharge 0 Seed 50 r Head (ft) 96 Efficiency (%) NA Pu in Serie and Parallel For erie oeration at a given caacity/dicharge, the total head equal the u of the head added by each u. For arallel oeration, the total dicharge i ultilied by the nuber of u for a given head. Generating Pu Curve Pu in erie (Chin 000 Figure.9) 6

17 Pu in arallel (Chin 000 Figure.0) Pu in Parallel: For a yte coniting of three u in arallel, one-third of the total flow,, goe through each u, and the head added by each u i the ae a the total head, H added by the u yte: H 0. and the characteritic curve of the u yte i therefore: H 0.0 Multile Pu Exale.4 (Chin 000) A u erforance curve i: h 0. What i the u erforance curve for a yte having three of thee u in erie and a yte having three of thee u in arallel Pu in Serie: For the u in erie, the ae flow () goe through each u and each u add one-third of the total head, H : H 0. The characteritic curve of the u yte i therefore: H 6 0. Variable Seed Pu Equation for Variable Seed Pu: H H P P N N N N N N where Pu dicharge H head (total dicharge/dynaic head) P ower inut N u eed (revolution/tie tyically in inute) 7 i i

18 Variable Seed Pu Pu Selection Exale A u tation i to be deigned for an ultiate caacity of 00 g at a total head of 80 ft. The reent requireent are that the tation deliver 750 g at a total head of 60 ft. One u will be required a a tandby. Pu Selection Exale 8

19 Pu Selection Exale Poibilitie are Pu A and/or Pu B. Pu Selection Exale One u will not uly the needed head at future condition. Pu B i better choice. Therefore, need to look at the three condition ioed on thi deign and ee if B can uly. Two u ut roduce 00 g at 80 ft TDH. One u ut roduce 600 g at 80 ft TDH. One u ut eet reent requireent of 750 g at 60 ft TDH. Pu Selection Exale Puing Station for Sewage Contant-eed u hould not be turned on and off too frequently ince thi can caue the to overheat. In all uing tation there ay be only two u, each of which ut be able to deliver the axiu anticiated flow. Lower flow accuulated in the wet well until a ufficient volue ha accuulated to run the u for hort eriod (the run tie). The wet well ay alo be ized to enure that the u will not tart ore often than a ecified tie eriod (the cycle tie). Exale 5- (McGhee 99) A all ubdiviion roduce an average watewater flow of 0,000 L/day. The iniu flow i etiated to be 5,000 L/day and the axiu 40,000 L/day. Uing a -inute running tie and a 5- inute cycle tie, deterine the deign caacity of each of two u and the required wet well volue. 9

20 5 in 0.5(40,000 L / day ) x 65L 440 in/ day out V V 0.5out 0.5out A iniu deth need to be aintained over the u uction. With an intake velocity of about 0.6 /ec ( ft/ec), a ubergence of about 00 ( ft) i needed. It i alo coon to rovide additional freeboard (about 600, or ft) above the axiu water level. Thi i a uitable wet well for thi exale: McGhee 99, Figure 5-. The required volue i deterined by the -inute running tie requireent in thi exale and will therefore be larger than 56L. V 5 in The cycle tie will be hortet when in i 0.5 out. Therefore, for a iniu 5 inute cycle tie: in V (40,000 L / day 5,000 L / day ) 56.5L 440 in/ day For a -inute running tie: V out in V in Suberible ewage u intallation: tc tr t f McGhee 99, Figure 5-4 and 5-5 Wet it-dry it ewage uing tation: V V out in in The total cycle tie i therefore: tf The filling tie with the u off i: tr The u running tie i the working volue of the wet well divided by the net dicharge, which i the uing rate inu the inflow: Each u ut be able to deliver the eak flow of 40,000 L/day. 0

Solved problems 4 th exercise

Solved problems 4 th exercise Soled roblem th exercie Soled roblem.. On a circular conduit there are different diameter: diameter D = m change into D = m. The elocity in the entrance rofile wa meaured: = m -. Calculate the dicharge

More information

ρ water = 1000 kg/m 3 = 1.94 slugs/ft 3 γ water = 9810 N/m 3 = 62.4 lbs/ft 3

ρ water = 1000 kg/m 3 = 1.94 slugs/ft 3 γ water = 9810 N/m 3 = 62.4 lbs/ft 3 CEE 34 Aut 004 Midterm # Anwer all quetion. Some data that might be ueful are a follow: ρ water = 1000 kg/m 3 = 1.94 lug/ft 3 water = 9810 N/m 3 = 6.4 lb/ft 3 1 kw = 1000 N-m/ 1. (10) A 1-in. and a 4-in.

More information

SOLUTIONS FOR TUTORIAL QUESTIONS COURSE TEP Ideal motor speed = 1450rev/min x 12/120 = 145 rev/min

SOLUTIONS FOR TUTORIAL QUESTIONS COURSE TEP Ideal motor speed = 1450rev/min x 12/120 = 145 rev/min SOLUTIONS FOR TUTORIL QUESTIONS COURSE TE 49 ) Maxiu pup diplaceent i c /rev. Motor diplaceent i c /rev. a) t full pup diplaceent: Ideal otor peed 4rev/in x / 4 rev/in The lo in otor peed due to the voluetric

More information

gravity force buoyancy force drag force where p density of particle density of fluid A cross section perpendicular to the direction of motion

gravity force buoyancy force drag force where p density of particle density of fluid A cross section perpendicular to the direction of motion orce acting on the ettling article SEDIMENTATION gravity force boyancy force drag force In cae of floating: their i zero. f k V g Vg f A where denity of article denity of flid A cro ection erendiclar to

More information

8 Pages, 3 Figures, 2 Tables. Table S1: The reagents used for this study, their CAS registry numbers, their sources, and their stated purity levels.

8 Pages, 3 Figures, 2 Tables. Table S1: The reagents used for this study, their CAS registry numbers, their sources, and their stated purity levels. Suleentary Material for The Feaibility of Photoenitized Reaction with Secondary Organic Aerool Particle in the Preence of Volatile Organic Coound Kurti T. Malecha and Sergey A. Nizkorodov* * nizkorod@uci.edu

More information

MAE 101A. Homework 3 Solutions 2/5/2018

MAE 101A. Homework 3 Solutions 2/5/2018 MAE 101A Homework 3 Solution /5/018 Munon 3.6: What preure gradient along the treamline, /d, i required to accelerate water upward in a vertical pipe at a rate of 30 ft/? What i the anwer if the flow i

More information

Rubble Mound Breakwater Design Example Given: Design Conditions. γ = 17 kn/m 3 φ = 30 c = 0

Rubble Mound Breakwater Design Example Given: Design Conditions. γ = 17 kn/m 3 φ = 30 c = 0 Rubble Mound Breakwater Deign Exaple Given: Deign Condition Water depth: 5.5 Beach lope: 1:0 Deign high water: 1.7 Deign wave H H 1/10.5 T 8 ec L o 100 Allowable overtopping: 0.4 /ec/ Aror unit: rough

More information

Conservation of Energy

Conservation of Energy Add Iportant Conervation of Energy Page: 340 Note/Cue Here NGSS Standard: HS-PS3- Conervation of Energy MA Curriculu Fraework (006):.,.,.3 AP Phyic Learning Objective: 3.E.., 3.E.., 3.E..3, 3.E..4, 4.C..,

More information

Determination of Flow Resistance Coefficients Due to Shrubs and Woody Vegetation

Determination of Flow Resistance Coefficients Due to Shrubs and Woody Vegetation ERDC/CL CETN-VIII-3 December 000 Determination of Flow Reitance Coefficient Due to hrub and Woody Vegetation by Ronald R. Copeland PURPOE: The purpoe of thi Technical Note i to tranmit reult of an experimental

More information

1-D SEDIMENT NUMERICAL MODEL AND ITS APPLICATION. Weimin Wu 1 and Guolu Yang 2

1-D SEDIMENT NUMERICAL MODEL AND ITS APPLICATION. Weimin Wu 1 and Guolu Yang 2 U-CHINA WORKHOP ON ADVANCED COMPUTATIONAL MODELLING IN HYDROCIENCE & ENGINEERING epteber 9-, Oxford, Miiippi, UA -D EDIMENT NUMERICAL MODEL AND IT APPLICATION Weiin Wu and Guolu Yang ABTRACT A one dienional

More information

4 CAVITATION AND PRESSURE IMPULSES INDUCED BY THE PROPELLER**

4 CAVITATION AND PRESSURE IMPULSES INDUCED BY THE PROPELLER** 4 CAITATION AND PRESSURE IMPULSES INDUCED BY THE PROPELLER** The chief ai in thi ection i: to how how the cavitie on a roeller blade working in a wake change volue during one revolution to how how the

More information

Momentum. Momentum. Impulse. Impulse Momentum Theorem. Deriving Impulse. v a t. Momentum and Impulse. Impulse. v t

Momentum. Momentum. Impulse. Impulse Momentum Theorem. Deriving Impulse. v a t. Momentum and Impulse. Impulse. v t Moentu and Iule Moentu Moentu i what Newton called the quantity of otion of an object. lo called Ma in otion The unit for oentu are: = oentu = a = elocity kg Moentu Moentu i affected by a and elocity eeding

More information

Bernoulli s equation may be developed as a special form of the momentum or energy equation.

Bernoulli s equation may be developed as a special form of the momentum or energy equation. BERNOULLI S EQUATION Bernoulli equation may be developed a a pecial form of the momentum or energy equation. Here, we will develop it a pecial cae of momentum equation. Conider a teady incompreible flow

More information

15 N 5 N. Chapter 4 Forces and Newton s Laws of Motion. The net force on an object is the vector sum of all forces acting on that object.

15 N 5 N. Chapter 4 Forces and Newton s Laws of Motion. The net force on an object is the vector sum of all forces acting on that object. Chapter 4 orce and ewton Law of Motion Goal for Chapter 4 to undertand what i force to tudy and apply ewton irt Law to tudy and apply the concept of a and acceleration a coponent of ewton Second Law to

More information

Figure 1 Siemens PSSE Web Site

Figure 1 Siemens PSSE Web Site Stability Analyi of Dynamic Sytem. In the lat few lecture we have een how mall ignal Lalace domain model may be contructed of the dynamic erformance of ower ytem. The tability of uch ytem i a matter of

More information

Application of Newton s Laws. F fr

Application of Newton s Laws. F fr Application of ewton Law. A hocey puc on a frozen pond i given an initial peed of 0.0/. It lide 5 before coing to ret. Deterine the coefficient of inetic friction ( μ between the puc and ice. The total

More information

Homework #6. 1. Continuum wave equation. Show that for long wavelengths the equation of motion,, reduces to the continuum elastic wave equation dt

Homework #6. 1. Continuum wave equation. Show that for long wavelengths the equation of motion,, reduces to the continuum elastic wave equation dt Hoework #6 Continuu wave equation Show that for long wavelength the equation of otion, d u M C( u u u, reduce to the continuu elatic wave equation u u v t x where v i the velocity of ound For a, u u i

More information

s s 1 s = m s 2 = 0; Δt = 1.75s; a =? mi hr

s s 1 s = m s 2 = 0; Δt = 1.75s; a =? mi hr Flipping Phyic Lecture Note: Introduction to Acceleration with Priu Brake Slaing Exaple Proble a Δv a Δv v f v i & a t f t i Acceleration: & flip the guy and ultiply! Acceleration, jut like Diplaceent

More information

Fluids Lab 1 Lecture Notes

Fluids Lab 1 Lecture Notes Fluid Lab Lecture Note. Bernoulli Equation. Pitot-Static Tube 3. Aireed Meaurement 4. Preure Nondimenionalization Reference: Anderon 3.-3.5, Denker 3.4 ( htt://www.av8n.com/how/ ) Bernoulli Equation Definition

More information

Topic 7 Fuzzy expert systems: Fuzzy inference

Topic 7 Fuzzy expert systems: Fuzzy inference Topic 7 Fuzzy expert yte: Fuzzy inference adani fuzzy inference ugeno fuzzy inference Cae tudy uary Fuzzy inference The ot coonly ued fuzzy inference technique i the o-called adani ethod. In 975, Profeor

More information

ROOT LOCUS. Poles and Zeros

ROOT LOCUS. Poles and Zeros Automatic Control Sytem, 343 Deartment of Mechatronic Engineering, German Jordanian Univerity ROOT LOCUS The Root Locu i the ath of the root of the characteritic equation traced out in the - lane a a ytem

More information

MICRO-HYDRO INSTALLATION SIZING CALCULATIONS Jacques Chaurette eng. January 17, 2008

MICRO-HYDRO INSTALLATION SIZING CALCULATIONS Jacques Chaurette eng. January 17, 2008 MICRO-HYDRO INSTALLATION SIZING CALCULATIONS Jacque Chaurette eng. January 7, 008 Calculation for micro-hydro ine jet impact elocity are baed on the ame ort of calculation done for pump ytem, except there

More information

Physics 6A. Practice Midterm #2 solutions. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Physics 6A. Practice Midterm #2 solutions. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Phyic 6A Practice Midter # olution or apu Learning Aitance Service at USB . A locootive engine of a M i attached to 5 train car, each of a M. The engine produce a contant force that ove the train forward

More information

Name: Answer Key Date: Regents Physics. Energy

Name: Answer Key Date: Regents Physics. Energy Nae: Anwer Key Date: Regent Phyic Tet # 9 Review Energy 1. Ue GUESS ethod and indicate all vector direction.. Ter to know: work, power, energy, conervation of energy, work-energy theore, elatic potential

More information

Introduction to Laplace Transform Techniques in Circuit Analysis

Introduction to Laplace Transform Techniques in Circuit Analysis Unit 6 Introduction to Laplace Tranform Technique in Circuit Analyi In thi unit we conider the application of Laplace Tranform to circuit analyi. A relevant dicuion of the one-ided Laplace tranform i found

More information

ME 3560 Fluid Mechanics

ME 3560 Fluid Mechanics Sring 018 ME 3560 Fluid Mechanic Chater III. Elementary Fluid Dynamic The Bernoulli Equation 1 Sring 018 3.1 Newton Second Law A fluid article can exerience acceleration or deceleration a it move from

More information

Tuning of High-Power Antenna Resonances by Appropriately Reactive Sources

Tuning of High-Power Antenna Resonances by Appropriately Reactive Sources Senor and Simulation Note Note 50 Augut 005 Tuning of High-Power Antenna Reonance by Appropriately Reactive Source Carl E. Baum Univerity of New Mexico Department of Electrical and Computer Engineering

More information

THE BICYCLE RACE ALBERT SCHUELLER

THE BICYCLE RACE ALBERT SCHUELLER THE BICYCLE RACE ALBERT SCHUELLER. INTRODUCTION We will conider the ituation of a cyclit paing a refrehent tation in a bicycle race and the relative poition of the cyclit and her chaing upport car. The

More information

Lecture 10. Erbium-doped fiber amplifier (EDFA) Raman amplifiers Have replaced semiconductor optical amplifiers in the course

Lecture 10. Erbium-doped fiber amplifier (EDFA) Raman amplifiers Have replaced semiconductor optical amplifiers in the course ecture 1 Two tye of otical amlifier: Erbium-doed fiber amlifier (EDFA) Raman amlifier Have relaced emiconductor otical amlifier in the coure Fiber Otical Communication ecture 1, Slide 1 Benefit and requirement

More information

Related Rates section 3.9

Related Rates section 3.9 Related Rate ection 3.9 Iportant Note: In olving the related rate proble, the rate of change of a quantity i given and the rate of change of another quantity i aked for. You need to find a relationhip

More information

Social Studies 201 Notes for November 14, 2003

Social Studies 201 Notes for November 14, 2003 1 Social Studie 201 Note for November 14, 2003 Etimation of a mean, mall ample ize Section 8.4, p. 501. When a reearcher ha only a mall ample ize available, the central limit theorem doe not apply to the

More information

(British) (SI) British Metric L T [V] = L T. [a] = 2 [F] = F = 2 T

(British) (SI) British Metric L T [V] = L T. [a] = 2 [F] = F = 2 T Hydraulics ecture # CWR 40 age () ecture # Outline: Review of terminology in fluid mechanics: Energy or work Hydraulic head Bernoulli s aw, Conductivity (examle) ransient & turbulent Friction head loss

More information

PHYSICS 211 MIDTERM II 12 May 2004

PHYSICS 211 MIDTERM II 12 May 2004 PHYSIS IDTER II ay 004 Exa i cloed boo, cloed note. Ue only your forula heet. Write all wor and anwer in exa boolet. The bac of page will not be graded unle you o requet on the front of the page. Show

More information

18 Problem 1. 7 d Sketch a cross section of a switched reluctance machine and explain the principle of operation.

18 Problem 1. 7 d Sketch a cross section of a switched reluctance machine and explain the principle of operation. Exam Electrical Machine and Drive (ET4117) 9 November 01 from 14.00 to 17.00. Thi exam conit of 3 roblem on 3 age. Page 5 can be ued to anwer roblem 4 quetion a. The number before a quetion indicate how

More information

USE OF INTERNET TO DO EXPERIMENTS IN DYNAMICS AND CONTROL FROM ZACATECAS MEXICO IN THE LABORATORY OF THE UNIVERSITY OF TENNESSEE AT CHATANOOGAA.

USE OF INTERNET TO DO EXPERIMENTS IN DYNAMICS AND CONTROL FROM ZACATECAS MEXICO IN THE LABORATORY OF THE UNIVERSITY OF TENNESSEE AT CHATANOOGAA. USE OF INTERNET TO DO EXPERIMENTS IN DYNAMICS AND CONTROL FROM ZACATECAS MEXICO IN TE LABORATORY OF TE UNIVERSITY OF TENNESSEE AT CATANOOGAA. Jim enry *, Joé Alberto González Guerrero, Benito Serrano Roale..

More information

CIRCLE YOUR DIVISION: Div. 1 (9:30 am) Div. 2 (11:30 am) Div. 3 (2:30 pm) Prof. Ruan Prof. Naik Mr. Singh

CIRCLE YOUR DIVISION: Div. 1 (9:30 am) Div. 2 (11:30 am) Div. 3 (2:30 pm) Prof. Ruan Prof. Naik Mr. Singh Nae: CIRCLE YOUR DIVISION: Div. 1 (9:30 a) Div. (11:30 a) Div. 3 (:30 p) Prof. Ruan Prof. Nai Mr. Singh School of Mechanical Engineering Purdue Univerity ME315 Heat and Ma Tranfer Exa # edneday, October

More information

AP CHEM WKST KEY: Atomic Structure Unit Review p. 1

AP CHEM WKST KEY: Atomic Structure Unit Review p. 1 AP CHEM WKST KEY: Atoic Structure Unit Review p. 1 1) a) ΔE = 2.178 x 10 18 J 1 2 nf 1 n 2i = 2.178 x 10 18 1 1 J 2 2 6 2 = 4.840 x 10 19 J b) E = λ hc λ = E hc = (6.626 x 10 34 J )(2.9979 x 10 4.840 x

More information

Research on an Axial Magnetic-Field-Modulated Brushless Double Rotor Machine

Research on an Axial Magnetic-Field-Modulated Brushless Double Rotor Machine Energie 213, 6, 4799-4829; doi:1.339/en694799 Article OPEN ACCESS energie ISSN 1996-173 www.di.co/journal/energie Reearch on an Axial Magnetic-Field-Modulated Bruhle Double Rotor Machine Ping Zheng *,

More information

Practice Problems Solutions. 1. Frame the Problem - Sketch and label a diagram of the motion. Use the equation for acceleration.

Practice Problems Solutions. 1. Frame the Problem - Sketch and label a diagram of the motion. Use the equation for acceleration. Chapter 3 Motion in a Plane Practice Proble Solution Student Textbook page 80 1. Frae the Proble - Sketch and label a diagra of the otion. 40 v(/) 30 0 10 0 4 t () - The equation of otion apply to the

More information

CHAPTER 5. The Operational Amplifier 1

CHAPTER 5. The Operational Amplifier 1 EECE22 NETWORK ANALYSIS I Dr. Charle J. Kim Cla Note 9: Oerational Amlifier (OP Am) CHAPTER. The Oerational Amlifier A. INTRODUCTION. The oerational amlifier or o am for hort, i a eratile circuit building

More information

The Realm of Hydrogeology

The Realm of Hydrogeology The Real of Hydrogeology In class exercise Stagnant Flow Plot hydraulic head and ressure vs. deth for (also indicate the hydrostatic line) Stagnant flow (no flow) Steady downward flow Steady uward flow

More information

Lecture 2 Phys 798S Spring 2016 Steven Anlage. The heart and soul of superconductivity is the Meissner Effect. This feature uniquely distinguishes

Lecture 2 Phys 798S Spring 2016 Steven Anlage. The heart and soul of superconductivity is the Meissner Effect. This feature uniquely distinguishes ecture Phy 798S Spring 6 Steven Anlage The heart and oul of uperconductivity i the Meiner Effect. Thi feature uniquely ditinguihe uperconductivity fro any other tate of atter. Here we dicu oe iple phenoenological

More information

4.5 Evaporation and Diffusion Evaporation and Diffusion through Quiescent Air (page 286) bulk motion of air and j. y a,2, y j,2 or P a,2, P j,2

4.5 Evaporation and Diffusion Evaporation and Diffusion through Quiescent Air (page 286) bulk motion of air and j. y a,2, y j,2 or P a,2, P j,2 4.5 Evaporation and Diffuion 4.5.4 Evaporation and Diffuion through Quiecent Air (page 86) z bul otion of air and j z diffuion of air (a) diffuion of containant (j) y a,, y j, or P a,, P j, z 1 volatile

More information

3. Internal Flow General Concepts:

3. Internal Flow General Concepts: 3. Internal Flow General Concet: ρ u u 4 & Re Re, cr 2300 μ ν π μ Re < 2300 lainar 2300 < Re < 4000 tranitional Flow Regie : Re > 4000 turbulent Re > 10,000 fully turbulent (d) 1 (e) Figure 1 Boundary

More information

3.185 Problem Set 6. Radiation, Intro to Fluid Flow. Solutions

3.185 Problem Set 6. Radiation, Intro to Fluid Flow. Solutions 3.85 Proble Set 6 Radiation, Intro to Fluid Flow Solution. Radiation in Zirconia Phyical Vapor Depoition (5 (a To calculate thi viewfactor, we ll let S be the liquid zicronia dic and S the inner urface

More information

Example 1: Example 1: Example 2: a.) the elevator is at rest. Example 2: Example 2: c.) the elevator accelerates downward at 1.

Example 1: Example 1: Example 2: a.) the elevator is at rest. Example 2: Example 2: c.) the elevator accelerates downward at 1. Exaple 1: 60 kg, v 1 100 N (wet), v 2 220 N (eat), a? Exaple 1: wo force parallel to the ground act upon a box with a a of 60 kg. One force i directed wet and ha a trength of 100 N. he other force i directed

More information

WATER DISTRIBUTION NETWORKS

WATER DISTRIBUTION NETWORKS WATER DISTRIBUTION NETWORKS CE 370 1 Components of Water Supply System 2 1 Water Distribution System Water distribution systems are designed to adequately satisfy the water requirements for a combinations

More information

Chapter 10: Flow Flow in in Conduits Conduits Dr Ali Jawarneh

Chapter 10: Flow Flow in in Conduits Conduits Dr Ali Jawarneh Chater 10: Flow in Conduits By Dr Ali Jawarneh Hashemite University 1 Outline In this chater we will: Analyse the shear stress distribution across a ie section. Discuss and analyse the case of laminar

More information

DIFFERENTIAL EQUATIONS

DIFFERENTIAL EQUATIONS Matheatic Reviion Guide Introduction to Differential Equation Page of Author: Mark Kudlowki MK HOME TUITION Matheatic Reviion Guide Level: A-Level Year DIFFERENTIAL EQUATIONS Verion : Date: 3-4-3 Matheatic

More information

Physics 30 Lesson 1 Momentum and Conservation of Momentum in One Dimension

Physics 30 Lesson 1 Momentum and Conservation of Momentum in One Dimension Phyic 30 Leon 1 Moentu and Conervation of Moentu in One Dienion I. Phyic rincile Student often ak e if Phyic 30 i harder than Phyic 0. Thi, of coure, deend on the atitude, attitude and work ethic of the

More information

Rupro, breach model used by Cemagref during Impact project

Rupro, breach model used by Cemagref during Impact project PAQUIER 1 Rupro, breach model ued by Cemagref during Impact project A PAQUIER Cemagref, France andre.paquier@cemagref.fr SUMMARY For embankment dam, piping and overtopping failure are the mot frequent

More information

_10_EE394J_2_Spring12_Inertia_Calculation.doc. Procedure for Estimating Grid Inertia H from Frequency Droop Measurements

_10_EE394J_2_Spring12_Inertia_Calculation.doc. Procedure for Estimating Grid Inertia H from Frequency Droop Measurements Procedure or Etiating Grid Inertia ro Frequency Droop Meaureent While the exion or inertia and requency droop are well known, it i prudent to rederive the here. Treating all the grid generator a one large

More information

Practice Problem Solutions. Identify the Goal The acceleration of the object Variables and Constants Known Implied Unknown m = 4.

Practice Problem Solutions. Identify the Goal The acceleration of the object Variables and Constants Known Implied Unknown m = 4. Chapter 5 Newton Law Practice Proble Solution Student Textbook page 163 1. Frae the Proble - Draw a free body diagra of the proble. - The downward force of gravity i balanced by the upward noral force.

More information

Theoretical Research on the Calculation of Bending Moment at Arc Vault in Steel Corrugated Plate

Theoretical Research on the Calculation of Bending Moment at Arc Vault in Steel Corrugated Plate Theoretical Reearch on the Calculation of Bending oment at Arc Vault in Steel Corrugated Plate Baijian Li, Liangheng Zhu, Xinha Fu, Xu Luo. School of Civil Engineering,Tranortation, South China Univerity

More information

Sound Wave as a Particular Case of the Gravitational Wave

Sound Wave as a Particular Case of the Gravitational Wave Oen Journal of Acoutic, 1,, 115-1 htt://dx.doi.org/1.436/oja.1.313 Publihed Online Seteber 1 (htt://www.scirp.org/journal/oja) Sound Wave a a Particular ae of the Gravitational Wave Vladiir G. Kirtkhalia

More information

Christian Linde Olsen Griffith University, Faculty of Engineering and Information Technology, Gold Coast Campus.

Christian Linde Olsen Griffith University, Faculty of Engineering and Information Technology, Gold Coast Campus. 1 Abtract Rubble Mound Breakwater Chritian Linde Olen Griffith Univerity, Faculty of Engineering and Information Technology, Gold Coat Campu. 1. Abtract The paper deal with the deign of a rubble mound

More information

FUNDAMENTALS OF POWER SYSTEMS

FUNDAMENTALS OF POWER SYSTEMS 1 FUNDAMENTALS OF POWER SYSTEMS 1 Chapter FUNDAMENTALS OF POWER SYSTEMS INTRODUCTION The three baic element of electrical engineering are reitor, inductor and capacitor. The reitor conume ohmic or diipative

More information

Physics 6A. Practice Midterm #2 solutions

Physics 6A. Practice Midterm #2 solutions Phyic 6A Practice Midter # olution 1. A locootive engine of a M i attached to 5 train car, each of a M. The engine produce a contant force that ove the train forward at acceleration a. If 3 of the car

More information

Practice Midterm #1 Solutions. Physics 6A

Practice Midterm #1 Solutions. Physics 6A Practice Midter # Solution Phyic 6A . You drie your car at a peed of 4 k/ for hour, then low down to k/ for the next k. How far did you drie, and what wa your aerage peed? We can draw a iple diagra with

More information

TOTAL HEAD, N.P.S.H. AND OTHER CALCULATION EXAMPLES Jacques Chaurette p. eng., June 2003

TOTAL HEAD, N.P.S.H. AND OTHER CALCULATION EXAMPLES Jacques Chaurette p. eng.,   June 2003 TOTAL HEAD, N.P.S.H. AND OTHER CALCULATION EXAMPLES Jacques Chaurette p. eng., www.lightmypump.com June 2003 Figure 1 Calculation example flow schematic. Situation Water at 150 F is to be pumped from a

More information

Determination of Flow Resistance Coefficients Due to Shrubs and Woody Vegetation

Determination of Flow Resistance Coefficients Due to Shrubs and Woody Vegetation December 000 Determination of Flow Reitance Coefficient Due to hrub and Woody Vegetation by Ronald R. Copeland PURPOE: The purpoe of thi Technical Note i to tranmit reult of an experimental invetigation

More information

RADIATION THERMOMETRY OF METAL IN HIGH TEMPERATURE FURNACE

RADIATION THERMOMETRY OF METAL IN HIGH TEMPERATURE FURNACE XVII IMEKO World Congre Metrology in the 3rd Millennium June 22 27, 2003, Dubrovnik, Croatia RADIATION THERMOMETRY OF METAL IN HIGH TEMPERATURE FURNACE Tohru Iuchi, Tohru Furukawa and Nobuharu Sato Deartment

More information

CE 6403 APPLIED HYDRAULIC ENGINEERING UNIT - V PUMPS

CE 6403 APPLIED HYDRAULIC ENGINEERING UNIT - V PUMPS CE 6403 APPLIED HYDRAULIC ENGINEERING UNIT - V PUMPS Centrifugal pups - Miniu speed to start the pup - NPSH - Cavitations in pups Operating characteristics - Multistage pups - Reciprocating pups - Negative

More information

PHY 171 Practice Test 3 Solutions Fall 2013

PHY 171 Practice Test 3 Solutions Fall 2013 PHY 171 Practice et 3 Solution Fall 013 Q1: [4] In a rare eparatene, And a peculiar quietne, hing One and hing wo Lie at ret, relative to the ground And their wacky hairdo. If hing One freeze in Oxford,

More information

Second Law of Motion. Force mass. Increasing mass. (Neglect air resistance in this example)

Second Law of Motion. Force mass. Increasing mass. (Neglect air resistance in this example) Newton Law of Motion Moentu and Energy Chapter -3 Second Law of Motion The acceleration of an object i directly proportional to the net force acting on the object, i in the direction of the net force,

More information

Chapter 7 Energy Principle

Chapter 7 Energy Principle Chater 7: Energy Princile By Dr Ali Jawarneh Hashemite University Outline In this chater we will: Derive and analyse the Energy equation. Analyse the flow and shaft work. Derive the equation for steady

More information

NCAAPMT Calculus Challenge Challenge #3 Due: October 26, 2011

NCAAPMT Calculus Challenge Challenge #3 Due: October 26, 2011 NCAAPMT Calculu Challenge 011 01 Challenge #3 Due: October 6, 011 A Model of Traffic Flow Everyone ha at ome time been on a multi-lane highway and encountered road contruction that required the traffic

More information

4 Conservation of Momentum

4 Conservation of Momentum hapter 4 oneration of oentu 4 oneration of oentu A coon itake inoling coneration of oentu crop up in the cae of totally inelatic colliion of two object, the kind of colliion in which the two colliding

More information

Social Studies 201 Notes for March 18, 2005

Social Studies 201 Notes for March 18, 2005 1 Social Studie 201 Note for March 18, 2005 Etimation of a mean, mall ample ize Section 8.4, p. 501. When a reearcher ha only a mall ample ize available, the central limit theorem doe not apply to the

More information

12.4 Atomic Absorption and Emission Spectra

12.4 Atomic Absorption and Emission Spectra Phyic Tool box 12.4 Atoic Abortion and iion Sectra A continuou ectru given off by a heated olid i caued by the interaction between neighbouring ato or olecule. An eiion ectru or line ectru i eitted fro

More information

= s = 3.33 s s. 0.3 π 4.6 m = rev = π 4.4 m. (3.69 m/s)2 = = s = π 4.8 m. (5.53 m/s)2 = 5.

= s = 3.33 s s. 0.3 π 4.6 m = rev = π 4.4 m. (3.69 m/s)2 = = s = π 4.8 m. (5.53 m/s)2 = 5. Seat: PHYS 500 (Fall 0) Exa #, V 5 pt. Fro book Mult Choice 8.6 A tudent lie on a very light, rigid board with a cale under each end. Her feet are directly over one cale and her body i poitioned a hown.

More information

( ) Zp THE VIBRATION ABSORBER. Preamble - A NEED arises: lbf in. sec. X p () t = Z p. cos Ω t. Z p () r. ω np. F o. cos Ω t. X p. δ s.

( ) Zp THE VIBRATION ABSORBER. Preamble - A NEED arises: lbf in. sec. X p () t = Z p. cos Ω t. Z p () r. ω np. F o. cos Ω t. X p. δ s. THE VIBRATION ABSORBER Preable - A NEED arie: Lui San Andre (c) 8 MEEN 363-617 Conider the periodic forced repone of a yte (Kp-Mp) defined by : 1 1 5 lbf in : 1 3 lb (t) It natural frequency i: : ec F(t)

More information

Physics 20 Lesson 28 Simple Harmonic Motion Dynamics & Energy

Physics 20 Lesson 28 Simple Harmonic Motion Dynamics & Energy Phyic 0 Leon 8 Siple Haronic Motion Dynaic & Energy Now that we hae learned about work and the Law of Coneration of Energy, we are able to look at how thee can be applied to the ae phenoena. In general,

More information

S E V E N. Steady-State Errors SOLUTIONS TO CASE STUDIES CHALLENGES

S E V E N. Steady-State Errors SOLUTIONS TO CASE STUDIES CHALLENGES S E V E N Steady-State Error SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Steady-State Error Deign via Gain 76.39 a. G( (50)(.3). Syte i Type. Step input: e( ) 0; Rap input: e( ) v 76.39.59 ;

More information

BASIC INDUCTION MOTOR CONCEPTS

BASIC INDUCTION MOTOR CONCEPTS INDUCTION MOTOS An induction motor ha the ame phyical tator a a ynchronou machine, with a different rotor contruction. There are two different type of induction motor rotor which can be placed inide the

More information

1.1 Speed and Velocity in One and Two Dimensions

1.1 Speed and Velocity in One and Two Dimensions 1.1 Speed and Velocity in One and Two Dienion The tudy of otion i called kineatic. Phyic Tool box Scalar quantity ha agnitude but no direction,. Vector ha both agnitude and direction,. Aerage peed i total

More information

Automatic Control Systems. Part III: Root Locus Technique

Automatic Control Systems. Part III: Root Locus Technique www.pdhcenter.com PDH Coure E40 www.pdhonline.org Automatic Control Sytem Part III: Root Locu Technique By Shih-Min Hu, Ph.D., P.E. Page of 30 www.pdhcenter.com PDH Coure E40 www.pdhonline.org VI. Root

More information

PHYSICS 151 Notes for Online Lecture 2.3

PHYSICS 151 Notes for Online Lecture 2.3 PHYSICS 151 Note for Online Lecture.3 riction: The baic fact of acrocopic (everda) friction are: 1) rictional force depend on the two aterial that are liding pat each other. bo liding over a waed floor

More information

Useful concepts associated with the Bernoulli equation. Dynamic

Useful concepts associated with the Bernoulli equation. Dynamic Useful concets associated with the Bernoulli equation - Static, Stagnation, and Dynamic Pressures Bernoulli eq. along a streamline + ρ v + γ z = constant (Unit of Pressure Static (Thermodynamic Dynamic

More information

Lecture 2 DATA ENVELOPMENT ANALYSIS - II

Lecture 2 DATA ENVELOPMENT ANALYSIS - II Lecture DATA ENVELOPMENT ANALYSIS - II Learning objective To eplain Data Envelopent Anali for ultiple input and ultiple output cae in the for of linear prograing .5 DEA: Multiple input, ultiple output

More information

Analysis the Transient Process of Wind Power Resources when there are Voltage Sags in Distribution Grid

Analysis the Transient Process of Wind Power Resources when there are Voltage Sags in Distribution Grid Analyi the Tranient Proce of Wind Power Reource when there are Voltage Sag in Ditribution Grid Do Nhu Y 1,* 1 Hanoi Univerity of ining and Geology, Deartment of Electrification, Electromechanic Faculty,

More information

DEVELOPMENT OF A STRUCTURED THERMOCLINE THERMAL ENERGY STORAGE SYSTEM

DEVELOPMENT OF A STRUCTURED THERMOCLINE THERMAL ENERGY STORAGE SYSTEM DEVELOPMENT OF A STRUCTURED THERMOCLINE THERMAL ENERGY STORAGE SYSTEM Brad M. Brown Matt N. Straer R. Paneer Selvam Univerity of Arkana Department of Civil Engineering 4190 Bell Engineering Center Fayetteville,

More information

two equations that govern the motion of the fluid through some medium, like a pipe. These two equations are the

two equations that govern the motion of the fluid through some medium, like a pipe. These two equations are the Fluid and Fluid Mechanic Fluid in motion Dynamic Equation of Continuity After having worked on fluid at ret we turn to a moving fluid To decribe a moving fluid we develop two equation that govern the motion

More information

5. Dimensional Analysis. 5.1 Dimensions and units

5. Dimensional Analysis. 5.1 Dimensions and units 5. Diensional Analysis In engineering the alication of fluid echanics in designs ake uch of the use of eirical results fro a lot of exerients. This data is often difficult to resent in a readable for.

More information

AP Physics Momentum AP Wrapup

AP Physics Momentum AP Wrapup AP Phyic Moentu AP Wrapup There are two, and only two, equation that you get to play with: p Thi i the equation or oentu. J Ft p Thi i the equation or ipule. The equation heet ue, or oe reaon, the ybol

More information

The Influence of the Load Condition upon the Radial Distribution of Electromagnetic Vibration and Noise in a Three-Phase Squirrel-Cage Induction Motor

The Influence of the Load Condition upon the Radial Distribution of Electromagnetic Vibration and Noise in a Three-Phase Squirrel-Cage Induction Motor The Influence of the Load Condition upon the Radial Ditribution of Electromagnetic Vibration and Noie in a Three-Phae Squirrel-Cage Induction Motor Yuta Sato 1, Iao Hirotuka 1, Kazuo Tuboi 1, Maanori Nakamura

More information

ME 322 Worksheet Winter 2007 Introduction to Compressible Flow

ME 322 Worksheet Winter 2007 Introduction to Compressible Flow ME 3 Workheet Winter 007 Introduction to Compreible Flow 1. A two-liter cylindrical tank, 10 cm in diameter, ha a piton that fit perfectly. The piton doe not leak, and there i no friction between the piton

More information

ME 315 Exam 3 8:00-9:00 PM Thursday, April 16, 2009 CIRCLE YOUR DIVISION

ME 315 Exam 3 8:00-9:00 PM Thursday, April 16, 2009 CIRCLE YOUR DIVISION ME 315 Exam 3 8:00-9:00 PM Thurday, Aril 16, 009 Thi i a cloed-book, cloed-note examination. There i a formula heet at the back. You mut turn off all communication device before tarting thi exam, and leave

More information

Pumping Stations Design For Infrastructure Master Program Engineering Faculty-IUG

Pumping Stations Design For Infrastructure Master Program Engineering Faculty-IUG umping Stations Design For Infrastructure Master rogram Engineering Faculty-IUG Lecture : umping Hydraulics Dr. Fahid Rabah Water and environment Engineering frabah@iugaza.edu The main items that will

More information

μ + = σ = D 4 σ = D 3 σ = σ = All units in parts (a) and (b) are in V. (1) x chart: Center = μ = 0.75 UCL =

μ + = σ = D 4 σ = D 3 σ = σ = All units in parts (a) and (b) are in V. (1) x chart: Center = μ = 0.75 UCL = Our online Tutor are available 4*7 to provide Help with Proce control ytem Homework/Aignment or a long term Graduate/Undergraduate Proce control ytem Project. Our Tutor being experienced and proficient

More information

Chapter 13. Root Locus Introduction

Chapter 13. Root Locus Introduction Chapter 13 Root Locu 13.1 Introduction In the previou chapter we had a glimpe of controller deign iue through ome imple example. Obviouly when we have higher order ytem, uch imple deign technique will

More information

Physics 30 Lesson 3 Impulse and Change in Momentum

Physics 30 Lesson 3 Impulse and Change in Momentum Phyic 30 Leon 3 Ipule and Change in Moentu I. Ipule and change in oentu According to Newton nd Law of Motion (Phyic Principle 1 on the Data Sheet), to change the otion (i.e. oentu) of an object an unbalanced

More information

CHAPTER 4 COMPARISON OF PUSH-OUT TEST RESULTS WITH EXISTING STRENGTH PREDICTION METHODS

CHAPTER 4 COMPARISON OF PUSH-OUT TEST RESULTS WITH EXISTING STRENGTH PREDICTION METHODS CHAPTER 4 COMPARISON OF PUSH-OUT TEST RESULTS WITH EXISTING STRENGTH PREDICTION METHODS 4.1 General Several tud trength rediction method have been develoed ince the 1970. Three o thee method are art o

More information

Lecture 17: Frequency Response of Amplifiers

Lecture 17: Frequency Response of Amplifiers ecture 7: Frequency epone of Aplifier Gu-Yeon Wei Diiion of Engineering and Applied Science Harard Unierity guyeon@eec.harard.edu Wei Oeriew eading S&S: Chapter 7 Ski ection ince otly decribed uing BJT

More information

Simulation of Wound Rotor Synchronous Machine under Voltage Sags

Simulation of Wound Rotor Synchronous Machine under Voltage Sags Simulation of Wound Rotor Synchronou Machine under oltage Sag D. Aguilar, G. azquez, Student Member, IEEE, A. Rolan, Student Member, IEEE, J. Rocabert, Student Member, IEEE, F. Córcole, Member, IEEE, and

More information

24P 2, where W (measuring tape weight per meter) = 0.32 N m

24P 2, where W (measuring tape weight per meter) = 0.32 N m Ue of a 1W Laer to Verify the Speed of Light David M Verillion PHYS 375 North Carolina Agricultural and Technical State Univerity February 3, 2018 Abtract The lab wa et up to verify the accepted value

More information

SERIES COMPENSATION: VOLTAGE COMPENSATION USING DVR (Lectures 41-48)

SERIES COMPENSATION: VOLTAGE COMPENSATION USING DVR (Lectures 41-48) Chapter 5 SERIES COMPENSATION: VOLTAGE COMPENSATION USING DVR (Lecture 41-48) 5.1 Introduction Power ytem hould enure good quality of electric power upply, which mean voltage and current waveform hould

More information

ME 375 FINAL EXAM Wednesday, May 6, 2009

ME 375 FINAL EXAM Wednesday, May 6, 2009 ME 375 FINAL EXAM Wedneday, May 6, 9 Diviion Meckl :3 / Adam :3 (circle one) Name_ Intruction () Thi i a cloed book examination, but you are allowed three ingle-ided 8.5 crib heet. A calculator i NOT allowed.

More information

Numerical Investigation of Heat Transfer and Fluid Flow around the Rectangular Flat Plane Confined by a Cylinder under Pulsating Flow

Numerical Investigation of Heat Transfer and Fluid Flow around the Rectangular Flat Plane Confined by a Cylinder under Pulsating Flow Journal of Alied Fluid Mechanic, Vol. 9, No.,. 1569-1577, 216. Available online at www.jafmonline.net, ISSN 1735-3572, EISSN 1735-365. DOI: 1.18869/acadub.jafm.68.235.21 Numerical Invetigation of Heat

More information

Section J8b: FET Low Frequency Response

Section J8b: FET Low Frequency Response ection J8b: FET ow Frequency epone In thi ection of our tudie, we re o to reiit the baic FET aplifier confiuration but with an additional twit The baic confiuration are the ae a we etiated ection J6 of

More information