Final senior design project report

32
DynaValve LLC Utilizing Hydraulics to Actuate Engine Valves Submitted by: DynaValve LLC. Submitted to: Dr. Robert Woods Dr. Raul Fernandez

description

Mechanical engineering senior design project

Transcript of Final senior design project report

Page 1: Final senior design project report

DynaValve LLC

Utilizing Hydraulics to Actuate Engine Valves

Submitted by:

DynaValve LLC.

Submitted to:

Dr. Robert Woods

Dr. Raul Fernandez

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Monday, December 9, 2013

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Using Hydraulics to Actuate Engine Valves 1

Executive Summary

In a typical internal combustion engine, the cam drives the motion of mechanical components, which in turn actuate intake and exhaust valves. This report provides analysis and evaluation of an engine where the mechanical components are replaced by hydraulic components. Mathematical and engineering analysis was performed to determine the system’s frequency, hydraulic fluid flow rate, and hydraulic line parameters.

While the system was initially modeled as a high-order differential equation, analysis proved that resonance would not be achieved under normal operating conditions. Therefore, the system was modeled using much more manageable equations. Other calculations were performed to determine minimum hydraulic line diameter, minimum fluid velocity to purge air from the system, and minimum line wall thickness. All calculations can be found in the appendix.

Results of data analyzed show that a 1/8 inch diameter stainless steel tube should be used as the medium of transportation for the hydraulic fluid, which in this case will be engine oil. The baseline pressure in the system will be supplied and maintained by an oil pump and check valves, respectively. Oil will be pulled from the existing engine oil reservoir, removing the need for an additional fluid supply that would take up valuable engine space. Should any air enter the system (which could cause performance issues), the velocity of the fluid will be sufficiently high to remove it.

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Introduction

The year was 1886. Karl Benz had just been granted the patent for the first modern automobile that used an internal combustion engine. It was a massive engineering feat that would make travel much more time-efficient, but even Benz couldn’t have foreseen how his invention would change the world. Today there are roughly 1 billion cars worldwide. Ever since the idea for the first internal combustion engine was proposed, efforts have been made to improve its operation. Recently, with the emergence of environmental awareness and recognition of the world’s dependence on oil and natural gas, the efficiency of automobile engines has been at the forefront of the modern engineering landscape.

This project focuses on the feasibility of replacing mechanical components with hydraulic components. In a typical internal combustion engine like the one shown in Figure 1, a push rod is in contact with a cam at one end and a rocker arm at the other. As the cam rotates, it causes vertical motion of the push rod, which in turn causes the rocker arm to rotate about a pivot point. As the rocker arm rotates, it causes the valves that control intake and exhaust to open and close. In this project, the goal is to replace the pushrod and rocker arm with an actuator and hydraulic tubes. A pump will supply pressure to the incompressible hydraulic fluid, which will need to supply at least the same amount of force as the push rod- rocker arm design in order to effectively control valve operation.

Figure 1. Traditional Push Rod/Rocker Arm Setup vs. Hydraulic Actuator System [1]

The primary objective of this project is to evaluate the effectiveness of such a hydraulic system. Factors such as system pressure, hydraulic line size, and fluid flow properties will have to be analyzed in order for the system to perform at the highest level. The system should be designed in such a way that the performance of other engine components does not suffer. One way to accomplish this is to use an existing engine pump to control the hydraulic pressure rather than adding a separate pump. A further objective is to make this system easy to install and maintain. Adding a large number of separate components would presumably overcomplicate the assembly, making it more likely that one or more of the system’s parts will fail or have performance issues.

The idea of using hydraulics in an internal combustion engine is not a new one. Hydraulic lifters have become a popular alternative to their mechanical counterparts. A patent filed by General Motors in

Cam

Hydraulic Actuator

Hydraulic Line

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1978 proposed the use of hydraulic actuators as a replacement for valve springs [2]. Another patent filed by Richard Beaumont in 1985 suggested using a hydraulically driven distributor valve to control timing and operation of the engine [3]. However, most of these past applications revolve around a traditional in-line cylinder arrangement, while this project can apply to either that traditional arrangement or the more unorthodox rotary arrangement.

One might ask why the mechanical system that has been an integral component of the internal combustion engine since its inception should be changed. As mentioned earlier, hydraulics are already present in many automobile systems, so including them in the engine would not be a completely foreign concept. Hydraulic components are valuable because they have a large mechanical advantage, meaning that they can produce a large force output from a relatively low force input. They can be particularly advantageous in automobile engines because power can be transmitted through small tubes that will not take up valuable space in the compact area of the engine. This is especially beneficial to this project because if multiple hydraulic lines are used, one cam should be able to drive the motion of multiple valves. In terms of maintenance, fewer cams represent fewer possibilities of moving parts that could malfunction.

Overall Design

For this project it was required to design, model and build a hydraulic system to actuate the valves of an engine. For this design, we chose to use engine oil as the hydraulic fluid for the system. One of the problems associated with this was that oil could leak out as the engine ran and would need to be replaced.

Another problem encountered was the possibility of air getting into the system, which could cause detrimental effects to the engine’s performance. Thus, it was necessary to devise a method to purge the system of this air and maintain only engine oil within the line.

The final major problem encountered was the dynamics of the system and the line in particular. These variables were to be modeled as an infinite ordered differential equation with an approximation by a fourth ordered differential equation.

These problems and solutions will be discussed below. Also, dimensioning of the system and the necessary flow rate from the oil pump will be explained.

Hydraulic Line Material

For this design, stainless steel tubes will be used for the hydraulic line, as rubber hoses will have too much pliability in the radial direction at high pressure. This pliability could cause a delay or phase shift in the system and could cause the failure or even damage to the engine.

These stainless steel tubes are made by several different companies including Matchless, Aeroquip, and Eaton. They can be purchased at several places both in person and online. The websites that sell these lines include Grainger, hosewarehouse.com and McMaster-Carr. Compared to the rubber hose line, stainless steel tubes are relatively expensive, costing an average of $50 per foot compared to the approximately $20 per foot of a rubber hose [4].

Although it would be cheaper to use this rubber hose, performance standards require the use of steel braided tubing.

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System Pressure

The pressure of the engine oil within the system will obviously change during the course of the engine cycle. The maximum pressure will be the pressure imposed by the valve spring at the valve’s maximum lift. The minimum pressure within the entire system will be the pressure at which the engine oil leaves the oil pump of the engine. In most oil pumps the average pressure is around 60 psi.

The following graph shows the force that is applied by the spring throughout its displacement. In this example, the spring constant of the valve spring is taken to be 380 pounds per inch and the lift of the valve is taken to be 0.42 inches [5].

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450

50

100

150

200

250

300

Valve Displacement (in.)

Forc

e (lb

.)

Figure 2. Spring Force vs. Spring Displacement

As can be seen from the graph in Figure 2, the force that is induced within the spring varies from a minimum value of 80 pounds to a maximum of 240 pounds.

In the following figure, the pressure within the hydraulic actuating system is plotted versus the displacement of the spring. In this example the diameter of the actuator is taken to be 0.82 inches.

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0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450

50

100

150

200

250

300

350

400

450

500

Deflection (in.)

Pres

sure

(psi.

)

Figure 3. Pressure of the System vs. Displacement of the Valve

As can be seen in Figure 3, the pressure within the system varies from a minimum of 60 psi at the cam dwell to a maximum of 430 psi at the maximum displacement of the spring and valve.

With these numbers in mind we must design the system, including the hydraulic line and actuators, to be able to withstand these pressures.

Hydraulic Transmission Line Modeling

Ongoing research is being conducted regarding the improvement of the line dynamics of a hydraulic transmission line. Fluid line transmission requires a very precise and accurate line dynamics model. In combining the transmission line and the actuator valve dynamics, the system stability has to be studied and analyzed before it can serve its intended purpose. In this project, the aim is to accommodate the valve actuation system with a very unorthodox arrangement to couple a remote cam actuator with a valve actuation follower using a hydraulic fluid line. In modeling fluid transmission lines, a lot of concerns arise that need to be addressed in order to develop an optimal design. Surging is a crucial problem that creates several problems if not controlled properly [6].

Figure 4. Free Body Diagram for Line [9]

Pressure surge in a pipe occurs due to a sudden valve closing at an infinitesimal time “t”, causing a very large pressure to build up at the valve closure point. The sudden valve closure causes the fluid flow

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at the valve to halt abruptly, and the fluid following behind will collide with a static fluid, creating a shock wave which could possibly lead to the splitting of the pipe. The shock wave will then propagate backwards at the speed of sound, creating a loud banging noise and changing the dimensions of the pipe diameter along its path. When the shock wave meets a dead end due to the closed inlet valve, it will be reflected back as another round of waves travelling towards its origin, causing an interference with the valve’s natural frequency. The interference will then create total system instability, leading to uncontrolled oscillation of the pipe. The pressure surge is directly related to the pipe’s elastic property and the compressibility of the working fluid, which is why this design utilizes a rigid steel pipe. The properties of the steel pipe are such that instability will never be attained unless near impossible frequencies are generated. The main interest in this control system design process is the actuator valve. The opening and closing of the valve is guided by the pressure difference. Due to the fact that the actuator valve gates open and close instantaneously, controlling the pressure and flow propagation is needed. A distributed line modeling technique is ideal for the total transmission line system design. The system is highly exposed to unmeasured disturbances, noise, time delays and lags, making the distributed line modeling the first choice to model the system. Using this technique is similar to finite element analysis in that each of the segments of the capacitance and resistance of the line are considered to be connected to one another throughout the length of the line.

Figure 5. Dynamic Model of Transmission Line [8]

Using the distributed line model (DLM) is different from the simplified lumped parameter model. DLM is a nonlinear differential equation, unlike the finite order rational polynomial associated with the lumped parameter model. The first approach was finding out the natural frequency of the pipe based on its length. Based on the engine assembly, the initial length assumed was 9 inches. Dividing the speed of sound propagation (Co) by the line length (L) showed that the resonant frequency was approximately 400 rad/sec. The second step taken was finding an applicable diameter for the transmission line. The most conservative modeling technique, which is the distributed line model, was used to find the dissipation number for the line .The dissipation number (Dn) is a unit less normalized parameter, defined as the ratio of the viscous frequency to the characteristic frequency. This normalized value is the most accurate modeling technique, and it is used as the reference point when comparing different frequency response results.

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0 0.5 1 1.5 2 2.5 3 3.5-1

-0.5

0

0.5

1

1.5

2

2.5

Normalized Time, v t

8Dn2 Z

/(R

Lsinh

)

Normalized Impulse Response

Figure 6. Normalized Frequency Plot

Using the Dn helps to find the normalized transfer function which is easily applicable and will be good for our line properties. The transfer function generated using the normalized dissipation number of the line is

T ( s )= 0.0029 s3+0.0046 s2−0.078 s+10.0002 s5+0.002 s4+0.039 s3+0.019 s2+s

(Eq. 1)

The applicable range of the dissipation number used for the system is 0.001< Dn < 0.5.The diameter of the line is then decided based on the frequency response of the system. The system

parameters frequency response, which was based on the dissipation number, indicated that the diameter of the line has a very large range of values which led the design team to decide the diameter to be as small as 0.125 inches. The resonance frequency, calculated after the diameter was applied on a lumped parameter modeling technique, was approximately 61000 rad/sec, indicating that the small diameter value decided upon is more than enough to have an acceptable pressure and flow parameters.

The transfer function based on lumped parameter model indicates that the system is stable for the set variables.

T ( s )=−2.27 x 10−12 s3−7.63 x10−6 s2−0.0068+3.93 x107

s4+2300 s3+3.73 x109+8.54 x 1012+4.94 x 1015 (Eq.2)

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102

103

104

105

106

107

108

0

90

180

270

360

Phas

e (d

eg)

Bode Diagram

Frequency (rad/sec)

-400

-350

-300

-250

-200

-150

System: GFrequency (rad/sec): 6.09e+004Magnitude (dB): -157

Mag

nitud

e (d

B)

Figure 7. Frequency Response for Transmission Line

Determination of the Diameter of the Tube

The minimum diameter of the hydraulic tube connecting two lifters is determined considering fluid dynamics and the strength of the tube. Several approaches can be used to determine the diameter of the hydraulic tube. For the purposes of this the design, the damping ratio was used to calculate the diameter. The damping ratio of the system is a function of the dimension of the tube and properties of fluid. The fluid considered in this analysis is standard engine oil. The equation used for diameter determination is

ζ =16

νCo

∗l

d2

(Eq. 3)

Where:

ζ = Damping factor

ν = Kinematic viscosity

Co = speed of sound in fluid

l = length of the tube

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d = diameter of the tube

The range of the damping ratio for the system to be stable is 0.1 < ζ < 0.5. The design length of the tube is 9 in. The minimum diameter of the tube is less than two hundredths of an inch. Working with such small dimensions would be very difficult and extremely expensive to manufacture. As such, a diameter of 0.125 (1/8) inch was used.

Determination of Thickness of the Tube

Even though the tube can be modeled as a thin walled tube, the required thickness to withstand the pressure is calculated. Tube thickness is calculated using the equivalent bulk modulus of the system. The bulk modulus of the fluid is the property that indicates the “springiness” of the fluid and is defined as the pressure needed to cause a given decrease in volume. It is a measure of the fluid’s resistance to compressibility. Typical oil will decrease about 0.5% in volume for every 1000 psi increase in pressure. The bulk modulus of the tube should also be considered [10]. If there is any entrapped air in the tube, the bulk modulus of the gas should be considered, but since the system will be purged, air can be neglected. The equivalent bulk modulus of the system is given by

1βe

= 1βc

+ 1β l

(Eq. 4)

Where,

βe = equivalent bulk modulus

βl = bulk modulus of fluid

βc = bulk modulus of the tube

The bulk modulus of the tube gives the measure of stiffness of the tube. The hydraulic tube is considered to be a thin-walled cylinder and the bulk modulus of the tube is given as

βc=t . Ed

(Eq. 5)

t = thickness

E = Young’s modulus of the tube

d = outer diameter of the tube

Substituting βc in equation 2, the following equation is developed.

βe

βl

= 1

1+β l . Dt . E

(Eq. 6)

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The obtained minimum thickness required is 4∗10−4 in. As expected, minimum thickness is very small, so using a commercially available standard stainless tube serves the purpose.

Purging the System

As in any hydraulic system, keeping the air out becomes an issue. In most designs, the air is bled out manually, effectively eliminating the problem. For example, in a brake system for automobiles, the air is bled out of the system by manually pumping the brakes and then relieving the pressure at the brake caliper. While this system works effectively for brakes, it will not be applicable for the hydraulic actuating system used for lifting an engine valve. This is because there is a constant possibility for air to get into the hydraulic line, whereas brake systems have a very small chance of developing air once bled correctly [11].

The air could be brought into the system in two main ways. The first possibility for air to get in is at the initial start-up. For example, if the engine has not been started for a long period of time, it is likely that the oil will leak out of the line and drain to the bottom of the oil pan. The oil leakage could occur from the clearance between the piston and the cylinder wall. Therefore air will be in the system as soon as the engine is fired. This will cause the engine to not start due to the compressibility of the air in the line causing the valves to not open. The possibility is illustrated in Figure 10.

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Figure 10. Oil Leakage from Piston Clearance

Another possibility for air to get into the system is during maneuvering of the vehicle. If the car is being steered in a direction that causes the oil to slosh to one side, it is possible for air to be pumped in from the oil pump. Figure 11 shown below illustrates how this can happen.

Figure 11. Engine Oil Sloshes

Due to the various possibilities of air being developed in the system, a self-purging design must be developed to ensure proper performance. A good way to keep the system purged is to allow a constant

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flow of oil through the hydraulic line while in use. However, it is necessary for the constant flow to shut off while actuating the valve with the camshaft. Without proper shutoff, the system will release all of the displaced oil out of the outlet hole shown in Figure 12.

Figure 12. Purging Design

The design of the purging system is shown in Figure 12. The figure was developed in SolidWorks by modifying a 16 valve engine head downloaded from GrabCad [12]. The idea is to pump oil into the valve actuator and have it flow towards the camshaft. A hole is then placed appropriately so that it shuts off when the cam is actuated and opens when the cam is in the dwell portion of rotation. To prevent backflow towards the oil pump, a check valve should be placed at the inlet of the oil. The outlet hole will allow air to be pushed out and removed from the system before the cam is used to lift the valve. For convenience, the design is also shown below in Figure 13 with a larger scale.

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Figure 13. Detailed Purging Design

To ensure that the air is removed on every cycle the velocity of the oil must be high enough to push the bubble out in one cycle. The bubble will have approximately 50% of the time it takes for one full rotation of the cam to be removed. During the other 50% of the cam rotation, the outlet hole will be closed off. This idea is illustrated in Figure 14.

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Figure 14. Outlet Hole Shutoff

To make the best use of this time, the line should be designed in a way shown below in Figure 15. This particular design allows the buoyancy force to move the bubble to the top of the line. Therefore, even when the outlet hole is closed off, the bubble will still be moving towards the outlet for the fall period of the cam. This places the bubble at portion B of the hydraulic line when the dwell time is initiated. At this time the outlet hole is opened and the flow of oil is developed. The bubble will have to travel a vertical distance of “h” to be removed from the system.

Figure 15. Design of the Hydraulic Line

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To minimize the velocity needed, the value of h should be kept to a minimum. This will reduce the flow rate needed to keep the system running at its optimum level. The velocity needed to initiate movement of the bubble can be found from equation 7[13].

12

ρoil A Cd v2min=ρoil gV (Eq. 7)

This equation shows that the buoyancy force is the opposing force whereas the drag force is the force pushing the bubble out. A free body diagram of the air bubble is shown in Figure 16.

Figure 16. Free Body Diagram of Air Bubble

The minimum velocity needed to initiate movement was found to be approximately 1 ft/s, and the full derivation is shown in the Appendix in section (C). In addition, the average velocity needed to push the bubble completely out can be found from the kinematic equation 8 [14], assuming that the cam is rotating at 2000 rpm and that the overall height is 2 inches. The average velocity was found to be 10 ft/s. The complete derivation is shown in the Appendix in section (C).

v=∆ x∆ t

(Eq. 8)

There are also other advantages of this design that increase the reliability of the engine. For example, since the oil is being removed at the camshaft end of the system, the oil can be used to lubricate the cam and actuator. In addition, there is no need to worry about the oil leaking out after delayed usage of the engine since the oil will be replaced upon startup.

Calculating Diameter Using Secondary Criterion

As a secondary criterion for transmission line size calculations, hydraulic line manufacturers recommend that the maximum velocity of fluid flowing through the line be less than 20 feet per second [15]. Thus, to determine what size the line must be to achieve an acceptable value, calculations regarding volumetric flow rate and velocity must be made.

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Using standard equations governing cam velocity, it was found that the maximum velocity of the cam is approximately 55 inches per second. The cam pushes the actuator which then transfers the energy of the cam into the hydraulic fluid, through the system and then to the valve actuator, effectively opening the valve. The area of the actuator that the cam pushes through is about a quarter inch. Using the equation:

Q=AV (Eq. 9)

the volumetric flow rate of the fluid flowing through the actuator is then 14 cubic inches per second.Now that the flow rate in the actuator is known, the flow rate throughout the system can be

assumed to be the same (with some small amount of error). To solve for the diameter of the line necessary to maintain a maximum velocity of twenty feet per second, equation 9 can be solved for area and area can then be solved for the diameter.

The diameter decided upon in the previous section would be the preferred diameter of the line using the secondary criterion. Thus, an eighth of an inch is the desired diameter for the system. Using the maximum 20 feet per second as the velocity flowing through the line, the diameter is found to be approximately 0.15 inches. Noting that the flow rate through the line will be nominally smaller than the flow rate through the actuator, the diameter that was solved for is acceptably close to the eighth of an inch diameter that dynamics calls for.

By using this secondary criterion, the diameter of an eighth of an inch was verified and is the best choice for use within the system.

Conclusion

In conclusion, this report presents a design for a hydraulic actuating system for a valve train that functions proficiently provided the person implementing stays within an acceptable range of such constraints as line length, line diameter and speed of engine. All of these constraints provide ample room for implementation and are applicable for many different engines and applications from automotive to agriculture to even aeronautical.

The preliminary test engine parameters, such as pressure within the system and the forces associated with this pressure, were analyzed. Accordingly, the entire design of the system revolved around these pressures and forces, and the system was designed to sustain them and function properly with them in mind.

During the design of this system, careful considerations were given to the dynamic effects associated with the transmission line of the system. These concerns, in the end, were unfounded as the dynamic effects such as surge and other timing issues can be neglected due to the very high natural frequencies associated with a system like this. Thus, a much simpler model was introduced and a suitable response diagram was attained.

Using principles of the bulk modulus, the necessary thickness of the line was calculated. This is the minimum thickness that will prevent timing problems due to the flex from the transmission line pressures. Also, purging the system of any air will be accomplished by maintaining a constant fluid flow throughout the system. This requires implementation of minimal additional components foreign to the standard automobile seen in practice. It does possibly require some upgrades to parts such as the oil pump of the automobile and the oil reservoir already mounted to the engine.

After careful consideration of all of the characteristics and properties associated with a hydraulic actuating system, an acceptable design has been developed. This design includes the use of an eighth of an inch line that is approximately nine inches long and an actuator with a piston that is approximately six tenths of an inch in diameter. This system should be run at a rate not exceeding three thousand rotations per minute and the hydraulic fluid used within the system should be standard SAE grade oil. Much

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beyond these constraints, it cannot be guaranteed that the dynamics of the system can be neglected and that this simple model can be used.

References

[1] DOE-HDBK-1018/1-93, “Diesel Engine Fundamentals,” Diesel Engines, nuclearpowertraining.com, Web 25, September 2013, from http://nuclearpowertraining.tpub.com/h1018v1/css/h1018v1_31.htm

[2] Trenne, Myron (To General Motors Corporation), “Hydraulic Valve Actuator System,” U.S. Patent 05/901,452, May 1, 1978.

[3] Beaumont, Richard. “Internal Combustion Hydraulic Engine,” U.S. Patent 06/733,074. May 13, 1985.

[4] “Hose Master- Flexible Metal Hose.” Grainger.com. Accessed December 2013. [http://m.grainger.com/mobile/product/HOSE-MASTER-Flexible-Metal-Hose-2ZV57]

[5] “Edelbrock Valve Springs and Retainers.” Edelbrock.com. Accessed September 2013. [http://edelbrock.com/automotive_new/mc/valvetrain/springs_retainers.shtml]

[6] Hullender, D.A., 1990. “Effects of Fluid Transmission Lines in Pressure Measurement” Instrumentation and Control – Fundamentals and Applications, Section 10.5, pp. 434-438, John Wiley & Sons, Inc., Somerset, NJ

[7] Craig, Kevin. "Hydraulic Transmission Lines." EDN. N.p., n.d. Web. 18 Nov. 2013. <http://www.edn.com/electronics-blogs/mechatronics-in-design/4423778/Hydraulictransmission-lines>.

[8] Shinners, Stanley M. Advanced modern control system theory and design. New York: Wiley, 1998. Print.

[9] "Keep an Eye on Hydraulic Transmission Lines." N4SAcom. N.p., n.d. Accessed 18 Nov. 2013. [http://n4sa.com/arkiv/107629.]

[10] Woods, R.L. and Lawrence, K., 1997, Modeling and Simulation of Dynamic Systems, Prentice Hall, Upper Saddle River, NJ.

[11] “Service Manual Procedure – Brake Bleeding Procedure.” Dodge Rem Service Manual. Accessed October 2013. [dodgeram.org http://dodgeram.org/tech/repair/Brakes/beeding.htm. ]

[12] Murarik, Peter. “Valve Train dohc.” GRABCAD.com. Accessed December 2013. [https://grabcad.com/library/valve-train-dohc-1/files.]

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[13] Munson, Bruce R., “Buoyancy, Flotation, and Stability,” Fundamentals of Fluid Mechanics, 2009

[14] Giancoli, Douglas C., “Kinematics in One Dimension,” Physics for Scientist and Engineers with Modern Physics, 2009

[15] “Line Sizing and Fluid Velocity,” RHM Fluid Power Inc, Westland, MI [http://www.rhmfp.com/tech-tips/161-line-sizing-and-fluid-velocity. Accessed 10/3/2013.]

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Appendix A

Normalized Frequency Program

function [NGZoverSinh,GZoverSinh] = ZoverSinh(Den,Beta,Mu,d,L )format shortgr=0.00625;Mu=4.6e-5;Nu=.015;Den= 870;L=0.25;d=2*r;Bf=1.152e9;Bl=1.73e7;Beta=(Bf * Bl)/(Bf + Bl);Dn=Nu*L*sqrt(Den/Beta)/r^2;% Dissipation numberRL=128*Mu*L/(pi*d^4);% Steady state flow resistanceWv=Nu/r^2;% Normalizing frequency, rad/sec if Dn>0.5 Dn 'Dn is greater than 0.5 which is outside the range for the computed coefficients' else if Dn<0.001 Dn 'Dn is less than 0.001 which is outside the range for the computed coefficients' elseif Dn<=0.2a3=0.0454*Dn^3.5523;a2=-0.1012*Dn^2.0585;else a3=-0.2167*Dn^4+0.2569*Dn^3-0.1077*Dn^2+0.0225*Dn; a2=-0.8047*Dn^4+1.7023*Dn^3-1.005*Dn^2+0.2079*Dn-0.0173;endif Dn<=0.1 a1=-5436.6*Dn^5+1213.3*Dn^4-91.209*Dn^3+2.9331*Dn^2-.0036*Dn+3e-5;else a1=1.1585*Dn^4-1.8786*Dn^3+0.0199*Dn^2+0.1218*Dn-0.0067;endb4=0.0047*Dn^4.0812;b3=0.035*Dn^3.6798;b2=0.1912*Dn^2.0661;b1=0.6685*Dn^1.6288; fprintf('Normalized Transfer Function, 8Dn^2Z/[RLSinh]')NGZoverSinh=tf([a3 a2 a1 1],[b4 b3 b2 b1 1 0]);NGZoverSinhzpk=zpk(NGZoverSinh);[Normalized_poles]=roots([b4 b3 b2 b1 1 0]);figure[y,t]=impulse(NGZoverSinh);plot(t,y,'r','LineWidth',2)xlabel('Normalized Time, \omega_v t')

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ylabel('8D_n^2Z/(R_Lsinh\Gamma)')title('Normalized Impulse Response')figurefprintf('Un-normalized transfer function, Z/Sinh')GZoverSinh=tf([a3/Wv^3 a2/Wv^2 a1/Wv 1]*RL/(8*Dn^2),[b4/Wv^5 b3/Wv^4 b2/Wv^3 b1/Wv^2 1/Wv 0]);GZoverSinhzpk=zpk(GZoverSinh)[Poles]=roots([b4/Wv^5 b3/Wv^4 b2/Wv^3 b1/Wv^2 1/Wv 0])[Y,T]=impulse(GZoverSinh);plot(T,Y,'k','Linewidth',2)xlabel(' Time, seconds')ylabel('Z/sinh\Gamma, (N/m^2)/(m^3/s)')title('Impulse Response, \Delta P_b/(\Delta Q_a) or -\Delta P_a/(\Delta Q_b)') end end

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Using Hydraulics to Actuate Engine Valves 21

Appendix B

Lumped Parameter Program to Find Out the Resonant Frequency Based on the Set Parameters

p=870; %Kg/m^3M=0.05;%kgb=115; % N.s/m DAMPING Vis=4.6e-9;Bf=1.21e9;Bl=1.73e7;L=.25;%mCd=1;K=66548.2;%N/md=0.003175;%mr=d/2;Ps=0.687e6;%paPi=3.14;V=L*pi*r^2;Av=5.29e-4;%m^2Ac=5.29e-4;%m^2Be = (Bf * Bl)/(Bf + Bl); C1 = 32*Vis/(p*d^2);C2 = Pi*d^2 / 8*L(Cd)^2*(Av)^2;C3 = pi*d^2 /4*p*L;B1 = b / M;B2 = K / M;B3 = Ac / M; D1 = 4*Be/ (Pi*d^2);D2 = 4*Be*Ac /(Pi*d^2); A=[0 1 0 0;-B2 -B1 0 B3;0 0 -C1 -C3;0 -D2 D1 0];B=[0;0;C3;0];C=[1,0,0,0];D=[0];

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Using Hydraulics to Actuate Engine Valves 22

Appendix C

Diameter of the Tube

Figure 8. Diameter of Transmission Line

Using, ζ =16

νCo

∗l

d2

Here, ν

Co = 0.0045*10−6

l = 9 in

ζ = 0.1

d2 = 16

νCo

∗l

ζ

= (16)(0.0045*10−6)*(0.250.1

)

d = 0.0017 in = 0.42 mm

Therefore, 0.42 mm is the minimum diameter of the tube.

Diameter

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Using Hydraulics to Actuate Engine Valves 23

Appendix D

Determination of Thickness of the Tube

Figure 9. Thickness of Transmission Line

Assuming, β l=200∗103 psi

E = 28∗106 psi

βe

βl

=0.75

Equivalent bulk modulus 1βe

= 1βc

+ 1β l

βe

βl

= 1

1+β l

βc

βe

βl

= 1

1+β l . Dt . E

t=4∗10−4in.

The obtained thickness is the minimum thickness of the tube.

Thickness

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Using Hydraulics to Actuate Engine Valves 24

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Using Hydraulics to Actuate Engine Valves 25

Appendix E

Buoyant Forces and Drag Force

Fb=ρoil gV

Fd=12

ρoil A Cd v2min

Fb=Fd

12

ρoil A Cd v2min=ρoil gV

vmin=√ 2 gVA Cd

Where: g=32.174ft

s2=386.09 ¿

s2

Cd=1

r=.0625

V= 43

π r3=.001¿3

A=π r2=.0123

r=.0625

vmin=1fts

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Using Hydraulics to Actuate Engine Valves 26

Appendix F

Average Velocity Needed to Push Bubble Out

v=∆ x∆ t

∆ x=2

∆ t=50 % of camrotation time (dwell portion of cam)

Assuming 2000 rpm

2000 rpm=200060

=33.3 r ev per sec

Therefore each cam rotation takes .03 sec.

∆ t=50 % of camrotation time (dwell portion of cam)

∆ t=.5∗.03 sec=.015 sec

v=∆ x∆ t

= 2.015

=11 fps