FLUID FLOW PATTERNS N THE HEAT ’ TRANSFER PASSAGES Q52”: AN INTERHAL } COMBUSTION meme ' Thai; for the Dogm of u s; MICHsGAN STATE COLLEGE -' - " A. F. M. ABDUL AZEZ... . ‘1953‘ ' ‘0‘. J S . ‘ e \:3 )IJz‘A .s ' o ' t\. . ‘. .,' '3’,“ I n , ‘ " I ‘- This is to certify that the thesis entitled Fluid Flow Patterns in the Heat Transfer Passages of an Internal Combusion Engine presented by A.F.M. Abdul A218 has been accepted towards fulfillment of the requirements for JéL—degree in_!s.!¢__ afiwid‘if Major professor Date ML FLUID FLou PATIERNS IN THE HEAT TRANSFER PASSAGES OF AN INTERNAL COMBUSTION ENGINE BY A. F; M. ABDUL gggz, A THESIS Submitted to the School of Graduate Studies of Michigan State College Of Agriculture and Applied Science in partial fulfillment of the requirements for the degree of MASTER OF SCIENCE Department of Mechanical Engineering E22? 4A- A .,.I hang: it 00‘ 0‘. A ACKNOWLEDGEMENTS The author is very grateful to his major professor Dr. Louis L. Otto for his most valuable suggestions in desi- gning the apparatus and adopting test methods, for his wise guidance in the experimental proceedure, and for his keen interest and constant supervision in the whole process without which this investigation would have never been a success. Further it was only his activqhelp which made it possible to finish the work within a limited time. The author, therefore, wiehes to express his heartfelt thanks to him and dedicates the work to his name. The author cannot but express his sincere apprecia- tion of the close co-operation and immense help rendered by Mr. Ray T. Pearson in carrying out all the machine shOp work in preparation of the investigation. The author also wishes to thank his friend Mr. A.S.M. Nurullah in preparing and checking the test results. Lastly the author is indebted to Michigan State College for the materials used in connection with the investigation. 5 3’3 52- .3 I3 ( .‘I W E, ii VITA The author was born in a small village of the Province of Bengal in India (now in Pakistan) on January 31, 1932. His father was a poor farmer and had no means to educate his sons. The author passed hi3.#th grade from the village Free Primary school and obtained a scholarship from the government. He then was admitted into a high school three miles from his house, where he enjoyed a free studentship and prosecuted his studies upto the end of the eigth grade. That was the year of l9#4 and theyear of famine in Bengal when he was promoted to the ninth grade. His father being extremely unable to give him a single penny for his education wanted to take him back to his farm work. But the author was determined to carry on with education. He therefore fled away frOm home in search of a charitable High School. Fortunately he met Mr. N. M. Das, the .head master of Nandina High School, a village high School. It was for his kind help that theauthor could pass the high school final examination. He passed the high school final examination with distinction and obtained a ;scholarship from the government., With this scholarship he went; to the Dacca iii Intermediate College located in East Pakistan and studied there Physics, Chemistry, Mathematics, Botany, English and Vernacular for two years. During this time he got a stipend from the Muslim Education Fund awarded by the Director of Public Instruction, East Bengal. He passed the Intermediate Science final examination with much credit and received a scholarship from thegovernment. At this time he awarded a stipend from his district. With these helps the author' was admitted into the Engineering College, Dacca, where also he received a scholarship. In 1952 he received his Bachelor of Science degrre in Mechanical Engineering frOm this college after four years of studies. The author had an ambition to become a teacher. So he wanted to go on for higher studies. He received a grant for passage to the United States from the Shakoor Educational Trust Fund in East Pakistan, and took a loan from the Multi, purpose Society of Kishorganj, East Pakistan. With these he came to the Michigan State College and became a Graduate student in Mechanical Engineering majoring in Automotive Engineering. The author is now a candidate for the degree of Master of Science in Mechanical Engineering. TABLE OF CONTENTS 1. Summary 2. 12. 13. 1h. 15. The Problem for Investigation Choice and Design of Testing Equipments Design of Drive Motor Pulley Methods adopted for Solution Test Conditions Proceedure for Pressure Test Low-speed Pressure Readings Medium-speed Pressure Readings High-speed Pressure Readings Sample Calculation for Measurements of Pressure by Piezometric Tubes Sample Calculation for measurements of Pressure by U-Tubes Results of Pressures Results of Vacuum Relation between Speed of the Motor and Speed of the Pump 10 13 18 21 2h 27 33 35 39 1+1 1+6 iv TABLE OF ILLUSTRATIONS Location of Pressure Points In the Engine Pressure Points in the Pump and the Radiator Pressure Vs. Speed Pressure Vs. “Distance Rate of Circulation Vs. Speed Power demanded by the Pump Vs. Speed Section showing Area of Water Flow through the Jacket of the Block Section of Minimum Area of the Flow of Water through the Jacket of the Hench 19 20 43 41+ 55 53 56 57 vi SUMMARY This work is an attempt to investigate and to establish relationshop among power demanded by the pump in pumping cooling water through the engine, rate of water circulation in the cooling system, pressures at different significant points and velocities_at critical sections, sections of minimum cross sectional area, in the cooling system for an Internal Combustion Engine used in automotive vehicles. The engine for investigation was a Buick straight- eight-eylinder engine of current production. For the measurements of pressures in the cooling system water was circulated by the pump driven by a motor through a V-belt. Low pressure readings were taken-in.the , piezometric tubes to be able to record small changes in pressures and high pressure readings were taken with the mercury U-tubes. To determine the rate of circulation through the cooling system and power demanded by the pump, the pump was disconnected from the engine and was couple to a dynamometer. A venturimeter was put in the water lineoof pump discharge. Rate of water circulation was calculated from the venturi- meter readings and the power demanded from the dynamometer readings. Necessary arrangements were made to create the condition for this test identical to that for actual operation. Velocities were calculated for four sections, (i) At the section of pump discharge, (ii) At the minimum section through the flow area of the of the water jacket of the engine block. (iii)At the section on the flow way from the block to the head of the engine. (iv) At the section of minimum flow area of the water jacket of the engine head. Results of pressure distribution have been so arranged in the table, (Table No. 1) that at a quick glance at the table it is very easy to visualize what is happening in the pressure distribution pattern with increase in pump speed as well as with increase in distance from the outlet of the pump.Also graphs have been drawn to show the relation; ship between pressure and pump speed and between pressure and positions of the pressure points in the system. This second graph, (Graph No. 2) giges a remarkable way of showing the pressure distribution pattern in thesystem. Graphs have been drawn to show the relation between rate of circulation and pump speed and between power demand df the pump and the pump speed. These graph give to a different scale, the rate of circulation and power demand of the pump with respect to the engine speed. The table No.3 and the graph No.3 give an idea how the rate of circulation and power demand of theypump are related to thespeed FOR INVESTIGATION In the Internal Cembustion Engines power is obtained by burning fuel inside the cylinder. The combustion results in a very high temperature and in the liberation of a large amount of heat. approximately :0 to 35 per cent of this heat travels into and through the cylinder walls. If the cylinder walls are not prOperly cooled, the result will be too much heating of the cylinder walls, the lubricating oil will be burnt and the gradually increasing friction between the cylinder walls and the pistons will quickly damage the wearing surfaces. Further inefficient cooling will .cause thermal stresses in the material of the cylinder walls. Therefore the cooling should be adequate, proper andtefficient. In present times all of the automotive engines are cooled by the pump circulation system. In this system some power from the engine output is used to drive the pump. So the rate of pumping should be such as to cause correct rate of water circulation, neither more nor less for proper and efficient cooling. If the water rate is greater than this more power is needed to drive the pump, which lowers the mechanical efficiency of the engine. 0n the other hand if sufficient amount of water is not circulated, the engine will wear out too rapidly. For this reason the present investigation seeks to establish relationships among power demanded by the pump, rate of circulation through the cooling system, and pressure and velocities at significant points in the cooling system. When the water pump rate is established an attempt is made to find the water velocities through thepngine at different speeds of the pump. The pump is driven through a V—belt by the engine crank shaft so water velocities through the engine can be related to the R.P.M.of the engine. The water velocities through the engine at different speeds will help to find the film coefficient of heat transfer between cylinder walls and cooling water. This film coefficient of heat transfer can be utilised to evaluate the rate of heat transfer to the cooling water to determine whether or not the cooling water flow rate is correct. The knowledge of thi Optimum amount oflwater flow at any speed of the engine is a basis for the design of the pump. An investigation is made to find at various points the pressures in the cooling water through the heat transfer passages. This will give a knowledge of the pressure distribution in the cooling water ~along the heat transfer passages and hence the nature and magnitude of the drop in pressure at any point from the inlet to the outlet of the water jacket. The point of maximum pressure drop will be seen . Thiswill help the designers to improve their water jacket to produce less drop in pressure. Tne less drOp in pressure means less resistance in the flow path, which saves power to the pump. The pressure at a few points in thebuction line is taken. This will help to check for 'cavitaion. So our problem is to find pressure distribution in the cooling water along the heat transfer passages, to find rate of water pumping with respect to engine speed and to find power needed to drive the water pump with respect to the rate of 2 water pumping. CHOICE AND DESIGN OF TESTING EQUIPMENTS Necessity wasto choose or to design devices for: (i) Driving the pump (ii) recording low speed pressures (iii) Recording high speed pressures (iv) Measuring rate of water circulation and (v) Determining power demand of thegpump For driving the pump a variable speed D.C motor of the following specification, provided with starting and speed control panel was used. A pulley was designed for motor . The pump was.driven through a V-belt. Volts - 220 D.C. Max. R.P.M. at full load - 1650 Current at full load - 11.9 amps. For measuring rate of water circulation a venturimeter of the following dimensions was used. The venturimeter was inserted in thedischarge line from the pump with a gate valve for control. ghroat diameter = 1 inch -Jiameter at entrance = 2 inches For the determination of the power demand of the pump a General Electric Co. dynamometer was used. The dynamo- meter csnstant was LB PULL x R.P.M. H.i’. = 5000 For recording low and high speed pressures straight piezomet- ric tuges and mercury U-tubes were designed and made. With the help of LIna-Mes the pressure at the point of maximum pressure, and vacuum at the point of maximum vacuum were deter- mined prior to the test for both the highest speed of the low range and highest speed of the high range. At low speed . .. At high Speed Max. Pressure in lbs/sq.in 5-5 13 Max. vacuum in ins of Hg 4 12 Minimum length required for 3.5 x 3u.14.7 = 8.1 ft. 13 x 30/14.7 = 26.b inches (a) Piexometric tube (b) Mercury U-tube (c) Water U-tube for vacuum 4 x 34/30 = k.52 ft Let DESIGN OF DRIVE MOTOR PULLEY Maximum R.P.M. of the crankshaft = #400 = N1 Diameter of the crankshaft pulley = 5.27 ins = Dl Diameter of the pump pulley = 5.8 ins = D2 So maximum R.P.M. of the pump N2 is N2 - D1 1 2 4400 Or N2 = 5.27 x 5.8 = 3940 R.P.M. of the motor at full load = 1050 = N) D be the.diimeter of the motor pulley in inches 5 II. = N2 D2 N3 Or D 5940 x 5.5 5 = 1050 = 15.5 lO METHODS ADOPTED FOR SOLUTION The problem of the present investigation was split up into three parts. They are:- (i) To find the pressure distribution in the cooling water along the heat transfer passages (ii) To find the rate of pumping water with respect to the engine speed. (iii) To find the power demand of the pump to the required flow to water. The method adopted for solution of the first part was as follows:- The engine was dismantled, water flow passages were observed and an engine head of the same type of our investiga- tion model was dissected to observe theflow passages in order tapping to locate the represententative pressure points at the best available static zones. The pressure tapping points were connected either to the straight piezometric tubes in case of low pressures to read very small drops between points or ll to the U manometer tubes in case of high pressures. A few pressure taps were made in the suction line to r ead vacuum. The pump was run at the various speeds and the readings were taken at the corresponfing speeds. To get the solution of the second part, ie., to find the rate of pumping water with respect of the engine speed, the pump was disconnected from the engine under investigation and was tested separately where.it was possible to measure the flow rate of water with the help of a venturimeter. The pump was driven directly by a dynamometer through a coupling. A venturimeter was inserted in the line from the pump discharge to the radiator inlet. A gate valve in this line was used to check and control the flow of water in order to produce a pre- determined pressure at the . discharge point of theypump. This predetermined pressure at the pump discharge is the pressure which was obtained under actual conditions ofpumping water through the system at various speeds. Readings of the venturimeter were taken and rates of water circulation calculated for various required speeds. At low speeds the Venturimeter did not show any indication, and hence flow rate was established by extrapolating the curve drawn from the high Speed readings. 12 To find the power needed to drive the pump at various speeds, the pump was connected to a dynamometer of known constant and by creating the same operating conditions as were under actual test the reading of the scale was taken. 15 TEST CONDITIONS For the.determination of the pressure distribution pattern in the cooling system there were two objectives:- (a) Pressure response with respect to the increase in engine speed in an wide range of engine speed. This was achieved by selecting quite large number speeds from the idling to the t0p speed of the engine. (b) Pressure drOp in the system at the same speed with respect to the distance covered from the pump outlet to the jacket outlet. In order to achieve this -end observations of flow passages were made to know their way and nature. Thirty significant points for pressure tapping in the discharge line and four in the suction line were decided and made at the best available static pressure zones. Similar conditions under which the water pump was run to force water into the cooling system were set up when the pump was tested‘separately.for the.rate of circulation of water. The idea of putting the pump under similar conditions isto get from the pump in the test condition the same flow . rate as was obtained in the actual condition at the various 14 specified speeds. This could be done by equating the energy of water at a given point of the pump for both the test and the actual conditions. Let the subscripts a and t stand for actual and test conditions respectively. The energy of water at a given point is by the well known Bernoulli Lquation N + Z ‘3 w {\JI < 0"; + where V = velocity of water at that point in ft.sec. P a pressure of water at point in lbs/sqin elevation of that point referred to any x; t] N datum in ft. w a specific weight of water at that temperature in pounds per cubic foot In the actual condition the energy is and in the test condition 2 £+Pt+. 2g w t at the same point. If the two energies are to be equal l5 va Pa Vt Pt n + -- + Z r: -' + + a a t 2g w 2g w If the flow rates are equal Va must be equal to Vt as the crosbsectional areas remain the same. So the equation stands If we refer the elevation for both the cases to a convenient point which will be common for both the case, we can make This was done by referring to the pump shaft. The height of the pump shaft from the zero datum of manometer board in the test condition was so set that the heighqof the pump shaft from the zero datum of the new manometer board was equal. So ultimately the equation is w w By regulating the check valve inserted in the water line P was made equal to Pa for a given speed. First, the inlefi‘side of thepump was connected 16 through a valve to the Water supply tank and outlet side through another Valve. It was arranged to regulate both the valves in order to get equal Pa's and Pés for the suction as well as discharge side pressures. But the resistance offered by the valve in the suction line was so large that with the available water tank the necessary suction could not be obtained even with the valve wide open. So the inlet side was then connected to the radiator in the same position as it was in the engine. It was argued that since the resistances in the suction line remained same, if the valve in the discharge line is regulated to bring about the same pressure (to made Pa = Pt) for the discharge side only suction vacuums will automatically adjust to corresponding vacuums at different speeds. For the determination of flow of water with respect to engine speed the pump could be driven at any speed. But the pump was driven at same speeds as it was driven when taking pressure readings for only at those speedsthe pressures were known for the discharge side. When the flow rate, the resistance to the flow, and the pump operating head are same for both the actual and test conditions the power demanded by the pump must be the same. 17 So the test to know the power to drive the pump has three conditions to fulfil. (a) To create same flowzx This was achieved to perform the second part of the investigation. (b) To ;ut in same reSistance to flow path:- Since the same radiator with the same hose length in the same condition was used, the resistance in the suction side was unchanged. The valve in the discharge line acted as a resistor By regulating the valve the resistance in the line was adjusted until.Pt was equal to Pa for a given speed. It was argued that since the speed of the pump and the discharge pressure for the test were same as for the actual condition this resis- tance must be equal to the resistance to the flow path offered in the actual investigation. (c) The pump Operating head is the difference bet- ween the pressure head and the suction head. As in both the cases these heads were same the pump operating head was same 0 18 PROCEDURE A. For pressure readings:- nll the pressure and vacuum points were connected either to the piezometric and water filled U-tubes, in case of low speed tests or to the mercury U-tubes, in case of high speed tests with the help of brass nipple and plastic tubes. The radiator was filled with water at room temperature to maintain the water level upto the filling point. The pump was driven at various speeds by means of a variable speed D.C. shunt wount motor. Readings were noted from the graph attached to the manometer board against the tubes. Parallax error was avoided. For every Speed the reading of the reference pressure point was taken. I9 more... A n :9: a 3.5m l‘ .XUOmx ofimmrw as» we onmm Luxuo mt» Co wu o» d 36$“ mwen£1x (who as» us once; as wuwwoado_umsn Woueuoa was ha on m EOPm MFODESF mumvo 05.... be my.» Mom cpsmmuxl 3.5.. . mu£QOm assumesm of» O» nsoamuxLOu whenisz .uFmvru er.» p3 muFaom okswmokm we romance; a man 20 Pressure points in the pump and in the radiator 35 O .. 52 Radiator Fig. 2 LOW SPEED PRESSURE READINGS Speed of the motor in R.P.M. Height of water level in the piezomter tubes in inches 1 1 1 13 22.65 33.3 10.8 18.65 28.5 10.7 18.20 27.4 10.75 18.60 28.2 10.7 18.#o 27.8 10.7 18.55 27.9 10.65 18.50 27.6 10.8 18.A 27.7 10 6 18.3 27.5 10.6 18.5 27 6 10.5 18.10 27.25 10.5 18.15 27.4 10 A 17.9 26.8 10.5 18.1 27.5 10.5 18.1 27.3 21 LOW SPEED PRESS Speed of the motor in R.P.M. URE READINGS 42.6 36.9 38.9 58.2 Height of water level in the 33 Closed end Open end Closed end Open end (Contd.) 395 470 16 24 16.74 25.2 15.35 22.9 15.4 22.9 15.2 22.5 15.2 22.6 14.8 22 14.75 21.9 14.4 21.35 14.4 21.3 14.1 20.7 14.2 20.9 13.4 19.7 13.7 20.2 U-tubes 24.3 25 19.7 18.7 24.5 24.8 20.6 20.2 22 LOW-SPEED PRESSURE READINGS (Contd.) Speed of the motor in R.P.M. 280 395 470 34 Closed end 52.6 55.4 56.? Open end 49.6 46.0 46.3 35 Closed end 50.2 51.8 53.3 Open end 44.8 42.2 ‘40.? Vacuum in inches of water 32 2.6 4.6 6.3 33 ' 2.3 3.9 4.6 34 3.0 0.6 10.4 35 5.4 9.0 12.6 4.1 2.3 19.4 20.3 23 MEDIUM—SPEED PRESSURE READINGS Speed of the motor in R.P.M. 24 Height of the water level in the pie- zometer tube water Closed limb level ......................... in the hg H O press.' _ llimb l Hg in the open limb Hg in the closed limb H O in the press. limb O\U1-F \'l 45-5 24.5 29.4 90 87.8 89 89.9 88.8 89.1 88.1 88.5 19.1 6.5 14 12.5 12.2 13.8 13 10 11.8 12.2 8.8 12.3 15.1 9.7 17.2 12.7 10.2 15.4 9.6 9.6 4.7 9.1 11.3 6 7» Open limb Hg H20 17.0 17.2 18.3 16.1 16.3 15.4 16.5 17 16.4 15.7 15.4 MEDIUM-SPEED PRESSURE READINGS 25 I..— n_-_._.._._1n~.2"_ll'k!~. "416- - . ' m! (Contd.) l 2 3 4 5 6 7 I; """"""" {38:5 """ i33'm’52mi53 """ £53” 13 07.4 15.6 10.8 16.1 17.2 14 86 13.5 9.3 10.3 15.5 15 86.7 14.5 9.4 14.3 16 16 87 13 0.5 12.2 15 15.2 17 86.7 13.9 8.5 11.3 15.1 18 75.5 15.7 8.2 13.5 13.8 19 75.4 20 72.3 20 72.3 21 72.3 22 70.6 23 71.3 14 8.9 14 14.1 2L4 69.3 25 69.6 25 67.5 27 67.6 ‘5 10.4 14.4 15.4 17.5 28 65.9 11 8.3 10.3 12.7 12.9 29 66.2 9-5 7.9 12.3 12.7 26 1 2 3 4 5 6 7 30 62.4 13.5 8.2 10.9 12.1 13.1 31 64.1 13.2 11.4 11.6 15.6 Pressure points Water levels in the -U—tubes i Closed limb Open limb 3 * L. 32 36 7.5 33 32.2 12.4 34 64 - 30 35 74 19 Speed = 880 Pressure Readings of the U tubes . Water Closed limb Open limb p01nts level in the Hg 820 Hg H20 press ................... 1299-----------------_-..---..-_--.. 32 5.3 3.8 4.8 33 15.5 14.1 34 10.1 5.8 HIGH—SPEED PRESSURE READINGS Speed a 1440 E .............. é ......... 2----fr ......... 2-..--é Pr°°sur° ..... BSESEB§§-EB-E§2-9:EEES§ ....... points Water Closed limb ' Open limb level --------------------------- in the Hg H 0 Hg H O . 2 2 press. limb o 1.1 1 16 4.3 9 20.8 2 18.8 8.8 19.9 21.9 3 19 6.6 8 19.6 4 19 5.6 8.2 18.4 5 19.5 6.4 14 18.4 6 18.2 7.1 11 20 7 19.5 6.1 19.5 8 5.6 5.6 19 9 5.4 15.6 18.5 10 18.4 4.8 7.3 17.5 11 19.5 7.3 10.1 20.2 12 19.5 6.1 7.6 18.9 113 19.5 7.6 12.3 20.4 .-----------‘---------—---------------------—----. 5.3 5.9 20.4 18 7.2 12.6 15.8 1505 I”. HIGH-SPEED PRESSURE READINGS (Contd.) Speed = 1630 l ................... é ......... 2 ..... E .......... 2-----2 Pressure Readings of the U-tubes P°1nt° Water Closed limb Open limb level in the Hg 320 Hg HZO press. limb 0 1.1 l 2 15-5 23-5 2 7.7 17.5 23.1 3 3.5 15 2115 21.8 4 3.6 15 20.6 5 4.3 12 22.1 6 5 16.5 22.3 7 4 4 22.2 8 3.3 3.3 21.1 9 3.7 14 20.8 .10 2.7 14 19.8 .11 5.2 16.5 22.3 .12 4 lo 21 13 5.5 14 22.6 29 18.5 8.2 4.6 13.7 16.6 16 12 17 30 31 HIGH-SPEED PRESSURE READINGS (Contd.) Spged = 1090 F l 2 3 4 5 8 i ----------------------------------------------------- E J Pressure points headings of the U-tubes ‘ Water Closed limb Open limb i 4 level ; J in the Hg H20 Hg H20 Ly press. limb O 1.1 1 ' 1.5 13.5 24 2 6.9 15.8 23.9 3 4.1 14 ' 22 22.2 4 3.3 14 21 5 4 12 22.2 6 4.6 12 22.4 7 3.8 3.8 22.3 8 303 3.3 2101" 9 3.7 12.5 21 10 2.2 11 20 ll 5 12.5 22.6 12 4 12.5 21.4 13 5.2 11 22.9 21.1 21.6 20.9 21.1 18.7 18.7 32 14.0 Let So Then 33 SAMPLE CALCULATION FOR MEASUREMENTS OF PRESSURE BY PIELOMETRIC TUBES the height of water column in the piezometer tube with reference to zero datum in inches = H the height of water column in the piwzometer tube to which all other readings to be referred for datum = h the pressure in inches of water above adopted master datum plane = H - h Pressure in inches of Hg = ( H - h ) x Pressure in lbs/sq.in 34 x 12 ( H - h ) x .00362 Let us take a numerical example, Speed = 010 Pressure point = 1 H = 59 h = .8 -l ."A Pressure Pressure Pressure in inches of water above adopted master datum plane = H - h = 59 - .8 = 58.2 58.2 in inches of Hg = . 1306 -; 4026 in lbs/aq.in 50.2 x .00302 .211 34 Fig. j Blue - water Red - Mercury .-:;m—-e _ .' ~ "g--. .‘_- e. . iv .‘ ~ "A uni/4:4..‘Aifmmzta l '. 8 y w V V Case 1 Gene 2 Case 3 SAMPLE CALCULATION FOR MEASUREMENTS OF PRESSURE BY U-TUBES To find pressure at the;p1ane 0 0 Case 1 Pressure at l = 0 gauge & Case 2 Pressure at 2 = H ine of Hg Preeeure at 3 H ins of Hg ( weight of air negligible ) 36 Pressure at O O = H ins of Hg + h ins of H20 = H y-g- ins of Hg 13.6 Case 3. :— Pressure at 1 = 0 w ins of H 0 Pressure at 2 2 = 1- ins of Hg 13.6 Pressure at 3 w/13.6 + H ins of Hg Pressure at 4 " " w + h Pressure at O O = + H ins of Hg 13.6 Numerical Example Speed = 1440 and Pressure Point N0 = 14 Pressure point No 9 Case 1 Case 2 H = 16,8 - 6.1 H = 18.5 - 5.h h = 19.3 h = 15.6 h h Pressure = H + —- Pressure = H +-— 13.6 13.6 37 19.5 15.6 Pressure = 12.7 + g- = 13.1 + f 13.6 13.6 = 14.14 ins of Hg = 14.24 ins of Hg Case 3. Pressure point 28 i II 15.9 f 15.4 = .5 :1: H 1.505 " 505 h=19 19.5 13.6 Pressure + 10 = 11.45 ins of Hg These are the pressures at the zero datum. The pressure above adopted master plane ( >7 ’lfiz r.‘} is found by subtracting from this pressure the pressure at the plane 0 O. .V-A. ‘V‘I— . - 1' .' ‘1 _ ’3 L} '3“ r 111 - m n ‘ , r‘ . M _ $1: ‘ H ' T ~ . L fl) : ") l1 ; 1' i 1 ‘ J .x‘ ‘ I" ,. A g P ‘1 “.9... bh—B—O-I F - Ku Fig. 4 Case 4 Pressure Point 1 Speed = 1440 Pressure Pressure Pressure Pressure Pressure Pressure at at at at at at :08 a u'ins of HZO =1 +Hinsong 13.6 .. + H -92.- 1306 ins 0: H8 + H - E + h/13.6 ins of Hg 13.6 38 R.P.M. of the pump RESULTS IN INCHES OF Hg Table No 1. 3350 3800 3940 Pressure . points 1 2 crummtw 10 11 12 13 14 15 .884 .720 .715 .717 .717 .715 .717 .711 1.26 1.28 1.29 1.27 1.28 1.27 1.94 1.97 2.00 1.97 1.98 1.95 1.96 1.95 1.95 " 1.93 1.94 1.89 " 1.93 3.37 3.36 3.36 ’ 3.31 6.62 6.45 ‘ 6.54 6.6 6.54 6.55 6.46 3.32 '. 3.28 3.31 7.2 7.1 A 7.13 8 7.09 7.1 6.95 6.8 17.2522.8 14.2 18.0 14.2318.4 14.2 18.5 14.2 18.5 14.2418.4 14.3 18.5 14.3 18.4 14.2 18.3 14.1 18.1 14.1 18.2 14 18 14.1 18.1- 14. 17.9 14 17.9 23.41 18.07 18.67 18.75 18.80 18.76 18.81 18.69 18.60 18.5 18.55 18.44 18.46 18.39 18.27 39 '1', '5 -V ’.. ‘F.‘ R.P.M of the Pump IN RESULTS INCHES 0F (Contd.) Hg ~ 2040 3350 3800 3940 Pressure points 16 17 18 19 20 21 22 23 24 26 27;. 28.: 29 50 31 25 .52 1.26 1.27 1.1 1.16 1.05 1.06 1.04 1.04 1.01 .985 .985 .96 .97 .91. 1.93 1.93 1.69 1.7 1.61 1.61 1.58 1.58 1.54 1.49 1.49 1.44 1.45 1.37 5.52 5.52 2.88 5.00 2.78 2.78 2.72 2.72 2.65 2.58 2.58 2.51 2.51 2.38 .94 1.41 2.44 6.4 6.37 5.55 5-51+ 5.52 5.52 5.18 5.24 5.11 4.96 4.96 4.84 4.86 4.58 4.71 .562 1.01 1.54 2.65 5.11 6.9 13.9 17.8 18.18 6.75 13.9 17.8 18.1 6.10 12.8 15.8 16.16 5.71 11.8 14.7 15.89 5.3 11.3 13.7 14.31 5.2 11.1 13.1 14.28 5.2 11.1 12 7 14.19 5.02 10.8 11.5 14.0 5.1 10.7 12.9 13.88 40 41 RESULTS VACUUM IN INCHES OF as “I Table 2 R.P.M. of L. thePump 655 921 1090 1420 1955 2040 3350 5800 5940 Pressure Points33 .169 .586 .338 1.46 1.4 5.7 5.91 6.0 52 .185 .558 .46 2.07 1.52 6.45 54 .22 .651 .765 1.45 2.5 4.5 8.45 9.08 .665 .911 1.5 4.05 4.0 9.4 11.6 12 35 .396 SAMPLE CALCULATION FOR VACUUM Case 1. :- U-tubes filled with water Let the closed limb read Hw inches of water and the Open limb read hw inches of water Hw- hw Vacuum in inches of Hg 2 -—————- 13.6 Case 2. :- U-tubes filled with mercury Let the closed limb read Hm inches of Hg and the open limb read hm inches of Hg Vacuum in inches of Hg = Hm- hm 42 3‘»me JP' '~ ‘. 25 15 Pressure in ins. Hg 10 Graph 1. 43 No No No No No 1. 2. 3. 4. 5. Pressure Vs. Speed at the pump discharge at entrance to the block at exit from the block at entrance to the head at exit from the head r __ 1? O 1000 2000 Speed of the pump in R.P.M. 3000 4000 12.29 .9.82 7.36 Pressure in 1bs/sq.in 4.91 44 Pressure Vs. Distance along the flow path N0 1. at a pump speed 2040 No 2. at a pump speed 1420 N0 3. at a pump speed 653 _.__minhfl-_- __- m ”‘l‘ . __-_ .L_____ _____ ,7 .. .. _- 0=e P 30: Distances in inches - « Block&——Pwe-w~wuun- F1. [7. E q 11 i in" lPressure Ein ins Hg _ -._______1. Graph 2. ! 2 -._ ...__ _.. -._--.._ .-..._.1..-_. -11 .1 - - _ h_ "—._—v ntrance ressure point exit pressure point I - T — -- -- TH... —— — ~—--—-———-+ .30 1 0 10 20 50 Relation between speed of the motor and speed of theiPUmp Speed of the motor in R.P.M. = N3 280 595 470 610 840 880 Speed of the pump in R.P.M. = N 653 921 1090 1420 1953 2040 2 Speed of themotor . in R.P.M. 1440 1640 1690 Speed of the pump ‘ in R.P.M. 3350 3000 3940 Sample Calculation E2 _5 N3 D2 _ 13,5 x 280 0r N2 ‘ 5.8 = 653 The speed of the motor has been determined with.a tachometer. 4S .wiq u _IAM. THE TEST T0 The original (a) to (b) to (c) to 4? DATA SET UP ' FOR KNOW RATE OF WATER FLOW AND DRIVE POWER levels of Hg in the U-tubes adjust pressure a 16.3 inches adjust vacuum = .45 " .record venturimeter readings = 19.1 " Pump Speed 1420 1955 2040 3350 5940 The pressure should read in the U-manometer L1 Epsopen limb for Vacuum Pressure 1.17 18 2.48 20.54 2.65 20.62 5.15 24.65 6.45 28.00 SAMPLE CALCULATION TO SHOW "What the pressure should read " “nu UDTAINED Speed of the pump = 3350 Pressure at the discharge point, point 2 = 17.2 ins of Hg If there were no water pressure in the pressure limb of the manometer the pressure shoul show in the Open limb 24.9 ins and in the closed limb 16.5 - —L—l'2°2 7.7 ins So the pressure exerted by water ( water was continous upto the Hg level in the closed limb ) the height of 7.7 ins of H o 2 = .565 ins of Hg So when the pressure snows 17.2 ins of Hg, Actual pressure is 17.2 + .565 = 17.765 ins of Hg And therefore when the pressure should be 2 17.2 ins of Hg the pressure should show 17.2 leo2 17.765 : 16e7 Therefore the pressure snould read in the open limb = 24.65 ins DATA OBTAINED FROM THE TEST T0 KNUH RnTE OF WATER FLCH AND DRIVE POWER AND RESULTS Table no 3 LB PULL X R.P.M. Dynamometer H.P. = 5000 Pump speed Venturimeter Dynamometer Rate of H.P. Readings Readings Flow Demanded ............. 5’1--_-i°2--------1:35----_------Effii‘.3°_---------- 1420 19.6 18.6 2.63 .056 .720 1953 20.64 1736 2.91 .066 1.136 2040 21.01 17.19 2.98 .094 1.216 3350 26.97 11.23 3.96 .191 2.656 3600 31.4 6.8 4.28 .238 3.264 50 SAMPLE CALCULATION 1. Rate of f\ow:- Venturimeter equation = V = in ft./sec. 5 = in ft.of water Example Speed of the pump 3940 P1 ' P2 = 52.5 - 5.7 = 26.8 ins of Hg w 26.8 x 13.6 = ft of 320 12 1 l 2 2 V A 2 1 v1 A2 I: 50 S HPLE CALCULATION 1. Rate or 1‘0an- Venturimeter equation = Zg w V = in fto/SeCO z in ft.of water l’r‘ Example Speed of the pump 3940 l 2 = 32.5 - 5.7 = 26.8 ins of Hg w 26.8 x 13.6 = ft Of H20 12 30.4 ft. of H20 1 l 2 2 V A 2 1 Or--=-— v1 A2 2 Orv = e7851+XD 2 1 x Vl 2 1 0 D1 = Dia at entrance = 2 inches 9 9 D2 = Dia at throat = 1 inch v: - vi 15 vi Therefore : 28 28 And 15 vi = 30.4 23 1+ So Q = AlVl = .7854 x-- x 11.4 144 3 e249 CftheC 2. For H.P. demanded Lb Pull x R.P.M. H.P. = 5000 4.41 x 3940 5000 3.48 52 PROCEDURE B. For finding rate of water circulation and power demand of the pump The pump was disconnected from the engine and was fitted to a dynamometer for direct drive through a coupling. Necessry water connections were made and a venturimeter was put in the discharge.1ine. The dynamometer was run at various desired speeds. at every Speed the pressures at the discharge point, point 1, were made equal to those for actual conditions by regulating the valve. The venturimeter leads were connected to an U-tube with plastic tubes. headings of venturimeter and dynamometer scale were taken every speed. Graph 3 I ‘3 }—..—.___..4..]__._.____._.. ----..-.---.-.-______+ ! ! i . j Flow rate Vs. Speed I F10. .2E rate in * ; cuft/sec i “L 01 T....._-,_..-_- _. --_-_---- -._, 4.8 H.P.demand Vs. Speed ” .9-. -~ “ —-. - v -—-- ‘~‘.—‘.—-. 4 ”/- O ‘23:".-- . _| . ._ .. . - fl . _ - ..---__ O 1000 2000 3000 4360 Speed of the pump in R.P.A. '53 54 VELOCITIES Velocities were calculated for four sections in the flow passages of the cooling System. These four sections were critical sections in the sense described velow:- The first section was the section at the pump discharge, it was a circular path of diameter 1.25 inches. The second section ( Fig. 7 ) was the section of minimum area of water flow in the jacket around the engine block cut across the diameter of the.cylinder. The area was calculated from the given dimensions. The third one was the section for the transfer of water from the block to the head of the engine. This consisted of five circular paths, one is of 1 inch diameter and the rest four are of diameter % inch each. The last sec- tion was the section of flow of water through the water jacket of the head. This was also a section of minimum areaaand was found by cutting the head at various sections. The area at this section was irregular ( Fig. 8 ) and it was found by making its impression on the graph paper. Area Area Area Area of section of section of section at section RESULTS ll 17654 x .03 .00856 sq.ft .028 sq.ft .7854 x 1.252 1 144 0.0109 sq.ft + 4 x .7854 x Velocities of water .52 144 (calculated from the graph) 27e8 55 Fig. 7 Section showing the area of water flow through jacket of the block 94 4 /////]W A //////7////f//Y//////// W/M/f/f/ffl///// 3... Not to scale through the wat m' “mum area of flow of water er jacket of the head \ ' /////// // Fig. \\\\\\\\\\\a\t \\\\§\\§»\\\\\\\\\P “ ”I\\\\ \\X\\\\\\U n-Y 8 Relation between Pump Speed and Engine Pump Speed Engine Speed Pump Speed Engine Speed 921 1010 3350 3670 1090 1195 3800 4160 Speed 1420 1953 1564 2140 3940 4330 Sample Calculation Diameter of the crank shaft pulley for driving the pump Diameter of the pump pulley EngineSpeed 715 Pump Speed x 5.8 653 x'-- 5-27 = 5.27 inches = 5.8 inches 5.8 5-27 ( when pump speed = 653 ) 59 CONCLUSION This investigation reveals some informations about " FLUID FLOW PATTERNS IN THE HEAT TRANSFER PASSAGES OF AN INTERNAL COMBUSTION ENGINE " which were before unknown to the students and teachers of Internal Combustion Engines. These may now fill up the vacant place in the literature. A look to the graphs Pressure Vs. Speed and Pressure Vs. Distance of pressure points in the system from the pump outlet snows that maximum pressure drops take place at exit from the pump to the entrance of the blocx, and at the exit from the block to the entrance of the heau. This is because of entrance and exit losses. The pressure dr0p along the block is negligible compared to the dr0ps described above, while there is a considerable amount of drop along the head. The reasons are that the flow path along the head offers large resistance due to its zigzag course, and the plain path along the block offers a very small resistance. The velocity of water through the water jacket must be turbulent, this is" necessary for high rate of neat transfer through the heat transfer surfaces of the head. 60 From the graph 1 it is seen that the pressure has the steepest gradient at qpump speed between 2000 to 3500 R.P.M., this is because with the increase of rate of water circulation pressure increases. But after that, gradient changes and tends to become less and less steep, this indicates some hindrance wich may be a larger friction factor at high speed. The maximum vacuum in the cooling system is 12 inches of Hg or the minimum absolute pressure is 18 inches of Hg when the barometer reads 30 inches of Hg. At this pressure the boiling point of water is 187.450 F. So if the thermostat is designed for 1600 F or 1800 F there is no chance for v cavitation. The graph No 3 , rate of water circulation Vs. Speed can be used to check for correct rate of water circulation through the cooling system, Horse Power demanded by the Pump Vs. Speed can be used to figure out Mechanical Efficiency. The loss of power to drive the pump is an important factor for small ecars which are designe for 30 Or 40 H.P. The future designers of water pump and heat transfer passages may be benefited by this investigation. 61 BIBLIOGRAPHY P. M. Heldt, hater Cooling, page 507, Automotive Engines Ninth edition. P.M. Heldt, Nyack, N.Y. 1935 Ben G. Elliot, Engine Cooling, page 208, Automobile.Power Plants, First edition McGraw-Hill Book Company, Inc., S. Goldstein,.Velocity and pressure measurements, page 248, Modern Developments in Fluid Mechanics, V01 1 Clarendon Press, Oxford. Hunter Rouse, Elementary Mechanics of Fluids, John Wiley & Sons, Inc. New York. L.C. Lichty, Engine Cooling, page 428, Internal Combustion Engines. McGraw-Hill Book Company, Inc. New York. King, Wisler, Woodburn, Hydraulics, John Wiley & Sons Inc. New York. F. T. Morse, Power Plsnt Engineering and Design, D. Van Nostrand Company, Inc. New York. 1.1.7154 t " . c '7' (£1. ':J~'- a ~ ." .- ‘ '5'“. sin—1 -. s o O ”1.” ‘. ‘. f "".l'_-.'- ,N' "+8? I J.r_. I” ‘ ' 1 A) ' c 'v {35: " ?