- Length: 14 pages
- Sources: 5
- Subject: Education - Mathematics
- Type: A2 Coursework
- Paper: #15285074

The Table 1 provides the layout of the spreadsheet, which reveals the time on the first row and the system variables on the subsequent columns on the right. The table 2 defines the Constant separately, which can be easily changed in order to investigate the system response.

Table 1: Layout of the Spreadsheet

A

B

C

D

E

F

G

H

I

1

Time (sec)

Fuel (%)

T

TF

TN

(Nm)

Vel

Pos

rpm

2

0

=B2*L$1

0

=C2-D2

=E2/L$3

0

0

=G2*60/(2*3.14159)

3

=$A2+0.1

=G2*L$2

=G2+(A3-A2) *0.5*(F2+F3)

=H2+(A3-A2)*0.5*(G2+G3)

4

Table 2: The Engine Parameters

K

L

Engine Parameter

1

k1=

1

2

k2=

0.1

3

J=

0.1

1. The report constructs the spreadsheets using respective equations as being defined in Fig 2.

In the row 1 of the spreadsheet, the report defines the time and the system variables to evaluate the system model as being revealed in Fig 4 from left to right.

2. The system responses are built by filling in the row 2, and the initial time is placed in cell A2 which contain 0.

3. The report uses 100% fuel input throughout the stimulation process. Thus, cell B3 contains the value 100.

4. By entering equation for the applied torque in Cell 2 using Equation 4, where constant k2 is defined in the spreadsheet.

5. The report assumes that engine is initial at rest (t=0), where the speed of the engine is equal to zero, making the torque frictional to be equal to zero. However, the report does not use the equation in the first entry because angular speed has not yet been evaluated. Thus, 0 value is placed in cell D2.

6. The resultant torque (TN) is the frictional torque TF and is subtracted from the applied torque, T, which eventually represents summing block within the system model. Thus, an equation is provided in cell E2, which subtract D2 from C2.

7. Moreover, the report obtains the angular acceleration from Equation 1 using TN and the moment of engine's inertial.

8. At t=0, the angular sped is zero.

9. The report converts the angular speed in Column G. from radian per second to (Colum H) revolution per minute using conversion factor 30 / ?.

10. The report further defines system parameter as t=0 in row 3. The value 0.1 seconds defines time and time interval, and cell A3 represent the time interactions.

11. The value B3 is set to value 100.

12. The equation is not used in cell D2 to define TF because it could reference a null value. Thus, Equation 3 is used for TF where k1 is defined in the spreadsheet.

13. Thus, the Equations define in E2 and F2 are placed in the next row.

14. To obtain value for ?, the report performs numerical integration of ?. The time interval is taken from the first column and multiplies by ? appropriate value, and the equation starts from the third row.

15. Thus, the equation in cell H2 is copied down in the next row.

The arrows in Table 1 represent that the cell is copied to represent the time period. The dollar sign in the equation shows that Excel cannot change the cell references in row and column.

Table 2: Spreadsheet for the Engine Stimulation

The table 2 presents the spreadsheet of the engine stimulation. The results of the stimulation is presented in the graphical form and the time taken to reach 3000 rpm with the 100% fuel input is approximately 0.4 seconds as being revealed in Fig 5.

The report uses the spreadsheet to determine the time that the engine takes to reach 3000 rpm with continuous 100% fuel input. Based on the diagram of fuel vs. RPM, it is revealed that when the RPM reaches 9549.00, the rpm remains constant and its level is off over 9549 and this is point where equilibrium is approached, "which is a consequence of the frictional torque being related to the square of angular velocity" (Golten, . & Verwer, 2003 P. 9).

Fig 5: Engine Speed vs. Time

The output of the JavaScript is in the following link:

http://azizautomotivedesign.webs.com/

The Appendix 1 contains the code of the Javascript of the Engine Speed vs. Fuel.

Fig: Output of the Javascript

http://azizautomotivedesign.webs.com/

Overview of the graph demonstrating the engine speed vs. Time, which reveals that the velocity of the engine increase with 100% fuel consumption however, when the engine speeds reach 9549, this is the point the engine reaches the equilibrium.

Part Two

This section demonstrates the strategy to add a control system. The strategy is to add different control systems to test the engine performances. The system in Fig 4 only uses the nonlinear with ?2 .Moreover; the system diagram has a loop without the feedback control. However, the report intends to adjust the fuel input of the system manually. With feedback control, the report compares the output with reference value to obtain an error, which is used to obtain the desired response. Typically, the error margin is used to achieve a desired output.

Fig 6: Close Loop System (Negative Feedback)

+ Error

Desired Output Actual Output

"The control loop has ability to implement a variety of control algorithms to achieve the desired output in an effective and efficient manner" (Golten, & Verwer, 2003 P. 10). The negative feedback is achieved by subtracting feedback from the input to achieve a desire measure of error in the output. The process is called negative feedback because the feedback is deducted from the input reference. The report implements the control block to complete the system since the block in Fig 4 only represents the process block. Thus, the report uses both Excel and JavaScript to test the performance of different speed control system using on/off control system

(i) on/off control

Regulation of engine speed through manual process of adjustment of fuel input may lead to trial and error approach, which may take time to achieve a desired result. However, there is a need to introduce controller and adjust the fuel input to achieve a desired results. The report starts from the simplest form of control type on/off controller. The principle to determine error in the output speed is to apply either 0% (off) and 100% (on) to fuel to decrease or increase the engine speed to remain as close to 800 rpm as possible. The accuracy of controller is generally being determined by sampling time. The report will determine the control loop system of the diagram in fig 6. The report attempts to use the output speed and set the desired speed how close the system could achieve a correct speed. The graphical illustration in Fig 7 reveals the on/off controller, which reveals whether the output speed is greater or less than the set point.

Fig 7: The Function of on/Off Controller

0% Error

The on/of controller illustration reveals whether the engine speed is greater than the set point, the error is negative and set at 0% if fuel is less than set point and when the error is positive and the fuel is set at 100%. The spreadsheet is used to demonstrate the process and use a conditional function and is extended to include controller.

The conditional function start with if statement as follows:

IF X and Y THEN Z

However, the following rule describes the control process of the engine model

"IF (engine speed is too low) and (engine speed increasing) THEN (throttle 100%)

IF (engine speed is too low) and (engine speed decreasing) THEN (throttle 100%)

IF (engine speed is too high) and (engine speed increasing) THEN (throttle 0%)

IF (engine speed is too high) and (engine speed decreasing) THEN (throttle 0%)." ( Golten, . & Verwer, 2003 P. 11).

The table below illustrates the four conditional statements:

Increasing

Decreasing

Too low

1

1

Too high

0

0

Thus, the algorithms are simplified as follows:

IF (error >0) THEN (throttle 100%) ELSE (throttle 0%)

As being revealed in the diagram below, when fuel input is set to 50%, the time taken for the engine to achieve 3000-rpm changes 1 minute.

The output of the JavaScript is presented below in the following link:

http://azizautomotivedesign.webs.com/engine-speed-fuel-at-50

The table below reveals the layout f the spreadsheet for on/off controller.

Table: Layout of the Spreadsheet

A

B

C

D

D

E

F

G

H

I

1

Time (sec)

Error

Fuel (%)

T (Nm)

TF

(Nm)

TN

(Nm)

Acc (rad/s2)

Vel (rad/s2)

Pos

rpm

2

0

=$L$11-$12

=if ($B2>0,100, 0)

=$L$8*$B2

0

=$C2-D2

=$E2/$L$9

0

0

=$G2*30/PI ()

3

=$A2+0.1

=$L$7*POWER ($G2,2)

=$G2+($A3-$A2)*$F2

=H2+(A3-A2)*0.5*(G2+G3)

4

Table…