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HYDRAULIC MACHINERY.pdf

This document provides an overview of hydraulic machinery and pumps. It discusses various types of pumps including positive displacement pumps like piston pumps and rotary pumps like gear pumps. It introduces concepts like net positive suction head (NPSH), cavitation, and pump selection criteria. Dimensionless parameters for pumps are presented, including the shape factor which relates to optimal pump efficiency. Methods for calculating pump power requirements, head-discharge curves, and applying affinity laws are outlined.

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Hydraulic Machinery Pumps, Turbines...
Hydraulic Machinery Overview ➢ Types of Pumps ➢ Dimensionless Parameters for Turbomachines ➢ Power requirements ➢ Head-discharge curves ➢ Pump Issues ➢Cavitation ➢NPSH ➢Priming ➢ Pump selection
Types of Pumps ➢Positive displacement ➢piston pump ➢Diaphragm pump ➢peristaltic pump ➢Rotary pumps ➢gear pump ➢two-lobe rotary pump ➢screw pump ➢ Jet pumps ➢ Turbomachines ➢axial-flow (propeller pump) ➢radial-flow (centrifugal pump) ➢mixed-flow (both axial and radial flow)
Reciprocating action pumps ➢ Piston pump ➢ can produce very high pressures ➢ hydraulic fluid pump ➢ high pressure water washers diaphragm pump
Peristaltic Pump ➢ Fluid only contacts tubing ➢ Tubing ___ and roller _______ with respect to the tubing determine flow rate ➢ Tubing eventually fails from fatigue and abrasion ➢ Fluid may leak past roller at high pressures ➢ Viscous fluids may be pumped more slowly ID velocity
Rotary Pumps ➢Gear Pump ➢fluid is trapped between gear teeth and the housing ➢Two-lobe Rotary Pump ➢(gear pump with two “teeth” on each gear) ➢same principle as gear pump ➢fewer chambers - more extreme pulsation trapped fluid
Rotary Pumps ➢ Disadvantages ➢precise machining ➢abrasives wear surfaces rapidly ➢pulsating output ➢ Uses ➢vacuum pumps ➢air compressors ➢hydraulic fluid pumps ➢food handling
Screw Pump ➢ Can handle debris ➢ Used to raise the level of wastewater ➢ Abrasive material will damage the seal between screw and the housing ➢ Grain augers use the same principle
Positive Displacement Pumps ➢What happens if you close a valve on the effluent side of a positive displacement pump? ➢What does flow rate vs. time look like for a piston pump? 0 0.5 1 1.5 2 2.5 0 0.5 1 1.5 revolutions total flow 1st piston 2nd piston 3rd piston 3 pistons Thirsty Refugees
Jet Pump “eductor” ➢ A high pressure, high velocity jet discharge is used to pump a larger volume of fluid. ➢ Advantages ➢no moving parts ➢self priming ➢handles solids easily ➢ Disadvantage ➢inefficient ➢ Uses ➢deep well pumping ➢pumping water mixed with solids http://spaceflight.nasa.gov/shuttle/upgrades/ojp.html
(inefficient process) Turbomachines impeller casing housing ( ) ( ) 2 1 2 1 z t t T Q r V rV r é ù = - ë û ➢ Demour’s centrifugal pump - 1730 ➢ Theory ➢conservation of angular momentum ➢conversion of kinetic energy to potential energy in flow expansion ___________ ________ ➢ Pump components ➢rotating element - ___________ ➢encloses the rotating element and seals the pressurized liquid inside - ________ or _________
Pressure Developed by Centrifugal Pumps 2 2 p V h g = velocity head ➢ Centrifugal pumps accelerate a liquid ➢ The maximum velocity reached is the velocity of the periphery of the impeller ➢ The kinetic energy is converted into potential energy as the fluid leaves the pump ➢ The potential energy developed is approximately equal to the ________ ____ at the periphery of the impeller ➢ A given pump with a given impeller diameter and speed will raise a fluid to a certain height regardless of the fluid density
Radial Pumps Impeller Vanes Casing Suction Eye Impeller Discharge centrifugal Flow Expansion diameter rotational speed ➢ also called _________ pumps ➢ broad range of applicable flows and heads ➢ higher heads can be achieved by increasing the _______ or the ________ ______ of the impeller 2 2 p V h g =
Axial Flow propeller ➢also known as __________ pumps ➢low head (less than 12 m) ➢high flows (above 20 L/s)
Dimensionless Parameters for Turbomachines ➢ We would like to be able to compare pumps with similar geometry. Dimensional analysis to the rescue... ➢ To use the laws of similitude to compare performance of two pumps we need ➢exact geometric similitude ➢all linear dimensions must be scaled identically ➢roughness must scale ➢homologous - streamlines are similar ➢constant ratio of dynamic pressures at corresponding points ➢also known as kinematic similitude 3 D Q  same
Kinematic Similitude: Constant Force Ratio   VD gl V 2 gl V   l V 2 V c viscous gravity surface-tension elastic ➢ Reynolds ➢ratio of inertial to _______ forces ➢ Froude ➢ratio of inertial to ________ force ➢ Weber ➢ratio of inertial to _______ ______ forces ➢ Mach ➢ratio of inertial to _______ forces
Turbomachinery Parameters 2 2 C V p p   − = 2 C p H h g V = impeller V D  = 2 2 C p H impeller h g D  = 2 2 3 , , , p flow H impeller impeller flow flow h g D Q C f Re D D D D      = =       impeller (Impeller is better defined) 3 , , , , , , flow p impeller flow flow D Q C f Re F W M D D D     =       Where is the fluid?
Shape Factor ➢Related to the ratio of flow passage diameter to impeller diameter ➢Defined for the point of best efficiency ➢What determines the ideal shape for a pump? ) ( f S =   , , , p Q  Exercise
Impeller Geometry: Shape Factor 0.18 0.37 1.25 2.33 3.67 axial: high _______, low _______ mixed radial mixed ( ) 3 4 p Q S gh w = pressure flow pressure flow S 500 1000 3400 6400 10000 N ( ) 3 4 sp p N Q N h = *N in rpm, Q in gpm, H in ft * Impeller diameter Nsp = 2732S Radial: high _______, low ____
Use of Shape Factor: Specific Speed ➢ The maximum efficiencies for all pumps occurs when the Shape Factor is close to 1! ➢Flow passage dimension is close to impeller diameter! ➢Low expansion losses! ➢ There must be an optimal shape factor given a discharge and a head. ➢ Shape factor defined for specific cases ➢Double suction ➢Treat like two pumps in parallel ➢Multistage (pumps in series) ➢Use Q and H for each stage ( ) 3 4 p Q S gh w = Why multistage?
Additional Dimensionless Parameters 2 2 p H h g C D w = 5 3 D P CP  = 3 Q Q C D  = 1 2 3 4 Q H C S C = impeller power Alternate equivalent way to calculate S. D is the _______ diameter P is the _____ (defined at max efficiency) w p P Qh g =
Head-Discharge Curve Theoretical head- discharge curve Actual head- discharge curve Q ➢ circulatory flow - inability of finite number of blades to guide flow ➢ friction - ____ ➢ shock - incorrect angle of blade inlet ___ ➢ other losses ➢ bearing friction ➢ packing friction ➢ disk friction ➢ internal leakage V2 V2 p h 2 2 p H h g C D w = 3 Q Q C D  =
Pump Power Requirements w p P Qh g = s w P P P e = m s m P P e = p m P m Qh P e e g = water pump shaft motor Water power Subscripts w = _______ p = _______ s = _______ m = _______
Impeller Shape vs. Power Curves S 1 - O.33 2 - 0.81 3 - 1.5 4 - 2.1 5 - 3.4 Discharge (% of design) Power (% of design) axial radial Implications http://www.mcnallyinstitute.com/
Affinity Laws ➢With speed, , held constant: 3 2 1 2 1         =   P P 2 1 1 2 2 p p h h w w æ ö = ç ÷ è ø 2 1 2 1   = Q Q  With diameter, D, held constant: 5 1 1 2 2 P D P D æ ö = ç ÷ è ø 2 1 1 2 2 p p h D h D æ ö = ç ÷ è ø 3 1 1 2 2 Q D Q D æ ö = ç ÷ è ø 2 2 p H h g C D w = 3 Q Q C D  = H Q P  =  5 3 D P CP  = homologous Q C = held constant
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0 0.02 0.04 0.06 0.08 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 D=0.366 m Dimensionless Performance Curves 2 2 p H h g C D w = 3 D Q CQ  = Efficiency (defined at max efficiency) 4 3 2 1 H Q C C S =  Curves for a particular pump  ____________ of the fluid! Independent ( ) ( ) 0.5 0.75 0.087 4.57 0.026 = shape
Pump Example ➢Given a pump with shape factor of 4.57, a diameter of 366 mm, a 2-m head, a speed of 600 rpm, and dimensionless performance curves (previous slide). ➢What will the discharge be? ➢How large a motor will be needed if motor efficiency is 95%? Exercise 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0 0.02 0.04 0.06 0.08 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 D=0.366 m CH = Hg  2 D2 3 D Q CQ  = Efficiency 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0 0.02 0.04 0.06 0.08 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 D=0.366 m CH = Hg  2 D2 3 D Q CQ  = Efficiency
Pumps in Parallel or in Series ➢Parallel ➢Flow ________ ➢Head ________ ➢Series ➢Flow ________ ➢Head ________ ➢Multistage adds same same adds
Cavitation in Water Pumps ➢ water vapor bubbles form when the pressure is less than the vapor pressure of water ➢ very high pressures (800 MPa or 115,000 psi) develop when the vapor bubbles collapse 0 1000 2000 3000 4000 5000 6000 7000 8000 0 10 20 30 40 Temperature (C) Vapor pressure (Pa)
Net Positive Suction Head ➢ NPSHR - absolute pressure in excess of vapor pressure required at pump inlet to prevent cavitation ➢given by pump manufacturer ➢determined by the water velocity at the entrance to the pump impeller ➢ NPSHA - pressure in excess of vapor pressure available at pump inlet ➢determined by pump installation (elevation above reservoir, frictional losses, water temperature) ➢ If NPSHA is less than NPSHR cavitation will occur
Net Positive Suction Head 1 2 2 2 s s v R p V p NPSH g g g = + - s = suction Total head -pv! NPSHR increases with Q2! Elevation datum Absolute pressure 2 2 eye eye v R p V p NPSH g g g = - + At cavitation! z How much total head in excess of vapor pressure is available?
NPSHA 2 2 atm v s s v L p p p V p z h g g g g g - D - - = + - 2 2 atm s s reservoir L p p V z h g g g + = + + 2 2 atm s s L p p V z h g g g - D - = + atm v L A p p z h NPSH g g - D - - = Subtract vapor pressure 2 2 1 1 2 2 1 2 2 2 L p V p V z z h g g g g + + = + + +
NPSHr Illustrated Pv Pressure in excess of vapor pressure required to prevent cavitation NPSHr NPSHr can exceed atmospheric pressure!
NPSH problem Determine the minimum reservoir level relative to the pump centerline that will be acceptable. The NPSHr for the pump is 2.5 m. Assume you have applied the energy equation and found a head loss of 0.5 m. 18°C ? Exercise
Pumps in Pipe Systems 60 m 1 km Pipe diameter is 0.4 m and friction factor is 0.015. What is the pump discharge? hp = z2 − z1 + hl hp = f(Q) 1 m 1 m 2 p h a bQ = - often expressed as p1  + V1 2 2g + z1 + hp = p2  + V2 2 2g + z2 + hl
Pumps in Pipe Systems 0 20 40 60 80 100 120 0 0.2 0.4 0.6 0.8 Discharge (m3/s) Head (m) system operating point Static head Head vs. discharge curve for ________ pump What happens as the static head changes (a tank fills)? p h 2 2 p H h g C D w = Could you solve this with a dimensionless performance curve?
Priming CH = p 2 D2 p = CH 2 D2 2 2 p H h g C D w = density 1.225 kg/m3 ➢ The pressure increase created is proportional to the _______ of the fluid being pumped. ➢ A pump designed for water will be unable to produce much pressure increase when pumping air ➢Density of air at sea level is __________ ➢Change in pressure produced by pump is about 0.1% of design when pumping air rather than water!
Priming Solutions ➢ Applications with water at less than atmospheric pressure on the suction side of the pump require a method to remove the air from the pump and the inlet piping ➢ Solutions ➢foot valve ➢priming tank ➢vacuum source ➢self priming foot valve to vacuum pump priming tank
Self-Priming Centrifugal Pumps ➢Require a small volume of liquid in the pump ➢Recirculate this liquid and entrain air from the suction side of the pump ➢The entrained air is separated from the liquid and discharged in the pressure side of the pump
Variable Flows? ➢How can you obtain a wide range of flows? ➢__________________________ ➢__________________________ ➢__________________________ ➢__________________________ ➢__________________________ ➢Why is the flow from two identical pumps usually less than the 2x the flow from one pump? Valve Multiple pumps (same size) Multiple pumps (different sizes) Variable speed motor Storage tank
RPM for Pumps ➢60 cycle ➢Other options ➢variable speed ➢belt drive number of poles sync full load rad/sec 2 3600 3500 367 4 1800 1750 183 6 1200 1167 122 8 900 875 92 10 720 700 73 12 600 583 61 14 514 500 52 16 450 438 46 18 400 389 41 20 360 350 37 22 327 318 33 24 300 292 31 26 277 269 28 28 257 250 26 30 240 233 24
Estimate of Pump rpm ➢The best efficiency is obtained when S=1 ➢Given a desired flow and head the approximate pump rpm can be estimated! ( ) 3 4 p Q S gh w = ( ) 3 4 p gh Q w » Pump for flume in DeFrees Teaching Lab… Q = 0.1 m3/s, hp = 4 m. Therefore  = 50 rads/s = 470 rpm Actual maximum rpm is 600!
Pump Selection ➢ Material Compatibility ➢ Solids ➢ Flow ➢ Head ➢ NPSHa ➢ Pump Selection software ➢ A finite number of pumps will come close to meeting the specifications!
Pump Selection Chart Model X Model M http://www.pricepump.com/
End of Curve Operation ➢ Right of the BEP (Best Efficiency Point) ➢is sufficient NPSH available for the pump to operate properly? ➢fluid velocities through the suction and discharge nozzles of the pump could be extremely high, resulting in increased pump and system noise (and wear) ➢ Left of BEP operation ➢high thrust loads on the pump bearings and mechanical face seals result in premature failure. ➢The pump is oversized, resulting in lower efficiency and higher operating and capital costs.
Gould’s Pump Curves BEP = 1836 L/s ( ) 3 4 p Q S gh w = Splitcase double suction 890 rpm = 93.2 rad/s S=0.787 Check the Power!
Pump Installation Design ➢Why not use one big pump? ➢Can the system handle a power failure? ➢Can the pump be shut down for maintenance? ➢How is the pump primed? ➢Are there enough valves so the pump can be removed for service without disabling the system?
Pump Summary ➢ Positive displacement vs. turbomachines ➢ Dimensional analysis ➢Useful for scaling ➢Useful for characterizing full range of pump performance from relatively few data points ➢ Turbomachines convert shaft work into increased pressure (or vice versa for turbines) ➢ The operating point is determined by where the pump and system curves intersect ➢ NPSH
Water problem? Early in my college days I took a break and spent 17 months in Salvadoran refugee camps in Honduras. The refugee camps were located high in the mountains and for several of the camps the only sources of water large enough to sustain the population of 6-10,000 were located at much lower elevations. So it was necessary to lift water to the camps using pumps. When I arrived at the camps the pumps were failing frequently and the pipes were bursting frequently. Piston pumps were used. The refugees were complaining because they needed water. The Honduran army battalion was nervous because they didn’t want any refugees leaving the camp. There was only one set of spare parts (valve springs and valves) for the pump and the last set of parts only lasted a few days. The pump repair crew didn’t want to start using the pump until the real cause of the problem was fixed because spare parts have to be flown in from Miami.
Water in Colomoncagua
Waiting for water
Water problem: proposed solutions? piston pump (80 L/min) 2 km pipeline (2” galvanized and then 3” PVC) with rise of 100 m
Shape Factor Solution ➢Create a dimensionless grouping p  p Q2 3 p Q2 3 4 3 Eliminate ______ Eliminate _______ Eliminate ______ S =  Q p        3 4 ( ) 3 4 p Q S gh w = mass length time ) ( f S =   , , , p Q  2 2 2 2 3 M L T L L M T L         =           ( ) 2 2/3 2/3 2 4/3 3 1 L T T T L       =        
Pump Curve Solution 2 2 p H h g C D w = CQ = Q D3 s rev s rev / 8 . 62 2 60 min 1 min 600 =                           =   ( )( ) ( ) ( ) 037 . 0 366 . 0 / 8 . 62 / 8 . 9 2 2 2 2 = = m s s m m CH 068 . 0 = Q C 3 D C Q Q = ( )( )( ) s m m s Q / 21 . 0 366 . 0 / 8 . 62 068 . 0 3 3 = = ( )( )( ) ( )( ) 3 3 9800 / 0.21 / 2 5.55 0.78 0.95 N m m s m P kW = = p m P m Qh P e e g =
Pump Curve Solution 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0 0.02 0.04 0.06 0.08 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 D=0.366 m 2 2 p H h g C D w = CQ = Q D3 Efficiency
NPSH solution 18°C ? A R NPSH NPSH = atm v l R p p z h NPSH g - D = - - Pa pv 2000 = 3 / 9789 m N =  101300 atm p Pa = ( ) ( ) m m m N Pa Pa z 5 . 2 5 . 0 / 9789 2000 101300 3 − − − =  m z 14 . 7 =  atm v A L p p NPSH z h g - = - D -
Implications of Power Curves ➢You are going to start a radial flow pump powered by an electric motor. You want to reduce the starting load on the motor. What can you do? ➢What would you do if you were starting an axial flow pump? ➢How could reducing the head on a radial flow pump result in motor failure? Close the effluent valve Open the effluent valve An effluent pipe break would increase the flow and increase the power requirement
Find Q ( ) ( ) 2 1 2 1 z t t T Q r V rV r é ù = - ë û 2 2 z t T QV r r = p1  + V1 2 2g + z1 + hp = p2  + V2 2 2g + z2 + hl 2 2 z t T Q V r w r w = 2 2 t z p V r T h Q g w w g = = 2 2 2 2 2 2 t V r V z g g w = + work Dimensional analysis Neglect head loss Datum is reservoir level Let A = 10 cm2 2 2 t z V r T VA g w w g =
How could we lift water more efficiently? 2 2 2 2 2 t V r V z g g w = + D 2 2 2 2 2 t Q A V r g z AV w = - D = Solve for Q=AV Decrease V without decreasing Q! ( vt  r cs2 p z h T z Q z w g = D D 2 2 2 2 t p z A z V r g z z h T g w w D - D D = 2 2 z t T Q V r w r w =
Lost energy 2 2 2 2 2 t V r V z g g w = + D 2 2 2 2 2 t V r V z g g w = + D
0.1 1 10 100 1000 0.0001 0.001 0.01 0.1 1 10 Flow (m3/s) Pumping head (m) Axial Mixed Radial Positive displacement 20 40 60 10 200 400 600 100 2000 4000 6000 1000 1 2 4 6 Selection of Pump Type Power (kW) p h
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REFERENCES 1. L. Nelik. Centrifugal and Rotary Pumps: Fundamentals with Applications, CRC Press, Boca Raton, 1999. 2. Sulzer Pumps. Centrifugal Pump Handbook, 3rd edition, Sulzer Pumps Ltd., Winterthur, Switzerland, Elsevier, 2010. 3. S. L. Dixon and C. A. Hall. Fluid Mechanics and Thermodynamics of Turbomachinery, 6th Edition, Elsevier Inc., Amsterdam, 2010. 4. M. P. Singh and G. M. Lucas. Blade Design and Analysis for Steam Turbines, McGraw-Hill Companies, Inc., 2011. 5. M. P. Boyce. Gas Turbine Engineering Handbook, 3rd edition, GPM, Houston, Texas, 2005. 6. C. Soares. Microturbines: Applications for Distributed Energy Systems, Elsevier, Amsterdam, 2007. 7. A. S. Leyzerovich. Wet-Steam Turbines for Nuclear Power Plants, PennWell Corporation, Tulsa, Oklahoma, 2005. 8. S. Yedidiah. Centrifugal Pump User's Guidebook: Problems and Solutions, Chapman & Hall, New York, 1996. 9. M. L. Adams Jr. Power Plant Centrifugal Pumps Problem Analysis and Troubleshooting, CRC Press, Boca Raton, 2017. 10. Sinaga, Nazaruddin. Energy Efficiency As Research and Business Opportunity, Proceeding, International Workshop on Improvement of UNDIP Research Ability and Networking to Stimulate Sustainable Energy, Grand Candi Hotel, Semarang, October 2009. 11. Cahyono, Sukmaji Indro, G. H. Choe, and Nazaruddin Sinaga. Numerical Analysis Dynamometer (Water Brake) Using Computational Fluid Dynamic Software. Proceedings of the Korean Solar Energy Society Conference, 2009. 12. Nazaruddin Sinaga, Abdul Zahri. Simulasi Numerik Perhitungan Tegangan Geser Dan Momen Pada Fuel Flowmeter Jenis Positive Displacement Dengan Variasi Debit Aliran Pada Berbagai Sudut Putar Rotor, Jurnal Teknik Mesin S-1, Vol. 2, No. 4, Tahun 2014. 13. Septianto, Fajar, A. Widodo dan N. Sinaga. Analisa Penurunan Efisiensi Motor Induksi Akibat Cacat Pada Cage Ball Bantalan, Jurnal Teknik Mesin S-1, Vol. 4, No. 4, Tahun 2015.
14. Fatichuddin, Mochamad dan N.Sinaga. Pengaruh Komposisi Air Terhadap Kebutuhan Daya Kompresor Pada Sistem Pembangkit Listrik Biogas dari Limbah Tandan Kosong Kelapa Sawit, Jurnal Ilmiah Momentum, Vol. 12 No. 2, Oktober 2016. 15. Nazaruddin Sinaga. Perancangan Awal Conventer Kit LPG Sederhana untuk Konversi Mesin Bensin Skala Kecil, Eksergi, Jurnal Teknik Energi POLINES, Vol. 13, No. 1, Januari 2017. 16. Nazaruddin Sinaga. Kaji Numerik Aliran Jet-Swirling Pada Saluran Annulus Menggunakan Metode Volume Hingga, Jurnal Rotasi Vol. 19, No. 2, April 2017. 17. Nazaruddin Sinaga. Analisis Aliran Pada Rotor Turbin Angin Sumbu Horisontal Menggunakan Pendekatan Komputasional, Eksergi, Jurnal Teknik Energi POLINES, Vol. 13, No. 3, September 2017. 18. Muchammad, M., Sinaga, N., Yunianto, B., Noorkarim, M.F., Tauviqirrahman, M. Optimization of Texture of The Multiple Textured Lubricated Contact with Slip, International Conference on Computation in Science and Engineering, Journal of Physics: Conf. Series 1090-012022, IOP Publishing, Online ISSN: 1742-6596 Print ISSN: 1742-6588. 19. Nazaruddin Sinaga, Mohammad Tauiviqirrahman, Arif Rahman Hakim, E. Yohana. Effect of Texture Depth on the Hydrodynamic Performance of Lubricated Contact Considering Cavitation, Proceeding of International Conference on Advance of Mechanical Engineering Research and Application (ICOMERA 2018), Malang, October 2018. 20. Syaiful, N. Sinaga, B. Yunianto, M.S.K.T. Suryo. Comparison of Thermal-Hydraulic Performances of Perforated Concave Delta Winglet Vortex Generators Mounted on Heated Plate: Experimental Study and Flow Visualization, Proceeding of International Conference on Advance of Mechanical Engineering Research and Application (ICOMERA 2018), Malang, October 2018. 21. Nazaruddin Sinaga, K. Hatta, N. E. Ahmad, M. Mel. Effect of Rushton Impeller Speed on Biogas Production in Anaerobic Digestion of Continuous Stirred Bioreactor, Journal of Advanced Research in Biofuel and Bioenergy, Vol. 3 (1), December 2019, pp. 9-18. 22. Nazaruddin Sinaga, Syaiful, B. Yunianto, M. Rifal. Experimental and Computational Study on Heat Transfer of a 150 KW Air Cooled Eddy Current Dynamometer, Proc. The 2019 Conference on Fundamental and Applied Science for Advanced Technology (Confast 2019), Yogyakarta, Januari 21, 2019.
23. Nazaruddin Sinaga. CFD Simulation of the Width and Angle of the Rotor Blade on the Air Flow Rate of a 350 kW Air-Cooled Eddy Current Dynamometer, Proc. The 2019 Conference on Fundamental and Applied Science for Advanced Technology (Confast 2019), Yogyakarta, Januari 21, 2019. 24. Anggie Restue, Saputra, Syaiful, and Nazaruddin Sinaga. 2-D Modeling of Interaction between Free-Stream Turbulence and Trailing Edge Vortex, Proc. The 2019 Conference on Fundamental and Applied Science for Advanced Technology (Confast 2019), Yogyakarta, January 21, 2019. 25. B. Yunianto, F. B. Hasugia, B. F. T. Kiono, N. Sinaga. Performance Test of Indirect Evaporative Cooler by Primary Air Flow Rate Variations, Prosiding SNTTM XVIII, 9-10 Oktober 2019, 1-7. 26. E. Yohana, N. Sinaga, I. Haryanto, E. Dharmawan. Taperless Type Blade Design with NACA 5513 Airfoil for Wind Turbine 500 TSD, IOP Conference Series: Earth and Environmental Science 448 (2020) 012006, doi:10.1088/1755-315/448/1/012006.

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HYDRAULIC MACHINERY.pdf

  • 1.
  • 2.
    Hydraulic Machinery Overview ➢Types of Pumps ➢ Dimensionless Parameters for Turbomachines ➢ Power requirements ➢ Head-discharge curves ➢ Pump Issues ➢Cavitation ➢NPSH ➢Priming ➢ Pump selection
  • 3.
    Types of Pumps ➢Positive displacement ➢pistonpump ➢Diaphragm pump ➢peristaltic pump ➢Rotary pumps ➢gear pump ➢two-lobe rotary pump ➢screw pump ➢ Jet pumps ➢ Turbomachines ➢axial-flow (propeller pump) ➢radial-flow (centrifugal pump) ➢mixed-flow (both axial and radial flow)
  • 4.
    Reciprocating action pumps ➢Piston pump ➢ can produce very high pressures ➢ hydraulic fluid pump ➢ high pressure water washers diaphragm pump
  • 5.
    Peristaltic Pump ➢ Fluidonly contacts tubing ➢ Tubing ___ and roller _______ with respect to the tubing determine flow rate ➢ Tubing eventually fails from fatigue and abrasion ➢ Fluid may leak past roller at high pressures ➢ Viscous fluids may be pumped more slowly ID velocity
  • 6.
    Rotary Pumps ➢Gear Pump ➢fluidis trapped between gear teeth and the housing ➢Two-lobe Rotary Pump ➢(gear pump with two “teeth” on each gear) ➢same principle as gear pump ➢fewer chambers - more extreme pulsation trapped fluid
  • 7.
    Rotary Pumps ➢ Disadvantages ➢precisemachining ➢abrasives wear surfaces rapidly ➢pulsating output ➢ Uses ➢vacuum pumps ➢air compressors ➢hydraulic fluid pumps ➢food handling
  • 8.
    Screw Pump ➢ Canhandle debris ➢ Used to raise the level of wastewater ➢ Abrasive material will damage the seal between screw and the housing ➢ Grain augers use the same principle
  • 9.
    Positive Displacement Pumps ➢Whathappens if you close a valve on the effluent side of a positive displacement pump? ➢What does flow rate vs. time look like for a piston pump? 0 0.5 1 1.5 2 2.5 0 0.5 1 1.5 revolutions total flow 1st piston 2nd piston 3rd piston 3 pistons Thirsty Refugees
  • 10.
    Jet Pump “eductor” ➢ Ahigh pressure, high velocity jet discharge is used to pump a larger volume of fluid. ➢ Advantages ➢no moving parts ➢self priming ➢handles solids easily ➢ Disadvantage ➢inefficient ➢ Uses ➢deep well pumping ➢pumping water mixed with solids http://spaceflight.nasa.gov/shuttle/upgrades/ojp.html
  • 11.
    (inefficient process) Turbomachines impeller casing housing () ( ) 2 1 2 1 z t t T Q r V rV r é ù = - ë û ➢ Demour’s centrifugal pump - 1730 ➢ Theory ➢conservation of angular momentum ➢conversion of kinetic energy to potential energy in flow expansion ___________ ________ ➢ Pump components ➢rotating element - ___________ ➢encloses the rotating element and seals the pressurized liquid inside - ________ or _________
  • 12.
    Pressure Developed by CentrifugalPumps 2 2 p V h g = velocity head ➢ Centrifugal pumps accelerate a liquid ➢ The maximum velocity reached is the velocity of the periphery of the impeller ➢ The kinetic energy is converted into potential energy as the fluid leaves the pump ➢ The potential energy developed is approximately equal to the ________ ____ at the periphery of the impeller ➢ A given pump with a given impeller diameter and speed will raise a fluid to a certain height regardless of the fluid density
  • 13.
    Radial Pumps Impeller Vanes Casing Suction EyeImpeller Discharge centrifugal Flow Expansion diameter rotational speed ➢ also called _________ pumps ➢ broad range of applicable flows and heads ➢ higher heads can be achieved by increasing the _______ or the ________ ______ of the impeller 2 2 p V h g =
  • 14.
    Axial Flow propeller ➢also knownas __________ pumps ➢low head (less than 12 m) ➢high flows (above 20 L/s)
  • 15.
    Dimensionless Parameters for Turbomachines ➢We would like to be able to compare pumps with similar geometry. Dimensional analysis to the rescue... ➢ To use the laws of similitude to compare performance of two pumps we need ➢exact geometric similitude ➢all linear dimensions must be scaled identically ➢roughness must scale ➢homologous - streamlines are similar ➢constant ratio of dynamic pressures at corresponding points ➢also known as kinematic similitude 3 D Q  same
  • 16.
    Kinematic Similitude: Constant ForceRatio   VD gl V 2 gl V   l V 2 V c viscous gravity surface-tension elastic ➢ Reynolds ➢ratio of inertial to _______ forces ➢ Froude ➢ratio of inertial to ________ force ➢ Weber ➢ratio of inertial to _______ ______ forces ➢ Mach ➢ratio of inertial to _______ forces
  • 17.
    Turbomachinery Parameters 2 2 C V p p   − = 2 C p H hg V = impeller V D  = 2 2 C p H impeller h g D  = 2 2 3 , , , p flow H impeller impeller flow flow h g D Q C f Re D D D D      = =       impeller (Impeller is better defined) 3 , , , , , , flow p impeller flow flow D Q C f Re F W M D D D     =       Where is the fluid?
  • 18.
    Shape Factor ➢Related tothe ratio of flow passage diameter to impeller diameter ➢Defined for the point of best efficiency ➢What determines the ideal shape for a pump? ) ( f S =   , , , p Q  Exercise
  • 19.
    Impeller Geometry: Shape Factor 0.18 0.37 1.25 2.33 3.67 axial:high _______, low _______ mixed radial mixed ( ) 3 4 p Q S gh w = pressure flow pressure flow S 500 1000 3400 6400 10000 N ( ) 3 4 sp p N Q N h = *N in rpm, Q in gpm, H in ft * Impeller diameter Nsp = 2732S Radial: high _______, low ____
  • 20.
    Use of ShapeFactor: Specific Speed ➢ The maximum efficiencies for all pumps occurs when the Shape Factor is close to 1! ➢Flow passage dimension is close to impeller diameter! ➢Low expansion losses! ➢ There must be an optimal shape factor given a discharge and a head. ➢ Shape factor defined for specific cases ➢Double suction ➢Treat like two pumps in parallel ➢Multistage (pumps in series) ➢Use Q and H for each stage ( ) 3 4 p Q S gh w = Why multistage?
  • 21.
    Additional Dimensionless Parameters 2 2 p H hg C D w = 5 3 D P CP  = 3 Q Q C D  = 1 2 3 4 Q H C S C = impeller power Alternate equivalent way to calculate S. D is the _______ diameter P is the _____ (defined at max efficiency) w p P Qh g =
  • 22.
    Head-Discharge Curve Theoretical head- dischargecurve Actual head- discharge curve Q ➢ circulatory flow - inability of finite number of blades to guide flow ➢ friction - ____ ➢ shock - incorrect angle of blade inlet ___ ➢ other losses ➢ bearing friction ➢ packing friction ➢ disk friction ➢ internal leakage V2 V2 p h 2 2 p H h g C D w = 3 Q Q C D  =
  • 23.
    Pump Power Requirements wp P Qh g = s w P P P e = m s m P P e = p m P m Qh P e e g = water pump shaft motor Water power Subscripts w = _______ p = _______ s = _______ m = _______
  • 24.
    Impeller Shape vs.Power Curves S 1 - O.33 2 - 0.81 3 - 1.5 4 - 2.1 5 - 3.4 Discharge (% of design) Power (% of design) axial radial Implications http://www.mcnallyinstitute.com/
  • 25.
    Affinity Laws ➢With speed,, held constant: 3 2 1 2 1         =   P P 2 1 1 2 2 p p h h w w æ ö = ç ÷ è ø 2 1 2 1   = Q Q  With diameter, D, held constant: 5 1 1 2 2 P D P D æ ö = ç ÷ è ø 2 1 1 2 2 p p h D h D æ ö = ç ÷ è ø 3 1 1 2 2 Q D Q D æ ö = ç ÷ è ø 2 2 p H h g C D w = 3 Q Q C D  = H Q P  =  5 3 D P CP  = homologous Q C = held constant
  • 26.
    0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0 0.02 0.040.06 0.08 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 D=0.366 m Dimensionless Performance Curves 2 2 p H h g C D w = 3 D Q CQ  = Efficiency (defined at max efficiency) 4 3 2 1 H Q C C S =  Curves for a particular pump  ____________ of the fluid! Independent ( ) ( ) 0.5 0.75 0.087 4.57 0.026 = shape
  • 27.
    Pump Example ➢Given apump with shape factor of 4.57, a diameter of 366 mm, a 2-m head, a speed of 600 rpm, and dimensionless performance curves (previous slide). ➢What will the discharge be? ➢How large a motor will be needed if motor efficiency is 95%? Exercise 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0 0.02 0.04 0.06 0.08 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 D=0.366 m CH = Hg  2 D2 3 D Q CQ  = Efficiency 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0 0.02 0.04 0.06 0.08 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 D=0.366 m CH = Hg  2 D2 3 D Q CQ  = Efficiency
  • 28.
    Pumps in Parallelor in Series ➢Parallel ➢Flow ________ ➢Head ________ ➢Series ➢Flow ________ ➢Head ________ ➢Multistage adds same same adds
  • 29.
    Cavitation in WaterPumps ➢ water vapor bubbles form when the pressure is less than the vapor pressure of water ➢ very high pressures (800 MPa or 115,000 psi) develop when the vapor bubbles collapse 0 1000 2000 3000 4000 5000 6000 7000 8000 0 10 20 30 40 Temperature (C) Vapor pressure (Pa)
  • 30.
    Net Positive SuctionHead ➢ NPSHR - absolute pressure in excess of vapor pressure required at pump inlet to prevent cavitation ➢given by pump manufacturer ➢determined by the water velocity at the entrance to the pump impeller ➢ NPSHA - pressure in excess of vapor pressure available at pump inlet ➢determined by pump installation (elevation above reservoir, frictional losses, water temperature) ➢ If NPSHA is less than NPSHR cavitation will occur
  • 31.
    Net Positive SuctionHead 1 2 2 2 s s v R p V p NPSH g g g = + - s = suction Total head -pv! NPSHR increases with Q2! Elevation datum Absolute pressure 2 2 eye eye v R p V p NPSH g g g = - + At cavitation! z How much total head in excess of vapor pressure is available?
  • 32.
    NPSHA 2 2 atm v ss v L p p p V p z h g g g g g - D - - = + - 2 2 atm s s reservoir L p p V z h g g g + = + + 2 2 atm s s L p p V z h g g g - D - = + atm v L A p p z h NPSH g g - D - - = Subtract vapor pressure 2 2 1 1 2 2 1 2 2 2 L p V p V z z h g g g g + + = + + +
  • 33.
    NPSHr Illustrated Pv Pressure inexcess of vapor pressure required to prevent cavitation NPSHr NPSHr can exceed atmospheric pressure!
  • 34.
    NPSH problem Determine theminimum reservoir level relative to the pump centerline that will be acceptable. The NPSHr for the pump is 2.5 m. Assume you have applied the energy equation and found a head loss of 0.5 m. 18°C ? Exercise
  • 35.
    Pumps in PipeSystems 60 m 1 km Pipe diameter is 0.4 m and friction factor is 0.015. What is the pump discharge? hp = z2 − z1 + hl hp = f(Q) 1 m 1 m 2 p h a bQ = - often expressed as p1  + V1 2 2g + z1 + hp = p2  + V2 2 2g + z2 + hl
  • 36.
    Pumps in PipeSystems 0 20 40 60 80 100 120 0 0.2 0.4 0.6 0.8 Discharge (m3/s) Head (m) system operating point Static head Head vs. discharge curve for ________ pump What happens as the static head changes (a tank fills)? p h 2 2 p H h g C D w = Could you solve this with a dimensionless performance curve?
  • 37.
    Priming CH = p 2 D2 p =CH 2 D2 2 2 p H h g C D w = density 1.225 kg/m3 ➢ The pressure increase created is proportional to the _______ of the fluid being pumped. ➢ A pump designed for water will be unable to produce much pressure increase when pumping air ➢Density of air at sea level is __________ ➢Change in pressure produced by pump is about 0.1% of design when pumping air rather than water!
  • 38.
    Priming Solutions ➢ Applicationswith water at less than atmospheric pressure on the suction side of the pump require a method to remove the air from the pump and the inlet piping ➢ Solutions ➢foot valve ➢priming tank ➢vacuum source ➢self priming foot valve to vacuum pump priming tank
  • 39.
    Self-Priming Centrifugal Pumps ➢Requirea small volume of liquid in the pump ➢Recirculate this liquid and entrain air from the suction side of the pump ➢The entrained air is separated from the liquid and discharged in the pressure side of the pump
  • 40.
    Variable Flows? ➢How canyou obtain a wide range of flows? ➢__________________________ ➢__________________________ ➢__________________________ ➢__________________________ ➢__________________________ ➢Why is the flow from two identical pumps usually less than the 2x the flow from one pump? Valve Multiple pumps (same size) Multiple pumps (different sizes) Variable speed motor Storage tank
  • 41.
    RPM for Pumps ➢60cycle ➢Other options ➢variable speed ➢belt drive number of poles sync full load rad/sec 2 3600 3500 367 4 1800 1750 183 6 1200 1167 122 8 900 875 92 10 720 700 73 12 600 583 61 14 514 500 52 16 450 438 46 18 400 389 41 20 360 350 37 22 327 318 33 24 300 292 31 26 277 269 28 28 257 250 26 30 240 233 24
  • 42.
    Estimate of Pumprpm ➢The best efficiency is obtained when S=1 ➢Given a desired flow and head the approximate pump rpm can be estimated! ( ) 3 4 p Q S gh w = ( ) 3 4 p gh Q w » Pump for flume in DeFrees Teaching Lab… Q = 0.1 m3/s, hp = 4 m. Therefore  = 50 rads/s = 470 rpm Actual maximum rpm is 600!
  • 43.
    Pump Selection ➢ MaterialCompatibility ➢ Solids ➢ Flow ➢ Head ➢ NPSHa ➢ Pump Selection software ➢ A finite number of pumps will come close to meeting the specifications!
  • 44.
    Pump Selection Chart ModelX Model M http://www.pricepump.com/
  • 45.
    End of CurveOperation ➢ Right of the BEP (Best Efficiency Point) ➢is sufficient NPSH available for the pump to operate properly? ➢fluid velocities through the suction and discharge nozzles of the pump could be extremely high, resulting in increased pump and system noise (and wear) ➢ Left of BEP operation ➢high thrust loads on the pump bearings and mechanical face seals result in premature failure. ➢The pump is oversized, resulting in lower efficiency and higher operating and capital costs.
  • 46.
    Gould’s Pump Curves BEP= 1836 L/s ( ) 3 4 p Q S gh w = Splitcase double suction 890 rpm = 93.2 rad/s S=0.787 Check the Power!
  • 47.
    Pump Installation Design ➢Whynot use one big pump? ➢Can the system handle a power failure? ➢Can the pump be shut down for maintenance? ➢How is the pump primed? ➢Are there enough valves so the pump can be removed for service without disabling the system?
  • 48.
    Pump Summary ➢ Positivedisplacement vs. turbomachines ➢ Dimensional analysis ➢Useful for scaling ➢Useful for characterizing full range of pump performance from relatively few data points ➢ Turbomachines convert shaft work into increased pressure (or vice versa for turbines) ➢ The operating point is determined by where the pump and system curves intersect ➢ NPSH
  • 49.
    Water problem? Early inmy college days I took a break and spent 17 months in Salvadoran refugee camps in Honduras. The refugee camps were located high in the mountains and for several of the camps the only sources of water large enough to sustain the population of 6-10,000 were located at much lower elevations. So it was necessary to lift water to the camps using pumps. When I arrived at the camps the pumps were failing frequently and the pipes were bursting frequently. Piston pumps were used. The refugees were complaining because they needed water. The Honduran army battalion was nervous because they didn’t want any refugees leaving the camp. There was only one set of spare parts (valve springs and valves) for the pump and the last set of parts only lasted a few days. The pump repair crew didn’t want to start using the pump until the real cause of the problem was fixed because spare parts have to be flown in from Miami.
  • 50.
  • 51.
  • 52.
    Water problem: proposed solutions? pistonpump (80 L/min) 2 km pipeline (2” galvanized and then 3” PVC) with rise of 100 m
  • 53.
    Shape Factor Solution ➢Createa dimensionless grouping p  p Q2 3 p Q2 3 4 3 Eliminate ______ Eliminate _______ Eliminate ______ S =  Q p        3 4 ( ) 3 4 p Q S gh w = mass length time ) ( f S =   , , , p Q  2 2 2 2 3 M L T L L M T L         =           ( ) 2 2/3 2/3 2 4/3 3 1 L T T T L       =        
  • 54.
    Pump Curve Solution 22 p H h g C D w = CQ = Q D3 s rev s rev / 8 . 62 2 60 min 1 min 600 =                           =   ( )( ) ( ) ( ) 037 . 0 366 . 0 / 8 . 62 / 8 . 9 2 2 2 2 = = m s s m m CH 068 . 0 = Q C 3 D C Q Q = ( )( )( ) s m m s Q / 21 . 0 366 . 0 / 8 . 62 068 . 0 3 3 = = ( )( )( ) ( )( ) 3 3 9800 / 0.21 / 2 5.55 0.78 0.95 N m m s m P kW = = p m P m Qh P e e g =
  • 55.
    Pump Curve Solution 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 00.02 0.04 0.06 0.08 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 D=0.366 m 2 2 p H h g C D w = CQ = Q D3 Efficiency
  • 56.
    NPSH solution 18°C ? A R NPSHNPSH = atm v l R p p z h NPSH g - D = - - Pa pv 2000 = 3 / 9789 m N =  101300 atm p Pa = ( ) ( ) m m m N Pa Pa z 5 . 2 5 . 0 / 9789 2000 101300 3 − − − =  m z 14 . 7 =  atm v A L p p NPSH z h g - = - D -
  • 57.
    Implications of PowerCurves ➢You are going to start a radial flow pump powered by an electric motor. You want to reduce the starting load on the motor. What can you do? ➢What would you do if you were starting an axial flow pump? ➢How could reducing the head on a radial flow pump result in motor failure? Close the effluent valve Open the effluent valve An effluent pipe break would increase the flow and increase the power requirement
  • 58.
    Find Q ( )( ) 2 1 2 1 z t t T Q r V rV r é ù = - ë û 2 2 z t T QV r r = p1  + V1 2 2g + z1 + hp = p2  + V2 2 2g + z2 + hl 2 2 z t T Q V r w r w = 2 2 t z p V r T h Q g w w g = = 2 2 2 2 2 2 t V r V z g g w = + work Dimensional analysis Neglect head loss Datum is reservoir level Let A = 10 cm2 2 2 t z V r T VA g w w g =
  • 59.
    How could welift water more efficiently? 2 2 2 2 2 t V r V z g g w = + D 2 2 2 2 2 t Q A V r g z AV w = - D = Solve for Q=AV Decrease V without decreasing Q! ( vt  r cs2 p z h T z Q z w g = D D 2 2 2 2 t p z A z V r g z z h T g w w D - D D = 2 2 z t T Q V r w r w =
  • 60.
    Lost energy 2 2 2 2 2 t Vr V z g g w = + D 2 2 2 2 2 t V r V z g g w = + D
  • 61.
    0.1 1 10 100 1000 0.0001 0.001 0.010.1 1 10 Flow (m3/s) Pumping head (m) Axial Mixed Radial Positive displacement 20 40 60 10 200 400 600 100 2000 4000 6000 1000 1 2 4 6 Selection of Pump Type Power (kW) p h
  • 62.
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  • 63.
    REFERENCES 1. L. Nelik.Centrifugal and Rotary Pumps: Fundamentals with Applications, CRC Press, Boca Raton, 1999. 2. Sulzer Pumps. Centrifugal Pump Handbook, 3rd edition, Sulzer Pumps Ltd., Winterthur, Switzerland, Elsevier, 2010. 3. S. L. Dixon and C. A. Hall. Fluid Mechanics and Thermodynamics of Turbomachinery, 6th Edition, Elsevier Inc., Amsterdam, 2010. 4. M. P. Singh and G. M. Lucas. Blade Design and Analysis for Steam Turbines, McGraw-Hill Companies, Inc., 2011. 5. M. P. Boyce. Gas Turbine Engineering Handbook, 3rd edition, GPM, Houston, Texas, 2005. 6. C. Soares. Microturbines: Applications for Distributed Energy Systems, Elsevier, Amsterdam, 2007. 7. A. S. Leyzerovich. Wet-Steam Turbines for Nuclear Power Plants, PennWell Corporation, Tulsa, Oklahoma, 2005. 8. S. Yedidiah. Centrifugal Pump User's Guidebook: Problems and Solutions, Chapman & Hall, New York, 1996. 9. M. L. Adams Jr. Power Plant Centrifugal Pumps Problem Analysis and Troubleshooting, CRC Press, Boca Raton, 2017. 10. Sinaga, Nazaruddin. Energy Efficiency As Research and Business Opportunity, Proceeding, International Workshop on Improvement of UNDIP Research Ability and Networking to Stimulate Sustainable Energy, Grand Candi Hotel, Semarang, October 2009. 11. Cahyono, Sukmaji Indro, G. H. Choe, and Nazaruddin Sinaga. Numerical Analysis Dynamometer (Water Brake) Using Computational Fluid Dynamic Software. Proceedings of the Korean Solar Energy Society Conference, 2009. 12. Nazaruddin Sinaga, Abdul Zahri. Simulasi Numerik Perhitungan Tegangan Geser Dan Momen Pada Fuel Flowmeter Jenis Positive Displacement Dengan Variasi Debit Aliran Pada Berbagai Sudut Putar Rotor, Jurnal Teknik Mesin S-1, Vol. 2, No. 4, Tahun 2014. 13. Septianto, Fajar, A. Widodo dan N. Sinaga. Analisa Penurunan Efisiensi Motor Induksi Akibat Cacat Pada Cage Ball Bantalan, Jurnal Teknik Mesin S-1, Vol. 4, No. 4, Tahun 2015.
  • 64.
    14. Fatichuddin, Mochamaddan N.Sinaga. Pengaruh Komposisi Air Terhadap Kebutuhan Daya Kompresor Pada Sistem Pembangkit Listrik Biogas dari Limbah Tandan Kosong Kelapa Sawit, Jurnal Ilmiah Momentum, Vol. 12 No. 2, Oktober 2016. 15. Nazaruddin Sinaga. Perancangan Awal Conventer Kit LPG Sederhana untuk Konversi Mesin Bensin Skala Kecil, Eksergi, Jurnal Teknik Energi POLINES, Vol. 13, No. 1, Januari 2017. 16. Nazaruddin Sinaga. Kaji Numerik Aliran Jet-Swirling Pada Saluran Annulus Menggunakan Metode Volume Hingga, Jurnal Rotasi Vol. 19, No. 2, April 2017. 17. Nazaruddin Sinaga. Analisis Aliran Pada Rotor Turbin Angin Sumbu Horisontal Menggunakan Pendekatan Komputasional, Eksergi, Jurnal Teknik Energi POLINES, Vol. 13, No. 3, September 2017. 18. Muchammad, M., Sinaga, N., Yunianto, B., Noorkarim, M.F., Tauviqirrahman, M. Optimization of Texture of The Multiple Textured Lubricated Contact with Slip, International Conference on Computation in Science and Engineering, Journal of Physics: Conf. Series 1090-012022, IOP Publishing, Online ISSN: 1742-6596 Print ISSN: 1742-6588. 19. Nazaruddin Sinaga, Mohammad Tauiviqirrahman, Arif Rahman Hakim, E. Yohana. Effect of Texture Depth on the Hydrodynamic Performance of Lubricated Contact Considering Cavitation, Proceeding of International Conference on Advance of Mechanical Engineering Research and Application (ICOMERA 2018), Malang, October 2018. 20. Syaiful, N. Sinaga, B. Yunianto, M.S.K.T. Suryo. Comparison of Thermal-Hydraulic Performances of Perforated Concave Delta Winglet Vortex Generators Mounted on Heated Plate: Experimental Study and Flow Visualization, Proceeding of International Conference on Advance of Mechanical Engineering Research and Application (ICOMERA 2018), Malang, October 2018. 21. Nazaruddin Sinaga, K. Hatta, N. E. Ahmad, M. Mel. Effect of Rushton Impeller Speed on Biogas Production in Anaerobic Digestion of Continuous Stirred Bioreactor, Journal of Advanced Research in Biofuel and Bioenergy, Vol. 3 (1), December 2019, pp. 9-18. 22. Nazaruddin Sinaga, Syaiful, B. Yunianto, M. Rifal. Experimental and Computational Study on Heat Transfer of a 150 KW Air Cooled Eddy Current Dynamometer, Proc. The 2019 Conference on Fundamental and Applied Science for Advanced Technology (Confast 2019), Yogyakarta, Januari 21, 2019.
  • 65.
    23. Nazaruddin Sinaga.CFD Simulation of the Width and Angle of the Rotor Blade on the Air Flow Rate of a 350 kW Air-Cooled Eddy Current Dynamometer, Proc. The 2019 Conference on Fundamental and Applied Science for Advanced Technology (Confast 2019), Yogyakarta, Januari 21, 2019. 24. Anggie Restue, Saputra, Syaiful, and Nazaruddin Sinaga. 2-D Modeling of Interaction between Free-Stream Turbulence and Trailing Edge Vortex, Proc. The 2019 Conference on Fundamental and Applied Science for Advanced Technology (Confast 2019), Yogyakarta, January 21, 2019. 25. B. Yunianto, F. B. Hasugia, B. F. T. Kiono, N. Sinaga. Performance Test of Indirect Evaporative Cooler by Primary Air Flow Rate Variations, Prosiding SNTTM XVIII, 9-10 Oktober 2019, 1-7. 26. E. Yohana, N. Sinaga, I. Haryanto, E. Dharmawan. Taperless Type Blade Design with NACA 5513 Airfoil for Wind Turbine 500 TSD, IOP Conference Series: Earth and Environmental Science 448 (2020) 012006, doi:10.1088/1755-315/448/1/012006.