A fluid machine is a device either for converting the energy held by a fluid into mechanical energy or vice versa.

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1 FLUID MACHINE A fluid machine is a device either for converting the energy held by a fluid into mechanical energy or vice versa. Fluid machine may be divided into two groups; 1. Positive displacement group * Reciprocating pump, etc Part one : Introduction of Pump 1

2 . Rotodynamic group * Pelton wheel, etc Depend on energy movement; fluid machine could be divided into three categories 1. Pump. Turbine 3. Jack Part one : Introduction of Pump

3 PUMP INTRODUCTION Rotodynamic pump is essentially a turbine in reverse ; which mean that mechanical energy is transferred from the rotor to the fluid. It is classified according to the direction of the fluid path through them. 1. Radial / centrifugal flow Part one : Introduction of Pump 3

4 . Axial flow 3. Mixed-flow type In general usage, the word PUMP is applied to a machine dealing with a liquid. A machine in which the working fluid is a gas is more usually termed as fan, blower or compressor. Part one : Introduction of Pump 4

5 HEAD OF PUMP Part one : Introduction of Pump 5

6 CENTRIFUGAL PUMP This type of pumps is the converse of the radial-flow (Francis) turbine. Whereas the flow in the turbine in inwards, the flow in the pumps is outwards. The rotor (impeller) rotates inside a spiral casing. The inlet pipe is axial, and fluid enters the eye, that is the center of the impeller with little, if any, whirl component of velocity. Part two : Centrifugal Pump 1

7 From there it flows outwards in the direction of the blades, and having received energy from the impeller, is discharged with increased pressure and velocity into the casing. It then has a considerable tangential (whirl) component of velocity which is normally much greater than that required in the discharge pipe. The kinetic energy of the fluid leaving the impeller is largely dissipated in shock losses unless arrangements are made to reduce the velocity gradually. Part two : Centrifugal Pump

8 Velocity triangle Inlet ; Tangential velocity of impeller U = ω 1 r 1 Absolute velocity vector at α 1 to tangent V 1 Relative velocity to impeller blades V r 1 = V1 U1 Components velocity of V1 V w1 : whirl velocity V f 1: radial flow velocity Inlet blade angle β 1 Part two : Centrifugal Pump 3

9 Outlet ; Tangential velocity of impeller U = ω r Absolute velocity vector at α to tangent V Relative velocity to impeller blades = V r V U Components velocity of V V w : whirl velocity V f : radial flow velocity Inlet blade angle β Part two : Centrifugal Pump 4

10 Velocity triangle for centrifugal pump: Part two : Centrifugal Pump 5

11 Calculation is done base on Euler s Turbine Equation. The one-dimensional theory simplifies the problem very considerably by making the following assumptions: 1. The blades are infinitely thin and the pressure difference across them is replaced by imaginary body forces acting on the fluid and producing torque.. The number of blades in infinitely large. Thus, v = 0 θ 3. No variation of velocity in the meridional plane (z-axis). Thus, In reality, v = f ( r, θ, z) Part two : Centrifugal Pump 6

12 Torque = Rate of change of angular momentum Angular momentum = (Mass) x (Tangential velocity) x (Radius) Specific energy, Y = ge = P m& (unit : J/kg) Euler s Head ; H E 1 = g ( v u v u ) w w1 1 (unit : m) Part two : Centrifugal Pump 7

13 Relation of u, v w and H E H E = 1 g ( v u cosα v u α ) 1 1 cos 1 α = 90 o 1 = w1 0 v = v v and 1 f H E = v u g w Part two : Centrifugal Pump 8

14 relation of β and HE from ; H E v u g w = = C1 + CQ 1 tan β Euler s head is depends on the value of β Part two : Centrifugal Pump 9

15 velocity triangle and the position of blades Blade condition with Euler s head value. β = 90 o has the highest Part two : Centrifugal Pump 10

16 Relation of β and with Bernoulli equation. H E Euler s head : g V H H H H w P V P E + = + = Reaction degree of pump = + = + = tan β U V gh V H H f E w P E Part two : Centrifugal Pump 11

17 LOSSES IN PUMP 3 major types of losses 1. Losses of hydraulic power a. Circulatory flow b. Friction c. Shocking in impeller. Loss of volume 3. Loss of mechanical energy Part three : Losses and Efficiency of Pump 1

18 a. Circulatory Flow SF : Slip Factor SF = V V w w = H H ideal actual = H E Part three : Losses and Efficiency of Pump

19 b. Friction losses h f = k 1 Q h f : Friction losses k 1 : Constant Q : Flow rate c. Shock losses h sh = k ( Q Q ) o k : Shock losses Q : Designed flow rate Q o : Actual flow rate Part three : Losses and Efficiency of Pump 3

20 EFFICIENCY OF PUMP Overall Efficiency : = o gqh P i m Mechanical Efficiency : mech = ( V U V U ) 1 g( Q + Q) g w w1 1 P i Manometric Efficiency : mano = V gh m wu Vw 1U 1 Volumetric Efficiency : Q v = Q + Q Part three : Losses and Efficiency of Pump 4

21 Part four : Reaction Turbine Francis Turbine 6

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36 TURBINE Turbine is a prime mover to subtract energy from fluid. Energy from water will be changed to mechanical energy. Turbines are subdivided into impulse and reaction machines. In impulse turbines, the total head available is first converted into the kinetic energy. The fluid energy which is reduced on passing through the runner in entirely kinetic, it follows that the absolute velocity at outlet is smaller than the absolute velocity at inlet (jet velocity). The fluid pressure is atmospheric throughout and the velocity is constant except for a slight reduction due to friction. Example : Pelton wheel Part four : Reaction Turbine Francis Turbine 1

37 In the reaction turbines, the fluid passes first through a ring of stationary guide vanes in which only part of the available total head is converted into kinetic energy. The guide vanes discharge directly into the runner along the whole of its periphery, so that the fluid entering the runner has pressure energy as well as kinetic energy. The pressure energy is converted into kinetic energy in the runner. Therefore, the relative velocity is not constant but increases through the runner. There is, therefore, a pressure difference across the runner. Example : Francis turbine Part four : Reaction Turbine Francis Turbine

38 FRANCIS TURBINE Main parts : 1. spiral. guide vanes 3. runner / impeller 4. draft tube Part four : Reaction Turbine Francis Turbine 3

39 Net head : Pi Vi H input = H = + + ρ g g Z i Part four : Reaction Turbine Francis Turbine 4

40 Velocity triangle : Part four : Reaction Turbine Francis Turbine 5

41 Part four : Reaction Turbine Francis Turbine 6

42 Euler s head : H E = V w 1U g 1 Hydraulic efficiency : η h = V w 1U 1 V gh w U Mechanical efficiency : η mech = ρ gq [ 1 ( V U V U )] g w1 P o 1 w Overall efficiency : η = o P o ρgqh Part four : Reaction Turbine Francis Turbine 7

43 Main parts : PELTON WHEEL 1. jet nozzle. runner / impeller 3. bucket Part four : Impulse Turbine Pelton Wheel 1

44 Velocity triangle : at inlet : V r 1 V1 U1 V w = = and 1 1 V at outlet : V = k V r r1 V w = U k Vr1 cos(180 ) Power : P = ρqu( V 1 U )[1 + k cos(180 )] Part four : Impulse Turbine Pelton Wheel

45 EFFICIENCY OF PELTON WHEEL 1. hydraulic efficiency η h = U ( V1 U )[1 + k cos(180 )] V 1 velocity at maximum hydraulic efficiency : U = V 1 Part four : Impulse Turbine Pelton Wheel 3

46 maximum hydraulic efficiency : 1+ k cos(180 ) ηh = power at max hydraulic efficiency : P max V 1 = ρq [1 + 4 k cos(180 )] Part four : Impulse Turbine Pelton Wheel 4

47 . mechanical efficiency η mech = P ρqu( V 1 U )[1 + k cos(180 )] 3. volumetric efficiency η v = Q a Q 4. overall efficiency η = o P ρgqh Part four : Impulse Turbine Pelton Wheel 5

48 Hydraulic efficiency : η h = V w 1U 1 V gh w U Mechanical efficiency : η mech = ρ gq [ 1 ( V U V U )] g w1 P o 1 w Overall efficiency : η = o P o ρgqh Part four : Impulse Turbine Pelton Wheel 6

49 DEGREE OF REACTION (ºR) / DARJAH TINDAKBALAS (ºR) Darjah tindakbalas ini menyatakan kejatuhan tekanan statik terhadap jumlah tenaga yang dipindahkan oleh penggerak. Pada turbin, tindakbalas hanyalah sebahagian sahaja tenaga tekanan yang ditukarkan kepada tenaga kinetik, dan tenaga yang selebihnya masih kekal dalam bentuk tenaga tekanan. Tenaga tekanan diubah menjadi tenaga kinetik semasa air melalui bilah. Persamaan Bernoulli diantara bahagian masuk dan bahagian keluar di bilah, boleh dinyatakan seperti berikut: P1 V1 + ρg g H H E L P V = + + H ρg g :Turus Euler (m) : Kehilangan oleh penggerak (m) E + H L (i) Jika kehilangan oleh penggerak boleh diabaikan, maka H L = 0. Turus Euler (Euler head, H E ) boleh dinyatakan seperti berikut: P1 P V1 V H E = + ρg g (ii) di mana, P 1 P = Kejatuhan tekanan statik merentasi penggerak ρg V V g 1 = Kejatuhan turus halaju merentasi penggerak Persamaan (ii) di atas jelas menunjukkan bahawa turus Euler yang dihasilkan oleh penggerak bergantung kepada tenaga tekanan dan tenaga kinetik.

50 Darjah tindakbalas dinyatakan sebagai: P1 P Kejatuhan tekanan statik ρg R = = Jumlah tenaga yang dipindahkan (iii) H E Jika tekanan di masukkan adalah sama dengan tekanan di keluaran, P1 = P. Oleh itu, darjah tindakbalas menjadi sifar. Ini bermakna turbin tersebut ialah turbin dedenyut (impulse turbine Pelton wheel). Daripada persamaan (ii), jika tiada kehilangan, bermakna V1 = V. Oleh itu turus Euler akan menjadi: P1 P H E = (iv) ρ g Masukkan persamaan (iv) ke dalam persamaan (iii), didapati nilai darjah tindakbalas menjadi satu (100%). Ini bermakna turbin tersebut adalah turbin tindakbalas sepenuhnya (reaction turbine Francis turbine). Daripada persamaan (ii), P1 P V1 V = H E ρg g (v) Masukkan persamaan (v) ke dalam persamaan (iii). Darjah tindakbalas boleh dinyatakan sebagai: V1 V H E g R = (vi) H E

51 Diketahui bahawa, H = ( u v u v ) E 1 g Dengan v w = 0 1 w1 w Maka, persamaan (vi) boleh diringkaskan menjadi V1 V R = 1 u 1v w 1

52 TUTORIAL FOR PUMP AND TURBINE QUESTION 1 (a) Dengan menggunakan lakaran yang jelas, dapatkan satu ungkapan untuk turus manometer Hm sebuah pam rotodinamik yang mengepam air merentasi satu perbezaan tekanan (p h -p s )/ρg dengan p h dan p s masing-masing adalah turus tekanan di paip hantar dan paip sedut. (b) Sebuah pam empar dipacu oleh sebuah motor elektrik pada kelajuan 1450 pusing per minit, ppm. Diameter, lebar bilah dan sudut bilah pendesak di bahagian keluar masing-masing ialah 600mm, 400mm dan 30º. Diameter, lebar bilah dan sudut bilah dibahagian masuk ialah 300mm, 80mm dan 0º. Jangka tekanan yang dipasang searas pada paip hantar dan paip sedut masing-masing menunjukkan bacaan positif 13.5 bar dan negative 0.5 bar. Dengan menganggap diameter paip sedut dan paip hantar adalah sama besar, tentukan : QUESTION i. Turus manometer H m ii. Kecekapan manometric η mano iii. Kuasa yang perlu dibekalkan oleh motor jika kecekapan mekanikal ialah 98%. (a) Terangkan apa yang dimaksudkan : i. Turus manometric pam ii. Turus Euler pam iii. Kecekapan manometric pam iv. Kecekapan keseluruhan pam (b) Pam empar dengan diameter pendesak di bahagian masukan adalah 30cm dan dikeluaran 60cm. Tebal pendesak ialah 1cm. Tebal bilah menempati 10% daripada luasan aliran dikeluaran. Bilah melengkung ke belakang dengan sudut bilah dimasukan ialah 30º dan dikeluaran ialah 40º diukur terhadap tangent. Kadar aliran air yang dipam adalah 0.5m 3 /s. Dengan menganggap bahawa air masuk ke pendesak tanpa berpusar dan halaju aliran adalah malar, kirakan : i. Kelajuan pam dalam putaran per minit, ppm. ii. Kuasa keluaran P o jika kecekapan manometric adalah 85%. iii. Beza tekanan merentasi pendesak dalam unit bar.

53 QUESTION 3 (a) Biasanya saiz paip sedutan pam empar lebih besar daripada saiz paip hantaran. Jelaskan dengan ringkas tentang kenyataan ini. (b) Apakah kelebihan pam empar berbanding dengan pam aliran paksi. (c) Pendesak sebuah pam empar berdiameter 00mm pada bahagian masuk dan 400mm pada bahagian keluar. Lebar pendesak pada bahagian masuk ialah 15mm dan pada bahagian keluar ialah 8mm. Bilah melengkung ke belakang dengan sudut 38º terhadap tangent. Jika pam beroperasi pada kelajuan 1500 ppm dengan kadar alir 15 liter setiap saat, tentukan perubahan tekanan di dalam pendesak. Abaikan semua kesusutan tenaga. QUESTION 4 (a) Sketch forward, radial and backward centrifugal pump impeller blades complete with their outlet velocity triangles. Which blade is the most suitable for pumping liquid and why? (b) Centrifugal pump supplies water at the rate of 400 litres per second and the pressure difference across the pump is 00kN/m. Diameter and width of the impeller at outlet are 40cm and 10cm respectively. Blade thickness occupied 10% of the circumference. Impeller inlet diameter is half of the outlet diameter. Assume losses in casing and impeller are negligible, zero whirl at inlet and diameter of suction and delivery pipes are equal. If the blades are radial, find : i. The pump power input in horse power if overall efficiency is 80%. ii. The impeller speed in rpm. iii. Inlet blade angle if flow velocity is constant.

54 TUTORIAL FOR FRANCIS TURBINE QUESTION 1 1. Terangkan dua perbezaan utama diantara turbin dedenyut (impulse) dan turbin tindakbalas (reaction). Sebuah turbin Francis paksi tegak (radial) beroperasi dengan kadar aliran air ialah liter per, dengan kelajuan 48 ppm (rpm). Halaju air pada bahagian masuk bekas sesiput ialah 9.0 m/s dan turus tekanan pada bahagian ini pula ialah 60 m air. Garis penengah bahagian masuk bekas sesiput terletak 3.35 m di atas aras air pada larian ekor (draft tube). Diameter bahagian masuk pelari (runner) ialah.4 m dan lebarnya 0.3 m. Jika kecekapan keseluruhan 90%, tentukan; i. Kuasa turbin ii. Sudut bilah pandu iii. Sudut pelari pada bahagian masuk QUESTION 1. Turbin Francis mempunyai halaju tentu tanpa dimensi dalam julat (pelari halaju rendah) ke (pelari halaju tinggi). Bagi pelari di kedua-dua hujung julat ini, lukiskan segitiga halaju di bahagian masukkan pelari dengan lengkap. Jelaskan semua tatatanda dan symbol yang digunakan dalam lakaran anda.. Sebuah turbin Francis aliran ke dalam (backward) beroperasi di bawah turus 45 m. Diameter di masukan pelari ialah 90 cm dan di keluaran ialah 60 cm. Pelari ini mempunyai lebar yang malar sebesar 1 cm. Sudut bilah di keluaran ialah 15º (diukur daripada garisan tangent ke bulatan pelari). Halaju aliran adalah malar pada nilai 3.5 m/s dan air keluar meninggalkan pelari tanpa pusaran. Jika kecekapan hidraulik ialah 90%, kirakan; i. Kelajuan turbin ii. Kadar alir iii. Sudut bilah pandu iv. Sudut bilah pelari di masukkan v. Kuasa yang terhasil

55 QUESTION 3 1. Dari analisis hidrodinamik, kecekapan hidraulik sebuah turbin tindakbalas boleh diungkapkan seperti berikut; Σh η h = 1 H dengan Σh L dan H B masing-masing semua kesusutan turus di dalam turbin dan turus bersih. Jika Σh L minimum, jelas sekali η h adalah maksimum. Dapatkan satu ungkapan bagi Vw (halaju pusaran di bahagian keluar) yang memaksimumkan nilai kecekapan hidraulik, jika jumlah kesusutan turus di dalam turbin boleh dinyatakan seperti berikut; Vr V Σ hl = k1 + k g g dengan Vr dan V masing-masing ialah halaju relative dan halaju mutlak di bahagian keluar rotor dan k1 dan k adalah pemalar. L B. Pada sebuah turbin tindakbalas, turus kesusutan pada rotor ialah 5% dari turus kinetic halaju relative keluar rotor, dan kesusutan pada draf tiub ialah 0% dari turus kinetik halaju mutlak keluar rotor. Halaju aliran serentas turbin adalah malar, iaitu 7.5 m/s. Halaju keliling di bahagian keluar rotor ialah U =15 m/s. Lakarkan rajah segitiga halaju masalah ini. Seterusnya pada kecekapan maksimum dan turus bersih sebanyak 90 m, tentukan; i. Halaju pusaran di keluaran ii. Sd\udut bilah pandu di keluaran iii. Sudut bilah di keluaran iv. Nilai kecekapan hidraulik turbin berkenaan.

56 QUESTION 4 1. Describe briefly the difference between impulse and reaction turbine. What do you understand by Degree of Reaction for turbines.. A francis turbine is required to give 180 kw output power under a head H of 0 m. overall efficiency is 80% and the hydraulic efficiency is 85%. The turbine speed is 600 rpm. If a flow velocity is constant at V f = 0.3 gh and peripheral velocity at inlet U1 = 0.8 gh and the water discharge without whirl, determine; i. The diameter of the runner at inlet ii. The guide vane angle iii. The runner blade angle at inlet iv. The width of the runner at inlet assuming blade thickness occupied 15% of the circumference.

57 TUTORIAL FOR PELTON WHEEL QUESTION 1 1. Tunjukkan dengan jelas beserta gambarajah bahawa kecekapan hidraulik roda pelton dapat dinyatakan sebagai berikut; U ( V η = h j U ) 1+ k cos(180 ) V j dengan V j adalah halaju jet, k adalah nisbah halaju relative, U adalah halaju tangent roda dan θ adalah sudut balikan sauk. Seterusnya tunjukkan bahawa kecekapan hidraulik maksimum diberikan sebagai; η h maksimum 1+ k cos(180 ) = V j. Sebuah turbin roda pelton mempunyai tekanan di muncung 950 kpa dengan pekali halaju muncung adalah C v =0.95. Diameter roda adalah 3 m dan jumpah jet adalah empat. Turbin ini dikehendaki menghasilkan kuasa keluar sebesar 6 MW. Sudut balikan sauk adalah 165º dan halaju relative yang keluar berkurang 10% berbanding yang masuk (V r /V r1 = 0.9). Jika kecekapan keseluruhan adalah 85% dan kecekapan hidraulik 93%, kirakan; i. Nisbah halaju tangent bilah terhadap halaju jet (U/V j ). Jika ada dua jawapan, pilih yang munasabah dan nyatakan sebabnya. ii. Kelajuan putaran roda (rpm) iii. Jumlah kadar alir iv. Diameter jet v. Kecekapan hidraulik maksimum yang boleh dicapai oleh roda ini.

58 QUESTION 1. Bermula dengan takrif pekali turus K H dan pekali kuasa K P, terbitkan rumus bagi laju tentu tanpa dimensi untuk turbin hidraulik. Jelaskan maksud semua tatatanda dan symbol yang digunakan serta nyatakan unit SI untuk setiap parameter yang terlibat bagi menghasilkan laju tentu tanpa dimensi.. Sebuah roda pelton dwi-jet diperlukan untuk menjana 5510 kw kuasa pada laju tentu tanpa dimensi Air dibekalkan menerusi paip empis air sepanjang 1000 m daripada aras takungan yang berada 350 m di atas kedudukan nozel. Pekali halaju nozel, Cv=0.97, nisbah laju = 0. 46, kecekapan keseluruhan 85% dan pekali darcy, f=0.04. Jika kehilangan geseran di dalam paip empis air berjumlah 5% daripada turus kasar, tentukan; i. Kelajuan turbin (rpm) ii. Diameter nozel iii. Jejari minimum bulatan sauk iv. Diameter paip empis air QUESTION 3 1. Terangkan prinsip operasi turbin roda pelton. Nyatakan 3 komponen utam turbin roda pelton dan peranan setiap komponen ini.. Sebuah roda pelton dengan jet air, dibekalkan dengan tenaga hidraulik melalui talian paip sepanjang km dari sebuah takungan air. Aras air dalam takungan terletak 380 m di atas pusat jet. Turbin berputar dengan kelajuan 500 rpm dan mengeluarkan 5 MW. Jika kesusutan turus geseran di dalam paip 8% daripada turus turbin dan factor geseran f=0.006, Cv=0.95, nisbah halaju bilah dengan halaju jet ialah 0.48, factor geseran untuk paip ialah 0.9 dan kecekapan keseluruhan turbin 88%, tentukan; i. Diameter paip ii. Diameter roda pelton

59 QUESTION 4 1. Dari analisis hidrodinamik terhadap sebuah roda pelton boleh ditunjukkan bahawa U kecekapan hidraulik ada fungsi daripada nisbah = serta sudut pesongan = ( 180 ), iaitu untuk keadaan unggul. η = (1 )(1 + cos ) h i. Tunjukkan η h adalah maksimum pada = 0. 5 ii. Dalam keadaan operasi sebenarnya, η h maksimum terjadi untuk, Berikan tiga sebab. iii. Sekiranya Vr=kVR1, kecekapan maksimum masih juga terjadi pada = 0.5. Berikan alas an mengapa hal ini terjadi. V j. Sebuah turbin pelton dengan jet menghasilkan MW kuasa pada putaran 400 rpm. Diameter roda ialah 1.5 m. Turus kasar diukur pada muncung nozel ke permukaan air di empangan ialah 00 m. Kecekapan penghantaran kuasa melalui paip penstock dan nozel ialah 90%. Halaju relative yang keluar dari sauk telah berkurangan sebanyak 10%. Bilah memesongkan jet air pada sudut 165º. i. Lakarkan rajah segitiga halaju roda pelton ini ii. Tentukan turus bersih, H B iii. Tentukan turus Euler, H E iv. Tentukan kecekapan hidraulik v. Tentukan diameter jet

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